Category: Class 11

MP Board Class 11th Chemistry Solutions Chapter 10 s-ब्लॉक तत्त्व

s-ब्लॉक तत्त्व NCERT अभ्यास प्रश्न

प्रश्न 1.
क्षार धातुओं के सामान्य भौतिक तथा रासायनिक गुण क्या हैं ?
उत्तर:
भौतिक गुण –
1. क्षार धातुएँ चाँदी के समान सफेद, नरम तथा हल्की धातुएँ होती हैं।
2. इनके धनत्व बहुत कम होते हैं (बड़े आकार के कारण)। यह वर्ग में नीचे जाने पर घटता है। यद्यपि पोटैशियम, सोडियम से हल्की धातु है।
3. क्षार धातुओं के गलनांक तथा क्वथनांक निम्न होते हैं, क्योंकि इन धातुओं में मात्र एक संयोजी इलेक्ट्रॉन के उपस्थिति के कारण इनके बीच दुर्बल धात्विक बंध होते हैं।
4. क्षार धातुएँ, Be तथा Mg इनके लवण ज्वाला में विशिष्ट रंग प्रदान करते हैं, क्योंकि इनके संयोजी इलेक्ट्रॉन आसानी से निम्न से उच्च ऊर्जा स्तर तक उत्तेजित हो जाते हैं –

रासायनिक गुण:
क्षार धातुएँ अपनी निम्न आयनन एन्थैल्पी के कारण अत्यधिक क्रियाशील होती है। इनकी क्रियाशीलता वर्ग में नीचे जाने पर घटती है –

1. वायु के साथ अभिक्रियाशीलता:
क्षार धातुएँ वायु की उपस्थिति में ऑक्साइड बनाने के कारण मलिन हो जाती है। लीथियम मोनोऑक्साइड, सोडियम परॉक्साइड तथा अन्य धातुएँ सुपर ऑक्साइड बनाती है।

• 4Li+ O₂ → 2Li₂O (ऑक्साइड)
• 2Na + O₂ → NaO₂ (परॉक्साइड)
• M+ O₂ → MO₂ (सुपरऑक्साइड) (M = K, Rb, Cs)

लीथियम वायु की नाइट्रोजन के साथ सीधे संयोग करके लीथियम नाइट्राइड (Li₃N) बनाती है।

2. जल के साथ क्रिया:
क्षार धातुएँ जल के साथ क्रिया करके हाइड्रॉक्साइड तथा डाइहाइड्रोजन बनाती हैं।
2M + 2H₂O → 2M⁺ + 2OH^– + H₂, (M = क्षार धातु) लीथियम का जल के साथ क्रिया सोडियम की तुलना में कम होता है। क्षार धातुएँ प्रोटॉन दाता स्पीशीज, जैसे – एल्कोहॉल, ऐल्काइन और गैसीय NHS से भी क्रिया करती हैं।

3. डाइहाइड्रोजन के साथ क्रिया:
क्षार धातुएँ डाइहाइड्रोजन के साथ सुगमतापूर्वक क्रिया करके आयनिक या लवणीय हाइड्राइड बनाती हैं।

(M = Li,Na, K आदि)

(iv) हैलोजन के साथ क्रिया:
क्षार धातुएँ हैलोजनों के साथ सुगमतापूर्वक क्रिया करके आयनिक हैलाइड बनाती हैं।

(M = क्षार धातु)

(v) अपचायक प्रकृति:
क्षार धातुएँ अपने संयोजी इलेक्ट्रॉन को सरलता से त्याग देती है। अतः ये प्रबल अपचायक होती हैं।
Na → Na⁺ +e^–
लीथियम सर्वाधिक प्रबल तथा सोडियम दुर्बलतम अपचायक है।

(vi) दव अमोनिया में विलेयता:
सभी क्षार धातुएँ अमोनिया में विलेय हो जाती है तथा गहरे नीले रंग का विलयन बनाती है।

(vii) सभी क्षार धातुएँ परस्पर तथा अन्य धातुओं के साथ मिलकर मिश्रधातु बनाती हैं। मर्करी के साथ अमलगम बनाती है तथा इस प्रकार की क्रियाएँ अत्यधिक ऊष्माक्षेपी होती हैं।

प्रश्न 2.
क्षारीय मृदा धातुओं के सामान्य अभिलाक्षणिक गुणों में आवर्तिता की विवेचना कीजिए।
उत्तर:
वर्ग – 2 के तत्व Be, Mg, Ca, Sr, Ba तथा Ra को संयुक्त रूप से क्षारीय धातु कहते हैं (Be के अतिरिक्त)।
सामान्य गुण (A) परमाण्विक गुण:
1. इलेक्ट्रॉनिक विन्यास – इनके सामान्य इलेक्ट्रॉनिक विन्यास को [उत्कृष्ट गैस] ns² द्वारा दर्शाते हैं।
2. परमाण्विक तथा आयनिक त्रिज्या – इन तत्वों की परमाण्विक तथा आयनिक त्रिज्याएँ एक ही आवर्त में निकटवर्ती क्षार धातुओं की अपेक्षा कम होती है। एक ही वर्ग में परमाण्विक तथा आयनिक त्रिज्याएँ, परमाणु संख्या में वृद्धि के साथ – साथ बढ़ती है।
3. आयनन एन्थैल्पी – क्षारीय मृदा धातुओं की प्रथम आयनन एन्थैल्पी, इनके छोटे आकार के कारण –

क्षारीय मृदा धातुओं की द्वितीय आयनन एन्थैल्पी, संगत क्षार धातुओं की अपेक्षा कम होती है।
4. जलयोजन एन्थैल्पी – क्षारीय मृदा धातुओं की जलयोजन एन्थैल्पी क्षार धातु आयनों के समान वर्ग में नीचे की ओर जाने से क्रमशः घटती है।
Be²⁺ > Mg²⁺ > Ca²⁺ > Sr²⁺ > Ba²⁺
क्षारीय मृदा धातु आयनों की जलयोजन एन्थैल्पी, संगत क्षार धातुओं की अपेक्षा कम होती है।

(B) भौतिक गुण:
1. ये धातुएँ चाँदी के समान सफेद, चमकीली तथा नरम (परन्तु क्षार धातुओं की अपेक्षा कठोर) होती है।
2. इन धातुओं के छोटे आकार के कारण इनके गलनांक तथा क्वथनांक संगत क्षार धातुओं की अपेक्षा कम होते हैं। यह क्रम यद्यपि नियमित नहीं है।
3. ज्वाला के प्रति रंग – Be तथा Mg के अतिरिक्त सभी क्षारीय मृदा धातुएँ ज्वाला के साथ एक विशिष्ट रंग देते हैं। इनके विभिन्न रंग इलेक्ट्रॉनों को उत्तेजित तथा उत्तेजनहीन करने में प्रयुक्त ऊर्जाओं के विभिन्नता के कारण होता है।

4. घनत्व – इन धातुओं के घनत्व Be से Ca तक घटते हैं, तदुपरान्त बढ़ते हैं, ये धातुएँ अपने छोटे आकार के कारण, संगत क्षार धातुओं की अपेक्षा सघन, भारी तथा कठोर होती हैं।

(C) रासायनिक गुण:
क्षारीय मृदा धातुएँ क्षार धातुओं की अपेक्षा कम क्रियाशील होती हैं। इन धातुओं की क्रियाशीलता वर्ग में नीचे की ओर जाने पर क्रमशः घटती है।
1. वायु तथा जल के साथ क्रिया – Be तथा Mg वायु (ऑक्सीजन) तथा जल के प्रति अक्रिय होते हैं क्योंकि इनकी सतह पर ऑक्साइड की परत जम जाती है। यद्यपि, बेरीलियम चूर्ण वायु में आसानी से जल जाता है।

इसी प्रकार मैग्नीशियम अधिक विद्युत् धनात्मक है तथा वायु में चमकीले प्रकाश के साथ जलकर Mgo तथा Mg,N, बनाता है। Ca, Sr तथा Ba वायु के साथ शीघ्रता से क्रिया करके ऑक्साइड तथा नाइट्राइड बनाते हैं।
2. हाइड्रोजन के साथ क्रिया – Be के अतिरिक्त ये सभी धातुएँ गर्म करने पर हाइड्रोजन के साथ योग करके MH, प्रकार के हाइड्राइड बनाती है। (M = Be, Mg, Ca, Sr, Ba)
3. हैलोजन के साथ क्रिया – ये धातुएँ हैलोजन के साथ उच्च ताप पर क्रिया करके MX₂ प्रकार के हैलाइड बनाती है। M + X₂ → MX₂,
(X = F, CI, Br)
4. अम्लों के साथ क्रिया-ये धातुएँ अम्लों के साथ शीघ्रता से क्रिया करती हैं तथा H, गैस मुक्त करती हैं।
M + 2HCl → MCl₂ + H₂
5. अपचायक प्रकृति- क्षार धातुओं की भाँति, क्षारीय मृदा धातुएँ भी प्रबल अपचायक प्रकृति की होती है। यद्यपि इनकी अपचायक क्षमता क्षार धातुओं की अपेक्षा कम होती है। वर्ग में नीचे जाने पर इनकी अपचायक क्षमता क्रमशः घटती है।

6.  द्रव NH₃ में विलेयता:
क्षार धातुओं की भाँति, क्षारीय मृदा धातुएँ द्रव अमोनिया में घुलकर गहरा नीला काला विलयन बनाती हैं।
M + (x + y)NH₃ → [M(NH₃)_x]²⁺ + 2[e(NH₃)_y]^–

प्रश्न 3.
क्षार धातुएँ प्रकृति में क्यों नहीं पाई जाती हैं ?
उत्तर:
क्षार धातुएँ अत्यधिक रासायनिक सक्रियता के कारण प्रकृति में मुक्त अवस्था में नहीं पाई जाती हैं। ये भूपर्पटी में हैलाइड, सल्फेट, कार्बोनेट, सिलिकेट, बोरेट, ऑक्साइड आदि अयस्कों के रूप में पाई जाती हैं।

प्रश्न 4.
Na₂O₂में सोडियम की ऑक्सीकरण अवस्था ज्ञात कीजिए।
उत्तर:
माना Na₂O₂में Na की ऑक्सीकरण अवस्था x है। H₂O₂ के समान Na₂O₂ में परॉक्साइड बंध (-O-O-) होता है। अत: Na₂O₂ में Na की ऑक्सीकरण अवस्था –
x × 2 + (-1) × 2 = 0
2x – 2 = 0
⇒ x = \(\frac { 2 }{ 2 }\) = +1

प्रश्न 5.
पोटैशियम की तुलना में सोडियम कम अभिक्रियाशील क्यों है ? बताइए। ‘
उत्तर:
पोटैशियम की आयनन एन्थैल्पी (∆iH) (419 kJmol^-1) सोडियम की आयनन एन्थैल्पी (496 kJmol^-1) की अपेक्षा कम होती है तथा पोटैशियम का मान इलेक्ट्रोड विभव, (-2.925 V), सोडियम के संगत मान (-2.714 V) की अपेक्षा अधिक ऋणात्मक है। यही कारण है कि पोटैशियम, सोडियम की अपेक्षा अधिक ऋणात्मक है। अत: पोटैशियम, सोडियम से अधिक क्रियाशील होता है।

प्रश्न 6.
निम्नलिखित के सन्दर्भ में क्षार धातुओं एवं क्षारीय मृदा धातुओं की तुलना कीजिए –

1. आयनन एन्थैल्पी
2. ऑक्साइडों की क्षारकता
3.  हाइड्रॉक्साइडों की विलेयता।

उत्तर:

1. आयनन एन्थैल्पी – अधिक नाभिक आवेश एवं छोटे परमाणु आकार के कारण, क्षारीय मृदा धातुओं की आयनन एन्थैल्पी संगत क्षार धातुओं से अधिक होती है।
2. ऑक्साइडों की क्षारकता – क्षार धातुओं के आयनन एन्थैल्पी कम अथवा वैद्युत धनात्मक गुण संगत क्षारीय मृदा धातुओं से अधिक होते हैं। अतः क्षारीय धातु ऑक्साइड संगत क्षारीय मृदा धातु ऑक्साइडों से अधिक क्षारीय होते हैं।
3. हाइड्रॉक्साइडों की विलेयता – क्षारीय मृदा धातुओं के हाइड्रॉक्साइडों की विलेयता संगत क्षारीय हाइड्रॉक्साइडों की अपेक्षा कम होती है।

प्रश्न 7.
लीथियम किस प्रकार मैग्नीशियम से रासायनिक गुणों में समानताएँ दर्शाता है ?
उत्तर:
लीथियम तथा मैग्नीशियम समान आकार के कारण गुणों में अत्यधिक समानता प्रदर्शित करते हैं –
(परमाण्विक त्रिज्या : Li = 152pm, Mg = 160pm; आयनिक त्रिज्या Li⁺ = 76pm, Mg²⁺ = 72pm)। इनके समान गुण निम्न हैं –
1. लीथियम तथा मैग्नीशियम दोनों अपने संबंधित वर्गों के तत्वों की अपेक्षा हल्के तथा कठोर हैं।
2. लीथियम तथा मैग्नीशियम दोनों जल के साथ धीरे-धीरे क्रिया करते हैं।
3. लीथियम तथा मैग्नीशियम दोनों के ऑक्साइड तथा हाइड्रॉक्साइड जल में बहुत कम विलेय हैं। इनके हाइड्रॉक्साइड गर्म करने पर अपघटित होते हैं।

4. Li तथा Mg दोनों वायु की नाइट्रोजन से सीधे क्रिया करके नाइट्राइड (Li₃N तथा Mg₃N₂) बनाते हैं।
5. Li तथा Mg दोनों गर्म करने पर कार्बन के साथ संयुक्त होते हैं।

• 2Li + 2C → Li₂C₂
• Mg + 2C → MgC₂

6. Li एवं Mg दोनों का ऑक्सीजन में गर्म करने पर मोनोऑक्साइड बनाते हैं।

• 4Li + O₂ → 2Li₂o
• 2Mg + O₂ → 2MgO

7. Li₂SO₄ , MgSO₄ की भाँति फिटकरी नहीं बनाता है।
8. LiCl, MgCl दोनों प्रस्वेद्य है तथा जलीय विलयन में हाइड्रेट (LiCl – 2H₂O तथा MgCl₂6H₂O) के रूप में क्रिस्टलीय होते हैं।
9. LiCO₃ तथा MgCO₃ दोनों गर्म करने पर आसानी से अपघटित होकर ऑक्साइड तथा CO₂ बनाते हैं।
10. Li तथा Mg दोनों द्वारा ठोस हाइड्रोजन कार्बोनेट नहीं बनाए जा सकते हैं।
11. LiNO₃ तथा Mg(NO₃)₂ दोनों गर्म करने पर अपघटित होकर नाइट्रोजन डाइऑक्साइड बनाते हैं।

प्रश्न 8.
क्षार धातुएँ तथा क्षारीय मृदा धातुएँ रासायनिक अपचयन विधि से क्यों प्राप्त नहीं किए जा सकते ? समझाइए।
उत्तर:

• क्षार धातुएँ तथा क्षारीय मृदा धातुएँ स्वयं में प्रबल अपचायक हैं। अतः ये धातुएँ अपने ऑक्साइंड तथा अन्य यौगिकों द्वारा रासायनिक अपचयन विधि द्वारा प्राप्त नहीं की जा सकती है।
• इन धातुओं की प्रकृति अत्यधिक विद्युत् धनात्मक होती है। अतः इनके लवणों में जलीय विलयन से इन्हें अन्य धातुओं द्वारा विस्थापित नहीं किया जा सकता है।

प्रश्न 9.
प्रकाश वैद्युत सेल में लीथियम के स्थान पर पोटैशियम एवं सीज़ियम क्यों प्रयुक्त किए जाते हैं?
उत्तर:
पोटैशियम तथा सीज़ियम की आयनन एन्थैल्पी लीथियम की अपेक्षा अधिक कम होती है। अतः ये धातुएँ प्रकाश में रखने पर इलेक्ट्रॉन आसानी से उत्सर्जित करती है, परन्तु लीथियम ऐसा नहीं कर पाती है। यही कारण है Li की अपेक्षा K तथा Cs का प्रयोग प्रकाश वैद्युत सेल में किया जाता है।

प्रश्न 10.
जब एक क्षार धातु को दव अमोनिया में घोला जाता है, तब विलियन विभिन्न रंग प्राप्त कर सकता है। इस प्रकार के रंग-परिवर्तन कारण बताइए।
उत्तर:
क्षार धातुओं का अमोनिया में तनु विलयन का रंग गहरा नीला होता है क्योंकि अमोनीकृत इलेक्ट्रॉन प्रकाश के अदृश्य क्षेत्र में ऊर्जा अवशोषित करते हैं। यदि विलयन की सान्द्रता 3M से अधिक बढ़ा दी जाये तो रंग ताँबे – काँस्य जैसा हो जाता है।

प्रश्न 11.
ज्वाला को बेरीलियम एवं मैग्नीशियम कोई रंग प्रदान नहीं करते हैं, जबकि अन्य क्षारीय मृदा धातुएँ ऐसा करती हैं, क्यों?
उत्तर:
Be तथा Mg परमाणु अपने छोटे आकार तथा अधिक प्रभावी नाभिकीय आवेश के कारण अपने इलेक्ट्रॉनों को अधिक प्रबलता सं बाँधे रखते हैं। अतः इन्हें उच्च उत्तेजन ऊर्जा की आवश्यकता होती है तथा ये बुन्सन ज्वाला द्वारा उत्तेजित नहीं हो पाते हैं। जबकि अन्य क्षारीय मृदा धातुओं के उच्च ऊर्जा स्तरों में इलेक्ट्रॉनों का सरलतापूर्वक उत्सर्जन हो सकता है। अतः ज्वाला में विशिष्ट रंग देते हैं।

प्रश्न 12.
सॉल्वे प्रक्रम में होने वाली विभिन्न अभिक्रियाओं की विवेचना कीजिए।
उत्तर:
सॉल्वे अमोनिया प्रक्रम में, अमोनिया द्वारा संतृप्त ब्राइन (NaCl का सान्द्र विलयन) में CO₂ प्रवाहित की जाती है। इस प्रक्रम में शीघ्र विलेय सोडियम बाइकार्बोनेट बनता है।

इस प्रकार बने सोडियम बाइकार्बोनेट को छानकर, सुखाकर तथा गर्म करके सोडियम कार्बोनेट प्राप्त करते हैं।

कार्बोनेटिंग स्तंभ में प्रयुक्त CO₂ को कैल्सियम कार्बोनेट को गर्म करके प्राप्त करते हैं। इस क्रिया में बने Cao को जल में घोलकर बुझा हुआ चूना प्राप्त कर लेते हैं।

अमोनिया पुनः प्राप्ति स्तंभ में NH₄Cl तथा Ca(OH)₂ को गर्म करके NH₃ प्राप्त करते हैं।

प्रश्न 13.
पोटैशियम कार्बोनेट सॉल्वे विधि द्वारा नहीं बनाया जा सकता है, क्यों?
उत्तर:
पोटैशियम कार्बोनेट को सॉल्वे विधि से नहीं बनाया जा सकता, क्योंकि पोटैशियम कार्बोनेट जल में अत्यधिक विलेय होने के कारण अवक्षेपित नहीं होता है।

प्रश्न 14.
Li₂CO₃ कम ताप पर एवं Na₂CO₃ उच्च ताप पर क्यों विघटित होता है ?
उत्तर:
Li⁺ आयन आकार में छोटे होते हैं, जो छोटे ऋणायन ऑक्साइड O^2-के साथ, CO₃^2- की तुलना में स्थायी जालक बनाते हैं इसलिए Li₂CO₃ निम्न ताप पर Li₂O में विघटित होता है। जबकि Na⁺ आयन बड़े आकार का होता है, जो बड़े ऋणायन CO₃^2- के साथ O^2-आयन की तुलना में स्थायी जालक बनाता है। अतः Na₂CO₃ अत्यधिक स्थायी है, जो उच्च ताप पर ही Na₂O में विघटित होता है।

प्रश्न 15.
क्षार धातुओं के निम्नलिखित यौगिकों की तुलना क्षारीय मृदा धातुओं के संगत यौगिकों से विलेयता एवं तापीय स्थायित्व के आधार पर कीजिए –
(a) नाइट्रेट
(b) कार्बोनेट
(c) सल्फेट।
उत्तर:
(a) क्षार तथा क्षारीय मृदा धातुओं में नाइट्रेट –
1. क्षार तथा क्षारीय मृदा धातुओं के नाइट्रेट जल में अति विलेय होते हैं।

2. क्षार धातुओं के नाइट्रेट (लीथियम नाइट्रेट के अतिरिक्त) उच्च ताप पर नाइट्राइड में अपघटित हो जाते हैं। अधिक ताप पर पुनः गर्म करने पर, ऑक्साइड प्राप्त होते हैं। क्षारीय मृदा धातुओं के नाइट्रेट (बेरियम नाइट्रेट के अतिरिक्त) गर्म करने पर संगत ऑक्साइड देते हैं तथा NO₂ तथा O₂ का मिश्रण मुक्त होता है।

(b) क्षार तथा क्षारीय मृदा धातुओं में कार्बोनेट –
1. क्षार धातुओं के कार्बोनेट 1273 K तक स्थायी होता है। इससे अधिक ताप पर ये पिघलकर संगत ऑक्साइड बनाते हैं।

अपेक्षाकृत कम स्थायी है तथा शीघ्रता से अपघटित हो जाती है।

सभी क्षारीय मृदा धातुओं के कार्बोनेट गर्म करने पर संगत धातु ऑक्साइड तथा CO₂ देते हैं।

क्षारीय मृदा धातुओं के कार्बोनेटों का तापीय स्थायित्व वर्ग में नीचे जाने पर क्रमशः घटता है। BeCO₂ सबसे कम स्थायी है।
2. सभी क्षार धातु कार्बोनेट सामान्यतः जल में विलेय होते हैं तथा इनकी विलेयता वर्ग में नीचे जाने पर क्रमशः बढ़ती है, क्योंकि इनकी जलयोजन ऊर्जा की अपेक्षा जालक ऊर्जा अधिक शीघ्रता से घटती है। क्षारीय मृदा धातुएँ कार्बोनेट जल में विलेय होती है तथा इनकी विलेयता वर्ग में नीचे जाने पर क्रमशः घटती है। यद्यपि ये CO₂ की उपस्थिति में अधिक विलेय होती है।

(c) क्षार तथा क्षारीय मृदा धातुओं में सल्फेट –
1. क्षार धातुओं के सल्फेट (Li₂SO₄ के अतिरिक्त) ताप स्थायी होते हैं, जबकि क्षारीय मृदा धातुओं के सल्फेट गर्म करने पर अपघटित हो जाते हैं। इनका ताप स्थायित्व वर्ग में नीचे की ओर जाने पर क्रमशः बढ़ता है।
2. क्षार धातु सल्फेट (Li₂SO₄ के अतिरिक्त) जल में विलेय होते हैं। क्षारीय मृदा धातुओं के सल्फेटों की विलेयता वर्ग में नीचे की ओर जाने पर क्रमशः घटती है। BeSO₄ तथा MgSO₄ जल में शीघ्र विलेय हैं, जबकि BaSO₄ जल में पूर्णतया अविलेय है।

प्रश्न 16.
सोडियम क्लोराइड से प्रारम्भ करके निम्नलिखित को आप किस प्रकार बनाएँगे –

1. सोडियम धातु
2. सोडियम हाइड्रॉक्साइड
3. सोडियम परॉक्साइड
4. सोडियम कार्बोनेट।

उत्तर:
1. सोडियम धातु:
इसे NaCl (40%) तथा CaCl₂(60%) के गलित मिश्रण द्वारा 273K पर डाउन्स सेल में वैद्युत-अपघटन द्वारा बनाया जाता है। कैथोड पर मुक्त Na को केरोसीन (मिट्टी के तेल)में एकत्रित करते हैं, जबकि Cl₂ एनोड पर मुक्त होती है।

कैथोड पर – Na⁺ + e →Na_(s)
एनोड पर – 2Cl^– → Cl_2(g) + 2e^–

2. सोडियम हाइड्रॉक्साइड:
इसे कास्टनर – कैलनर सेल में NaCl के जलीय संतृप्त विलयन (ब्राइन) के वैद्युत – अपघटन द्वारा प्राप्त करते हैं। इस सेल में मर्करी कैथोड तथा कार्बन एनोड का प्रयोग करते हैं। सोडियम धातु जो कैथोड पर मुक्त होती है, मर्करी के साथ संयोग करके सोडियम अमलगम बनाती है। Cl₂ गैस एनोड पर मुक्त होती है।

एनोड पर – 2Cl^– → Cl_2(g) + 2e^–
इस प्रकार प्राप्त सोडियम अमलगम की क्रिया जल से कराने पर सोडियम हाइड्रॉक्साइड तथा हाइड्रोजन गैस मुक्त होती है।
2Na – अमलगम + 2H₂O → 2NaOH + 2Hg + H₂

3. सोडियम परॉक्साइड:
गलित NaCl के वैद्युत – अपघटन से प्राप्त गलित सोडियम धातु को वायु की अधिकता में गर्म करने पर सोडियम ऑक्साइड प्राप्त होता है। इससे वायु की पुनः क्रिया कराने पर Na₂O₂ प्राप्त होता है।

4. सोडियम कार्बोनेट- इसे सॉल्वे अमोनिया प्रक्रम द्वारा बनाया जाता है। अमोनियाकृत सान्द्र ब्राइन विलयन (NaCl के जलीय विलयन) में CO₂ प्रवाहित करने पर, सोडियम हाइड्रोजन कार्बोनेट अवक्षेपित होता है। जिसे पुनः गर्म कराने पर सोडियम कार्बोनेट प्राप्त होता है।

प्रश्न 17.
क्या होता है, जब –

1. मैग्नीशियम को हवा में जलाया जाता है
2. बिना बुझे चूने को सिलिका के साथ गर्म किया जाता है
3. क्लोरीन को बुझे चूने से क्रिया कराया जाता है
4. कैल्सियम नाइट्रेट को गर्म किया जाता है।

उत्तर:

(iii) यह Cl₂ से क्रिया करके कैल्सियम हाइपोक्लोराइड बनाता है।

प्रश्न 18.
निम्नलिखित में से प्रत्येक के दो-दो उपयोग बताइए –
(a) कास्टिक सोडा
(b) सोडियम कार्बोनेट
(c) बिना बुझा चूना।
उत्तर:
(a) कास्टिक सोडा –

• इसका उपयोग साबुन, कागज, कृत्रिम रेशम आदि बनाने में होता है।
• वस्त्र उद्योग में सूती कपड़ों के मर्सरीकरण में इसका प्रयोग होता है।

(b) सोडियम कार्बोनेट –

• इसका प्रयोग जल के मृदुकरण, धुलाई तथा निर्मलन में करते हैं।
• इसका उपयोग काँच, साबुन, बोरेक्स कास्टिक सोडा के निर्माण में होता है।

(c) बिना बुझा चूना –

• इसका प्रयोग कास्टिक सोडा से धावन सोडा बनाने में करते हैं।
• इसका प्रयोग शर्करा के शुद्धिकरण में तथा रंजकों के निर्माण में करते हैं।

प्रश्न 19.
निम्नलिखित की संरचना बताइए –

1. BeCl(वाष्य)
2. BeCl₂(ठोस)।

उत्तर:
1. वाष्प अवस्था में यह सेतु बंधित क्लोराइड द्विलक की भाँति रहता है।

2. ठोस अवस्था में, यह क्लोरीन आबंध के साथ बहुलक श्रृंखला संरचना प्रदर्शित करता है।

प्रश्न 20.
सोडियम एवं पोटैशियम के हाइड्रॉक्साइड एवं कार्बोनेट जल मे विलेय है, जबकि मैग्नीशियम एवं कैल्सियम के संगत लवण जल में अल्प विलेय है। समझाइए।
उत्तर:

∆Hविलयन का मान जितना अधिक ऋणात्मक होता है, यौगिक की विलेयता उतनी ही कम होती है। सोडियम तथा पोटैशियम हाइड्रॉक्साइड तथा कार्बोनेटों की जल योजन ऊर्जा उनकी जालक ऊर्जा से अधिक होती है। अतः ये जल में विलेय होते हैं। मैग्नीशियम तथा कैल्सियम हाइड्रॉक्साइड की जालक ऊर्जा इनकी जलयोजन ऊर्जा से अधिक होती है। अतः ये जल में अल्प विलेय होते हैं।

प्रश्न 21.
निम्नलिखित की महत्व बताइए –
(a) चूना पत्थर
(b) सीमेंट
(c) प्लास्टर ऑफ पेरिस।
उत्तर:
(a) चूना पत्थर (CaCO3):

• इसे मैग्नीशियम कार्बोनेट के साथ आयरन जैसी धातुओं के निष्कर्षण में गालक के रूप में प्रयोग करते हैं।
• इसका प्रयोग ऐन्टासिड, टूथपेस्ट में अपघर्षक के रूप में, च्यूइंगम में संघटक तथा सौन्दर्य प्रसाधनों के रूप में भी करते हैं।

(b) सीमेंट:

• यह भवन निर्माण हेतु एक महत्वपूर्ण यौगिक है।
• इसका उपयोग काँक्रीट, प्रगलित काँक्रीट, प्लास्टीरंग, पुल-निर्माण, भवन-निर्माण आदि में किया जाता है।

(c) प्लास्टर ऑफ पेरिस:

• इसका उपयोग भवन निर्माण तथा टूटी हुई हड्डियों के प्लास्टर में होता है।
• इसका उपयोग दंत चिकित्सा, अलंकरण कार्य तथा मूर्तियों एवं अर्द्धप्रतिमाओं को बनाने में भी होता है।

प्रश्न 22.
लीथियम के लवण साधारणतया जलयोजित होते हैं, जबकि अन्य क्षार धातुओं के लवण साधारणतया निर्जलीय होते है, क्यों?
उत्तर:
क्षार धातुओं में सबसे छोटा आकार होने के कारण Li⁺ की जलयोजन कार्य सबसे कम होती है। यही कारण है कि लीथियम लवण सामान्यतः जलयोजित होते हैं तथा अन्य क्षार धातुओं के लवण साधारणतः निर्जलीय होते हैं।

प्रश्न 23.
LiF जल में लगभग विलेय होता है, जबकि LiCl न सिर्फ जल में, बल्कि एसीटोन में भी विलेय होता है। कारण बताइए, अन्य क्षार धातुओं के लवण साधारणतया निर्जलीय होते है। क्यों ?
उत्तर:
LiF, उच्च जालक ऊर्जा के कारण जल में लगभग विलेय होता है। परन्तु LiCl जल में विलेय होता है। LiCl अपनी विशिष्ट सहसंयोजी प्रकृति के कारण, एसीटोन में भी विलेय होता है। (चूँकि सहसंयोजी प्रकृति, ऋणायन के आकार के साथ बढ़ती है। अतः विलेयता का क्रम निम्न है –
LiF < LiCl< LiBr < Lil

प्रश्न 24.
जैव द्रवों में सोडियम, पोटैशियम, मैग्नीशियम एवं कैल्सियम की सार्थकता बताइए।
उत्तर:
सोडियम तथा पोटैशियम आयन अंतराकाशी द्रव में महत्वपूर्ण भूमिका निभाते हैं। ये आयन शिरा-संकेतों के संचरण में भाग लेते हैं। पोटैशियम आयन कोशिका द्रव में प्रचुरता में पाए जाने वाले धनायन हैं। जहाँ ये अनेक एन्जाइमों को सक्रिय करते हैं, ग्लूकोस के ऑक्सीकरण से ATP के निर्माण में भाग लेते हैं।

पौधे में, प्रकाश अवशोषण के लिए मुख्य रंजक क्लोरोफिल होता है। क्लोरोफिल में मैग्नीशियम होता है। शरीर में कैल्सियम का 99% भाग हड्डियों तथा दांतों में होता है। यह अंतर तांत्रिकीय पेशीय कार्यप्रणाली, अंतर तांत्रिकीय प्रेषण, कोशिका झिल्ली अखंडता तथा रक्त स्कंदन में महत्वपूर्ण भूमिका निभाता है।

प्रश्न 25.
क्या होता है, जब –
(a) सोडियम धातु को जल में डाला जाता है।
(b) सोडियम धातु को हवा की अधिकता में गर्म किया जाता है।
(c) सोडियम परॉक्साइड को जल में घोला जाता है।
उत्तर:
(a) H₂ गैस मुक्त होती है, जो अभिक्रिया में अत्यधिक ऊष्मा उत्पन्न करने के कारण आग पकड़ लेती है।
2Na_2(s) + 2H₂O_(l) → 2NaOH_(aq)+ H_2(g)

(b) Nazo की समान मात्रा के साथ Na20, प्राप्त होता है।

(c) HO₂ बनता है।
Na₂O_2(s) + 2H₂O_(l) → 2NaOH_(aq) + H₂O_2(l).

प्रश्न 26.
निम्नलिखित में से प्रत्येक प्रेक्षण पर टिप्पणी लिखिए –
(a) जलीय विलयनों में क्षार धातु आयनों की गतिशीलता Li⁺ < Na⁺ <K⁺ <Rb⁺ <Cs⁺ क्रम में होती है।
(b) लीथियम ऐसी एकमात्र क्षार धातु है, जो नाइट्राइड बनाती है।
(c) M²⁺_(aq) + 2e^– → M_(s); हेतु E° ( जहाँ, M = Ca, Sr या Ba) लगभग समान है।
उत्तर:
(a) आयन का आकार जितना कम होता है, जलयोजन उतना ही अधिक होता है तथा आयन का जलयोजन जितना अधिक होता है, उसकी आयनिक गतिशीलता उतनी ही कम होती है। अतः जलयोजन क्षमता का क्रम निम्न होगा –
Li⁺ < Na⁺ <K⁺ <Rb⁺ <Cs⁺
अतः आयनिक गतिशीलता विपरीत क्रम में बढ़ेगी –
Li⁺_(aq) < Na⁺_(aq) <K⁺_(aq) <Rb⁺_(aq) <Cs⁺_(aq)
(b) अपने छोटे आकार के कारण, क्षार धातुओं में केवल लीथियम नाइट्राइड बनाती है।
(c) M²⁺ + 2e^– M_(s); जहाँ, M = Ca, Sr, Ba के लिए E° का मान लगभग समान होता है। किसी भी इलेक्ट्रोड के लिए E का मान निम्नलिखित तीन कारकों पर निर्भर करता है –

• वाष्पन एन्थैल्पी
• आयनन एन्थैल्पी
• जलयोजन एन्थैल्पी

चूँकि इन तीनों कारकों का संयुक्त प्रभाव Ca, Sr तथा Ba के लिए लगभग समान रहता है। अतः इनके इलेक्ट्रोड विभव का मान लगभग स्थिर रहता है।

प्रश्न 27.
निम्नलिखित में से प्रत्येक प्रेक्षण पर टिप्पणी लिखिए –
(a) Na₂CO₃ का विलयन क्षारीय होता है।
(b) क्षार धातुएँ उनके संगलित क्लोराइडों के वैद्युत-अपघटन से प्राप्त की जाती है।
(c) पोटैशियम की तुलना में सोडियम अधिक उपयोगी है।
उत्तर:
(a) Na₂CO₃ एक दुर्बल अम्ल (H₂CO₃) तथा प्रबल क्षार का लवण है। अतः जल-अपघटन कराने पर यह क्षार (NaOH) देता है, जिसके कारण इसका जलीय विलयन क्षारीय होता है।
Na₂CO_3(s) + 2H₂O_(l) → 2NaOH_(aq) + H₂CO_3(aq)

(b)1. क्षार धातुएँ प्रबल अपचायक होती है, अतः इन्हें इनके ऑक्साइड तथा यौगिकों से निष्कर्षित नहीं किया जा सकता है।
2. अत्यधिक धनात्मक प्रकृति के कारण इनके लवणों के विलयन से इन्हें किसी अन्य धातु द्वारा विस्थापित नहीं किया जा सकता है।
3. क्षार धातुएँ अपने लवणों के जलीय विलयन के विद्युत्-अपघटन विधि द्वारा प्राप्त नहीं की जा सकती है, क्योंकि कैथोड पर Na धातु की अपेक्षा H, मुक्त होती है। यही कारण है कि क्षार धातुएँ अपने गलित क्लोराइडों के विद्युत्-अपघटन द्वारा प्राप्त होती है।

वैद्युत अपघटन में निम्नलिखित क्रियाएँ होती हैं –
कैथोड पर – 2cl^– → Cl₂ + 2e^–
एनोड पर – 2Na⁺ + 2e^– →2Na

(c) पोटैशियम की तुलना में सोडियम अधिक महत्वपूर्ण है क्योंकि यह क्रियाशील तो है परन्तु पोटैशियम की भाँति अतिक्रियाशील नहीं है। सोडियम के अन्य उपयोग हैं –

• नाभिकीय संयंत्रों में शीतलक के रूप में।
• पेट्रोल के लिए अपस्फोटन-रोधी पदार्थ, टेट्राएथिल लेड (TEL) के निर्माण में।
  4C₂H₂Cl + 4Na – Pb →(C₂H₅)₄Pb + 3Pb + 4NaCl

प्रश्न 28.
निम्नलिखित के मध्य क्रियाओं के संतुलित समीकरण लिखिए –

1. Na₂O₂ एवं जल
2. KO₂ एवं जल
3. Na₂O₂ एवं CO₂.

उत्तर:

1. Na₂O_2(s) +2H₂O_(l) → 2NaOH_(aq) + H₂O_2(aq)
2. 2KO_2(s) + H₂O_(l) →2KOH_(aq) + \(\frac { 3 }{ 2 }\)O_2(g)
   या 2KO_2(s) + H₂O_(l) →2KOH_(aq) + \(\frac { 3 }{ 2 }\)O_2(g)
3. Na₂O + CO₂ → Na₂CO₃ .

प्रश्न 29.
आप निम्नलिखित तत्वों को कैसे समझाएँगे –

1. BeO जल में अविलेय है, जबकि BeSO₄ विलेय है।
2. Bao जल में विलेय है, जबकि BaSO₄ अविलेय है।
3. एथेनॉल में dil. KI की तुलना में अधिक विलेय है।

उत्तर:
1. BeO जल में अविलेय है, क्योंकि BeO की जालक ऊर्जा का मान जलयोजन ऊर्जा से अधिक है, जबकि BeSO जल में घुलनशील है क्योंकि इसकी जलयोजन ऊर्जा जालक ऊर्जा से अधिक है।

2. दूसरे वर्ग के ऑक्साइड की विलेयता के लिए दोनों जालक ऊर्जा और जलयोजन ऊर्जा वर्ग में नीचे जाते समय घटती है, क्योंकि धनायन का आकार बढ़ता है। लेकिन जालक ऊर्जा, जलयोजन की अपेक्षा अधिक तीव्रता से घटती है। अत: BaO जल में घुलनशील है, क्योंकि जलयोजन ऊर्जा जालक ऊर्जा से अधिक है। परन्तु दूसरी ओर BaSO₄ जल में अविलेय है। जालक ऊर्जा की प्रबलता समान रहती है, क्योंकि ऋणायन का आकर इतना बड़ा है कि धनायन का आकार बढ़ने से कोई प्रभाव नहीं पड़ता। अत: BaSO₄ के लिए जालक ऊर्जा मान जलयोजन ऊर्जा से अधिक है।

3. एथेनॉल में Lil अधिक घुलनशील है, KI की तुलना में Lil में Li⁴ आयन का आकार छोटा है और घुलनशील है KI आयनिक यौगिक है अतः एथेनॉल में कम घुलनशील है।

प्रश्न 30.
इनमें से किस क्षार धातु का गलनांक न्यूनतम है-
(a) Na,
(b) K,
(c) Rb,
(d) Cs.
उत्तर:
(d) सीज़ियम का गलनांक न्यूनतम है (312K)।

प्रश्न 31.
निम्नलिखित में से कौन-सी क्षार धातु जलयोजित लवण देती है –
(a) Li,
(b) Na,
(c) K,
(d) Cs.
उत्तर:
(a) लीथियम अकेला ऐसा क्षार तत्व है, जो जलयोजित लवण देती है।

प्रश्न 32.
निम्नलिखित में कौन-सी क्षारीय मृदा धातु कार्बोनेट ताप के प्रति सबसे अधिक स्थायी है –
(a) MgCO₃
(b) CaCO₃
(c) SrCO₃
(d) BaCO₃
उत्तर:
(d) BaCO₃ ताप के प्रति अधिक स्थाई है, यह 1633K पर वियोजित होता है।

s-ब्लॉक तत्त्व अन्य महत्वपूर्ण प्रश्न

s-ब्लॉक तत्त्व वस्तुनिष्ठ प्रश्न

प्रश्न 1.
सही विकल्प चुनकर लिखिए –

प्रश्न 1.
प्लास्टर ऑफ पेरिस होता है –
(a) (CaSO₄)₂. H₂O
(b) CaSO₄.2H₂O
(c) CaSO₄. H₂O
(d) CaSO₄.
उत्तर:
(a) (CaSO₄)₂. H₂O

प्रश्न 2.
ब्लीचिंग पाउडर का सक्रिय घटक –
(a) Ca(OCl)₂
(b) Ca(OCl)Cl
(c) Ca(O₂Cl)₂
(d) CaCl₂O₂.
उत्तर:
(b) Ca(OCl)Cl

प्रश्न 3.
द्रव अमोनिया में सोडियम का विलयन निम्न के कारण अपचायक होता है –
(a) Na परमाणु
(b) NaH
(c) NaNH₂
(d) e^– (NH₃)_x
उत्तर:
(d) e^– (NH₃)_x

प्रश्न 4.
जल के साथ सोडियम की क्रिया तीव्र होती है, लीथियम की नहीं, क्योंकि लीथियम –
(a) का परमाणु भार अधिक है
(b) एक धातु है
(c) अधिक विद्युत् धनी है
(d) अधिक विद्युत् ऋणी है।
उत्तर:
(d) अधिक विद्युत् ऋणी है।

प्रश्न 5. विभिन्न क्लोराइडों में स्थायित्व का क्रम –
(a) LiCl > KCl > NaCl > CsCl
(b) CsCl > KCI > NaCl > LICl
(c) NaCl > KCl > LiCl > Csci
(d) KCl> CsCl> NaCl> LiCl.
उत्तर:
(a) LiCl > KCl > NaCl > CsCl

प्रश्न 6.
M ²⁺ की जलयोजन ऊर्जा अधिक होगी इससे –
(a) A³⁺
(b) Na⁺
(c) Be²⁺
(d) Mg³⁺
उत्तर:
(b) Na⁺

प्रश्न 7.
क्षारीय मृदा धातुओं का कौन-सा गुण परमाणु-क्रमांक बढ़ने के साथ बढ़ता है –
(a) आयनन ऊर्जा
(b) हाइड्रॉक्साइड की विलेयता
(c) सल्फेट की विलेयता
(d) ऋण-विद्युतता।
उत्तर:
(b) हाइड्रॉक्साइड की विलेयता

प्रश्न 8.
वायु को शुष्क करने के लिये किसका उपयोग ठीक है –
(a) CaCO₃
(b) Na₂CO₃
(c) NaHCO ₃
(d) CaO.
उत्तर:
(d) CaO.

प्रश्न 9. किस सल्फेट की विलेयता सबसे कम है –
(a) BaSO₄
(b) MgSO₄
(c) SrSO₄
(d) CaSO₄
उत्तर:
(a) BaSO₄

प्रश्न 10.
क्षारीय मृदा धातुओं के कार्बोनेटों में तापीय स्थायित्व का क्रम होगा –
(a) BaCO₃ > SrCO₃ > CaCO₃ > MgCO₃
(b) BaCO₃ > SrCO₃ > MgCO₃ > CaCO₃
(c) CaCO₃ > SrCO₃ > MgCO₃ > BaCO₃
(d) MgCO₃ > CaCO₃ > SrCO₃ > BaCO₃.
उत्तर:
(d) MgCO₃ > CaCO₃ > SrCO₃ > BaCO₃.

प्रश्न 11.
मैग्नीशियम का महत्वपूर्ण अयस्क है –
(a) मेलाकाइट
(b) केसीटेराइट
(c) कार्नेलाइट
(d) गेलेना।
उत्तर:
(c) कार्नेलाइट

प्रश्न 12.
क्षारीय मृदा धातुओं के सल्फेटों की घुलनशीलता समूह में नीचे जाने पर घटती है। इसका कारण है –
(a) गलनांक का बढ़ना
(b) जालक ऊर्जा का बढ़ना
(c) समन्वयन संख्या का बढ़ना
(d) उपर्युक्त सभी।
उत्तर:
(b) जालक ऊर्जा का बढ़ना

प्रश्न 13.
कौन-सा अयस्क लीथियम का नहीं है –
(a) पेंटालाइट
(b) ट्राइफिलाइट
(c) एल्बाइट
(d) स्पोड्यूमीन।
उत्तर:
(c) एल्बाइट

प्रश्न 14.
मैग्नीशियम उपस्थित है इसमें –
(a) हीमोग्लोबीन
(b) क्लोरोफिल
(c) विटामिन B₁₂
(d) विटामिन C.
उत्तर:
(b) क्लोरोफिल

प्रश्न 15.
CaCN₂ तथा C के मिश्रण को कहते हैं –
(a) बेराइट
(b) एनहाइड्राइट
(c) नाइट्रोलियम
(d) आइसलैंड स्पॉट।
उत्तर:
(c) नाइट्रोलियम

प्रश्न 2.
रिक्त स्थानों की पूर्ति कीजिए –

1. Ca²⁺ आयन की त्रिज्या K⁺ आयन से कम है क्योंकि …………. है।।
2. Be(OH)₂अम्ल तथा क्षार दोनों में विलेय है, क्योंकि इसकी प्रकृति ………….. है।
3. लीथियम समूह – 2 के ……………… तत्व से समानता रखता है।
4. द्रव अमोनिया में सोडियम धातु नीला रंग देती है यह ………….. के कारण है।
5. हाइड्रोलिथ का सूत्र …………… है।
6. s – ब्लॉक तत्व प्रबल …………… है।
7. क्षारीय मृदा धातु …………. रेडियोधर्मी गुण प्रदर्शित करता है।
8. क्षारीय धातु …………….. रेडियोधर्मी गुण प्रदर्शित करता है।
9. …………. क्षारीय मृदा धातु हाइड्रॉक्साइड उभयधर्मी है।
10. कार्नेलाइट का रासायनिक सूत्र …………… है।

उत्तर:

1. धनावेश अधिक
2. उभयधर्मी
3. Mg
4. e^– (NH₃)_x
5. CaH₂
6. अपचायक
7. रेडियम
8. फ्रान्सियम (Fr)
9. Be(OH)₂
10. KClMgCl₂ 6H₂O.

प्रश्न 3.
उचित संबंध जोड़िए –

उत्तर:

1. (e) CaO
2. (d) CaCN₂ तथा C
3. (b) MgCl₂
4. (a) Ca(OCI)Cl
5. (c) BaSO₄
6. (g) चूना पत्थर
7. (f) धुएँ का पर्दा

प्रश्न 4.
एक शब्द / वाक्य में उत्तर दीजिए –

1. क्षारीय मृदा धातुओं को बढ़ती हुई क्रियाशीलता के क्रम में व्यवस्थित कीजिए।
2. लीथियम अयस्क के दो नाम सूत्र सहित लिखिए।
3. मैग्नीशियम के दो अयस्कों के नाम लिखिए।
4. क्षार धातुओं को मिट्टी के तेल में रखा जाता है। क्यों?
5. ग्लोबल लवण का आण्विक सूत्र है।

उत्तर:

1. क्षारीय मृदा धातुओं की रासायनिक क्रियाशीलता समूह में ऊपर से नीचे आने (Be-Ra) पर बढ़ती है। Be < Mg < Ca < Sr < Ba < Ra
2. Li अयस्क के दो नाम व सूत्र –
   • स्पोड्यूमीन LiAl(SiO₃)₂
   • लेपिडोलाइट Li₂Al₂(SiO₃)₃.F(OH)₂
3. Mg के दो अयस्कों के नाम –
   • कार्नेलाइट
   • डोलोमाइट
4. अत्यधिक क्रियाशीलता के कारण
5. Na₂SO₄.10H₂O

s-ब्लॉक तत्त्व अति लघु उत्तरीय प्रश्न

प्रश्न 1.
एक ऐसे खनिज का नाम तथा सूत्र बताइये जिसमें Ca एवं Mg दोनों उपस्थित होते हैं।
उत्तर:
खनिज का नाम-डोलोमाइट। सूत्र – MgCO₃,CaCO₃

प्रश्न 2.
फ्लक्स का नाम लिखिए जिसको धात्विक प्रक्रमों में अम्लीय अशुद्धियों को दूर करने के लिये प्रयोग करते हैं।
उत्तर:

• लाइमस्टोन CaCO₃
• मैग्नेसाइट MgCO₃

प्रश्न 3.
प्लास्टर ऑफ पेरिस का उपयोग प्लास्टर चढ़ाने में किस गुण के कारण किया गया है ?
उत्तर:
प्लास्टर ऑफ पेरिस जल से क्रिया करके सीमेण्ट के समान कठोर हो जाता है। इस गुण के कारण इसका उपयोग टूटी हड्डियों को जोड़ने के लिये प्लास्टर चढ़ाने में किया जाता है।

प्रश्न 4.
IA तथा IIA समूह को s-ब्लॉक तत्व कहते हैं। क्यों?
उत्तर:
IA तथा IIA समूह में उपस्थित सभी तत्वों का इलेक्ट्रॉनिक विन्यास करने पर अंतिम इलेक्ट्रॉन S – उपकोश में प्रवेश करता है, इसलिये इसे s-ब्लॉक तत्व कहते हैं।
उदाहरण –

• Na₁₁ → 1s² 2s²2p⁶3s¹
• Ca₂₀ → 1s² 2s²2p⁶3s²3p⁶4s²

प्रश्न 5.
सॉरेल सीमेंट क्या है ? इसका उपयोग किन कार्यों में होता है ?
उत्तर:
MgCl₂के सान्द्र विलयन में MgO मिलाया जाये तो MgCl₂ 2MgO. xH₂O संघटन वाले मैग्नीशियम क्लोराइड का एक सफेद पेस्ट बनता है, जो जमकर कड़ा हो जाता है। इसे सॉरेल सीमेन्ट कहा जाता है। इसका उपयोग दाँतों की खोह भरने में तथा चीनी मिट्टी एवं पोर्सलेन के बर्तनों को जोड़ने में किया जाता है।

प्रश्न 6.
सोडियम कार्बोनेट को वायु में खुला छोड़ देने पर क्या परिवर्तन होता है ? समीकरण सहित समझाइये।
उत्तर:
Na₂CO₃ 10H₂O को वायु में खुला छोड़ देने पर इसका क्रिस्टलीय जल धीरे-धीरे निकल जाता है और यह मोनोहाइड्रेट Na₂CO₂ H₂O देता है जो 750°C पर गर्म करने पर निर्जल Na₂CO₃ देता है जिसे सोडा राख कहते हैं।

प्रश्न 7.
लीथियम हाइड्राइड का प्रयोग अन्य हाइड्राइडों के निर्माण में किया जा सकता है। बेरेलियम हाइड्राइड उनमें से एक है। इसके निर्माण के विभिन्न पद बताइए। इस प्रक्रम में प्रयुक्त रासायनिक समीकरण भी दीजिए।
उत्तर:
BeH₂ का निर्माण संगत जटिल क्षारीय धातुओं हाइड्राइडों जैसे लीथियम-ऐल्युमिनियम हाइड्राइड के अपचयन द्वारा करते हैं।

• 8LiH + Al₂Cl₆ → 2LiAlH₄ + 6LiCl
• 2BeCl₂ + LiAlH₄ → 2BeH₂ + LiCl + AlCl₂

प्रश्न 8.
कैल्सियम सल्फेट किस प्रकार बनाया जाता है ? इसके उपयोग लिखिए।
उत्तर:
प्रयोगशाला में Ca के ऑक्साइड, कार्बोनेट, क्लोराइड की अभिक्रिया तनु H₂SO₄ के साथ कराने पर कैल्सियम सल्फेट प्राप्त होता है।

• CaO+ H₂SO₄ → CaSO₄ + H₄O
• CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂

उपयोग:

• चाक बनाने में
• सीमेंट उद्योग में
• प्लास्टर ऑफ पेरिस बनाने में
• खाद के रूप में।

प्रश्न 9.
जिप्सम किसे कहते हैं ? इससे प्लास्टर ऑफ पेरिस कैसे बनाते हैं ?
उत्तर:
कैल्सियम सल्फेट CaSO₄ 2H₂O को जिप्सम कहते हैं । जिप्सम को 120 – 130°C तक गर्म करने पर इसमें से तीन चौथाई क्रिस्टल जल निकल जाता है तथा प्लास्टर ऑफ पेरिस प्राप्त होता है।

प्लास्टर ऑफ पेरिस जल अवशोषित करके पुनः जिप्सम में परिवर्तित हो जाता है।

प्रश्न 10.
अनबुझा चूना बनाते समय भट्टी का ताप 1000°C से अधिक गर्म नहीं करते।रासायनिक समीकरण सहित समझाइये।
उत्तर:
अनबुझा चूना बनाते समय भट्टी का ताप 1000°C से अधिक नहीं रखते क्योंकि इससे उच्च ताप पर CaO अशुद्धि के रूप में उपस्थित SiO₂ से मिलकर गलनीय सिलिकेट बना लेता है।

प्रश्न 11.
सोडियम को मिट्टी के तेल में रखा जाता है। क्यों?
उत्तर:
सोडियम अत्यन्त क्रियाशील तथा प्रबल धन विद्युती तत्व है। यह वायुमण्डल में उपस्थित 02, नमी तथा कार्बन डाइ-ऑक्साइड के साथ अभिक्रिया करके ऑक्साइड तथा हाइड्राक्साइड बनाता है। इसलिये सोडियम को मिट्टी के तेल में रखा जाता है जिससे वह वायु के संपर्क में न आ सके।

• 4Na + O₂ → 2Na₂O
• Na₂O+ H₂O → 2NaOH
• 2NaOH + CO₂ → Na₂CO₃ + H₂O

प्रश्न 12.
चूने के पानी का सूत्र लिखिए। इसमें CO₂ के प्रवाह से क्या परिवर्तन होगा?
उत्तर:
बुझे हुये चूने का जल में स्वच्छ विलयन चूने का पानी कहलाता है। इसका सूत्र Ca(OH)₂ होगा।

• चूने के पानी में CO₂ प्रवाहित करने से CaCO₃ बनने के कारण विलयन दूधिया हो जाता है।
  Ca(OH)₂ + CO₂ → CaCO₃ + H₂O
• चूने के पानी में CO₂ को देर तक प्रवाहित करने पर चूने के पानी का दूधियापन समाप्त हो जाता है।
  Ca(OH)₂ + CO₂ + H₂O → Ca(HCO₃)₂

प्रश्न 13.
फोटो रासायनिक सेल में किस धातु का उपयोग होता है और क्यों?
उत्तर:
फोटो रासायनिक सेल में पोटैशियम तथा सीजियम का उपयोग होता है, क्योंकि इनकी आयनन ऊर्जा अत्यन्त कम होती है।

प्रश्न 14.
सोडियम क्लोराइड से सोडियम का निष्कर्षण सामान्य अपचायक से क्यों नहीं किया जा सकता है ?
उत्तर:
सोडियम प्रबल अपचायक है। विद्युत् रासायनिक श्रेणी में इसका स्थान सबसे ऊँचा है। इससे प्रबल अपचायक उपलब्ध न होने से इसे सामान्य अपचायकों द्वारा अपचयित नहीं किया जा सकता है इसे केवल विद्युत् अपघटन द्वारा ही अपचयित किया जाता है।

प्रश्न 15.
Li और Be सहसंयोजी यौगिक बनाने की प्रवृत्ति रखते हैं। समझाइये।
उत्तर:
Li और Be परमाणु का आकार बहुत छोटा होता है और उनकी आयनन ऊर्जा अधिक होती है। अतः इनके संयोजकता कोश के इलेक्ट्रॉन नाभिक से दृढ़ता से जुड़े रहते हैं। साथ ही साथ आयनों की ध्रुवण क्षमता उच्च आवेश घनत्व के कारण अधिक होती है। इसलिये Li और Be सहसंयोजी यौगिक बनाने की प्रवृत्ति रखते हैं।

प्रश्न 16.
Li की तुलना में K और Cs का उपयोग फोटो रासायनिक सेल में करते हैं। क्यों?
उत्तर:
Cs तथा K के बड़े आकार के कारण इनकी आयनन ऊर्जा Li की तुलना में अत्यन्त कम है इसलिये सरलता से इलेक्ट्रॉन का त्याग कर सकते हैं। इसलिये फोटो रासायनिक सेल में Li की तुलना में K तथा Cs का उपयोग किया जाता है।

प्रश्न 17.
BeCl₂, को वायुमण्डल में रखने पर यह सफेद धूम्र देता है। क्यों?
उत्तर:
BeCl₂ आर्द्रता ग्राही होता है। यह वायुमण्डल में उपस्थित नमी को अवशोषित करके जल अपघटित हो जाता है और HCl गैस उत्पन्न करता है। HCl गैस बनने के कारण ही सफेद धूम्र प्राप्त होते हैं।
BeCl₂ + 2H₂O → Be(OH)₂ + 2HCl

प्रश्न 18.
प्रथम वर्ग में ऊपर से नीचे आने पर तत्वों की कठोरता बढ़ती जाती है, क्यों?
उत्तर:
प्रथम वर्ग में ऊपर से नीचे आने पर तत्वों के आकार में वृद्धि के साथ-साथ इनके घनत्वों में भी वृद्धि होती है और उनके परमाणुओं के मध्य आकर्षण बल भी बढ़ता जाता है। जिससे तत्वों की कठोरता बढ़ती जाती है।

s-ब्लॉक तत्त्व लघु उत्तरीय प्रश्न

प्रश्न 1.
s – ब्लॉक तत्व तथा p – ब्लॉक तत्वों को प्रतिनिधि तत्व कहते हैं। क्यों ?
उत्तर:
वे तत्व जिनके बाहरी कोश का इलेक्ट्रॉनिक विन्यास ns^1-2 तथा ns²np^1-6 होता है, प्रतिनिधि तत्व कहलाते हैं। क्योंकि इस समूह में उपस्थित प्रत्येक तत्व अपने समूह के गुणों का प्रतिनिधित्व करते हैं तथा प्रत्येक समूह के गुण दूसरे समूह के गुण से पूर्णतः भिन्न होते हैं।
(1) s – ब्लॉक तत्व – वे तत्व जिनका अंतिम इलेक्ट्रॉन s – उपकोश में प्रवेश करता है, s – ब्लॉक तत्व कहलाते हैं। इनका सामान्य इलेक्ट्रॉनिक विन्यास ns^1-2 होता है।
(2) p – ब्लॉक तत्व-वे तत्व जिनका इलेक्ट्रॉनिक विन्यास करने पर अंतिम इलेक्ट्रॉन p – उपकोश में प्रवेश करता है, p – ब्लॉक तत्व कहलाते हैं। इनका सामान्य इलेक्ट्रॉनिक विन्यास ns²np^1-6 होता है।

प्रश्न 2.
क्षार धातु प्रबल अपचायक होते हैं, क्यों?
उत्तर:
क्षार धातुओं के बड़े आकार के कारण इनकी आयनन ऊर्जा कम होती है। जिसके कारण ये संयोजी कोश के इलेक्ट्रॉन को आसानी से त्याग कर ऑक्सीकृत हो जाते हैं तथा इनके मानक इलेक्ट्रोड विभव का मान अधिक ऋणात्मक होता है। इसलिये ये इलेक्ट्रॉन का त्याग कर सरलता से M⁺ आयन बनाते हैं। इसलिये ये प्रबल अपचायक की तरह कार्य करते हैं।

प्रश्न 3.
BeSO₄ तथा MgSO₄ जल में शीघ्र विलेय है जबकि CaSO₄ SrSO₄ तथा BaSO₄ अविलेय है। क्यों?
उत्तर:
क्षारीय मृदा धातुओं की जालक ऊर्जा, सल्फेट आयन के वृहद् आकार के कारण लगभग समान होती है। अतः इनकी विलेयता जलयोजन ऊर्जा पर निर्भर करती है, जो वर्ग में नीचे जाने पर क्रमशः घटती है। Be²⁺ तथा Mg²⁺ आयनों की उच्च जलयोजन एन्थैल्पी, जालक एन्थैल्पी कारक को हीन कर देती है जिसके कारण इनके सल्फेट जल में विलेय होते हैं। दूसरी ओर Ca²⁺, Sr²⁺ तथा Ba²⁺ आयनों के लिए जलयोजन एन्थैल्पी कम होती है। अत: यह जालक एन्थैल्पी कारक को हीन नहीं कर पाती है। अतः इनके सल्फेट जल में अविलेय होते हैं।

प्रश्न 4.
जलीय विलयन में लीथियम की अपचायक क्षमता अधिक क्यों होती है ?
उत्तर:
किसी तत्व के जलीय विलयन में इलेक्ट्रॉनों की त्यागने की क्षमता का मापन इलेक्ट्रोड विभव द्वारा करते हैं। यह मुख्यतः निम्नलिखित तीन कारकों पर निर्भर करता है –

अपने आयनों के छोटे आकार के कारण, लीथियम की जलयोजन एन्थैल्पी सर्वाधिक होती है। यद्यपि Li की आयनन एन्थैल्पी, क्षार धातुओं से सर्वाधिक होती है परन्तु जलयोजन एन्थैल्पी भी आयनन एन्थैल्पी से अधिक होती है। अतः उच्च जलयोजन एन्थैल्पी के कारण, लीथियम जलीय विलयन में प्रबलतम अपचायक होता है।

प्रश्न 5.
प्रकाश वैद्युत सेल में लीथियम के स्थान पर पोटैशियम एवं सीजियम क्यों प्रयुक्त किए जाते हैं ?
उत्तर:
पोटैशियम तथा सीजियम की आयनन एन्थैल्पी लीथियम की अपेक्षा अधिक कम होती है। अतः ये धातुएँ प्रकाश में रखने पर इलेक्ट्रॉन आसानी से उत्सर्जित करती है, परन्तु लीथियम ऐसा नहीं कर पाती है। यही कारण है Li की अपेक्षा K तथा Cs का प्रयोग प्रकाश वैद्युत सेल में किया जाता है।

प्रश्न 6.
क्या कारण है कि क्षार धातुएँ M⁺ प्रकार का धनायन बनाती है, M⁺² प्रकार का धनायन नहीं?
उत्तर:
क्षार धातुओं के संयोजी कोश में 1 इलेक्ट्रॉन होता है। इनके बड़े आकार के कारण इनकी आयनन ऊर्जा अत्यन्त कम होती है। इसलिये ये सरलता से इलेक्ट्रॉन का त्याग कर M⁺ आयन बनाते हैं। इस M⁺¹ आयनिक अवस्था में इनका इलेक्ट्रॉनिक विन्यास अक्रिय गैसों के इलेक्ट्रॉनिक विन्यास के समान अत्यन्त स्थायी होता है। इसलिये इस अवस्था में ये रासायनिक दृष्टि से निष्क्रिय होते हैं और इसकी आयनन ऊर्जा अत्यन्त उच्च होती है। इसी कारण ये M⁺² आयन नहीं बनाते।

प्रश्न 7.
क्षार धातुओं में कौन-सी धातु प्रबल अपचायक है तथा क्यों ?
उत्तर:
किसी भी तत्व की अपचायक प्रवृत्ति उसके मानक इलेक्ट्रोड विभव पर निर्भर करती है। वे तत्व जिनका मानक इलेक्ट्रोड विभव ऋणात्मक होता है अपचायक की तरह कार्य करते हैं तथा मानक इलेक्ट्रोड विभव का मान जितना अधिक ऋणात्मक होता है वह तत्व उतना प्रबल अपचायक होता है। Li का मानक इलेक्ट्रोड विभव अत्यधिक उच्च ऋणात्मक मान दर्शाता है। इसलिये यह प्रबल अपचायक है।

प्रश्न 8.
क्षार धातु प्रकृति में मुक्त अवस्था में नहीं प्राप्त होते हैं, क्यों? अथवा, क्षार धातु सदैव आयनिक यौगिक का निर्माण करते हैं, क्यों?
उत्तर:
क्षार धातुओं के बड़े आकार के कारण इनकी आयनन ऊर्जा अत्यन्त कम होती है। इसलिये यह सरलता से इलेक्ट्रॉन का त्याग करके धनायन बना सकते हैं। अर्थात् प्रबल धनविद्युती तथा अत्यधिक क्रियाशील होने के कारण वायुमण्डल में उपस्थित ऋणविद्युती तत्व, जैसे-नमी, CO₂ के साथ सरलता से अभिक्रिया करके आयनिक यौगिकों का निर्माण करते हैं। इसलिये प्रकृति में मुक्त अवस्था में प्राप्त नहीं होते हैं।

प्रश्न 9.
क्षार धातु सरलता से ज्वाला परीक्षण देते हैं। क्यों ?
उत्तर:
क्षार धातुओं के बड़े आकार के कारण इनकी आयनन ऊर्जा के मान अत्यन्त कम होते हैं। इसलिये इन क्षारीय धातुओं तथा इनके यौगिकों को जब बुन्सन ज्वाला में गर्म किया जाता है तो संयोजी कोश का इलेक्ट्रॉन उत्तेजित होकर उच्च ऊर्जा स्तर में चला जाता है। कुछ समय पश्चात् यह इलेक्ट्रॉन अतिरिक्त ऊर्जा को दृश्य प्रकाश के रूप में प्रकीर्णित कर अपनी मूल स्थिति में वापस आ जाता है । इस प्रकार प्रकाश के प्रकीर्णन के कारण यह ज्वाला को विशिष्ट रंग प्रदान करते हैं।

प्रश्न 10.
Be तथा Mg ज्वाला परीक्षण नहीं देते हैं। क्यों ?
उत्तर:
Be तथा Mg में s – कक्षक पूर्ण कक्षक के रूप में होता है। इनके छोटे आकार तथा :-कक्षक के पूर्ण कक्षक होने की वजह से इनका स्थायित्व अधिक होता है जिसके कारण इनकी आयनन ऊर्जा उच्च होती है। जिसके कारण इलेक्ट्रॉन को बुन्सन ज्वाला द्वारा उत्तेजित करना संभव नहीं है। दूसरे शब्दों में, दृश्य प्रकाश द्वारा विकिरण संभव नहीं है। इसलिये Mg तथा Be ज्वाला परीक्षण नहीं देते हैं।

प्रश्न 11.
क्षार धातुएँ द्रव अमोनिया में घुलकर नीला विलयन बनाती है, जो प्रबल विद्युत् चालक होते हैं। समीकरण सहित कारण बताइये।
उत्तर:
क्षार धातुओं का द्रव अमोनिया में विलयन अमोनीकृत इलेक्ट्रॉन की उपस्थिति के कारण नीले रंग का होता है। इस विलयन की चालकता अमोनीकृत इलेक्ट्रॉन एवं अमोनिया युक्त धनायन दोनों की उपस्थिति के कारण होती है।
M + (x + y)NH₃ → [M(NH₃)_x]⁺ + [e(NH₃)_y]^–

प्रश्न 12.
Na क्षारीय है अथवा Na₂O, स्पष्ट कीजिये।
उत्तर:
Na₂O, Na की तुलना में अधिक क्षारीय है क्योंकि Na₂O जल से क्रिया करके NaOH बनाते हैं जबकि Na भी जल से अभिक्रिया करके NaOH बनाता है लेकिन पहले वह Nago बनाता है फिर NaOH

• 4Na + 2H₂O → 2Na₂O + 2H₂
• Na₂O + H₂O → 2NaOH

प्रश्न 13.
क्षार धातुएँ तथा क्षारीय मृदा धातुओं की तुलना निम्नलिखित बिन्दुओं के आधार परकीजिये –

1. N₂ के साथ क्रिया
2. कार्बोनेट पर ऊष्मा का प्रभाव
3. सल्फेटों की जल में विलेयता।

उत्तर:
क्षार धातुएँ तथा क्षारीय मृदा धातु में तुलना –

प्रश्न 14.
क्षार धातुओं को यदि वायुमंडल में खुला रखा जाये तो कुछ समय पश्चात् उनकी धात्विक चमक नष्ट हो जाती है। क्यों?
उत्तर:
प्रत्येक धातु में एक विशिष्ट चमक होती है जिसे धात्विक चमक कहते हैं । क्षार धातुओं के बड़े आकार के कारण इनकी आयनन ऊर्जा अत्यधिक कम होती है। इसलिये यह अत्यधिक क्रियाशील तथा प्रबल धनविद्युती होते हैं तथा वायुमण्डल में उपस्थित नमी, O₂ तथा CO₂ के साथ अभिक्रिया करके क्षारीय कार्बोनेट बनाते हैं। धातुओं की सतह पर ऑक्साइड तथा कार्बोनेट की पर्त बनने के कारण इनकी धात्विक चमक नष्ट हो जाती है।

प्रश्न 15.
क्षार धातु एवं क्षारीय मृदा धातुओं में प्रमुख अंतर लिखिये।
उत्तर:
क्षार धातु एवं क्षारीय मृदा धातुओं में प्रमुख अंतर –

क्षार धातु:

• ये + 1 संयोजकता दर्शाते हैं।
• इनके हाइड्रॉक्साइड प्रबल क्षार होते हैं।
• इनके कार्बोनेट, सल्फेट, फॉस्फेट जल में विलेय होते हैं।
• इनकी आयनन ऊर्जा का मान अपेक्षाकृत कम होता है।
• ये चमकदार, आघातवर्धनीय एवं तन्य होते हैं।

क्षारीय मृदा धातु:

• ये + 2 संयोजकता दर्शाते हैं।
• इनके हाइड्रॉक्साइड दुर्बल क्षार होते हैं, क्षार धातु की तुलना में।
• इनके यौगिक जल में अविलेय होते हैं।
• इनकी आयनन ऊर्जा का मान उच्च होता है।
• इनके ये गुण क्षार धातु की तुलना में अपेक्षाकृत कम होते हैं।

प्रश्न 16.
LiCl और RbCl में कौन प्रबल आयनिक होगा और क्यों ?
उत्तर:
LiCl की तुलना में RbCl प्रबल आयनिक यौगिक है। क्योंकि Li के छोटे आकार के कारण इसकी आयनन ऊर्जा अत्यधिक उच्च होती है। इसलिये यह सहसंयोजी यौगिक बनाता है जबकि Rb के बड़े आकार के कारण इसकी आयनन ऊर्जा अत्यन्त कम होती है। इसलिये यह सरलता से इलेक्ट्रॉन का त्याग करके धनायन बनाता है। इसलिये इसके यौगिक प्रबल आयनिक प्रवृत्ति दर्शाते हैं।

प्रश्न 17.
क्षार धातु और क्षारीय मृदा धातु में किसके कार्बोनेट जल में विलेय हैं और क्यों?
अथवा
क्षार धातु के कार्बोनेट जल में विलेय हैं जबकि क्षारीय मृदा धातु के कार्बोनेट जल में अविलेय हैं। क्यों?
उत्तर:
क्षार धातु के कार्बोनेट जल में विलेय है क्योंकि इनकी जलयोजन ऊर्जा, जालक ऊर्जा से अधिक होती है जबकि क्षारीय मृदा धातु के छोटे आकार तथा उच्च आवेश घनत्व के कारण इनकी जालक ऊर्जा उच्च तथा जलयोजन ऊर्जा से अधिक होती है। इसलिये क्षारीय मृदा धातु के कार्बोनेट जल में अविलेय है।

प्रश्न 18.
क्षारीय मृदा धातुओं के द्वितीय आयनन विभव का मान प्रथम आयनन विभव से अधिक है फिर भी क्षारीय मृदा धातु + 2 ऑक्सीकरण संख्या दर्शाते हैं + 1 नहीं, क्यों?
उत्तर:
क्षारीय मृदा धातु + 2 आयन का निर्माण करते हैं + 1 आयन का नहीं, क्योंकि

• इनके संयोजी कोश में 2 इलेक्ट्रॉन होते हैं ये तत्व स्थायी इलेक्ट्रॉनिक विन्यास प्राप्त करने के लिये 2e का त्याग कर M⁺² आयन का निर्माण करते हैं।
• इस द्विसंयोजी आयन के निर्माण के दौरान जालक ऊर्जा मुक्त होने लग जाती है जो द्वितीय आयनन ऊर्जा के मान को कम कर देती है। इसलिये ये सरलता से द्विसंयोजी आयन का निर्माण करते हैं।

प्रश्न 19.
BeCl₂ तथा अन्य क्षारीय मृदा धातुओं के क्लोराइडों में असमानता बताइए।
उत्तर:

• BeCl₂ सहसंयोजी यौगिक है जबकि अन्य क्षारीय मृदा धातुओं के क्लोराइड आयनिक है।
• BeCl₂ जल में अविलेय है जबकि अन्य क्षारीय मृदा धातुओं के क्लोराइड जल में विलेय है।
• BeCl₂ कार्बनिक विलायकों में विलेय है जबकि अन्य क्षारीय मृदा धातुओं के क्लोराइड कार्बनिक विलायकों में विलेय है।
• BeCl₂ के गलनांक, क्वथनांक निम्न हैं जबकि अन्य क्षारीय मृदा धातुओं के क्लोराइड के गलनांक, क्वथनांक उच्च हैं।

प्रश्न 20.
BaSO₄ की विलेयता CaSO₄ से कम है। क्यों?
उत्तर:
क्षारीय मृदा धातुओं के सल्फेटों की विलेयता समूह में ऊपर से नीचे आने पर कम होती है। क्योंकि जालक ऊर्जा तो लगभग समान रहती है। परन्तु समूह में ऊपर से नीचे आने पर परमाणविक त्रिज्या में वृद्धि के कारण जलयोजन ऊर्जा में कमी आती है। जलयोजन ऊर्जा का मान जालक ऊर्जा से कम होने लगता है जिसके कारण विलेयता में कमी आती है।

प्रश्न 21.
Be की आयनन ऊर्जा B से अधिक है। क्यों?
उत्तर:

Be में s – कक्षक पूर्ण कक्षक है। जबकि B में p कक्षक अपूर्ण कक्षक है। अर्धपूर्ण तथा पूर्ण कक्षक अपूर्ण कक्षक की तुलना में अधिक स्थायी होते हैं तथा इनमें से इलेक्ट्रॉन को निकालने के लिये अत्यधिक ऊर्जा की आवश्यकता होती है। इसलिये Be की आयनन ऊर्जा बोरॉन से अधिक है।

प्रश्न 22.
सोडियम कार्बोनेट को अमोनिया सोडा विधि से कैसे बनाया जाता है ? इसके सिद्धान्त को लिखिये।
उत्तर:
इस विधि में पहले NaCl के सान्द्र विलयन को NH₃ द्वारा संतृप्त करते हैं । जिससे अमोनियामय सोडियम क्लोराइड बनाता है।
इस अमोनियामय सोडियम क्लोराइड विलयन में CO₂ गैस प्रवाहित करते हैं। जिससे अमोनियम बाइकार्बोनेट बनता है जो सोडियम क्लोराइड से क्रिया करके सोडियम बाइ-कार्बोनेट बनाता है।

• NH₃ + CO₂ + H₂O → NH₄HCO₃
• NH₄HCO₃ + NaCl → NaHCO₃ + NH₄Cl

सोडियम बाइ-कार्बोनेट अल्प विलेय होने से अवक्षेप के रूप में नीचे बैठ जाता है। इसे छानकर निस्तापित करने पर सोडियम कार्बोनेट बना लेता है।
2NaHCO₃ → Na₂CO₃ + H₂O + CO₂

प्रश्न 23.
सोडियम कार्बोनेट से –

• सोडियम बाइ-कार्बोनेट
• सोडियम हाइड्रॉक्साइड
• सोडियम सिलीकेट कैसे प्राप्त करते हैं ?

उत्तर:

• सोडियम कार्बोनेट के जलीय विलयन में CO₂ गैस प्रवाहित करने पर सोडियम बाइकार्बोनेट का सफेद अवक्षेप प्राप्त होता है।
  NaCO₃ + H₂O + CO₂ → 2NaHCO₃
• सोडियम कार्बोनेट को चूने के पानी के साथ उबालने पर सोडियम हाइड्रॉक्साइड बनता है।
  Na₄CO₃ + Ca(OH)₂ → 2NaOH + CaCO₃
• सोडियम कार्बोनेट को सिलिका के साथ गर्म करने पर सोडियम सिलिकेट बनता है।
  Na₂CO₃ + SiO₂ → Na₂SiO₃ + CO₂

प्रश्न 24.
बेकिंग सोडा क्या है ? इसे बनाने की विधि, गुण तथा उपयोग लिखिए।
उत्तर:
परिभाषा-सोडियम हाइड्रोजन कार्बोनेट को बेकिंग सोडा कहते हैं तथा इसका सूत्र NaHCO, है।
बनाने की विधि –

• अमोनिया सोडा विधि में सोडियम बाइ-कार्बोनेट माध्यमिक यौगिक के रूप में मिलता है।
  NH₄HCO₃ + NaCl + NaHCO₃ + NH₄Cl
• सोडियम कार्बोनेट विलयन में CO₂ गैस प्रवाहित करने पर सोडियम बाइकार्बोनेट बनता है।
  NaCO₃ + H₂O + CO₂ → 2NaHCO₃

गुण:
सफेद क्रिस्टलीय पदार्थ जल में अल्प विलेय, जलीय विलयन दुर्बल क्षारीय।
उपयोग:

• बेकिंग पाउडर बनाने में
• पेट की अम्लीयता कम करने की दवा में
• आग बुझाने के यंत्र में।

प्रश्न 25.
सोडियम कार्बोनेट बनाने की ली-ब्लॉक विधि का संक्षिप्त विवरण देते हुये समझाइये कि ली ब्लॉक-विधि से सॉल्वे विधि अच्छी क्यों है ?
उत्तर:
ली-ब्लॉक प्रक्रम – यह प्रक्रम तीन पदों में पूर्ण होता है।
(1) नमक को सान्द्र H₂SO₄ के साथ गर्म करने पर सोडियम सल्फेट (साल्ट केक) बनता है।

(2) पिसे हुये साल्ट केक, चूने पत्थर और कोक के मिश्रण को गर्म करने पर सोडियम कार्बोनेट के साथ कैल्सियम सल्फाइड बनता है। CaCO₃,Na₂CO₃ और Cas के इस मिश्रण को काली राख कहा जाता है।

(3) बारीक पिसी काली राख को जल के साथ उबालने पर Na₂CO₃ विलेय हो जाता है। अविलेय Cas और CaCO₃ के छानकर अलग कर देते हैं। छनित को वाष्पित करके ठोस Na₂CO₃प्राप्त कर लेते हैं।

सॉल्वे विधि की ली-ब्लॉक से श्रेष्ठता –

• सॉल्वे विधि सस्ती है।
• सॉल्वे विधि में शुद्ध Na₂CO₃ बनता है।
• सॉल्वे विधि में कोई हानिकारक धूम नहीं निकलते।
• सॉल्वे विधि में बीच में NaHCO₃ भी बनता है जो एक उपयोगी यौगिक है।

प्रश्न 26.
सोडियम कार्बोनेट की हारग्रीव-बर्ड सेल विधि का वर्णन कीजिये।
उत्तर:
हारग्रीव बर्ड सेल में कार्बन का एनोड तथा छिद्रयुक्त कॉपर का कैथोड होता है तथा इन्हें ऐस्बेस्टॉस झिल्ली द्वारा पृथक् रखा जाता है। NaCl विलयन का वैद्युत अपघटन कराने पर सोडियम तथा क्लोरीन बनते हैं । सेल में ऐस्बेस्टॉस के बाहरी ओर भाप और CO₂भेजी जाती है, जो Na से अभिक्रिया करके Na₂CO₃ बनाते हैं। इन विलयन का वाष्पन करने पर Na₂CO₃.10H₂O प्राप्त होता है।

• 2NaCl – 2Na⁺ + 2Cl^–
• 2Na⁺ + 2e^– → 2Na
• कैथोड पर – 2Na + 2H₂O → 2NaOH + H₂.
• 2NaOH + CO₂ →Na₂CO₃ + H₂O

प्रश्न 27.
सोडियम हाइड्रॉक्साइड बनाने की नेल्सन सेल विधि का चित्र सहित वर्णन नमक का ग्रेफाइट का ऐनोड कीजिये।
उत्तर:
नेल्सन सेल इस्पात की टंकी में ऐस्बेस्टॉस बेलनाकार ऐस्बेस्टॉस की तह लगी इस्पात की की खोल छिद्रयुक्त नली लगाकर बनाया जाता है। इस्पात की छिद्र युक्त नली कैथोड का कार्य करती है। इस नली में NaCl विलयन भरकर इस्पात की नली में लटका इस्पात का कैथोड देते हैं। कार्बन की छड़ एनोड का कार्य करती है। विद्युत् अपघटन पर सोडियम आयन मुक्त होता है NaOH का विलयन जो ऐस्बेस्टॉस की तह को पार कर कैथोड पर मुक्त होने के बाद टंकी में आने वाली भाप से क्रिया कर NaOH बनाता है।

2NaCl ⇌ 2Na⁺ + 2Cl^–
कैथोड पर – 2Na⁺ + 2e →2Na
2Na + 2H₂O → 2NaOH + H₂

प्रश्न 28.
बुझा हुआ चूना बनाने की विधि, गुण तथा उपयोग लिखिए।
उत्तर:
(1) अनबुझे चूने पर पानी का छिड़काव करने पर बुझा चूना प्राप्त होता है।
CaO + H₂O → Ca(OH)₂

(2) कैल्सियम लवणों पर क्षार की अभिक्रिया कराने पर Ca(OH)₂ प्राप्त होता है।
Ca(NO₃)₂ + 2NaOH → Ca(OH)₂ + 2NaNO₃

गुण:
(1) इसे 400°C तक गर्म करने पर कैल्सियम ऑक्साइड बनता है।

(2) चूने के पानी में CO₂ प्रवाहित करने पर CaCO₃ का सफेद दूधिया अवक्षेप बनता है।
Ca(OH)₂ + CO₂ → CaCO₃ + H₂ O
(3) चूने के पानी में देर तक CO₂ प्रवाहित करने पर Ca(HCO₃ )₂ बनने के कारण दूधियापन समाप्त हो जाता है।
CaCO₃ + CO₂ + H₂O → Ca(HCO₃ )₂
(4) शुष्क बुझे चूने पर Cl₂ गैस प्रवाहित करने पर ब्लीचिंग पाउडर प्राप्त होता है।
Ca(OH)₂ + Cl₂ → CaoCl₂ + H₂ O
उपयोग:

• कॉस्टिक सोडा तथा विरंजक चूर्ण के निर्माण में।
• कोल गैस के शोधन में।
• दीवारों पर सफेदी करने में।
• अमोनिया के निर्माण में।

प्रश्न 29.
सोडा लाइम विधि से सोडियम हाइड्रॉक्साइड कैसे बनाते हैं ?
उत्तर:
सोडियम कार्बोनेट के 10 – 20% विलयन को बुझे हुये चूने की उचित मात्रा के साथ भाप के द्वारा 84-85°C ताप गर्म करने पर कैल्सियम कार्बोनेट एवं कास्टिक सोडा बनता है। कास्टिक सोडा विलयन में रहता है जबकि CaCO₃ का सफेद अवक्षेप प्राप्त होता है। विलयन को छानकर वाष्पन करने या 98% शुद्ध ठोस कास्टिक सोडा प्राप्त होता है।
Na₂ CO₃ + Ca[OH]₂ → 2NaOH + CaCO₃

s-ब्लॉक तत्त्व दीर्घ उत्तरीय प्रश्न

प्रश्न 1.
निम्नलिखित में से प्रत्येक प्रेक्षण पर टिप्पणी लिखिए –
(a) जलीय विलयनों में क्षार धातु आयनों की गतिशीलता Li ⁺ > Na⁺ > K⁺ > Rb⁺ > Cs⁺
(b) लीथियम ऐसी एकमात्र क्षार धातु है, जो नाइट्राइट बनाती है।
(c) M²⁺_(aq) + 2e^– →M_(s) हेतु E° ( जहाँ, M = Ca, Sr या Ba) लगभग स्थिरांक है।
उत्तर:
(a) आयन का आकार जितना कम होता है, जलयोजन उतना ही अधिक होता है तथा आयन का जलयोजन जितना अधिक होता है, उसकी आयनिक गतिशीलता उतनी ही कम होती है। अतः जलयोजन क्षमता का क्रम निम्न होगा –
Li ⁺ > Na⁺ > K⁺ > Rb⁺ > Cs⁺
अतः आयनिक गतिशीलता विपरीत क्रम में बढ़ेगी
(b) अपने छोटे आकार के कारण, क्षार धातुओं में केवल लीथियम नाइट्राइड बनाती है।

(c) M²⁺_(aq) + 2e^– →M_(s)
जहाँ, M = Ca, Sr, Ba के लिए E° का मान लगभग समान होता है। किसी भी M²⁺/ M इलेक्ट्रोड के लिए E° का मान निम्नलिखित तीन कारकों पर निर्भर करता है –

• वाष्पन एन्थैल्पी
• आयनन एन्थैल्पी
• जलयोजन एन्थैल्पी।

चूँकि इन तीनों कारकों का संयुक्त प्रभाव Ca, Sr तथा Ba के लिए लगभग समान रहता है। अतः इन इलेक्ट्रोड विभव का मान भी लगभग स्थिर रहता है।

प्रश्न 2.
लीथियम अपने समूह के अन्य सदस्यों से अपसामान्य व्यवहार दर्शाता है, क्यों?
उत्तर:
Li अपने समूह के अन्य सदस्यों से अपसामान्य व्यवहार दर्शाता है, क्योंकि –

• इसके परमाणु तथा आयन का आकार छोटा होता है।
• आयनन ऊर्जा उच्च है।
• इलेक्ट्रॉनबंधुता उच्च है।
• d – कक्षक अनुपस्थित है।

अपसामान्य व्यवहार:

• Li अन्य क्षार-धातुओं की तुलना में अधिक कठोर है।
• Li के गलनांक तथा क्वथनांक अन्य क्षार धातुओं की तुलना में उच्च है।
• Liसहसंयोजी यौगिक बनाता है जबकि समूह के अन्य सदस्य आयनिक यौगिक बनाते हैं।
• Li केवल ऑक्साइड बनाता है जबकि अन्य क्षार धातुएँ परॉक्साइड तथा सुपर ऑक्साइड भी बनाते हैं।
• लीथियम हाइड्रॉक्साइड समूह के अन्य धातुओं के हाइड्राइड की तुलना में अधिक स्थायी है।
• लीथियम हाइड्रॉक्साइड दुर्बल क्षार है जबकि अन्य क्षार धातुओं के हाइड्रॉक्साइड प्रबल क्षार है।
• Li नाइट्रोजन के साथ संयोग कर नाइट्राइड बनाता है जबकि समूह की अन्य धातु नाइट्रोजन से संयोग नहीं करती है।
• लीथियम नाइट्रेट गर्म करने पर विघटित होकर Li₂O देता है जबकि अन्य क्षार धातु के नाइट्रेट गर्म करने पर अपघटित होकर नाइट्राइट देते हैं।
  1. 4LiNO₃ → 2Li₂O + 4NO₂ + O₂
  2. 2NaNO₃ → 2NaNO₂ + O₂
• Li₂CO₃ गर्म करने पर विघटित हो जाता है जबकि अन्य क्षार धातु के कार्बोनेट गर्म करने पर अपघटित नहीं होते हैं।
  

प्रश्न 3.
विकर्ण संबंध क्या है ? Li तथा Mg में विकर्ण संबंध लिखिये।
उत्तर:
विकर्ण संबंध:
द्वितीय आवर्त के कुछ तत्व तृतीय आवर्त के कुछ तत्व के साथ विकर्ण में समानता दर्शाते हैं जिसे विकर्ण संबंध कहते हैं।
Liतथा Mg में विकर्ण संबंध:

• लीथियम की परमाणु त्रिज्या 1.34A तथा Mg की परमाणु त्रिज्या 1.36A लगभग बराबर है।
• Liव Mg⁺² की ध्रुवण क्षमता लगभग समान है।
• Li व Mg दोनों कठोर होते हैं।
• Li व Mg की ऋण विद्युतता (1.0 और 1.2) बराबर होती हैं।
• Li तथा Mg दोनों के गलनांक तथा क्वथनांक उच्च हैं।
• Li तथा Mg दोनों N, के साथ अभिक्रिया करके नाइट्राइड बनाते हैं।
• Li तथा Mg दोनों ऑक्सीजन के साथ संयोग करके मोनो ऑक्साइड देते हैं।
• Li तथा Mg दोनों जल को विघटित करके H₂ देते हैं।
• Li तथा Mg के कार्बोनेट गर्म करने पर CO₂ गैस देते हैं।
• LiOH तथा Mg (OH)₂ दोनों दुर्बल क्षार हैं।।

प्रश्न 4.
क्षार धातु तथा क्षारीय मृदा धातु में क्या समानता है ?
उत्तर:

• क्षार धातु तथा क्षारीय मृदा धातु प्रकृति में मुक्त अवस्था में प्राप्त नहीं होते।
• क्षार धातु तथा क्षारीय मृदा धातु के ऑक्साइड जल में विलेय होकर प्रबल क्षार का निर्माण करते हैं।
• क्षार धातु तथा क्षारीय मृदा धातु कोमल तथा चमकीली होती हैं।
• इन्हें वायु में रखने पर इनकी सतह मलिन हो जाती है।
• क्षार धातु तथा क्षारीय मृदा धातु (Be तथा Mg को छोड़कर) ज्वाला परीक्षण देते हैं।
• क्षार धातु तथा क्षारीय मृदा धातु प्रबल अपचायक है।
• क्षार धातु तथा क्षारीय मृदा धातु दोनों आयनिक यौगिक का निर्माण करते हैं।
• दोनों के नाइट्रेट तथा हैलाइड जल में विलेय हैं।

प्रश्न 5.
Be अपने समूह के अन्य सदस्यों की तुलना में अपसामान्य व्यवहार दर्शाता है, क्यों?
उत्तर:
Be अपने समूह के अन्य सदस्यों से अपसामान्य व्यवहार दर्शाता है, क्योंकि

• इसके परमाणु तथा आयन का आकार छोटा होता है।
• आयनन ऊर्जा अत्यधिक उच्च होती है।
• इलेक्ट्रॉनबंधुता उच्च होती है।
• d-कक्षक की अनुपस्थिति।

अपसामान्य व्यवहार:

• Be कठोर है जबकि इस समूह के अन्य सदस्य कोमल धातु होती है।
• अन्य क्षारीय मृदा धातु की तुलना में Be के गलनांक तथा क्वथनांक उच्च होते हैं।
• Be के यौगिक सहसंयोजी होते हैं। जबकि अन्य क्षारीय मृदा धातु के यौगिक आयनिक होते हैं।
• Be अम्लों से सरलता से H₂ मुक्त नहीं करता जबकि अन्य क्षारीय मृदा धातु शीघ्रता से H₂ मुक्त करती है।
• बेरीलियम कार्बाइड जल अभिक्रिया कराने पर मेथेन देता है, जबकि अन्य सदस्य एसीटिलीन देते हैं ।
  Be₂C + 2H₂O → 2BeO + CH₄
  CaC₂ + 2H₂0 → Ca(OH)₂ + C₂H,₂
• BeO उभयधर्मी है जबकि अन्य सदस्यों के ऑक्साइड क्षारीय होते हैं।
• Be व Mg को छोड़कर सभी सदस्य ज्वाला परीक्षण देते हैं।
• Be गर्म करने पर भी जल के साथ कोई अभिक्रिया नहीं दर्शाता जबकि अन्य क्षारीय मृदा धातु जल के साथ सरलता से अभिक्रिया दर्शाते हैं।
• Be हाइड्रोजन के साथ मंद गति से अभिक्रिया करता है, जबकि अन्य शीघ्रता से अभिक्रिया करते हैं।

प्रश्न 6.
Be व AI में विकर्ण संबंध लिखिये।
उत्तर:

• Be व AI की परमाणविक त्रिज्या तथा आयनिक त्रिज्या लगभग बराबर है।
• दोनों सहसंयोजी यौगिक बनाते हैं।
• दोनों धातुएँ दुर्बल विद्युती धनी प्रकृति के होते हैं।
• दोनों धातुओं की सान्द्र HNO, से क्रिया कराने पर ये निष्क्रिय होते हैं।
• दोनों शीघ्रता से हाइड्राइड नहीं बनाती।
• Be तथा AI दोनों के ऑक्साइड उभयधर्मी प्रकृति के होते हैं।
  1. BeO +2HCl→ BeCl₂ + H₂O
  2. BeO + 2NaOH → Na₂BeO₂ + H₂O
  3. Al₂O₃+6HCl → 2AlCl₃ + 3H₂O
  4.  Al₂0₃ + 2NaOH → 2NaAlO₂ + H₂O
• दोनों धातुओं के कार्बाइड जल से क्रिया करके मीथेन देते हैं।
  Be₂C + 2H₂O → 2BeO + CH₂
  Al₄C₃ + 6H₂O → 2AI₂O₃ + 3CH₄
• BeCl₂ तथा AlCl₃ द्विलक तथा बहुलक के रूप में मिलते हैं।
• Be तथा A1 के ऑक्साइड दुर्बल क्षार हैं।
• BeCl₂तथा AlCl₃ प्रबल लुईस अम्ल हैं।

प्रश्न 7.
सॉल्वे विधि द्वारा सोडियम कार्बोनेट का निर्माण किस प्रकार किया जाता है ? नामांकित रेखाचित्र खींचिए एवं समीकरण लिखिये।
उत्तर:
जब अमोनियामय NaCl विलयन में CO₂ गैस प्रवाहित करते हैं तो अमोनियम बाइकार्बोनेट बनता है। जो NaCl से अभिक्रिया करके सोडियम बाइकार्बोनेट बनाता है।

• NH₃ + CO₂ + H₂O → NH₄HCO₃
• NH₄HCO₃ + NaCl → NaHCO₃ + NH₄Cl

सोडियम बाइकार्बोनेट अल्प विलेय होने से अवक्षेपित होकर नीचे बैठ जाता है। इसे छानकर निस्तापित करने पर Na₂co₃ प्राप्त होता है।
2NaHCO₃ → Na₂CO₃ + H₂O + CO₂

उपकरण एवं विधि –
(1) अमोनिया संतृप्त स्तम्भ – इसमें ब्राइन को अमोनिया से संतृप्त करते हैं। अमोनियामय ब्राइन बनता है तथा Ca एवं Mg की अशुद्धि अवक्षेपित होकर नीचे बैठ जाती है।
(2) छन्ना – Ca तथा Mg के अवक्षेप अमोनियामय ब्राइन से पृथक् हो जाते हैं।
(3) शीतकारक-अमोनियामय ब्राइन को ठण्डा करते हैं।
(4) कार्बोनेटीकरण स्तम्भ – अमोनियामय ब्राइन में चूने के भट्टी से प्राप्त CO₂ प्रवाहित करने पर सोडियम बाइकार्बोनेट प्राप्त होता है।

• 2NH₃ + H₂O + CO₂ → (NH₄)₂ CO₃
• (NH₄)₂CO₃CO₂ + H₂O → 2NH₄HCO₃
• NH₄HCO₃ + NaCl → NaHCO₃ + NaCl

(5) निर्वात् छन्ना – अविलेय NaHCO₃ छनकर पृथक् हो जाता है। विलयन में NH₄Cl तथा NH₄HCO₃ शेष रहता है। इसे पुनः प्राप्ति स्तम्भ में भेजा जाता है।
(6) चूने की भट्टी – चूने के पत्थर से CO₂ गैस प्राप्त की जाती है।

CaO जल के साथ क्रिया कर Ca(OH)₂ बनाता है जो NH₄Cl के साथ अभिक्रिया कर पुन: NH₃ देता है।
CaO + H₂O + Ca(OH)₂

(7) अमोनिया पुनः प्राप्ति स्तम्भ – निर्वात् छन्ने से प्राप्त द्रव पर ऊष्मा तथा बुझे चूने की क्रिया से अमोनिया प्राप्त करते हैं।

• NH₄HCO₃ → NH₃ + H₂O+CO
• 2NH₄Cl + Ca(OH)₂ → CaCl₂ + 2NH₃ + 2H₂O

(8) NaHCO₄ का जारण – निर्वात् छन्ना से प्राप्त सोडियम बाइकार्बोनेट प्राप्त होता है। इसे बेलनाकार भट्टियों में गर्म करते हैं जिससे सोडियम कार्बोनेट प्राप्त होता है।
2NaHCO₂ → Na₂CO₃ + H₂O + CO₂

प्रश्न 8.
जब वर्ग – 1 की एक धातु को द्रव अमोनिया में घोला गया, तो निम्नलिखित प्रेक्षण प्राप्त हुए –
(a) प्रारंभ में नीला विलयन प्राप्त हुआ।
(b) विलयन को सान्द्र करने पर, नीला रंग-काँस्य-रंग में परिवर्तित हो गया। विलयन के नीले रंग की व्याख्या कीजिए। विलयन को कुछ समय तक रखने पर प्राप्त उत्पाद का नाम बताइए।
उत्तर:
(a) वर्ग-1 धातुओं को द्रव अमोनिया में घोलने पर निम्नलिखित अभिक्रिया प्राप्त हुई –

विलयन का नीला रंग अमोनीकृत इलेक्ट्रॉनों के कारण होता है, जो दृश्य प्रकाश क्षेत्र की संगत् ऊर्जा का अवशोषण करके विलयन को नीला रंग प्रदान करता है।

(b) सान्द्र विलयन में, धातु आयन स्तर बन जाने के कारण नीला रंग, काँस्य रंग में बदल जाता है। नीला विलयन कुछ समय तक पड़े रहने पर हाइड्रोजन को मुक्त करता है तथा ऐमाइड बनते हैं।

प्रश्न 9.
क्षार धातुओं के परॉक्साइड तथा सुपर ऑक्साइडों का स्थायित्व वर्ग में नीचे की ओर जाने पर घटता है। उदाहरण सहित व्याख्या कीजिए ?
उत्तर:
परॉक्साइड तथा सुपर ऑक्साइडों का स्थायित्व धातु आयन का आकार बढ़ने पर बढ़ता है।
KO₂ < RbO₂ < CsO₂
क्षार धातुओं का ऑक्सीजन के साथ क्रिया करके विभिन्न ऑक्साइड बनाने का कारण, क्षार धातु के धनायन के परितः प्रबल धनात्मक क्षेत्र का निर्माण होता है। Li* का आकार सबसे छोटा है, जिसके कारण यह O^2-आयन को पुन: O₂ से क्रिया नहीं करने देता है। Na⁺ का आकार Li से बड़ा है अतः इसका धनात्मक Li⁺ के क्षेत्र से क्षीण होता है। K⁺ Rb⁺ Cs⁺जैसे बड़े आयन O^2-₂ आयन को पुन: 0₂ से क्रिया करके सुपरऑक्साइड (O^–₂) बनाने देते हैं।

पुनः धातु आयनों का आकार बढ़ने के साथ-साथ परॉक्साइडों तथा सुपरऑक्साइडों के स्थायित्व में भी वृद्धि होती है। इसका प्रमुख कारण जालक ऊर्जा प्रभाव के फलस्वरूप बड़े ऋणायनों का बड़े धनायनों द्वारा स्थायित्व प्रदान करना है।

प्रश्न 10.
सोडियम हाइड्रॉक्साइड प्राप्त करने की कास्टनर केलनर सेल का चित्र सहित वर्णन कीजिए।
उत्तर:
इसमें एक लोहे की आयताकार टंकी होती है जो स्लेट की पट्टियों द्वारा तीन भागों में विभाजित रहती है। ये पट्टियाँ हौज पेंदे को छूती नहीं हैं बल्कि निचला भाग हौज के तले में रखे हुये पारे से ढंका रहता है। पारे की पर्त तीनों भागों को एक-दूसरे से पृथक् रखती है। पारे की पर्त यांत्रिक प्रबन्ध द्वारा इधर – उधर घूमती रहती है।

बाहरी कक्ष में NaCl का विलयन भरा रहता है, जिसमें ग्रेफाइट की छड़ लगी रहती है। ये एनोड का कार्य करती है। बीच ऐनोड वाले भाग में NaOH का तनु विलयन भरा रहता है, जिसमें लोहे की छड़ का बना कैथोड लटका – रहता है। विद्युत् धारा प्रवाहित करने पर पारा प्रेरण बीच का कक्ष एनोड का तथा बाहरी कक्ष कैथोड का कार्य करते हैं।

बाहरी कक्ष में एनोड पर Cl₂ मुक्त होती है। सोडियम कैथोड पर मुक्त होकर पारे के साथ सोडियम अमलगम बना लेता है। सोडियम अमलगम उत्क्रेन्द्रीय पट्टियों की सहायता से सेल के मध्य भाग में आता है । यहाँ पर सोडियम अमलगम ऐनोड का कार्य करता है और आयरन की छड़ कैथोड का। इस भाग में NaOH भरा रहता है। विद्युत् धारा प्रवाहित करने पर OH^– आयन एनोड पर विसर्जित होते हैं तथा अमलगम में उपस्थित सोडियम से क्रिया कर NaOH बनाते हैं तथा H₂ गैस मुक्त करते हैं।

अभिक्रिया-बाहरी कक्ष में –
2NaCl ⇄ 2Na⁺ + 2Cl^–

कैथोड पर –
2Na⁺ + 2e → 2Na
2Na + xHg → Hg_xNa₂
एनोड पर –

मध्य कक्ष में –
NaOH ⇄  Na⁺ + OH^–

कैथोड पर –
2NaHg + 2H₂O → 2NaOH + 2Hg + H₂
Na⁺ + e → Na
2Na + 2H₂O → 2NaOH + H₂

एनोड पर –
2OH^– → 2OH + 2e
Hg_x Na₂ + 2OH^– → 2NaOH + xHg

प्रश्न 11.
जब कैल्सियम के यौगिक –
(A) में जल मिलाया जाता है तो यौगिक (B) के विलयन का निर्माण होता है। इस विलयन में कार्बन डाइ-ऑक्साइड प्रवाहित करने पर यह यौगिक (C) बनने के कारण दुधिया हो जाता है। कार्बन डाइ-ऑक्साइड की अधिक मात्रा में प्रवाहित करने पर, यौगिक (D) के निर्माण के कारण यह दुधियापन लुप्त हो जाता है तथा यौगिक (A), (B), (C) तथा (D) को पहचानिए तथा बताइए कि अंतिम पद में दुधियापन क्यों समाप्त हो जाता है ?
उत्तर:
यौगिक (B) के विलयन में CO₂ प्रवाहित करने पर विलयन का दुधिया होना संकेत करता है कि यौगिक (B) बुझा हुआ चूना [Ca(OH)₂] है तथा यौगिक (C) कैल्सियम कार्बोनेट है। चूँकि यौगिक (B), यौगिक (A) में H₂O को मिलाने से प्राप्त होता है। अतः यौगिक (A) में बिना बुझा चूना (CaO) है। संगत अभिक्रियाएँ निम्नलखित हैं –


(iii) CO₂ को अधिकता में प्रवाहित करने पर, विलेय कैल्सियम बाइकार्बोनेट बनाने के कारण दुधियापन लुप्त हो जाता है।

प्रश्न 12.
कैल्सियम ऑक्साइड बनाने की विधि, रासायनिक गुण तथा उपयोग लिखिये।
उत्तर:
चूने के पत्थर को गर्म करके कैल्सियम ऑक्साइड बनाते हैं।

यह अभिक्रिया उत्क्रमणीय है। अतः चूना प्राप्त करने हेतु CO₂ को जल्दी-जल्दी-हटाते रहना आवश्यक है। अभिक्रिया का-ताप 900°C होना चाहिये क्योंकि अधिक ताप पर मिट्टी और चूने की अभिक्रिया से गलनीय सिलिकेट बन जाता है। भट्टी में बाजू से दो अँगीठियों में कोयला जलाया जाता है। ऊपर से धीरे-धीरे चूने का पत्थर डालते रहते हैं, जो नीचे आते-आते अपघटित हो जाता है। CO₂ गैस ऊपरी भाग से बाहर निकलती है। इसे द्रवित कर सिलेण्डरों में भर लिया जाता है।

रासायनिक गुण:
(1) नम वायु से नमी तथा CO₂ सोखकर Ca(OH)₂ तथा CaCO₃ बनाता है।

• CaO + H₂O Ạ Ca(OH)₂
• CaO + CO₂ CaCO₂

(2) यह एक प्रबल क्षारीय ऑक्साइड है जो अम्लों के साथ अभिक्रिया कर लवण बनाता है।

• CaO + 2HCl → CaCl₂ + H₂O
• Cao + SiO₂→ CaSiO₃

(3) अमोनियम लवणों के साथ गर्म करने पर NH, गैस बनती है।
2NH₄Cl + CaO → CaCl₂ + H₂O + 2NH₂
(4) कार्बन और कैल्सियम ऑक्साइड को गर्म करने से CaC₂ प्राप्त होता है।

उपयोग:

• धातुकर्म के गालक के रूप में
• ऐल्कोहॉल तथा गैसों को सुखाने में
• लाइम लाइट उत्पन्न करने के लिये
• कोल गैस को शुद्ध करने में
• कागज बनाने में।

प्रश्न 13.
वर्ग-1 के एक तत्व का आयन कोशिकाओं में शिरा-संकेतों के संचरण, शर्करा तथा एमीनो अम्लों के प्रवाह में सहायक है। यह तत्व ज्वाला परीक्षण में ज्वाला के साथ पीला रंग देता है तथा
ऑक्सीजन के साथ ऑक्साइड तथा परॉक्साइड बनाता है।तत्व की पहचान कीजिए तथा इसके परॉक्साइड निर्माण की रासायनिक समीकरणों को लिखिए। यह तत्व ज्वाला के साथ रंग क्यों देता है ?
उत्तर:
ज्वाला परीक्षण में पीले रंग की ज्वाला दर्शाता है कि धातु सोडियम ही होनी चाहिए। यह 0, के साथ क्रिया करके सोडियम परॉक्साइड Na₂O₂ तथा सोडियम ऑक्साइड, Nao का मिश्रण देता है।

सोडियम की आयनन एन्थैल्पी कम होती है। जब सोडियम धातु या इसके लवण को बुन्सन ज्वाला में गर्म किया जाता है, तब ज्वाला की ऊष्मा बाह्यतम इलेक्ट्रॉन को उच्च ऊर्जा स्तर पर उत्तेजित कर देती है तब ये इलेक्ट्रॉन पुनः अपनी तलस्थ / आद्य अवस्था में आते हैं तो दृश्य क्षेत्र में विकिरण उत्सर्जन के कारण ज्वाला को पीला रंग प्रदान करते हैं।

प्रश्न 14.
निम्नलिखित की महत्ता बताइए –
(a) चूना पत्थर
(b) सीमेंट
(c) प्लास्टर ऑफ पेरिस।
उत्तर:
(a) चूना पत्थर (CaCO₃):

• इसे मैग्नीशियम कार्बोनेट के साथ आयरन जैसी धातुओं के निष्कर्षण में गालक के रूप में प्रयोग करते हैं।
• इसका प्रयोग ऐन्टासिड, टूथपेस्ट में अपघर्षक के रूप में, च्यूइंगम में संघटक तथा सौन्दर्य प्रसाधनों में रूपक के रूप में भी करते हैं।

(b) सीमेंट:

• यह भवन निर्माण हेतु एक महत्वपूर्ण यौगिक है।
• इसका उपयोग कांक्रीट, प्रबलित कांक्रीट, प्लास्टरिंग, पुल-निर्माण, भवन-निर्माण आदि में किया जाता है।

(c) प्लास्टर ऑफ पेरिस:

• इसका उपयोग भवन निर्माण तथा टूटी हुई हड्डियों के प्लास्टर में होता है।
• इसका उपयोग दंत-चिकित्सा, अलंकरण कार्य तथा मूर्तियों एवं अर्द्धप्रतिमाओं को बनाने में भी होता है।

MP Board Class 11th Chemistry Solutions

Students get through the MP Board Class 11th Chemistry Important Questions Chapter 11 The p-Block Elements which are most likely to be asked in the exam.

MP Board Class 11th Chapter 11 The p-Block Elements

The p-Block Elements Class 11 Important Questions Very Short Answer Type

Question 1.
How would you explain the higher stability of BC13 in comparison to TiCl₃?
Answer:
Boron exhibits only +3 oxidation state, thus it forms a stable compound BCl₃. Down the group, inert pair effect gradually becomes more effective due to which+1 oxidation state of Thallium is more stable than its +3 oxidation state. That is why BCl₃ is more stable thanTiCl₃.

Question 2.
Why does Boron trifluoride act as a Lewis acid?
Answer:
Due to electron deficiency, BF₃ acts as a strong Lewis acid. It easily reacts with Lewis base and completes its octet towards boron.
.

Question 3.
Giving the examples of BCl₃ and CCl₄ compounds, explain their behaviour towards water.
Answer:
BCl₃ is an electron deficient molecule. It easily accepts an electron pair from water and forms boric acid (H₃BO₃) and HCl.
BCl₃ + 3H₂O → H₃BO₃ + 3HCl
CCl₄ is an electron sufficient molecule in which 4 does not undergo hydrolysis.

Question 4.
What is the reason that caustic base like NaOH is not stored in aluminium containers?
Answer:
On keeping caustic base like NaOH in aluminium containers, aluminium dis-solves in base to form sodium meta aluminate, therefore base is not stored in aluminium containers.
2Al + 2NaOH + 2H₂O → 2NaAlO₂ + 3H₂.

Question 5.
At normal temperature, aluminium does not react with water. Why?
Answer:
In presence of air, a transparent protective oxide layer is formed on the surface of aluminium. Due to this layer, at normal temperature, aluminium does not react with water.

Question 6.
Among iron and aluminium, aluminium is more reactive as compared to iron but iron easily gets rusted than aluminium. Why?
Answer:
In presence of air, a transparent non-porous protective layer of oxide is developed on aluminium surface due to which at normal temperature it does not react with oxygen and moisture present in air. Whereas a porous oxide layer is formed on iron surface, due to which reactivity of iron increases. Therefore iron easily gets rusted.

Question 7.
Write the formula of double salt or Alum.
Answer:
General formula of double salt is R₂SO₄.M₂(SO₄)₃ where R is a monovalent metal like Na, K, Rb, Cs or NH4 radical and M is a trivalent metal like Fe⁺³, Al⁺³ or Cr⁺³.
Example: K₂SO₄.Al₂(SO₄)₃.24H₂O (Potash Alum).

Question 8.
Name the ores of aluminium. Give their formulae.
Answer:
Ores of aluminium are as follows :
1. Bauxite: Al₂O₃.2H₂O
2. Diaspore: Al₂O₃.H₂o
3. Cryolite: Na₃AlF₆
4. Alunite: K₂SO₄.Al₂(SO₄)₃.2Al(OH)₃
5. Corundum: Al₂O₃
6. Felspar: K₂O.Al₂O₃.SiO₃.

Question 9.
What happens when (Give chemical equation):
(a) Aluminium chloride is heated, (b) Alum is heated.
Answer:
(a) On heating aluminium chloride :
2AlCl₃.6H₂O → Al₂O₃ +6HCl+ 3H₂O
Thus on heating anhydrous aluminium chloride cannot be obtained.

(b) On heating alum to 200°C a porous substance is obtained.
K₂SO₄. Al₂ (SO₄)₃.24H₂O → K₂O + Al₂O₃ + 4 SO₂ + 24H₂O

Question 10.
Aluminium is a strong reducing agent. Why?
Answer:
Elements which donate electrons and form cations are known as reducing agents. Reducing nature of an element depends on its standard electrode potential. Higher the negative value of standard electrode potential of an element higher is its reducing character. Standard electrode potential of AI is -1.67, therefore aluminium acts as a strong reducing agent.

Question 11.
Why is Gallium a liquid at room temperature?
Answer:
In solid state crystalline structure of gallium is such that its lattice energy is very less and at low-temperature metallic bond between its atoms start breaking. Therefore gallium is obtained in liquid state at low temperature.

Question 12.
Boron does not form trivalent ion. Why?
Answer:
Due to small size of boron, its ionization energy is high and value of third ionization is more than that of first and second ionization energy. Therefore, it is not possible to remove or donate three electrons. Therefore it does not form trivalent ion.

Question 13.
Is boric acid a protic acid ? Explain.
Answer:
Boric acid, is not a protic acid, because it does not give proton on ionization in water. It accepts electron from the hydroxyl ion of water and behaves as a Lewis acid and releases H⁺ ion at the end.
B(OH)₃ + HOH → [B(OH)₄] + H⁺.

Question 14.
Give the reactions representing amphoteric behaviour of aluminium.
Answer:
Al dissolves both in acid and base and liberate dihydrogen. This behaviour of it, is amphoteric.
2Al_(s) + 3H₂SO_4(aq) → Al₂ (SO₄)_3(aq) + 3H_2(g)
2Al_(s) + 2NaOH_(aq) + 6H₂O_(l) → 2Na+ [Al(OH)₄]^–_(aq) + 3H_2(g)
Sodium tetrahydroxoaluminate (III).

Question 15.
Write resonance structures of CO₃^-2 and HCO₃^–.
Answer:
Resonance structures of CO₃^-2

Question 16.
Why is melting and boiling point of boron extremely high?
Answer:
Crystals of boron are formed by the development of covalent bonds between its atoms. Two atoms combine to form icosahedron network which contain 20 triangular faces and 12 comers. It makes boron extremely hard, therefore melting and boiling point of boron is extremely high.

Question 17.
Normally, boron does not react with acid or base under what conditions does it react with acid or base ?
Answer:
Normally, boron does not react with acid or base but if the acid is strongly reducing then boron reacts with it at high temperature and forms boric acid. Similarly, it reacts with base at high temperature to form borate.

Question 18.
What is inorganic benzene? Le
Answer:
Borazine is known as inorganic benzene. Its chemical formula is B₃N₃H₆. Structure of borazine is like benzene. It has cyclic and planar hexagonal structure.

Question 19.
What is corundum? Borazine
Answer:
Al is found in more than one crystalline forms. Its hardest crystalline form is known as corundum which acts as an abrasive.

Question 20.
State the names of ores ot boron. Give their formulae.
Answer:
Ores of boron are as follows:

1. Borax – Na₂B₄O₇.1oH₂O
2. Kernite – Na₂B₄O₇.2H₂O
3. Colemanite – Ca₂B₆O₁₁ .2H₂O
4. Orthoboric acid – H₃BO₃

Question 21.
Justify that Tl⁺³ is an oxidizing agent whereas Al⁺³ is not?
Answer:
Due to inert pair effect, in boron family, stability of + I oxidation state increases on moving down the group whereas stability of +3 oxidation state decreases. Therefore, Tl⁺¹ is more stable than Tl⁺³ Thus,
Tl⁺³ +2e ^– → Tl⁺¹
By the reaction, it is clear that Tl⁺³ is getting reduced to Tl⁺¹, therefore Tl⁺³ is an oxidizing agent but in Al, Al+3 oxidation state is possible and not Al⁺¹, therefore it is not possible for Al⁺³ to be an oxidizing agent.

Question 22.
What is the hybridized state of carbon in
(i) CO₃^2-
(ii) Diamond,
(iii) Graphite?
Answer:
Hybridization of carbon in CO₃^2-, diamond and graphite are sp², sp³ and sp² respectively.

Question 23.
B-Cl bond has a dipole moment, but dipole moment of BCl₃ molecule is zero. Why?
Answer:
In BCl₃, boron is sp² hybridized, due to which geometry of BCl₃ molecule is trigonal planar. It has a symmetrical shape and resultant dipole moment of symmetrical molecules is zero (because all dipole moments become negligible due to symmetry of molecules).

Thus, dipole moment of BCl₃ is zero.

Question 24.
What happens when boron is treated with caustic soda?
Answer:
At ordinary temperature, boron does not show any reaction with base but it reacts with caustic soda NaOH or caustic potash (KOH) to form borate and liberate H₂ gas.

Question 25.
Explain disproportionation reaction with example.
Answer:
Gallium exhibits +1 and +3 oxidation states. +3 oxidation state of gallium is more stable, therefore compound of gallium of+1 oxidation state oxidizes to compound of +3 oxidation state.
3GaCl → 2Ga + GaCl₃.

Question 26.
Boric acid acts as a Lewis acid, not as a protic acid. Why?
Answer:
In boric acid, octet of central metal boron is not complete. Its valence shell has 6 electrons, so it needs a lone electron pair to complete its octet. Due to being an electron pair acceptor, boric acid acts as a Lewis acid. It liberates H⁺ ion on reacting with water.
H₃BO₃ or B(OH)₃ + H₂O ⇌ [B(OH)₄]^–+ H⁺

Question 27.
How is borax obtained from colemanite?
Answer:
On boiling colemanite with concentrated sodium carbonate solution, borax is obtained.
Ca₂B₆O₁₁ + 2Na₂CO₃ → Na₂B₄O₇ + 2NaBO₂ + 2CaCO₃
On concentrating the solution, crystals Of borax are obtained on passing carbon dioxide in the mother liquor borax is obtained.
4NaBO₂ + CO₂ → Na₂B₄O₇ + Na₂CO₃

Question 28.
What is the effect of heat on boric acid?
Answer:
On heating boric acid to 100°C temperature, metaboric acid is obtained which on heating to high temperature forms boric anhydride.

Question 29.
Cryolite is used in the extraction of aluminium by alumina. Why?
Answer:
Melting point of pure alumina is very high 2050°C, but in presence of cryolite and fluorspar, it melts at 870°C. This way, cryolite reduces the melting point of alumina and also acts as an electrolyte.

Question 30.
How will you justify that diamond and graphite are the allotropes of carbon?
Answer:
On combustion of diamond and graphite in the presence of air, C02 gas is re-leased which on passing through lime water turns it milky. This proves that diamond and graphite are the allotropes of carbon.
C_diamond + O₂ →CO₂
C_graphite + O₂ → CO₂

Question 31.
What will happen if a piece of diamond is dropped on burning charcoal?
Answer:
If a piece of diamond is dropped on burning charcoal, then it will bum completely and only CO₂ gas is obtained and after combustion no residue is left which proves that diamond is the purest form of carbon.

Question 32.
Write the uses of carbon monoxide.
Answer:

• It is the main constituent of water gas (CO + H₂) and producer gas (CO + N₂).
• It is used for the preparation of some metal carbonyls.
• Carbon monoxide is used as reducing agent.

Question 33.
Diamond is less abundant than graphite in nature. Why?
Answer:
Diamond is formed due to conversion of carbon in molten state at extremely high pressure into crystalline form. But this stage in nature is very rare, therefore diamond is less abundant than graphite.

Question 34.
What is dry ice? Write its use.
Answer:
Solid carbon dioxide is known as dry ice because its crystals appear as ice and they do not wet paper or clothe. It gets converted to solid state at -78.5°C without forming a liquid. It is used as a coolant to prevent food materials from decay and as an anaesthetic in surgery.

Question 35.
What is carborundum? Write its main use.
Answer:
Structure of silicon carbide is hard like diamond, it is known as carborundum. It is used for sharpening and grinding the metals.

Question 36.
Write name and structural formulae of organic compounds used as refrigerant, anaesthetic and solvent.
Answer:

Question 37.
Write the uses of graphite.
Answer:
Uses of graphite :

1. It is a good conductor of electricity. So it is used as an electrode in dry cells, electric arcs etc.
2. Pencil, black paint, black ink is prepared from it.
3. Due to its lubricating property, it is used at high temperature to maintain the machines smooth.

Question 38.
Write the names of various varieties of coal.
Answer:
On the basis of carbon content, coal is of the following varieties :

1. Peat: Contains 60% carbon,
2. Lignite: Contains 70% carbon,
3. Bituminous : Contains 70% carbon,
4. Anthracite : Contains 90% carbon.

Question 39.
Write the uses of diamond.
Answer:
Uses of diamond :

• As valuable jewellery,
• Used in cutting glass,
• In drilling rocks,
• In polishing gems.

Question 40.
Carbon dioxide is acidic in nature. Explain with equation.
Answer:
Aquaous solution of carbon dioxide is acidic.
CO₂ + H₂O → H₂CO₃ (Carbonic acid)

It turns blue litmus to red and reacts with base to form salt.
2NaOH + CO₂ → Na₂CO₃ + H₂O
Ca(OH)₂ + CO₂ → CaCO₃ + H₂O.

Question 41.
Why should we not sleep in a closed room with burning chulha (Angithi)?
Answer:
We should not bum chulha (angithi) in a closed room, because the gas coming out of it contains large amount of CO. It enters the body by the process of respiration, combines with haemoglobin and forms carboxyhaemoglobin which creates obstruction in the transportation of oxygen and blood in the body. Therefore, the person can become unconcious and can die also.

Question 42.
What are carbides?
Answer:
Binery compounds of carbon which it forms with less electronegative or highly electropositive elements than itself are called carbides. These are of various types, like :

1. Ionic carbide,
2. Metallic carbide,
3. Intermediate carbide,
4. Covalent carbide.

Question 43.
Write the use of silica gel.
Answer:
Silica gel is a porous, amorphous solid which contains 4% of moisture. It is used as a catalyst in petroleum industry, also in chromatography.

Question 44.
What is thixotropy?
Answer:
On shaking or agitating a liquids, its viscosity decreases temporarily. This property is known as thixotropy.
When SiCl₄ is hydrolysed at a high temperature, then the silica obtained this property.

Question 45.
What are the interstitial carbides?
Answer:
When carbon atoms occupy the interstitial sites of crystal lattice of transition metals, then such carbides formed are called interstitial carbides. They are extremely hard and the melting points are high.
Example: Tungsten carbide, iron carbide.

Question 46.
What are methanides and acetylides?
Answer:
1. Carbides which give methane on hydrolysis are called methanides.
Al₄C₃ +12H₂O →4Al(OH)₃ +3CH₄

2. Carbides which give acetylene on hydrolysis are called acetylides.
CaC₂ + 2H₂O → Ca(OH)₂ + C₂H₂.

Question 47.
What are silanes and germanes ?
Answer:
Hydrides of Si and Ge are known as silanes and germanes which are represented by M_nH_2n+2
where M = Si, Ge. In silanes, value of n is from 1 to 8 whereas in germanes value of n is 1 to 5.

Question 48.
What is activated charcoal?
Answer:
Charcoal is soft and porous. It absorbs coloured substances and smelling gases. If it is heated in steam to 1100°C, then its absorbing power increases and it is known as activated charcoal.

The p-Block Elements Class 11 Important Questions Short Answer Type

Question 1.
What happens when boric acid is heated?
Answer:
When boric acid is heated, it liberates three molecules of water at various temperatures and forms boron trioxide at the end.

Question 2.
Describe the shapes of BF₃ and BH₄^–. Assign the hybridization of boron in these species.
Answer:
In BF₃, three bond pairs are present in boron. Thus, it is sp² hybridized and of trigonal planar structure, whereas bond pair number in [BH₄]^– is 4 due to which it is sp³ hybridized and has tetrahedral structure.

Question 3.
What is alum? State its general formula and use.
Answer:
Initially, double salts of potassium sulphate and aluminium sulphate K₂SO₄.Al₂(SO₄)₃.24H₂O are known as alums. But presently, all double salts with general formula R₂SO₄.M₂(SO₄)₃.24H₂O called alums. Where R = monovalent metal like : Na, K, Rb, Cs, etc. M = Trivalent metal like:PAl, Cr, Fe etc.
Uses :

• In the purification of water,
• In dyeing leather,
• In paper industry,
• In fire extinguishers,
• In dyeing clothes,
• In clotting of blood.

Question 4.
Write the name, composition and uses of four alloys of aluminium.
Answer:
Alloys of aluminium :

1. Aluminium bronze: Cu (90%) + Al (10%)
   Uses: For making utensils, cheap ornaments, coins.
2. Magnelium : Mg (10%) + Al (90%)
   Uses : Used for making aeroplanes, tools.
3. duralumin : Al (95%) + Cu (4%) + Mn (0.5%) + Mg (0.5%)
   Uses : Used for making airships.
4. Nickeloy : Al (95%) + Cu (4%) + Ni (1%)
   Uses : For making utensils, coins etc.

Question 5.
Aluminium is a weaker conductor of electricity than copper still it is used in electric cables. Why?
Answer:
Copper is a better conductor of electricity than aluminium but aluminium is a light metal and its density is less than copper. This way according to weight aluminium is a better conductor as compared to copper. Therefore, aluminium is used in making electric wire and cable instead of copper.

Question 6.
Boron forms only covalent compounds. Why?
Answer:
Due to small size and high ionization energy of boron, its tendency to form cation is very less. Therefore boron cannot lose three electrons to form trivalent ion. To achieve a stable electronic configuration it shares its electrons with atoms of other elements to form stable compounds. Therefore, boron forms only covalent compounds.

Question 7.
Halides of boron acts as strong Lewis acids. Why?
Answer:
There are three electrons in the valence shell of boron. When it shares its electrons with three halogen atoms, then also it contains total 6 electrons in its valence shell, it still needs lone electron pair to complete its octet. Therefore, it is an electron-deficient compound and acts as an electron acceptor and accepts electrons from an electron-pair donor compound to form a co-ordinate bond and form an addition compound. Therefore it acts as a strong Lewis acid.

Question 8.
Aluminium cannot be obtained from its ore by the process of reduction. Why?
Answer:
Due to strong electropositivity, aluminium acts as a reducing agent, therefore aluminium can be easily oxidized, on the basis of ionization energy and electron affinity, it is clear that aluminium acts as electron donor, not as an electron acceptor. Therefore, it cannot be reduced. That is why aluminium cannot be obtained from its ore by reduction.

Question 9.
Atomic radius of gallium is less than aluminium. Why?
Answer:
In gallium, there are 10 electrons in d-orbital. Shape of d-orbital is such that its screening effect is less effective due to which attractive force of nucleus for outermost electrons is more. As a result, outermost electrons are strongly attracted by the nucleus due to which atomic radius of gallium decreases. Therefore atomic radius of gallium is less than aluminium.

Question 10.
Suggest a reason as to why CO is poisonous?
Answer:
Carbon monoxide reacts with haemoglobin to form a stable compound named carboxyhaemoglobin. It is 300 times more stable than oxyhaemoglobin. Due to formation of carboxyhaemoglobin the tendency of flow of oxygen in blood decreases which in turn leads to suffocation. Less amount of it may cause headache and more amount of it may lead to death.

Question 11.
State the reason of formation of covalent compounds of boron halides with ammonia and amine.
Answer:
There are three electrons in valence shell of boron. When it shares electrons with three halogen atoms to form boron trihalide then also the valence shell of boron contains total 6 electrons. They still require an electron pair to complete their octet. Therefore, they are electron deficient compounds and act as electron acceptor and accept a lone pair of electron from any electron-pair donor compound like ammonia or amine to form a co-ordinate bond and result in the formation of an addition compound.

Question 12.
Boron family generally exhibit +1 and +3 oxidation state. Why?
Answer:
B₅ → 1s²,2s²2p¹

General electronic configuration of all the members of boron family is ns² np¹. In normal state, there is one unpaired electron in p-orbital of its valence shell. But in excited state, one electron of 2s gets excited and goes to 2p-subshell. Thus, in excited state there are three unpaired electrons. Therefore, all the members of boron family exhibit +1 and +3 oxidation state.

Question 13.
Explain inert pair effect in boron family.
Answer:
Members of boron family exhibit +1 oxidation state in normal state and +3 in excited state. On moving down the group, stability of +1 oxidation state increases but that of +3 oxidation number decreases because the two electrons of 5-orbital of valence shell do not participate in bond formation. This is known as inert pair effect.

With the increase in atomic number, electrons enter into d-subshell and shape of d and d-subshells is such that their screening effect is minimum due to which attractive force of nucleus on valence electron increases. This increase in attractive force is more in 5-orbital electrons than in electrons of p-subshell. Therefore 5-orbital electrons do not participate in bond formation.

Question 14.
How is boric acid obtained from colemanite?
Answer:
Colemanite is dissolved in boiling water and on passing sulphur dioxide gas through it, boric acid and calcium bisulphite are formed. Calcium bisulphite is soluble whereas boric acid crystallizes.

Question 15.
What is borax glass?
Answer:
Anhydrous sodium tetraborate (Na₂B₄O₇) is known as borax glass. It is obtained on heating ordinary borax above its melting point. It is a colourless glass-like substance which absorbs moisture from the environment and gets converted to decahydrate form. It is soluble in hot water, its aqueous solution is alkaline due to hydrolysis. On heating, it swells up to a white opaque substance. Anhydrous substance gives borax glass at 740°C.
Na₂B₄O₇+ 2H₂O → H₂B₄O₇ + 2NaOH

Question 16.
Explain the effect of heat on borax.
Answer:
On strong heating borax, its water of Crystallization separates and finally it melts and gets converted to transparent bead.

Boric anhydride B₂O₃ reacts with metallic oxides to form metaborate which has a specific colour. This reaction is known as borax bead test and is used in the detection of basic radicals.

Question 17.
How is excessive content of CO₂ responsible for global warming?
Answer:
CO₂ is a greenhouse gas. 75% of the solar energy reaching the earth is absorbed by the earth’s surface and the remaining is reflected back into the atmosphere. But the heat released from the hot surface cannot be sent back to space because the C02 present in the atmosphere absorbs this extra heat due to which average temperature of the atmosphere increases. This is known as global warming.

Question 18.
Explain why is there a phenomenal decrease in ionization enthalpy from carbon to silicon?
Answer:
On moving from carbon to silicon in the periodic table, the atomic size increases, thus distance between the outermost electron and nucleus increases. Thus, these electrons feel very less attraction of the nucleus, due to which their removal is very easy. Since size of Si atom is considerably larger due to which the outermost electrons feel less attraction. Therefore, its ionization enthalpy (to remove 1 electron) is less.

Question 19.
How would you explain the lower atomic radius of Ga as compared to Al?
Answer:
Electronic structures of Al and Ga are as follows :
₁₃Al = 1 s²,2s²2p⁶,3s²3p¹
₃₁ Ga = 1s²,2s²,2p⁶,3s²,3p⁶3d¹⁰,4s²4p¹
In this screening effect of d-electrons is very less. Thus, on moving from Al to Ga, screening effect of 10 d-electrons is unable to make the nuclear charge ineffective. Thus, atomic radius of Ga, due to effective nuclear charge is less than that of aluminium.

Question 20.
What is borax bead test? Explain.
Answer:
Borax bead test: On heating borax it loses its water of crystallisation. On further heating it forms a glassy bead which contains sodium metaborate and boric anhydride.
Na₂B₄O₇.10H₂O → Na₂B₄O₇+10H₂O
Na₂B₄O₇ → 2NaBO₂ + B₂O₃

Example: Copper sulphate forms a bluish green metaborate.
CuSO₄ → CuO+SO₃
CuO + B₂O₃ →Cu(BO₂ )₂
Cupric metaborate bead of bluish green colour In reducing flame the colour of the bead is red.
2Cu(BO₂)₂ →CuBO₂ +B₂O₃ + CUO
Thus, Borax bead test is used in the detection of basic radicals.

Question 21.
How many types of borax are there? Describe briefly.
Answer:
Borax exists in the following three forms :
1. Prismatic borax: It is decahydrate, Na₂B₄O₇.10H₂O. It is the ordinary form and is obtained by the crystallization of solution at ordinary temperature.
2. Octahedral borax: It is borax pentahydrate, Na₂B₄O₇.5H₂O. It is formed by the crystallization of the solution above 60°C.

3. Borax glass: It is anhydrous sodium tetraborate, Na₂B₄O₇. It is obtained by heating ordinary borax above its melting point. It is a colourless glass like substance. It absorbs moisture from the air and gets converted to decahydrate form. Its aqueous solution is basic.
Na₂B₄O₇ +2H₂O → H₂B₄O₇ +2NaOH.

Question 22.
How is borate radical detected?
Answer:
In the laboratory, for the detection of acidic borate radical BO_3^-3, salt is heated with ethanol and concentrated sulphuric acid, due to which vapours of ethyl borate are obtained. These vapours bum with a green core flame. Actually, the salt first gets converted to boric acid, which reacts with ethanol to form triethyl borate.
H₃BO₃ +3C₂H₅OH →B(OC₂H₅)₃ + 3H₂O.

Question 23.
Explain the structure of aluminium chloride.
Answer:
Aluminium chloride is actually obtained in the form of dimer Al₂Cl₆. There are three electrons in the valence shell of Al, when these three electrons are shared with three chlorine atoms, AlCl₃ is formed but there are total 6 electrons in the valence shell of Al. To complete its octet, it requires one lone electron pair. In tfcis conation, Al of AlCl₃  electron pair from chlorine of AlCl₃ and complete its octet.

Question 24.
What is the reason that AlF₃ is insoluble in anhydrous HF but becomes soluble on the addition of NaF Why?
Answer:
Strength of hydrogen bond in HF is more, due to which HF is comparatively a weak acid and cannot release F ion. On adding NaF, due to complete ionization concentration ofF ion considerably increases by which due to formation of complex ion AlF₆^-3 by AlF₃ it dissolves.
HF → H⁺ + F^–
3NaF → 3Na⁺+ 3F^–
AlF₃ + 3F^– → AlF₆^-3
AlF₆^-3 + 3Na⁺ → Na₃ [AlF₆].

Question 25.
In some reactions thallium resembles aluminium, whereas in others it resembles with group-1 metals. Support this statement by giving some evidences.
Answer:
Both thallium and aluminium are the elements of group-13. General electronic configuration of their valence shell is ns² np¹. Aluminium exhibits only +3 oxidation state. Like aluminium, thallium also exhibits +3 oxidation states in some of its compounds.
Example: Tl₂O₃, TlC₃ etc. Like aluminium, thallium also forms octahedral ion like [AlF₆]^-3– and [TlF₆]^3-.
Like alkali metals of group-1, due to inert pair effect, thallium exhibits +1 oxidation state.

Example: TlCl, Tl₂O etc. Like alkali metal hydroxides, TlOH is also soluble in water and its aqueous solution is strongly basic. Tl₂SO₄, like alkali metal sulphates form alum and Tl₂CO₃, like alkali metal carbonates soluble in water.

Question 26.
Explain the difference in properties of diamond and graphite on the basis of their structure.
Answer:
Differences in properties of Diamond and Graphite

Diamond	Graphite
1. In this C is sp3 hybridized.	In this C is sp2 hybridized.
2. Its geometry is three-dimensional tetrahedral.	Its geometry is two-dimensional hexagonal layers.
3. It is hardest substance with high density and high boiling point.	It is a soft slippery substance with low density and high boiling point.
4. It is a bad conductor of electricity (due to absence of free electrons).	It is a good conductor of electricity (fourth electron is free).
5. It is used in cutting glass, jewellery and as an abrassive.	It is used as a lubricant, in the manufacture of electrodes, in making pencils, crucibles (due to high melting point) etc.

Question 27.
Explain back bonding with example.
Answer:
There are total 6 electrons in the valence shell of boron trifluoride. Being an electron acceptor, BF³ acts as a Lewis acid. It should be a strong Lewis acid but it acts as a weak Lewis acid.

BF³ is a symmetrically planar molecule due to sp² hybridized boron in BF³. In this molecule, one 2p_z orbital of boron remains completely vacant. On the other side, there are 2 electrons in 2p_z orbital of fluorine. In this situation, due to overlapping between 2p_z orbital of boron and 2p_z orbital of fluorine a bond can be formed. This is known as back bonding.

Question 28.
1. BCl₃ is stable but B₂Cl₆ does not exist whereas AlCl₃ is unstable.
2. AlCl₃ is unstable, Al₂Cl₆ is stable. What is the cause?
Answer:
1. BCl₃ is stable because of the presence of electrons in its valence shell but due to back bonding, the various resonating structure provide stability to BCl₃ due to resonance. But B does not contain vacant d orbital, therefore, it cannot accept the electron pair donated by chlorine, therefore formation of B₂Cl₆ is not possible.

2. In the valence shell of AlCl₃, there are total 6 electrons. Due to incomplete octet, AlCl₃ is unstable but in Al₂Cl₆ dimer, vacant d orbital of aluminium can accept the lone electron pair donated by chlorine to complete its octet. Therefore Al₂Cl₆ is stable.

Question 29.
Write the formula and two uses of the following compounds :
1. Borax,
2. Boric acid.
Answer:
1. Borax:
Formula: Na₂B₄O₇.10H₂O
Uses:

• Due to its resisting character, it is used in the preparation of medicinal soaps.
• In the manufacture of spectacle glass (boro glass).

2. Boric acid :
Formula: H₃BO₃
Uses :

• Boric acid is used in the manufacture of enamel and in brightening utensils.
• Due to its antiseptic nature, boric acid is used to wash eyes.

The p-Block Elements Class 11 Important Questions Short Answer Type – II

Question 1.
Rationalise the given statements and give chemical reactions :
(a) Lead (II) chloride reacts with Cl₂ to give PbCl₄.
(b) Lead (IV) chloride is highly unstable towards heat.
(c) Lead is known not to form an iodide Pbl₄.
Answer:
(a) Due to inert pair effect, Pb is more stable in +2 oxidation state than in +4 oxidation state.
Thus, lead (II) chloride does not give lead (IV) chloride on reacting with chlorine.

(b) Lead (IV) chloride dissociate on heating and gives lead (II) chloride and Cl₂ because lead is more stable in +2 oxidation state than in +4 oxidation state.
2PbCl_4(l) →PbCl_2(s) + Cl_2(g)

(c) Due to strong reducing tendency of Pb⁺⁴ ions and reducing tendency of F^– ion, Pbl₄ does not exist.

Question 2.
If starting material for the manufacture of silicones is RSiCl₃. Write the structure of the product formed.
Answer:
By the hydrolysis of alkyl trichlorosilane followed by condensation polymerization, chain polymer (silicone) is obtained.

Question 3.
What is catenation? In which element this property is maximum and why?
Answer:
The property of linking with atoms of its own kind is called catenation. This property is maximum in carbon because carbon atom is small in size and its electronegativity is high. Bond energy of C-C bond (355.3 kJ mol^-1) is maximum. Thus, it possess the tendency to form long carbon chains.

Question 4.
Give structure of Diamond.
Answer:
Diamond is a tetrahedral molecule, forming three- dimensional network. Each carbon has sp3 hybridization and is linked with four other carbon atoms by strong covalent bonds as shown in the Fig. It is clear from the figure that each carbon is situated at the centre of a regular tetrahedron and other four carbon atoms are present at the four corners of it. Experimental data proves this structure. In this structure, C-C bond length is 154 A (154 picometer).

Hence, in structure of diamond there is three-dimensional network of strong covalent bonds. It is for this reason that the diamond is the hardest substance known and it has a very high melting point (m.p. of diamond is 3843 K). All the valencies of carbon take part in the formation of C – C bonds and therefore, diamond is a bad conductor of electricity.

Question 5.
Give structure of Graphite.
Answer:
Structure: In graphite, each carbon is sp2 hybridized. Each carbon is linked with three other carbon atoms by single covalent bonds. The fourth electron of each carbon remains free and this accounts for good conductivity of graphite. The C – C bond length in graphite is 1.42Å (142 picometer). All the three sp² hybridized orbitals lie in the same plane (are planar). Hence, in graphite carbon atoms form a hexagonal ring and lie in the same plane.

These rings combine together to form a plane. In graphite, there are many such planes which lie one over the other at a distance of 3.4Å. Weak van der Waals’ forces are acting in between these planes. Due to this reason, it is possible to slip one plane of graphite over other and this is responsible for lubricating property of graphite and its soft nature.

Question 6.
Explain industrial method of preparation of Artificial graphite with chemical equation.
Answer:
Preparation of artificial graphite: Graphite is obtained by Acheson (Edward G. Acheson) method in electric arc furnace. Mixture of coke and sand is heated in electric furnace. The furnace is fitted with carbon electrodes which are connected together with a thin rod of carbon as shown in the fig.
Electric current is passed. The temperature reaches to 3000°C. Iron oxide or silica acts as catalyst in the reaction.

Question 7.
Explain the method of preparation of artificial diamond.
Answer:
A mixture of sugar, charcoal and iron oxide is taken in a crucible and heated in an electric furnace to a temperature of 3000°C. Then, it is placed in molten lead. Temperature of molten lead is less than iron due to which iron starts solidifying. As a result of pressure carbon starts separating in the form of small crystals of diamond. Iron is separated by dissolving in HCl. This way, artificial diamond is obtained.

Question 8.
Explain Labell and van’t Hoff law related to valency of carbon atom. Or, What do you understand by symmetrical tetrahedral structure of carbon?
Answer:
Atomic number of carbon is 6. On this basis, 2 electrons are in first shell and 4 electrons in the second shell. Thus, its valency is four. According to Labell and van’t Hoff, if we consider carbon atom to be situated at the centre of tetrahedron and the four arms of tetrahedron representing the four valencies of carbon, then the value of angle between any two valencies is 109°28’. By Henry’s experiments, carbon valencies are symmetrically arranged. It is in tetrahedral form in space. These are not in a plane.

Question 9.
Give the cause of lubricanting property of graphite.
Answer:
In graphite, carbon atoms link with each other by covalent bonds and form a hexagonal network. These rings mutually form a layer. In graphite, many such layers lie one over the other at a distance of 3.4 Å and van der Waals’ force are bonded to each other by weak due to which they can easily slide over each other. As a result, graphite is soft and its melting point is high. Therefore graphite is used as a lubricant for such machines which become hot when used.

Question 10.
Diamond is used in cutting tools. Why?
Answer:
In diamond, each carbon atom is in sp3 hybridized state and each carbon is linked with other carbon atoms by strong covalent bonds. Thus a three-dimensional tetrahedral structure is formed in diamond which is very strong. Therefore, diamond is hardest known element and therefore it is used in cutting tools.

Question 11.
Diamond has a specific shine (brilliance). Why? Or, Diamond is used in making jewellery, why?
Answer:
High refractive index and total internal reflection makes diamond shine and appear beautiful. Therefore, diamond is extremely shiny and is used in making costly jewellery.

Question 12.
Graphite is soft and diamond is hard. Why?
Answer:
In graphite, each carbon atom is in sp² hybridized state and each carbon atom in graphite link with nearest three atoms in the same plane and form a hexagonal network. Many such planes are loosely held over one another with weak van der Waals’ force due to which layers of graphite can slide over one another. Due to This property graphite is soft.

In diamond, each carbon atom is in sp³ hybridized state in which each carbon atom is joined to other four carbon atoms by covalent bonds and form a tetrahedral three-dimensional structure, therefore diamond is hard.

Question 13.
Diamond is a bad conductor of electricity, whereas graphite a good conductor. Give reason.
Answer:
Both diamond and graphite are the allotropes of carbon and the outermost shell of their atoms contain four electrons. In diamond carbon is sp³ hybridized state and the four electrons of each carbon atom is linked to nearest four carbon atoms by strong covalent bonds. Thus, no free electron is present in any carbon atom, therefore it is extremely hard and a bad conductor of electricity.

In graphite each carbon is in sp² hybridized state and only three electrons are joined to nearest three carbon atoms by strong covalent bonds and the fourth electron remains free. Therefore, flow of electrons in graphite is easy. As a result graphite is soft, and a good conductor of electricity.

Question 14.
Carbon monoxide has bad effect on our body. Or, Why is CO gas poisonous?
Answer:
CO combines with haemoglobin of blood and forms a stable compound carboxyhaemoglobin.
Haemoglobin + O2 ⇌ Oxyhaemoglobin
In which the ability to carry the oxygen of blood is destroyed due to which the person may become unconscious or even die due to suffocation.

Question 15.
Differentiate physical properties of Diamond and Graphite in a tabular form.
Answer:
Differences in physical properties of Diamond and Graphite

Physical properties	Diamond	Graphite
1. Appearance	Transparent	Opaque, Black
2. Hardness	Hard	Soft
3. Thermal conductivity	Very low (bad conductor)	Low
4. Electrical conductivity	Bad conductor	Good conductor
5. Density	3510 kg/m3	2250 kg/m3
6. Use	In jewellery and in cutting tools	As a lubricant, as electrodes.

Question 16.
What is supercritical liquid?
Answer:
Any gas can be liquefied by increasing pressure at a temperature lower than its criticaL temperature. The pressure at which a gas can be liquefied is known as its critical pressure. But due to sublime nature of CO₂, it cannot be converted to liquid state, therefore above critical pressure it can be converted to supercritical liquid. Critical temperature and critical pressure for CO₂ is 31°C and 729 atmospheric pressure.

Question 17.
Suggest reasons why the B – F bond lengths in BF₃ (130 pm) and BF₄ (143 pm) differ.
Answer:
In BF₃, boron is sp2 hybridized. It contains vacant 2p-orbital. In BF₃, each F contains completely filled 2p-orbital. Since both these orbitals are of similar energy, pπ-pπ back bonding is formed in which one electron pair from completely filled 2p-orbital is transferred to vacant 2p-orbital of boron. This type of bond ] formation is known as back bonding. Thus, some double bond behaviour is found in B – F bond. That is why, bond length of all the three B – F bonds is less.

In [BF₄]^– ion, boron is sp³ hybridized. It does not contain vacant 2p-orbital due to which back bonding is not found in it. In [BF₄]^– ion, all the 4 B – F bonds are complete single bonds. Double bonds are smaller than single bonds. Thus, B – F bond length in BF₃ (130 pm) is less than in [B₄]^–(143 pm).

Question 18.
Write the reactions of carbon monoxide which states that it is :
1. Combustible,
2. Unsaturated compound,
3. Reducing agent.
Answer:
1. Combustible: It liberates CO₂ gas on combustion in presence of oxygen.
CO+\(\frac{1}{2}\) O₂ → CO₂

2. Unsaturated compound: Being an unsaturated compound, carbon monoxide forms addition compounds.

3. Reducing agent: It reduces with metal oxide to metal.
ZnO + CO → Zn + CO₂ ↑

Question 19.
Write similarities and dissimilarities in carbon and silicon.
Answer:
Similarities :

• Both carbon and silicon are non-metals.
• Valence shell configuration of both is ns²np².
• Both show allotropy.
• Both form covalent compounds.
• Covalency of both is four.
• Both possess the property of catenation.
• Oxides of both are acidic.

Dissimilarities:

Carbon	Silicon
1. Except graphite, carbon is a bad conductor of electricity.	Silicon is a semiconductor.
2. Maximum covalency of carbon is four.	Maximum covalency of silicon is six.
3. Tendency of catenation is more in carbon.	Tendency of catenation is less in silicon.
4. Carbon forms multiple bonds.	Silicon does not form multiple bond.
5. CO is known.	SiO is unknown.
6. CCl4 does not hydrolyse.	SiCl4 undergoes hydrolysis.

Question 20.
Carbon and silicon show tetravalency whereas Ge, Sn and Pb are bivalent. Why?
Answer:
Due to the small size of carbon and silicon, their ionization energy is very high. Therefore, they do not form ionic compounds by losing electrons but for completing their octet, they share electrons to form covalent compounds and exhibit tetravalency.

Due to large size of Ge, Sn and Pb their ionization energy is very less. Therefore, they can lose electrons to form ionic compounds also and these elements exhibit +2 and +4 oxidation state. Due to inert pair effect on moving from top to bottom in a group, stability of +4 oxidation state decreases, but stability of +2 oxidation state increases. Therefore, they represent +2 oxidation state.

Question 21.
SnCl₄ is a liquid whereas SnCl₂ is a solid. Why? Or, Molecular mass of a compound of tin is 189 and of the other is 260, still the first compound is solid whereas the second is a liquid. Why?
Answer:
In SnCl₄, Sn exhibits +4 oxidation state and forms covalent compounds, therefore SnCl₄ is a liquid whereas in SnCl₂, Sn exhibits +2 oxidation state and forms ionic compounds. Therefore, SnCl₂ is a solid.

Question 22.
Si, like C does not form graphite-like structure. Why?
Answer:
Si does not form graphite-like structure because :

1. Si does not form sp² hybridized compounds whereas in graphite, carbon is sp² hybridized.
2. Atomic radius of Si is greater than carbon due to which electron affinity, ionization energy etc. is less than carbon due to which Si does not form π -bonds like carbon.

Question 23.
Maximum covalency of carbon is 4, whereas maximum covalency of other members of this group is 6. Why? Or, Carbon does not show higher oxidation state like Si. Why?
Answer:
All the members of carbon family contain 4 unpaired electrons in their valence shell in excited state. They require 4 more electrons to complete their octet. Therefore, they share electrons with other elements and form covalent bond. Therefore, their valency is 4.

In carbon, due to absence of d-orbital higher oxidation state is not possible whereas in other members, higher oxidation state is possible due to presence of vacant d-orbital because due to presence of vacant d-orbital, they can behave as electron acceptor and can accept a lone electron pair from any other electron donor group forming a co-ordinate bond. Therefore, their maximum oxidation state is 6.

Question 24.
CCl₄ does not hydrolyse whereas SiCl₄ hydrolysis. Why?
Answer:
Due to absence of vacant d-orbital in carbon, its maximum oxidation state is 4 and it cannot increase its oxidation state. Therefore, CCl₄ does not hydrolyse.
In Si, due to presence of vacant d-orbital, its maximum oxidation state is 6, therefore SiCl₄ can easily accept the lone electron pair donated by water and this way its oxidation state increases, due to which it easily hydrolyse.

Question 25.
CO₂ is a gas, whereas Si02 is a high melting point solid. Why ? Or, Explain the structure of CO₂ and SiO₂.
Answer:
CO₂ has a linear structure. In it carbon is in sp hybridized state and CO₂ molecules are attracted by weak van der waals’ force of attraction. Therefore, at normal temperature CO₂ is a gas.

SiO₂ is a solid. Its structure is like a three dimentional network. In it, each Si is linked with four oxygen atom in tetrahedral form. Single covalent bond is present between Si and O atom. This single covalent bond is stronger than van der Waals’ force.
Therefore, SiO₂ is a hard solid and its melting point is high.

Question 26.
Compare CO₂ and SiO₂.
Answer:
Comparison of CO₂ and SiO₂

Properties	CO2	SiO2
1. Physical state	Gas	Solid
2. Structure	Linear monomer	Three-dimensional lattice
3. Melting point	Low	High
4. Reactivity	More reactive	Less reactive
5. Hybridization	sp	sp3
6. Solubility in water	Soluble	Insoluble
7. Reaction with base	Form carbonate	Form silicate.

Question 27.
Carbon like other members of its group do not form complex compound. Why?
Answer:
Tendency of an element to form co-ordinate or complex compound depend on the following factors:
1. Small atomic radius,
2. High charge density,
3. Presence of d-orbital.
From the electronic configuration of carbon, it is clear that carbon does not contain vacant d-orbital, therefore it cannot accept electron pair by ligand and form complex compound. Whereas other members of the group contain vacant d-orbital due to which they can easily accept the electron pair donated by ligand and can form coordinate bond. Therefore, they easily form complex compound.

Question 28.
M²⁺ is strong reducing agent whereas M⁴⁺ ion represent covalent character.
Answer:
All the members of carbon family contain 4 electrons in their valence shell and in M⁴⁺ state its ionization energy is very high. Therefore, all the elements share their electrons with other elements and form covalent compounds.
Whereas less ionization energy is required to remove two electrons, therefore in +2 oxidation state, they form ionic compounds. Due to the tendency to donate electrons, they behave as reducing agents.

Question 29.
Normally compounds of tin and lead like SnCl₂ and PbCl₂ are used as reducing agents whereas SnCl, and PbCl₄ are used as oxidizing agents. Why?
Answer:
In Sn and Pb, +2 oxidation state is more stable than +4 oxidation state. Thus, Sn4+ and Pb4+ possess the tendency to form Sn2+ and Pb²⁺ respectively due to which it reduces others. Hence Sn⁴⁺ and Pb⁴⁺ is oxidising agent and Sn²⁺ and Pb²⁺ possess the tendency to form Sn⁴⁺ and Pb⁴⁺. That’s why it oxidizes others.

Question 30.
Explain the orbital structure of monoxide by carbon.
Answer:
In CO, both carbon and oxygen are in sp-hybridized state.

One sp-orbital of carbon overlaps with one sp-orbital of oxygen and form σ-bond. Other orbital of carbon and oxygen contain one-one electron pair which is non-bonding. One electron p_z-orbital of carbon undergoes lateral overlapping with one electron of p_z – orbital of oxygen to form a π-bond.
Now, there is no electron in p orbital of carbon, whereas there are two electron in p_y orbital of oxygen. These also undergo lateral overlapping to form a bond which represents co-ordination in Lewis structure.

Question 31.
What is Silica garden?
Answer:
In a crystal of saturated aqueous solution of sodium silicate if sand, crystal of copper sulphate, ferrous sulphate, nickel sulphate, cadmium nitrate, manganese sulphate, cobalt nitrate etc. are added then after two three days coloured plants appear to grow in the solution. This is known as Silica garden.

Question 32.
Why does silicon monoxide not form like carbon monoxide ?
Answer:
Carbon can form a π-bond with oxygen after the formation of a σ-covalent bond. Also another vacant 2p_z orbital of carbon can undergo overlapping with lone electron pair situated in 2p_z orbital of oxygen. Because carbon is so much electronegative that it can accept a lone electron pair by oxygen. Whereas Si is less electronegative and it also has a bigger size due to which it cannot form 3pπ – 2pπ. Therefore, SiO is not possible.

Question 33.
Write balanced equations for the preparation of water gas, carburetted water gas and producer gas.
Answer:
1. Water-gas: This gas is a mixture of carbon monoxide and hydrogen. It is prepared by passing steam over red hot coke.

2. Carburetted water gas : On passing water gas through hot bricks placed in oil, acetylene and ethylene are formed which mix with water gas to form carburetted water gas. It contains CO = 30%, H₂ = 35%, saturated hydrocarbons = 15.20%, hydrocarbon = 10%, N2 = 2.5 – 5%, C02 = 2%.
3. Producer gas: It is a mixture of carbon monoxide and nitrogen. It is obtained by passing limited amount of air on red hot coke.
2C + Air (O₂ +N₂) →2CO + N₂

Question 34.
How is silica gel obtained from silicon tetrachloride?
Answer:
Silicon tetrachloride is obtained by, the reaction of silicon with chlorine.
Si + 2Cl₂ →SiCl₄
By the hydrolysis of SiCl4, silicon tetrahydroxide is obtained.
SiCl₄ + 4HOH → Si(OH)₄ +4HCl

This silicon tetrahydroxide is actually silicic acid monohydrate.
Si (OH)₄ ⇌ H₂SiO₃ .H₂O
On heating, silicic acid breaks into silica.
H₂SiO₃.H₂O → SiO₂ + 2H₂O
This silica, is known as silica gel (SiO₂ xH₂O).

The p-Block Elements Class 11 Important Questions Long Answer Type

Question 1.
Discuss the pattern of variation in the oxidation state of (i) B to TI and (ii) C to Pb.
Answer:

Boron and aluminium exhibit only +3 oxidation state because due to absence of d and d-electrons, they do not show inert pair effect. Elements from Ga to Tl exhibit +1 and +3 oxidation state. On moving down the group, tendency of+1 oxidation state increases because tendency of ns² electrons of valence shell for bond formation goes on decreasing. This is known as inert pair effect. Tl⁺ is more stable than Tl⁺³.

Carbon and silicon exhibit only +4 oxidation state. Tendency to represent +2 oxidation state in heavier members increases in the order Ge < Sn < Pb. This is due to class tendency of ns² electrons of valence shell towards bond formation (inert pair effect). Sn forms compounds in both the states and compounds of lead are more stable in +2 oxidation state than in +4 state.

Question 2.
Write diagonal relationship in Be and Al.
Answer:
Elements placed diagonally in second and third period show resemblance in properties. This is known as diagonal relationship.
1. Both possess same electronegativity.
Be =1.5 Al = 1.5 .
2. Polarizing power of Be⁺² and Al⁺³ is nearly same.

3. Alkaline earth metals are soft, but beryllium is hard like aluminium.
4. Like Al, Be is also passive to concentrated nitric acid.
5. Like Aluminium carbide, Be₂C reacts with water to liberate methane.
Be₊₂C + 2H₊₂O →2BeO + CH₄
Al₄C₃ + 12H₂O → 4Al(OH)₃ +3CH₄

6. Both Be arid Al react with NaOH and librate hydrogen.
Be + 2NaOH → Na₂BeO₂ + H₂
2Al+ 2NaOH +2H₂O →2NaAlO₂ +3H₂

7. Oxides of both are amphoteric.
BeO + 2 HCl →BeCl₂ +H₂O
BeO+2NaOH → Na₂BeO₂ +H₂O
Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O
Al₂O₃ +2NaOH →2NaAlO₂ + H₂O
8. BeCl₂ and AlCl₃ exist in the form of dimer and polymer.
9. Hydroxides of both are insoluble in water and dissociate on heating.
Be(OH)₂ →BeO + H₂O
2Al(OH)₃ → Al₂O₃ +3H₂O

10. BeCl₂ and AlCl₃ are strong Lewis acids.
11. Both the metals react with halogens to form halides.
Be + Cl₂ →BeCl₂
2Al + 3Cl, → 2AlCl₃
12. Halides of both show covalent behaviour and are soluble in organic solvents.

Question 3.
Write similarities and dissimilarities in B and Al.
Answer:
Similarities :

1. Electronic configuration of valence shell of both is ns²np.
2. Covalency of both is 6.
3. Oxidation number of both is +3.
4. Both form M₂O₃ type of oxides.
5. Compounds of both act as strong Lewis acids.
6. Oxides of both are of amphoteric nature.

Dissimilarities:

Boron	Aluminium
1. Boron is a non-metal.	1. Aluminium is a metal.
2. Boron is a bad conductor of heat and electricity.	2. Aluminium is a good conductor of heat and electricity.
3. Its melting point is very high.	3. Its melting point is very low.
4. It does not react with dilute HCl and H2SO4.	4. It reacts with dilute HCl and H2SO4 to liberate H2. 2Al + 3H2SO4 → Al2(SO4)3 + 3H2
5. It reacts with cone. HNO3B + 3HNO3 → H3BO3 + 3NO2	5. It is passive towards cone. HNO3.
6. It reacts with metals to form borides.	6. 3Mg + 2B → Mg3B2 It reacts with metals to form alloys.
7. Maximum covalency of boron is 4.	7. Maximum covalency of Al is 6.
8. Its carbide is covalent.	8. Its carbide are ionic and hydrolyse to give methane.

Question 4.
Boron shows anomalous behaviour with other members of the group. Ex-plain.
Answer:
Boron shows anomalous behaviour with other members of the group because :

1. Atomic and ionic radius of boron is less.
2. Boron is a bad conductor of electricity, where as others are good conductors.
3. Boron forms covalent compounds where as other members of the group form ionic compound.
4. Compounds of boron are insoluble in water but soluble in organic solvents where as compounds of other members are soluble in water.
5. Boron does not form trivalent ion like other members.
6. Maximum covalency of boron is 4 where as maximum covalency of other members is 6.
7. Oxides of boron is acidic where as oxides of other members are either amphoteric or of basic nature.
8. Boron reacts with other metals to form borides where as other members react with metals to form alloys.
9. Boron forms more than one type of hydride, where as other members form only one hydride.

Question 5.
What are boranes? Write their characteristics and uses.
Answer:
Hydrides of boron are known as boranes. Boron forms hydrides in two series :
Nido borane series : Its general formula is B_nH_n+4. Its first member BH₅ does not exist. Second member B₂H₆ is most important which is known as diborane. Other important members are pentaborane B₅H₉, hexaborane B₆H₁₀.
Borane series : Its general formula is B_nH_n+6. Important member of this series is tetraborane B₄H₁₀, pentaborane B₅H₁₁, hexaborane B₆H₁₂.

Method of preparation:
1. Diborane is obtained by the reaction of BX₃ with lithium hydride at 450K temperature.

2. Diborane is obtained by the reduction of Boron trihalide with lithium aluminium tetrahydride.
4BCl₃ +3LiAlH₄→ 2B₂H₆+ 3LiCl+3AlCl₃

Characteristics :

• Diborane is a colourless gas where as other members are volatile solid.
• Diborane after combustion in presence of oxygen releases heat. Therefore, it is used as a rocket fuel.
• It is stable at low temperature. It starts dissociating at a higher temperature.

Uses :

1. In the form of a rocket fuel.
2. As a catalyst in polymerization reaction.

Question 6.
Show the structure of diborane. Or, Draw the structure of diborane.
Answer:
Structure of diborane: It is very interesting because it does not have sufficient valence electrons. So it is called electron-deficient compounds. Its structure is determined by electron diffraction studies.

Terminal hydrogens (4) and both boron atoms are in one plane and bridged hydrogens are perpendicular to this plane in both sides. The four-terminal B-H bonds are normal electron-pair bond while B-H-B bonds are electron deficient in which 2 electrons join 3 atoms. Thus, it is called 2 electrons 3 centre or (2e – 3c) bond.

Question 7.
Compare boron and carbon.
Answer:
Similarities:
1. Both boron and carbon are non-metals.
2. Both show allotropy,
3. form more than one type of hydride.
4. Compounds of both are covalent.
5. Com- both are soluble in organic solvents.
6. Crystalline form of boron is hard like
7. Both CO₂ and B₂O₃ dissolve in alkali and form carbonate and borate.
2NaOH +CO₂ →Na₂CO₃ +H₂O
2NaOH + B₂O₃ → Na₂B₂O₃ + H₂O.
Dissimilarities:

Boron	Carbon
1. Electronic configuration of boron is 1s2,2s22p1.	Electronic configuration of carbon is 1s2,2s22p2.
2. Covalent of boron is 3.	Covalent of carbon is 4.
3. Compounds of boron are electron deficient.	Carbon forms double or triple bond.
4.Boarn does not double or triple Bond.	Compounds of boron are not electron deficient.
5. Compounds of boron are Lewis acids.	Compounds of carbon are not Lewis acids.

Question 8.
Write diagonal relationship between boron and silicon.
Answer:
Elements placed diagonally in second and third period show similarity in properties. This is known as diagonal relationship.

Diagonal relationship :
1. Both boron and silicon does not exist in free state in nature. They are found as compounds.
2. Both have same density and electronegativity.

3. Both are non-metals and exhibit allotropy.
4. Both have high melting point.
5. Both are bad conductor of electricity.
6. Both does not form cation.
7. Both do not react with dilute HCl.

8. Oxides of both react with base to form borate and silicate.
B₂O₃ +6NaOH → 2Na₃BO₃+ 3H₂O
SiO₂ + 2NaOH → Na₂SiO₃ + H₂O

9. Both form various covalent hydrides.
10. Both form halide with halogen which get hydrolysed with water.
4BF₃ +2H₂O →HBO₂ +3HBF₄
3SiF₄+3H₂O → H₂SiO₃+2H₂SiF₆

11. Both react with nitrogen at high temperature to form nitride.
2B + N₂ →2BN
2Si + 4N₂ →2SiN₄

12. Both the elements react with metals at high temperature to form boride and silicate.
3Mg +2B → Mg₃B₂
2Mg + Si → Mg₂Si.

Question 9.
Compare the structure of BCl₃ and AlCl₃.
Answer:
BCl₃ is an electron-deficient compound which always exist as a monomer. In BCl₃, B is in sp² hybridized state. Therefore, its structure is trigonal and bond angle is 120°. Because atomic radius of boron is small and chlorine bridge is unstable, therefore it does not form dimer structure.

AlCl₃ always exist as a dimer. In this dimer structure, each Al atom accept lone electron pair from the chlorine atom linked to other Al and complete its octet and achieve stability.

Question 10.
Explain the structure of diborane and boric acid.
Answer:
Structure of diborane: In diborane structure, four-terminal hydrogen and two boron atoms are in one plane. There are two bridged hydrogen atoms above and below this plane. The 4 terminal B – H bonds are normal bonds where as two bridged (B – H – B) are of different types, they are known as banana bond (or three centred bond).

Structure of boric acid: Boric acid has layered structure, in which H₃BO₃ units are joined by hydrogen atoms.

Question 11.
Explain Goldschmidt aluminothermic process with a labelled diagram. Or, Explain the thermite welding process.
Answer:
Thermite process: It is also called Goldschmidt aluminothermic process. In this process, refractory crucible is filled with oxide of the metal to be reduced, aluminium powder and barium peroxide. After filling magnesium ribbon it is fixed in sand. Now, the magnesium ribbon is set on fire. A very high temperature is produced at which metal oxide get reduced by aluminium and collected in molten state. Reaction is extremely exothermic and the O₂ required is supplied by barium peroxide. Selection of reducing metal is done on the basis of nature of ore.

Cr₂O₃+2Al → Al₂O₃ +2Cr + Heat
Fe₂O₃+2Al →2Fe + Al₂O₃

Question 12.
What is Alum? Write its general formula, method of preparation, properties and uses. Or, Explain the method of preparation of common alum. Write any four uses of it.
Answer:
Alums: Double salts which can be represented by the general formula R₂SO₄. M₂ (SO₄)₃. 24H₂O are called Alums. Here R=A monovalent metal Na, K, Rb, Cs or NH₄ radical
M = Trivalent metal like Fe, Al or Cr.
Alums in which trivalent metal is Al, are named as the alums of monovalent metal or radical present in them like :
Potash alum K₂SO₄. Al₂(SO₄)₃.24H₂O
Ammonium alum (NH₄)₂SO₄. Al₂(SO₄)₃.24H₂O.
Methods of preparation: On crystallizing a mixture of equimolecular proportion of potassium sulphate and aluminium sulphate solution.
K₂SO₄ + Al₂ (SO₄)₃ + 24H₂O →K₂SO₄. Al₂ (SO₄)₃.24H₂O

Properties: 1. Colourless, octahedral crystals are formed whose aqueous solution is acidic due to hydrolysis. Solid alum is soluble in water but insoluble in alcohol. It’s one molecule contain 24 molecules of crystallized water.
2. On heating it melt at 92°C, on heating up to 200°C all the water of crystallization is lost and alum swells up and becomes porous. This type of alum is known as burnt alum.

Uses :

• To stop the flow of blood and in medicines. Blood contains negative charge and due to positive charge of Al³⁺ coagulation takes place which stops bleeding.
• In dyeing and printing cloth, in sizing of paper.
• In purifying water.
• In special foam extinguishers.

Question 13.
What happens when :
(a) Borax is heated strongly,
(b) Boric acid is added to water,
(c) Aluminium is treated with dilute NaOH,
(d) BF₃ is reacted with ammonia.
Answer:
(a) When borax is heated strongly, sodium metaborate and glass-like transparent beads of boric anhydride are obtained.

Sodium metaborate Boric anhydride Glass like transparent bead

(b) Boric acid is partially soluble in cold water where as readily soluble in hot water. It acts as a weak monobasic acid. It is not a protic acid but by accepting a hydroxide ion from water and donating a proton it behaves like a Lewis acid.
H-OH + B(OH)₃ →[B(OH)₄]^–+H⁺
(c) When aluminium is treated with dilute NaOH, dihydrogen is released.
2Al_(s) + 2NaOH_(aq) + 6H₂O_(l) → 2Na⁺[ Al(OH)₄]^–+ 3H_2(g)
(d) BF₃, being a Lewis acid accepts an electron pair from NH₃ and forms a complex compound.

Question 14.
Carbon shows anomalous behaviour in comparison to other members of its group. Why?
Answer:
Carbon shows anomalous behaviour in comparison to other members of the group because:
1. Atomic and ionic radius is less.
2. Ionisation energy is high.
3. High electron affinity.
4. Absence of d-orbital.

Anomalous behaviour:

• Melting and boiling points of carbon is higher than other members.
• Tendency of catenation of carbon is higher than the other members.
• Carbon forms multiple bonds where as other members do not form multiple bonds.
• Monoxide of carbon is known where as that of other members are not known.
• Maximum covalency of carbon is 4 where as maximum covalency of other members is 6.
• Carbon like other members, does not form complex compounds.
• Carbon forms more than one type of hydride where as other members form only one type of hydride.
• CCl₄ does not hydrolyse where as tetrahalides of other members easily hydrolyse.
• CO₂ is a gas where as dioxide of other members are solid.

Question 15.
Explain the following reactions :
(a) Silicon is heated with methyl chloride at high temperature in the presence of copper.
(b) Silicon dioxide is treated with hydrogen fluoride.
(c) CO is heated with ZnO.
(d) Hydrated alumina is treated with aqueous NaOH solution.
Answer:
(a) On heating silicon with methyl chloride at high temperature in the presence of copper mixture of mono, di, tri methyl chlorosilicon and tetramethylsilane is formed.

(b) On treating silicon dioxide with hydrogen fluoride, silicon tetrafluoride is obtained which dissolves in HF to form hydro fluorosilicic acid.
SiO₂ +4HF → SiF₄ +2H₂O
SiF₄ + 2HF → H₂SiF₆

(c) CO is a strong reducing substance but it does not reduce ZnO because ΔG° for CO → CO₂, is always higher than corresponding value of ZnO. Thus, no reaction takes place.
(d) Alumina dissolves in NaOH to form sodium metaluminate.

Question 16.
Write notes on the following :
1. Freon,
2. Silicones.
Answer:
1. Freon: Dichloro difluoro methane are called freons, freons are formed by the reaction of carbon tetrachloride with HF or SF3 in presence of SbCl₅. Freon is used as a refrigerant in refrigerators and A.C.

2. Silicones: Silicone is a synthetic polymer whose basic unit is R₂SiCl₂. Its molecular formula is like ketone, therefore these are named silicones. By the reaction of alkyl halide and Si in the presence of Cu at 575 K temperature dialkyl dihalide silane is obtained.

Properties :
1. Chemically inert,
2. Water repellant,
3. Insulators,
4. Unaffected by heat or heat resistant.

Use :
1. As waterproof paper,
2. As a lubricant.

Question 17.
Give reasons :
(i) Cone. HNO₃ can be transported in aluminium container.
(ii) A mixture of dilute NaOH and aluminium pieces is used to open drain.
(iii) Graphite is used as lubricant.
(iv) Diamond is used as an abrasive.
(v) Aluminium alloys are used to make aircraft body.
(vi) Water should not be kept in aluminium utensils overnight.
(vii) Aluminium wire is used to make transmission cables.
Answer:
(i) Al reacts with cone. HNO₃ and develops a protective layer of aluminium oxide on its surface which prevent its further reactions.

Thus, Al becomes inactive. That is why cone. HNO₃ is transported in aluminium container.
(ii) NaOH reacts with A1 to liberate dihydrogen gas. This hydrogen gas is used to open drains.
2Al_(s) + 2NaOH_(aq)+2H₂O_(l) → 2NaAlO_2(aq) + 3H_2(g)

(iii) Graphite has layered structure. These layers are joined mutually by weak van der Waals’ attractive forces. Thus, can easily slide over one another. That is why graphite is used as a dry lubricant.
(iv) In diamond, each sp3 hybridized atom is linked with four other carbon atoms by which three dimensional lattice is formed. It is difficult to break this large covalent bond. Thus, diamond is the hardest substance found on earth. That is why it is used as an abrasive.

(v) Alloy of aluminium like duration is light, strong and corrosion-resistant. That is why, it is used in making aircraft body.
(vi) Aluminium reacts with water and oxygen (present in water) and forms a thin layer of poisonous aluminium oxide only on the walls. Therefore, water should not be kept in aluminium utensils overnight.
2Al_(s) + O_2(g) + H₂O_(l) → Al₂O_3(s) + H_2(g)

(vii) Aluminium has high electrical conductivity. It is used for the preparation of trans¬mission cables. Again on the basis of weight its electrical conductivity is double than Cu.

Question 18.
(a) Classify the following oxides into neutral, basic, acidic and amphoteric oxides:
CO, B₂O₃, SiO₂, Al₂O₃, PbO₃, PbO₂, Tl₂O₃.
(b) Write the chemical reactions to show their nature.
Answer:

(ii) Due to amphoteric nature Al₂O₃ and PbO₂, react both with acids and alkalies.
Al₂O₃ + 3H₂SO₄ → Al₂ (SO₄)₃ + 3H₂O

(iii) Due to basic nature TlO₃ reacts with acids.
Tl₂O₃ + 6HCl → 2TlCl₃ + 3H₂O

Question 19.
Write a brief note on fullerenes.
Answer:
Fullerenes is a crystalline allotrope of carbon, but due to its ball like structure and presence of 60 carbon atom in its spherical crystal, such type of crystals are like dust particles. Formula of one unit of crystal is C₆₀, C₇₀, C₈₀. C₆₀ fullerene is also known as Buckminster fullerene. On evaporating graphite in an electric arc in helium or argon medium and condensing it, dust like powder is collected.

Properties: Fullerenes are like dust particles which are soft and spherical. Dissolves in organic solvents to give a coloured solution. Reacts with alkali metal like sodium to give Na₃C₆₀ compound. Fullerenes polymerize in presence of ultraviolet rays though their crystals do not break at 1375K temperature.

Structure: In fullerenes, there are 20, six carbon atom rings, 12 five carbon atom rings. All five atomic rings are joined to six atomic rings to which a spherically symmetrical shape is obtained. Therefore, C₆₀ fullerene is also known as Buckyball. It is a cage-like football, radius of each sphere is 700pm.

Uses :
1. As a lubricant,
2. Compounds formed with alkali metals as semiconductors.

Question 20.
What do you understand by the following :
(a) Inert pair effect,
(b) Allotropy,
(c) Catenation.
Answer:
(a) Inert pair effect: The property of 5-electrons not to participate in bond formation is called inert pair effect. This is because the energy required to unpair the ns² electrons is more than the energy released in the formation of the two bonds. Havier elements of group – 13, 14, 15 exhibit lower oxidation state than the number of electrons present in their valence shell. For example, in T1 +1 oxidation state is more stable than +3 oxidation state.

(b) Allotropy: The existence of an element in two or more forms which possess different physical properties but similar chemical properties, then these forms are known as allotropes and the phenomenon is known as allotropy. This is either due to difference in the number of atoms in the molecule (like O₂ and O₃) or due to difference in the arrangement of atoms in the molecule [like graphite, diamond and fullerene (crystalline allotrope of carbon)].

(c) Catenation: The property of self linking of atoms of the same type to form long, open or closed chain is called catenation. This property is maximum in carbon and goes on decreasing on moving down the group. In group-14, the order is as follows :
C >> Si >> Ge = Sn >> Pb

Question 21.
A certain salt X gives the following result:
(i) Its aqueous solution is alkaline to litmus.
(ii) It swells up to a glass material Y on strong heating.
(iii) When cone. H₂SO₄ is added to a hot solution of X, white crystal of an acid Z separates out.
Answer:
(i) Aqueous solution of salt is alkaline which represents that X is a strong base and a weak acid.
(ii) On strong heating, salt X swells up to a glassy solid Y. This represents that salt X is borax.
(iii) Hot aqueous solution of borax, reacts with cone.H₂SO₄ to form crystals of orthoboric acid.
Equations involved in the above reactions are as follows :

Question 22.
Write method of preparation, properties and use of carbon monoxide.
Answer:
Preparation of CO :
1. Carbon monoxide is obtained by the incomplete combustion of carbon or hydrocarbon.
2. CO is obtained by the dehydration of formic acid with cone. H₂SO₄.

Physical properties :

• Density of this gas is 1.25 gm/litre.
• It is a colourless, odourless, poisonous gas.
• Its boiling point is 81 K.
• Partially soluble in water.
• Strong reducing agent.

Chemical properties :
1. Combustibility: Carbon monoxide undergoes combustion in the presence of air to form C02.
2CO + O₂ → 2CO₂
2. Reducing property: Carbon monoxide is a strong reducing agent. It reduces metal oxides to metals.
Fe₂O₃ + 3CO → 2Fe + 3CO₂
CuO + CO → Cu + CO₂

3. It reacts with chlorine and forms a poisonous gas named phosgene.
CO + Cl₂ → COCl₂
4. Manufacture of metal carbonyl: Due to the presence of non-bonded electron pair on carbon monoxide, it acts as a ligand and manufactures metal carbonyls by forming a co-ordinate bond with transition metals.
Ni + 4CO →Ni (CO)₄
Fe + 5CO → Fe (CO)₅

Uses :

1. In industries carbon monoxide is used as a fuel (as water gas and producer gas).
2. As a reducing agent.
3. In industrial preparation of methanol and formic acid.
4. In the purification of nickel.

Question 23.
Write a note on zeolite.
Answer:
Zeolite is a type of complex silicate in which few silicon atoms are replaced by Al⁺³ ion. To balance the difference in valency of Si⁺⁴ and Al⁺³ ion some other ions like Na⁺, K⁺, Ca⁺², Mg⁺² etc. are present which keeps the molecule electrically neutral. Their general formula is M_x[(AlO₂)_x(SiO₂)_y ].mH₂O.
Example: Na₂[Al₂Si₃O₁₀].2H₂O
Ca[Al₂Si₇O₁₈].6H₂O

Zeolites have a honeycomb-like structure which has pores and channels of various size. Size of these pores is between 260 pm to 740 pm. Molecules or ions of appropriate size can be absorbed in these pores and by the channels liberation or absorption of molecules like water can take places. Therefore, these are also known as molecular sieves and they act as selecting size catalysts. Reactions catalyzed by them depends on the size of pores present in them and also on the size of reactants and products.

Structure: 24 units of tetrahedral SiO^-4₄ ions join to form one block of zeolite. This dense octahedral block is known as octahedral block or sodalite cage. These blocks of sodalite join mutually by four-membered ring to form a two dimensional or three-dimensional network. Due to this type of structure zeolites are extremely porous. If blocks of sodalite cage are joined by double six-membered ring then the network formed is known as fouge sight.

Question 24.
What are allotropes? Sketch the structure of two allotropes of carbon namely diamond and graphite. What is the impact of structure on the physical properties of two allotropes?
Answer:
When an element exists in two or more than two forms which show different physical properties but same chemical properties then these forms are called allotropes and the phenomenon is known as allotropy.
Crystalline carbon mainly exist in two allotropes
(i) Graphite and
(ii) Diamond. In 1985 a third allotrope fullerene of carbon was discovered by H. W. Croto, E. Smale and R. F. Kurl.

In diamond each carbon atom is sp3 hybridized and linked to four other carbon atoms in a tetrahedral geometry. In diamond a three dimensional lattice of carbon atoms is formed. In graphite each carbon is sp² hybridized and forms three sigma bonds with three nearest carbon atoms. It has layered structure and these layers are joined by weak van der Waals’ force of attraction.

Structure of two allotropes diamond and graphite of carbon and effect on their physical properties :
Diamond is a tetrahedral molecule, forming three- dimensional network. Each carbon has sp3 hybridization and is linked with four other carbon atoms by strong covalent bonds as shown in the Fig. It is clear from the figure that each carbon is situated at the centre of a regular tetrahedron and other four carbon atoms are present at the four corners of it. Experimental data proves this structure. In this structure, C-C bond length is 154 A (154 picometer).

Hence, in structure of diamond there is three-dimensional network of strong covalent bonds. It is for this reason that the diamond is the hardest substance known and it has a very high melting point (m.p. of diamond is 3843 K). All the valencies of carbon take part in the formation of C – C bonds and therefore, diamond is a bad conductor of electricity.

Structure: In graphite, each carbon is sp2 hybridized. Each carbon is linked with three other carbon atoms by single covalent bonds. The fourth electron of each carbon remains free and this accounts for good conductivity of graphite. The C – C bond length in graphite is 1.42Å (142 picometer). All the three sp² hybridized orbitals lie in the same plane (are planar). Hence, in graphite carbon atoms form a hexagonal ring and lie in the same plane.

These rings combine together to form a plane. In graphite, there are many such planes which lie one over the other at a distance of 3.4Å. Weak van der Waals’ forces are acting in between these planes. Due to this reason, it is possible to slip one plane of graphite over other and this is responsible for lubricating property of graphite and its soft nature.

(i) Due to its hardness, diamond is used as an abrasive and in making dyes where as being soft in nature, graphite is used as pencils and as a dry lubricant in machines.
(ii) Diamond is not a conductor of electricity where as graphite is a good conductor of electricity because in it an electron of each carbon atom remains in free state.
(iii) Diamond is transparent where as graphite is not transparent.

Question 25.
Write Lewis structure and resonance structure of CO₂.
Answer:
Structure: Lewis structure of CO₂ is as follows :

Standard heat of CO₂ is ΔH⁰f = -393.5kJ /mol^-1 and C – O bond length is 115 pm. By this, it is clear that stability of CO₂ is even more than the stability that should be obtained on the basis of Lewis structure of CO₂.

This is only possible when resonance structure of CO₂ is possible. Carbon dioxide is a resonance hybrid of the following resonating structures.

Orbital structure: In CO₂, carbon is in sp hybridized state and both the oxygen are in sp² hybridized state. In each oxygen in two sp² orbitals non-bonding electron pair are present. One sp² orbital overlaps with sp orbital of carbon to form σ-bond. p_z orbital of one oxygen undergoes lateral overlapping with the p_z orbital of carbon to form π-bond. p_y orbital of other oxygen laterally overlaps with p orbital of carbon to form π-bond.
This way, CO₂ molecule is linear and its dipole moment is zero.

Question 26.
1. What is the basic difference in the silicones formed by the hydrolysis of R₂SiCl₂ and RSiCl₃.
(2) [SiF₆]^-2 is known whereas [SiCl₆]^-2 is not. Why?
Answer:
1. By the hydrolysis of R₂SiCl2, linear polymer of silicones is formed.

2. Size of fluoride ion is small, therefore silicon atom can combine with 6 fluoride ions. Therefore [SiF₆]^-2 is known where as due to larger size of chloride, silicon cannot combine with six chloride ions therefore [SiCl₆]^-2 is not known.

Question 27.
Give one method for industrial preparation and one for laboratory preparation of CO and CO₂ each.
Ans. (a) Carbon monoxide :
(i) Industrial preparation: On passing steam over hot coke, CO is obtained.

(ii) Laboratory method: CO is obtained by the dehydration of formic acid in the presence of cone. H₂SO₄.

Formic acid (b) Carbon dioxide :
(i) Industrial preparation: CO₂ is obtained by heating limestone.

(ii) Laboratory method: CO₂ is obtained by the action of dilute HCl on CaC03.
CaCO_3(s) +2HCl_(aq) →CaCl_2(aq) +CO_2(g)+H₂O_(l)

Question 28.
Write short notes on :
(i) Sodium zeolite,
(ii) Sodium silicate,
(iii) Silicones.
Answer:
(i) Sodium zeolite: Sodium zeolite is also known as sodium permutit. Permutit is mixed silicate of sodium and aluminium. Its formula is : Na₂[Al₃Si₂O₈.xH₂O]
Sodium zeolite is used in converting hard water into soft water. Water containing calcium and magnesium ion is known as hard water. When hard water is passed through permutit kept in a column then calcium and magnesium ions are displaced by sodium. So-dium salts do not make water hard and this way soft water is obtained. If sodium permutit is represented by Na₂P then the following reaction can be written for softening of water.


(ii) Sodium silicate or water glass: It is bright like glass and soluble in water, there-fore it is known as water glass. It contains sodium metasilicate and silica in excess and its formula is Na₂SiO₃ SiO₂ orNa₂SiO₃.3SiO₂.
It is obtained by fusing a mixture of sodium carbonate and sand.
Na₂CO₃+2SiO₂ → Na₂SiO₃. SiO₂ + CO₂
The product obtained is hard, solid like glass and is soluble in water.

If sand, crystals of copper sulphate, ferrous sulphate, nickel sulphate, cadmium nitrate, manganese sulphate and cobalt nitrate etc. are added to a saturated solution of sodium silicate in a tube, then after two or three days coloured plants seem to grow in the solution and this is known as silica garden.

(iii) Silicones: These are synthetic resins of silicon and organic compounds. They are generally prepared from sand, sodium chloride and petroleum. It can be obtained in the form of gas, sticky liquid, solid like rubber or hard like stone. These are organosilicon polymers.

Question 29.
What is zeolite? Explain with example. Write any four uses of it.
Answer:
Zeolite:

Zeolite is a type of complex silicate in which few silicon atoms are replaced by Al⁺³ ion. To balance the difference in valency of Si⁺⁴ and Al⁺³ ion some other ions like Na⁺, K⁺, Ca⁺², Mg⁺² etc. are present which keeps the molecule electrically neutral. Their general formula is M_x[(AlO₂)_x(SiO₂)_y ].mH₂O.
Example: Na₂[Al₂Si₃O₁₀].2H₂O
Ca[Al₂Si₇O₁₈].6H₂O

The p-Block Elements Class 11 Important Questions Objective Type

1. Choose the correct answer:

Question 1.
Which statement is false for boric acid :
(a) It is mono basic acid
(b) It is formed by hydrolysis of boron halide
(c) Its shape is planar
(d) It is tribasic acid.
Answer:
(a) It is mono basic acid

Question 2.
Aqueous solution of Borax is :
(a) Basic
(b) Acidic
(c) Neutral
(d) Amphoteric.
Answer:
(a) Basic

Question 3.
Hybridization of boron atom in diborane is :
(a) sp
(b )sp²
(c) sp³
(d ) sp³d².
Answer:
(b )sp²

Question 4.
Cause of polymerization of boric acid is :
(a) Acidic nature
(b) H-bond
(c) Mono basic acid
(d) Its geometry.
Answer:
(b) H-bond

Question 5.
Main ore of aluminium is :
(a) Bauxite
(b) Dolomite
(c) Galena
(d) Felspar.
Answer:
(a) Bauxite

Question 6.
Three centred-two electron bond is present in :
(a) NH₃
(b) B₂H₆
(c) BCl₃
(d) Al₂Cl₃.
Answer:
(b) B₂H₆

Question 7.
Carborundum is:
(a) B₄C
(b) SiC
(c) Al₃C₄
(d) CaC₂.
Answer:
(b) SiC

Question 8.
Which halide is electron deficient:
(a) CCl₄
(b) NCl₃
(c) Cl₂O
(d) BCl₃.
Answer:
(d) BCl₃.

Question 9.
Stable allotrope of carbon is :
(a) Diamond
(b) Graphite
(c) Coal
(d) Anthracite.
Answer:
(b) Graphite

Question 10.
Best conductor of heat is :
(a) Cu
(b) B
(c) Ag
(d) Diamond.
Answer:
(b) B

Question 11.
Which does not possess electrical conductivity :
(a) K
(b) Graphite
(c) Diamond
(d) Na.
Answer:
(c) Diamond

Question 12.
Amphoteric oxide is:
(a) CaO
(b) CO₂
(c) SiO₂
(d) SnO₂.
Answer:
(c) SiO₂

Question 13.
Hardest known substance is :
(a) Coke
(b) Carborundum
(c) Corundum
(d) Diamond.
Answer:
(d) Diamond.

2. Fill in the blanks:

1. When alumina contain both Fe₂O₃ and SiO₂ as impurity it is concentrated by ………………… process.
Answer:
Hall’s

2. When bauxite contain excess of silica as impurity, then it is concentrated by ………………… process. In this process ………………… is mixed and is heatd to 1800°C in a current of nitrogen.
Answer:
Serpeck, coke

3. Bauxite ore containing silica as impurity is mixed with coke and heated to 1800°C in a current of nitrogen when ………………… is formed and volatile silicon is separated.
Answer:
Aluminium nitride (AIN)

4. Hydrolysis of AIN formed by Serpeck process form ………………… which is heated to obtain Al₂O₃.
Answer:
Neutral

5. Aqueous solution of some oxides of group 13 elements convert blue litmus to red and red litmus to blue. Solution of this type of oxides are called ………………… .
Answer:
Al₂Cl₆

6. Aluminium chloride exist in the form of a dimmer. Its chemical formula is ………………… .
Answer:
Al(OH)₃

7. Boron halide forms electron ………………… compound.
Answer:
Deficient

8. Solid CO₂ is known as ………………… .
Answer:
Dry ice

9. Main factor of depletion of ozone layer is ………………… .
Answer:
Chlorofluoro carbon

10. Carbon monoxide reacts with chlorine in the presence of sunlight to form a poison-ous gas ………………… .
Answer:
Phosgene

11. First scientist to form artificial diamond is ………………… .
Answer:
Moissan

12. Germanium is a ………………… .
Answer:
Metalloid

13. Graphite is a ………………… and diamond is a of electricity.
Answer:
Good conductor,’bad conductor

14. Silicon carbide is known as …………………. .
Answer:
Carborundum.

3. Match the following:

I.

‘A’	‘B’
1. B2O3	(a) Amphoteric
2. TiO2	(b) Acidic
3. Al2O3	(c) Basic
4. Ca2O3	(d) Basic
5. Ln2O3	(e) Amphoteric.

Answer:
1. (c) Basic
2. (c) Basic
3. (d) Basic
4. (a) Amphoteric
5. (e) Amphoteric.

II.

‘A’	‘B’
1. In the preparation of artificial silk (cellulose xanthate)	(a) CCl4
2. In cutting and drilling diamond	(b) C2H2
3. In used for extinguishing fire in fire extinguishes	(c) B4C
4. Is used in welding as oxyacetylene flame	(d) Graphite
5. Black lead is known as Plumbago	(e) CS2.

Answer:
1. (e) CS₂
2. (c) B₄C
3. (a) CCl₄
4. (b) C₂H₂
5. (d) Graphite.

4. Answer in one word/sentence:

1. +1 oxidation state of T1 is more stable as compared to +3.
Answer:
Inert pair effect

2. What are hydrides of Boron known as?
Answer:
Borane

3. Which is known as two electron-three centre bond.
Answer:
Diborane

4. What will happen on mixing excess of ammonia solution in copper sulphate solution?
Answer:
A complex compound cupric ammonium sulphate is formed

5. Write the formula of alum.
Answer:
K₂SO₄. Al₂(SO₄)₃.24H₂O

6. What is the effect of heat on borax?
Answer:
Transparent bead of sodium metaborate + boric anhydride is formed

7. What is the name of C₆O carbon crystal?
Answer:
Buckminster Fullerene

8. Which carbide is harder than diamond?
Answer:
Boron carbide

9. What is dry ice?
Answer:
Solid carbon dioxide

10. Name the compound which harm ozone layer.
Answer:
C.F.C. or freon

11. Due to which property of carbon, its number of compounds is large?
Answer:
Property of catenation

12. Name the good conductor allotrope of carbon.
Answer:
Graphite

13. What is water glass?
Answer:
Sodium silicate.

 

Students get through the MP Board Class 11th Chemistry Important Questions Chapter 9 Hydrogen which are most likely to be asked in the exam.

MP Board Class 11th Chapter 10 The s-Block Elements

The s-Block Elements Class 11 Important Questions Very Short Answer Type

Question 1.
Why are alkali metals not found in nature ?
Answer:
Alkali metals, due to high chemical reactivity is not found in free state in nature. It is found in earth’s crust in the form of ores like halide, sulphate, carbonate, silicate, borate, oxide etc.

Question 2.
Explain why is sodium less reactive than potassium?
Answer:
Ionisation enthalpy of potassium (Δ_iH₁) 419 kJ mol^-1 is less than ionisation enthalpy (496 kJ mol^-1) of sodium and standard electrode potential (- 2.925 V) of potassium is more negative than the corresponding value (- 2.714 V) of sodium. That is why sodium is less reactive than potassium.

Question 3.
Potassium carbonate cannot be prepared by Solvay process. Why?
Answer:
Potassium carbonate cannot be prepared by Solvay process because potassium bicarbonate being soluble in water do not get precipitated.

Question 4.
Why does Li₂CO₃ decompose at a lower temperature where as Na₂CO₃ at a higher temperature?
Answer:
Lithium due to its small size extremely polarize CO₃^2- ions and form more stable Li₂O and CO₂. Carbonates of alkali metals (except Lithium) are stable at high temperature. That is why, Li₂CO₃ decompose at lower temperature and Na₂CO₃ at a higher temperature.

Question 5.
Write name and formula of an ore in which both Ca and Mg are present.
Ans. Name of ore: Dolomite
Formula: MgCO₃.CaCO₃.

Question 6.
Write the name of flux which are used for the removal of acidic impurities in metallurgical processes.
Answer:

1. Limestone CaCO₃,
2. Magnesite MgCO₃.

Question 7.
Which property of Plaster of Paris is used in plastering?
Answer:
Plaster of Paris reacts with water and becomes hard like cement. Due to this property, it is used in plastering for joining crack bones.

Question 8.
IA and IIA groups are known as s-block elements. Why?
Answer:
On writing the electronic configuration of all the elements present in IA and IIA group, the last electron enters the s-subshell, therefore it is known as .y-block element.
Example: Na₁₁ → 1s²,2s², 2p⁶,3s¹
Ca₂₀ →1s²,2s²,2p⁶,3s², 3p⁶,4s².

Question 9.
What is the Sorrel cement? For which work is it used?
Answer:
If MgO is mixed in concentrated solution of magnesium chloride then a white paste of magnesium oxy chloride of the composition MgCl₂. 2MgO. xH₂O is formed which settles and become hard. It is known as Sorrel cement. It is used to fill the cavities in teeth and in joining clay and porcelain utensils.

Question 10.
What changes are obtained on keeping sodium carbonate open in air ? Explain with equation.
Answer:
On keeping Na₂CO₃.10H₂O open in air its water of crystallization slowly evaporates and it gets converted to monohydrate Na₂CO₃.H₂O which is an efflorescent. On heating it to 750°C anhydrous sodium carbonate is formed which is also called soda-ash.

Question 11.
Lithium hydride can be used for the preparation of other hydrides. Beryllium hydride is one among them. State the various steps of its preparation. Write the chemical equations used in this process.
Answer:
BeH₂ is manufactured by the reduction of complex alkali metal hydrides like Lithium aluminium hydride.
8LiH + Al₂Cl₆ → 2LiAlH₄+6LiCl
2BeCl₂ + LiAlH₄ →2BeH₂ + LiCl + AlCl₃.

Question 12.
Determine the oxidation state of sodium in Na₂O₂.
Answer:
Let oxidation state of Na in Na₂O₂ bex In Na₂O₂ (Na-O-O-Na) one peroxide bond is found in which oxidation state of O is -1.

or 2x + 2(-1) = 0
or x = + 1
Thus, oxidation state of sodium in Na₂O₂ is +1.

Question 13.
How is calcium sulphate prepared? Write its use.
Answer:
In the laboratory, on reacting oxide, carbonate, chloride of Ca with dilute H₂SO₄, calcium sulphate is obtained.
CaO + H₂SO₄ → CaSO₄ + H₂O
CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂

Uses :

1. In preparing chalk,
2. In cement industry,
3. In the preparation of Plaster of Paris,
4. As a fertilizer.

Question 14.
What is Gypsum? How is Plaster of Paris prepared from it?
Answer:
Calcium sulphate (CaSO₄.2H₂O) is called gypsum.
When heated to 120°C-130°C in absence of hydrocarbon fuel, it loses three fourth of its water content to form Plaster of Paris.

Plaster of Paris again absorbs water and get converted to gypsum.

Question 15.
Temperature of furnace is kept less than 1000°C while preparing quick lime. Explain the reason with chemical equation.
Answer:
The temperature of the furnace used to prepare lime is kept less than 1000°C because at high-temperature CaO will react with Silica (SiO₂) present as impurity and form fusible silicate.

Question 16.
Sodium is stored in kerosene oil. Why?
Answer:
Sodium is extremely reactive and strong electropositive element. It reacts with O₂ present in the atmosphere, moisture and carbon dioxide to form oxide and hydroxide. Therefore, sodium is stored in keroseiie oil due to which it cannot come in contact with air.
4Na + O₂ → 2 Na₂O
Na₂O + H₂O →2NaOH
2NaOH + CO₂ →Na₂CO₃ + H₂O.

Question 17.
Write the formula of lime water. What changes will take place on passing CO₂ in it?
Answer:
Solution of quick lime in water is known as lime water. Its formula is Ca(OH)₂.
1. Lime water turns milky on passing CO₂ through it due to the formation of CaCO₃.
Ca(OH)₂ + CO₂ → CaCO₃ + H₂O

2. On passing CO₂ in lime water for a long time milkiness of lime water disappears.
CaCO₃ +CO₂ +H₂O → Ca(HCO₃)₂.

Question 18.
Why can K₂CO₃ not be prepared by Solvay process?
Answer:
K₂CO₃ cannot be prepared by Solvay process because KHCO₃ analogous to NaHCO₃ obtained in between the reaction is soluble due to which it cannot be obtained by crystallization.

Question 19.
Which metal is used in photochemical cell and why?
Answer:
In photochemical cell, potassium and caesium is used because of their low ionization energy.

Question 20.
Why can extraction of sodium by sodium chloride not be done by normal reducing agent?
Answer:
Sodium is a strong reducing agent. It occupies the highest place in electrochemical series. Stronger reducing agents than this is not available, so it cannot be reduced by normal reducing agents. It can be reduced only by electrolysis.

Question 21.
Li and Be have the tendency to form covalent compounds. Explain.
Answer:
Due to small size of Li and Be atoms and their high ionization energy, the electrons of the valence shell are firmly bound to their nuclei. The polarizing power of their ions is also high due to high charge density, hence Li and Be form covalent compounds.

Question 22.
Write down Lewis structure of O₂ and write oxidation state of each oxygen atom. What is the average oxidation state of this ion?
Answer:
Lewis structure of O₂^– ion is
Uncharged oxygen atom contains six electrons, thus its oxidation state is zero. But oxygen with -1 charge has 7 electrons, thus its oxidation state is -1.
Oxidation state of each oxygen atom = –\(\frac{1}{2}\)
O₂^– = 2x = -1
⇒ x = –\(\frac{1}{2}\)

Question 23.
In Solvay process, can we obtain sodium carbonate by directly reacting ammonium carbonate solution with sodium chloride?
Answer:
No, because ammonium carbonate reacts with sodium chloride as follows :
(NH₄)₂CO₃ + 2NaCl⇌Na₂CO₃ + 2NH₄Cl Since the products obtained Na₂CO₃ and NH₄Cl both are more soluble due to which equilibrium does not displace in forward direction. That is why we cannot obtain sodium carbonate by directly reacting ammonium carbonate solution with sodium chloride.

Question 24.
All compounds of Alkali metals are soluble in water but why are Lithium compounds more soluble in organic compounds?
Answer:
Due to comparatively small size of Li⁺ and high polarizing power covalent character is produced in it (Fajan’s Rule). Compounds of other alkali metals are of ionic nature due to which they are soluble in water whereas Lithium compounds due to covalent nature is soluble in alcohol and other organic compounds.

Question 25.
In place of lithium, caesium and potassium is used in photoelectric ceil. Why?
Answer:
Ionization energy of lithium is more than that of potassium and caesium. The elements which possess low ionization energy liberate their electrons easily and are used in photoelectric cells. Thus, potassium and caesium are used in photoelectric cells not lithium.

Question 26.
Beryllium halide produce smoke in air. Why?
Answer:
Normally air contain moisture, thus Beryllium halide hydrolyse in water and release HCl due to which it produces smoke in air.
BeCl₂ + 2H₂O → Be(OH )₂ + 2HCl

Question 27.
On moving downwards in the first group the hardness of element increases. Why?
Answer:
In the first group on moving downwards along with the increase in size of the elements their density also increases and the force of attraction between its atoms increases due to which there is an increase in their hardness.

The s-Block Elements Class 11 Important Questions Short Answer Type

Question 1.
s-block elements and p-block elements are called Representative elements. Why?
Answer:
Elements in which electronic configuration of outer shell is ns^1-2 and ns²np^1-6 are called Representative elements because each element present in this group represent the properties of their group and properties of each group completely differ from the properties of the other group.

1. s-block elements: Elements whose last electron enter into s-subshell are s-block
elements. Their general electronic configuration is ns^1-2.

2. p- block elements: Elements in which last electron enter into p-subshell are called p-block elements. Their general electronic configuration is ns²np^1-6.

Question 2.
Alkali metals are good reducing agents. Explain, why?
Answer:
Due to less ionization energy of alkali metals. They have tendency to loose electron (get oxidized) and form positive ion (M⁺) also alkali metals have negative value of standard reduction potential. Therefore, alkali metals are good reducing agent.

Question 3.
BeSO₄ and MgSO₄ are easily soluble in water whereas CaSO₄, SrSO₄ and. BaSO₄ are insoluble. Why?
Answer:
Lattice energy of Alkaline earth metals due to large size of sulphate ion is nearly same. Thus, their solubility depend on hydration energy which decreases on moving down the group. High hydration enthalpy of Be²⁺ and Mg²⁺ ions dominates lattice enthalpy factor due to which their sulphates are soluble in water.

Alternatively hydration enthalpy for Ca²⁺, Sr²⁺ and Ba²⁺ ions is less. Therefore, they cannot dominate lattice enthalpy factor. Thus, their sulphates are insoluble in water.

Question 4.
Why is reducing ability of Lithium more in aqueous solution?
Answer:
In aqueous solution of an element, measure of ability to lose electron depends on electrode potential. It mainly depends on three factors :

Due to small size of their ions, hydration enthalpy of Lithium is maximum. Though, ionization enthalpy of Li is maximum among Alkali metals but its hydration enthalpy is more than ionization enthalpy. Thus, due to high hydration enthalpy, Lithium is strongly reducing in aqueous solution.

Question 5.
Why are Alkali metals and Alkaline earth metals not obtained by chemical reduction methods? Explain.
Answer:

• Alkali and Alkaline earth metals are itself strong reducing agents. Thus, these metals cannot be obtained by chemical reduction methods of their oxides and other compounds.
• These metals are highly electropositive in nature: Therefore, these cannot be displaced from aqueous solution of their salts by other metals.

Question 6.
What is the reason that alkali metals form M⁺ ion and not M²⁺ ion?
Answer:
Alkali metals consist of one electron in the outermost shell which it can loose easily and achieve stable inert gas configuration. They do not possess the tendency to form M²⁺ type of ion because the value of ionization energy necessary to remove the second electron is very high.

Question 7.
Which of the alkali metal is strongly reducing and why?
Answer:
Reducing tendency of an element depends on its standard electrode potential. Elements whose standard electrode potentials are negative behave as reducing agents and higher the negative value of the standard electrode potentials of an element, stronger is its reducing tendency. Lithium has the highest negative standard electrode potential, therefore, it is strongly reducing.

Question 8.
Alkali metals do not exist in free state in nature. Why? Or, Alkali metals always form ionic compounds. Why?
Answer:
Due to the large size of alkali metals, their ionization energy is less, therefore it can easily lose electron and form cation. Therefore, being strongly electropositive and extremely reactive it easily reacts with electronegative elements like: O, Moisture, CO₂ etc. present in the atmosphere and forms ionic compounds. Therefore, they do not exist in free state in nature.

Question 9.
Why do alkali metals give flame test?
Answer:
Alkali metals give their characteristic colour to the oxidizing flame. The valence electron gets excited by absorbing energy from the flame. On falling back from higher energy state to lower energy state it emits radiations of lower energy of the range of visible region. Different alkali metals give different colour because they are having different ionization energy. The alkali metals which have high I.E. will emit light having minimum frequency and vice-versa.

Question 10.
Why Be and Mg do not give flame test?
Answer:
Except beryllium and magnesium all the alkaline earth metals impart characteristic colours to Bunsen flame.
Due to small size of Be and Mg atoms, the energy required to excite the valence electrons is very high which is not obtained in Bunsen flame. That is why Be and Mg do not impart any colour to flame.

Question 11.
When an alkali metal is dissolved in liquid ammonia then the solution obtains various colours. State the cause of such type of colour.
Answer:
The colour of dilute solution of alkali metal in liquid ammonia is dark blue because ammoniated electrons absorb energy from visible range of light.
M + (x + y)NH₃→ [M(NH₃)_x]⁺ +[e(NH₃)_y]Ammoniated electron
If concentration of solution is increased, it gets converted to bronze colour.

Question 12.
State two-two use of each of the following :
(a) Caustic soda,
(b) Sodium carbonate,
(c) Quick lime.
Answer:
(a) Caustic soda :

• It is used in the preparation of soap, paper, artificial silk etc.
• In the textile industry, it is used for mercerising cotton fabrics.

(b) Sodium carbonate :

• It is used in softening of water and washing.
• It is used in the manufacture of glass, soap, borax, caustic soda etc.

(c) Quick lime:

• It is used in the preparation of washing soda from caustic soda. ,
• It is used in the purification of sugar and in the manufacture of dyestuffs.

Question 13.
Alkali metals form blue coloured solution on dissolving in ammonia, which is good conductor of electricity? Explain, why?
Answer:
Alkali metals dissolve in liquor ammonia to form deep blue coloured solution of high electrical conductivity.
M + (x +y)NH₃ → [M(NH₃)_x]⁺ +[e (NH₃)_y]^–.
The blue colour of the solution is due to ammoniated electrons. Positive ion and electrons are responsible for the conductivity.

Question 14.
Na is alkaline or Na₂O. Explain.
Answer:
Na₂O is alkaline because it reacts with water to form NaOH. Na also reacts with water to form NaOH but it forms Na₂O firstly and then NaOH. Thus, Na₂O is alkaline.
4Na + 2H₂O → 2Na₂O + 2H2
Na₂O + H₂O → 2NaOH

Question 15.
Differentiate Alkali metals and Alkaline earth metals on the basis of following points: ‘
(i) Reaction with nitrogen.
(ii) Effect of heat on carbonates.
(iii) Solubility of sulphates in water.
Answer:
Differences of Alkali metals and Alkaline earth metals

Question 16.
Draw the structure of the following :
(i) BeCl₂ (vapour),
(ii) BeCl₂ (solid).
Answer:
(i) In vapour state, it exists as a bridged chloride dimer.

(ii) In solid state, it forms a polymeric chain with chlorine bond.

Question 17.
Hydroxides and carbonates of sodium and potassium are soluble in water, whereas corresponding salts of magnesium and calcium are partially soluble. Explain.
Answer:
Solubility of a salt in water depends on its Hydration energy and Lattice energy.
ΔH_solution = ΔH_{Hyderation energy} +ΔH_{Lattice energy}

Higher the negative value of ΔH Solution, lesser is the solubility of the compound. Hydration energy of hydroxides and carbonates of sodium and potassium is higher than their Lattice energy. Thus, they are soluble in water.
Lattice energy of magnesium and calcium hydroxides is more than their hydration energy. Thus, they are partially soluble in water.

Question 18.
Salts of Lithium are generally hydrated, whereas salts of other alkali metals are generally anhydrous. Why ?
Answer:
Due to smallest size among alkali metals, hydration energy of Li⁺ is least. That is why lithium salts are generally hydrated and salts of other alkali metals are normally anhydrous.

Question 19.
LiF is nearly soluble in water, whereas LiCl is not only soluble in water but is also soluble in acetone. Give reason.
Ans. LiF, due to high lattice energy is nearly soluble in water, but due to high hydration energy of Li ions, LiCl is also soluble in water. But due to specific covalent character of LiCl, it is also soluble in acetone (Since covalent character increases with increase in size of anion. Thus, order of solubility is :
LiF < LiCl < LiBr < Lil

Question 20.
Alkali metals lose their metallic lustre, when kept in open air. Why?
Answer:
All metals possess a specific lustre known as metallic lustre. Due to large size of alkali metals, their ionization energy is very less. Therefore, they are highly reactive and strongly electropositive and react with moisture, O₂ and CO₂ present in the atmosphere and form basic carbonate. Due to formation of oxide and carbonate layer on their surface, they lose their metallic lustre.

Question 21.
Differentiate between Alkali metals and Alkaline earth metals.
Answer:
Differences between Alkali metals and Alkaline earth metals

Alkali metal	Alkaline earth metal
1. They show +1 valency.	They show +2 valency.
2. Their hydroxides are strong alkalines.	Their hydroxides are comparatively weak alkalines.
3. Their carbonates, sulphates and phosphate are soluble in water.	These compounds of alkaline earth metals are insoluble in water.
4. Magnitude of their ionization energy is comparatively low.	Magnitude of their ionization energy is high.
5. These are lustrous, malleable and ductile.	These are comparatively less.

Question 22.
Among LiCl and RbCl, which is strongly ionic and why?
Answer:
RbCl is stronger ionic than LiCl because due to small size of Li, its ionization energy is very high. Therefore, it forms covalent compounds, whereas due to larger size of Rb, its ionization energy is very less, due to which it easily loses electron and forms cation. Therefore, its compounds show strong ionic character.

Question 23.
Among alkali metals and Alkaline earth metals, whose carbonate is soluble in water and why ? Or, Carbonates of Alkali metals are soluble in water whereas carbonates of Alkaline earth metals are insoluble in water. Why?
Answer:
Carbonates of Alkali metals are soluble in water because their hydration energy is more than their lattice energy, whereas due to small size and high charge density of Alkaline earth metals, their lattice energy is high and is more than their hydration energy. Therefore, carbonates of Alkaline earth metals are insoluble in water.

Question 24.
Explain the importance of sodium,potassium, magnesium and calcium in biological fluids.
Answer:
Sodium and potassium ions play an important role in interstitial fluids. These ions participate in the transmission of nerve signals. Potassium ions are the most abundant cations within cell fluids, where they activate many enzymes, participate in the oxidation of glucose to produce ATP.

The main pigment for the absorption of light in plants is chlorophyll which contains magnesium.
About 99% of body calcium is present in bones and teeth. It also plays important roles in neuromuscular function, intemeuronal transmission, cell membrane integrity and blood coagulation.

Question 25.
Though value of second ionization energy of Alkaline earth metals is more than their first ionisation energy still these metals show +2 oxidation state and not +l. Why?
Answer:
The value of second ionization energy of Alkaline earth metal is more than its first ionization energy still their maximum compounds show +2 oxidation state due to the following reasons:

• Electronic configuration of bivalent ion is similar to inert gases.
• During the formation of bivalent ion they release large amount of hydration energy which compensates the high value of second ionization energy.
• Due to small size of M²⁺ ion their hydration energy is high which compensate the high value of second ionization energy.

Question 26.
State the dissimilarities in BeCl₂ and other alkaline earth metal chlorides.
Answer:

1. BeCl₂ is a covalent compound whereas chlorides of other alkaline earth metals are ionic.
2. BeCl₂ is insoluble in water whereas other chlorides of other alkaline earth metals are soluble in water.
3. BeCl₂ is soluble in organic solvents whereas chlorides of other alkaline earth metals are soluble in organic solvents.
4. Melting and boiling points of BeCl₂ is low whereas that of chlorides of other alkaline earth metals is high.

Question 27.
Solubility of BaS0₄ is less than CaSO₄. Why?
Answer:
Solubility of sulphates of alkaline earth metals decreases on moving down the group because though lattice energy nearly remains same, their hydration energy decreases down the group due in increase in atomic radius. Value of hydration energy goes on becoming less than lattice energy due to which solubility decreases.

Question 28.
Ionization energy of Be is more than B. Why?
Answer:

In Be, s-orbital is complete whereas in B,p -orbital is incomplete. Half filled or fully filled orbital is more stable than incomplete orbital and huge amount of energy is required to remove the electron from them. Therefore, ionization energy of Be is more than Boron.

Question 29.
How is sodium carbonate prepared by ammonia soda process? Write its principle.
Answer:
In this process, first concentration NaCl solution is saturated with NH₃ by which ammonical sodium chloride is formed.
CO₂ gas is then passed through this ammonical sodium chloride solution when ammonium bicarbonate is formed which reacts with sodium chloride to form sodium bicarbonate.
NH₃ + CO₂ + H₂O → NH₄HCO₃
NH₄HCO₃ + NaCl → NaHCO₃ + NH₄Cl
Sodium bicarbonate being partially soluble settles at the bottom as precipitate. It is filtered and ignited to form sodium carbonate.
2NaHCO₃→ Na₂CO₃ + H₂O + CO₂.

Question 30.
How are (1) sodium bicarbonate (2) sodium hydroxide and (3) sodium silicate obtained from sodium carbonate?
Answer:
1. On passing CO₂ gas through aqueous sodium carbonate solution, white precipitate of sodium bicarbonate is obtained.
Na₂CO₃ + H₂O + CO₂ → 2NaHCO₃
2. On boiling sodium carbonate with lime water, sodium hydroxide is formed.
Na₂CO₃ + Ca(OH)₂ → 2NaOH + CaCO₃
3. On heating sodium carbonate with silica, sodium silicate is formed.
Na₂CO₃ + SiO₂ →Na₂SiO₃ + CO₂.

Question 31.
What is baking soda? Write its method of preparation, properties and use.
Answer:
Definition: Sodium hydrogen carbonate is known as Baking soda and its formula is NaHCO₃.
Method of preparation :
1. In Ammonia Soda process, sodium bicarbonate is obtained as an intermediate compound.
NH₄HCO₃ + NaCl →NaHCO₃ + NH₄Cl

2. Sodium bicarbonate is formed by passing C02 gas through sodium carbonate solu¬tion.
Na₂CO₃ + H₂O + CO₂ →2NaHCO₃
Properties: It is a white crystalline substance, partially soluble in water. Its aqueous solution is weakly basic.

Uses :

• In the preparation of baking powder,
• As a medicine to reduce acidity of the stomach,
• In fire extinguishers.

Question 32.
Describe briefly Le-Blanc process of preparation of sodium carbonate and explain the cause of superiority of Solvay process over Le-Blanc process.
Answer:
Le-Blanc process: The names of the three steps for the formation of sodium carbonate are as follows :
(i) Preparation of salt cake: Sodium chloride when heated with cone. H₂SO₄ produces sodium sulphate (salt lake) and HCl gas.
(ii) Black ash: On heating the mixture of sodium sulphate, calcium carbonate and coke gives a mixture of sodium carbonate and calcium sulphide which is known as Black ash.

(iii) Production of black ash from common salt: Mixture is mixed with water and filtered. Sodium carbonate is soluble in water, therefore it remains infiltrate, CaS is removed as residue. The evaporation of filtrate gives solid sodium carbonate..
2 NaCl + H₂SO₄ →Na₂SO₄ + 2HCl
Na₂SO₄ + 4C → Na₂s + 4CO
Na₂S + CaCO₃ → (Na₂CO₃+CaS) Black ash

Superiority of Solvay process over Le-Blanc process :

• Solvay process is cheap.
• Pure Na₂CO₃ is formed in Solvay process.
• No chemical fumes are released in Solvay process,
• In Solvay process, NaHCO₃ is formed in the middle which is a useful compound.

Question 33.
Describe Hargreave-Bird cell of sodium carbonate.
Answer:
Hargreave-Bird cell contains a carbon anode and a porous copper cathode which are separated by an asbestos sheet. By the electrolysis of NaCl solution, sodium and chlorine are formed. In the cell, steam and CO₂ are passed at the outer side of asbestos, which reacts with Na to form Na₂CO₃. On evaporating this solution Na₂CO₃.10H₂O is obtained.

2 NaCl ⇌ 2Na⁺+2Cl^–
2Na⁺+2e^– → 2Na
At cathode: 2Na + 2H₂O →2NaOH + H₂
2NaOH + CO₂ →Na₂CO₃ + H₂O
At anode: 2Cl^– – 2e^– → 2Cl
Cl + Cl → Cl₂.

Question 34.
Explain Nelson cell process of preparation of sodium hydroxide with diagram.
Answer:
Nelson cell consists of a cylindrical asbestos sheet covered by porous steel lining arranged inside a steel tank. The porous steel lining acts as cathode. NaCl solution is filled in this chamber. Carbon rod acts as anode.
On electrolysis, sodium ions are released which pass the asbestos sheet and after releasing at cathode react with the steam entering the tank and form NaOH.

2NaCl ⇌ 2Na⁺+2Cl^–
At cathode : 2Na⁺ +2e^– → 2Na
2Na + 2H₂O →2NaOH + H₂
At anode : 2Cl^– – 2e^– → 2Cl
Cl + Cl → Cl₂.

Question 35.
Write method of preparation, properties and uses of slaked lime.
Answer:
Method of preparation :
1. On sprinkling water on quick lime, slaked lime is obtained.
CaO + H₂O → Ca(OH)₂
2. Ca(OH)₂ is obtained by the reaction of a base on calcium salt.
Ca(NO₃)₂ + 2 NaOH → Ca(OH)₂ + 2NaNO₃

Properties :
1. On heating it to 400°C, calcium oxide is formed.

2. On passing CO₂ through lime water, milky white precipitate of CaCO₃ is formed.
Ca(OH)₂ +CO₂ → CaCO₃ + H₂O
3. On passing CO₂ through lime water for a longer time, milky white colour disappears.
CaCO₃ + CO₂ + H₂O → Ca(HCO₃)₂
4. On passing Cl₂ gas through dry slaked lime, bleaching powder is obtained.
Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O

Uses :

• In the manufacture of caustic soda and bleaching powder.
• In the purification of coal gas.
• In whitewashing.
• In the manufacture of ammonia.

Question 36.
Describe the soda-lime method for preparation of sodium hydroxide with chemical equation.
Answer:
Soda-lime method is also known as causticizing process. In this method, sodium hydroxide is obtained by boiling sodium carbonate solution with slaked lime.
Na₂CO₃ + Ca(OH)₂ ⇌ 2NaOH + CaCO₃
This reaction is reversible but due to precipitation of calcium carbonate, the reaction proceeds only in forward direction.

A mixture of sodium carbonate and slaked lime is taken in iron tanks. The mixture is heated to 80-85°C by steam. Precipitated calcium carbonate is separated periodically. The filterate is treated with dilute hydrochloric acid for testing the presence of sodium carbonate. More lime is added when effervescence occurs. Dilute hydrochloric acid is added when no more effervescence occurs and the solution is filtered to separate precipitate of calcium carbonate. Filterate obtained on evaporation in iron containers give 98% pure sodium hydroxide.

The s-Block Elements Class 11 Important Questions Long Answer Type

Question 1.
Write notes on each of the following observations :
(a) Mobility of alkali metal ions in aqueous solution are :
Li⁺ < Na ⁺ < K⁺ < Rb⁺ < Cs⁺
(b) Lithium is the only alkali metal which forms nitride.
(c) E° for M²⁺_(aq) + 2e^– → M_(s) (Where M = Ca, Sr or Ba) is nearly constant.
Answer:
(a) Smaller the size of an ion, higher is its hydration and higher the hydration of an ion lower is its mobility. Thus, order of hydration ability is as follows: ‘
Li⁺ < Na ⁺ < K⁺ < Rb⁺ < Cs⁺
Thus, ionic mobility is in the opposite order.

(b) Due to its small size, among alkali metals only Lithium forms nitride.

(c) E° for M²⁺_(aq) + 2e^– → M_(s) where M = Ca, Sr or Ba, value of E° is nearly same.

For a M²⁺/M electrode, value of E° depends on three factors:

• Enthalpy of Vaporization,
• Ionization enthalpy,
• Hydration enthalpy.
  Since, the combined effect of all these three factors is nearly same for Ca, Sr and Ba, therefore value of their electrode potentials is also nearly same.

Question 2.
How is lithium different from other members of its group?
Answer:
Anomalous behaviour of lithium: Properties of lithium are different than other members of the group due to the given reasons :
(i) Size of lithium atom and ion is very small.
(ii) Due to small size, polarizing power of lithium-ion is high, due to which compounds possess covalent character.
(iii) In comparison to other alkali metals its electropositive nature is less and ionization energy is high.
Lithium differ from alkali metals of group I due to following properties :
(i) Lithium is comparatively harder than the other alkali metals.
(ii) Melting and boiling point pf lithium is comparatively high.
(iii) Lithium reacts with oxygen to form only normal oxide (Li₂O), whereas other metals form peroxide (M₂O₂) and superoxide (MO₂).
(iv) Lithium hydride (LiH) is more stable as compared to hydrides of other metals.
(v) Lithium hydroxide (LiOH) is a weak base and is partially soluble in water, but hydroxides of other metals of the group are more soluble in water.
(vi) Lithium forms nitride (Li₃N) with nitrogen, whereas other metals of the group do not form nitride.
(vii) Lithium nitrate when heated decomposes to nitrogen dioxide and oxygen.

Whereas sodium nitrate and potassium nitrate when heated strongly form correspond¬ing nitrate with the release of oxygen.

(viii) Lithium carbonate and lithium hydroxide decompose on heating strongly, but carbonates and hydroxides of other metals do not decompose on heating.

(ix) Fluoride, phosphate, oxalate and carbonate of Lithium are partially soluble in water, whereas corresponding salts of other metals are soluble in water.

Question 3.
What is diagonal relationship? Write diagonal relationship between Li and Mg.
Answer:
Diagonal relationship: Some elements of second period show similarity with some elements of third period diagonally.
This is known as diagonal relationship. Diagonal relationship between Li and Mg :

• Atomic radius of Lithium is 1.34 Å and atomic radius of Mg is 1.36 Å.
• Polarizing power of Li⁺ and Mg²⁺ is nearly same.
• Li and Mg both are hard.
• Electronegativity of Li and Mg is (1.0 and 1.2) same.
• Melting and boiling point of Li and Mg is high.
• Both Li and Mg react with N₂ to form nitride.
• Both Li and Mg react with oxygen to form monoxide.
• Li and Mg both dissociate water to give H₂.
• Carbonates of both Li and Mg give CO₂ on heating.
• Both LiOH and Mg(OH)₂ are weak bases.

Question 4.
What are similarities between Alkali metal and Alkaline earth metals?
Answer:

• Alkali metals and Alkaline earth metals are not obtained in free state in nature.
• Oxides of alkali metals and alkaline earth metals dissolve in water to form strong bases.
• Alkali metals and Alkaline earth metals are soft and bright.
• On keeping in air, their surface become dull.
• Alkali metals and Alkaline earth metals (except Be and Mg) give flame test.
• Alkali metals and Alkaline earth metals are strong reducing agents.
• Alkali metals and Alkaline earth metals both form ionic compounds.
• Nitrate and halide of both are soluble in water.

Question 5.
Be shows anomalous behaviour as compared to other members of its group. Explain.
Answer:
Anomalous behaviour of beryllium: Being the first member of family, beryllium shows anomalous behaviour as compared to other members of family. It is because of following reasons:

• Smallest size of the atom and the ion.
• High ionization energy, electronegativity and heat of hydration.
• Vacant d-orbitals are not present.

Beryllium shows the following different properties from other members of the group :
1. Beryllium is hard, whereas other metals are soft.

• Beryllium does not liberate hydrogen quickly from acids whereas other metals liberate quickly.
• Beryllium carbide form methane by reaction with water whereas other members give acetylene.
  Be₂C + 2H₂O → 2BeO + CH₄
  CaC₂ +2H₂O → Ca(OH)₂ + C₂H₂
• Beryllium oxide is amphoteric whereas oxides of other metals of the group are basic.
• Beryllium and magnesium do not impart any colour to the flame whereas all other members impart colour.
• Beryllium react with hydrogen slowly whereas others react quickly.
• Compounds of beryllium are generally covalent where other members of the group form ionic compounds.

Question 6.
Explain why :
(a) Solution of Na₂CO₃ is alkaline.
(b) Alkali metals are obtained by the electrolysis of their corresponding fused chlorides.
(c) Sodium is more useful than Potassium.
Answer:
(a) Na₂CO₃ is a salt of a weak acid (H₂CO₃) and a strong base (NaOH). Thus, on hydrolysis it gives base (NaOH) due to which its aqueous solution is alkaline.

(b)
1. Alkali metals are strongly reducing, thus they cannot be extracted by their oxides and compounds.
2. Due to high electropositive nature, these cannot be displaced by other metals from their salt solutions.
3. Alkali metals cannot be obtained by the electrolysis of aqueous solution of their salts because at cathode in place of Na metal H2 is liberated. Due to this, Alkali metals are obtained by the electrolysis of their molten chlorides.

At Cathode:2Cl^– → Cl₂ + 2e^–
At Anode : 2Na⁺ + 2e^– → 2Na

(c) As compared to potassium, sodium is more important because it is reactive but not very reactive as potassium. Other uses of sodium are :

• As a coolant in nuclear plants.
• In the manufacture of Tetraethyl lead (TEL) used as antiknock agent for petrol.
  4C₂H₅Cl + 4Na- Pb → (C₂H₅)₄ Pb + 3Pb+4NaCl (TEL).

Question 7.
Explain diagonal relationship between Be and Al.
Answer:
Diagonal relationship between Beryllium and Aluminium: Beryllium resembles aluminium of group 3 in the following properties :

• Both beryllium and aluminium form covalent compounds.
• Both metals are passive towards reaction with concentrated HNO₃, because they are covered with a layer of oxide on their surface.
• Both the metals are of weak electropositive nature.
• Both do not form hydride quickly.
• Carbides of both the metals react with water to form methane.
  Be₂C + 2H₂O → 2BeO + CH₄
  Al₄C₃ + 6H₂O → 2 Al₂O₃ + 3CH₄
  
• Oxides of both are soluble in water and form hydroxides which are amphoteric in nature and react with acid and base to form salt.
  BeO + 2 HCl → BeCl₂ + H₂O
  Al₂O₃ +6HCl → 2AlCl₃ +3H₂O
  BeO + 2NaOH >Na2Be02 + H₂O
  Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O
  
• Both Be and Al do not give colour to the flame.

Question 8.
How is sodium carbonate manufactured by Solvay process?
Answer:
Principle: In this process, first concentrated solution of sodium chloride (Brine) is saturated with NH₃ to form ammonical sodium chloride. On passing CO₂ gas to this, ammonium bicarbonate is formed which reacts with sodium chloride and forms sodium bicarbonate. Precipitate of sodium bicarbonate is filtered which on calcination gives sodium carbonate.’
NH₃+CO₂+H₂O → NH₄HCO₃
NH₄HCO₃ + NaCl →NaHCO₃ +NH₄Cl
2NaHCO₃ → Na₂CO₃ + H₂O + CO₂

CO₂ formed is used again for carbonation NH₄Cl formed on treatment with slaked lime gives NH₃ which is used for saturation of brine.
2NH₄Cl + Ca(OH)₂ →CaCl₂ +2H₂O + 2NH₃
The various steps and reactions involved are given below :

Saturating tank: In this tank brine (NaCl) is saturated with ammonia. NaCl solution is introduced from the top while ammonia is introduced from bottom so that ammonium brine is formed and which collects at bottom. Calcium and magnesium salts are present as impurity in sodium chloride solution. They get precipitated as hydroxides by ammonium hydroxide.

Filter: Ammonical brine is filtered to removed the precipitated hydroxides of calcium and magnesium.

Cooler: The filtrate is cooled by passing through condenser. Cooling is necessary because when ammonia dissolves in brine a lot of heat is produced.

Carbonating tower: The cold solution of brine saturated with ammonia is intro¬duced into the carbonating tank from top.
Carbonating tower is fitted with a number of compound diaphragms each made of a horizontal iron plate.

Following reactions take place here :
2NH₃ + CO₂ +H₂O → (NH₄)₂CO₃
(NH₄)₂CO₃ +2NaCl → Na₂CO₃ + 2NH₄Cl
Na₂CO₃ +H₂O + CO₂ → 2NaHCO₃

Vacuum filter: The precipitated sodium bicarbonate along with solution of traces of ammonium carbonate and ammonium chloride is passed through rotatory vacuum filter at the bottom of tower where socfium bicarbonate separates leaving behind mother liquor containing ammonium chloride.

Limekiln: Here limestone is burnt to produce quick lime and carbon dioxide.

Formed CaO reacts with water to form slaked lime.
CaO + H₂O →Ca(OH)₂

Ammonia recovery tower: The mother liquor obtained from rotatory filter pump containing ammonium chloride is introduced into the ammonia recovery tank from the top part while the milk of lime is introduced from the top of lower part and steam is introduced into the tower from the bottom, ammonium chloride reacts with milk of lime to produce ammonia.
2NH₄Cl + Ca(OH)₂ → 2NH₃ +CaCl₂ +2H₂O

Sodium bicarbonate obtained from rotatory filters is ignited in a specially constructed cylindrical vessels when sodium carbonate is formed and CO₂ evolved is collected and used again.
2NaHCO₃ →Na₂CO₃ + H₂O+CO₂.

Question 9.
When a metal of group-1 is dissolved in liquid ammonia the following observations are obtained : ,
(a) Initially blue solution is obtained.
(b) On concentrating the solution, blue colour gets converted to bronze. Discuss the blue colour of the solution. Write the name of the product formed on keeping the solution for some time.
Answer:
(a) On dissolving the metals of group-1 in liquid ammonia the following reaction is obtained:
The blue colour of the solution is due to ammoniated electrons which absorbs the corresponding energy of the visible range and provides blue colour to the solution.
M+(x+y)NH₃ → [M(NH₃)_x]⁺ + [e(NH₃)_y]^–

(b) In concentrated solution, due to formation of metal ion layer the blue colour gets converted to bronze colour. Blue colour releases hydrogen on being kept for some time and amide is formed.

Question 10.
Stability of peroxides and superoxides of Alkali metals decreases on moving down the group. Discuss with example.
Answer:
Stability of peroxides and superoxides of metal ion increases with increase in size.
Thus, KO₂ < RbO₂ < CsO₂
Cause of formation of various oxide by alkali metals with oxygen is the formation of a boundary of strong electropositive region around the metal cation. Size of Li⁺ is the smallest due to which it does not allow O^2- ion to react with O². Size of Na⁺ is bigger than Li thusn its positivity decreases by Li⁺ region. Bigger ions like K⁺, Rb⁺ and Cs⁺ allow O^2-₂ ion again to react with O₂ to form superoxide (O₂^–)

With the increase in size of metal ions, stability of peroxides and superoxides also increases. Its main cause is that due to lattice energy effect, larger anion and larger cation provide stability.

Question 11.
Explain Castner-Kellner cell with labelled diagram to obtain sodium hydroxide.
Answer:
It consists of a rectangular iron tank which is divided into three parts by state partitions. These partitions do not touch the bottom of the tank which is covered by mercury. Layers of mercury are separated from each other in all the three parts. Mercuiy layer can move here and there by mechanical arrangement. NaOH solution is filled in the outer compartment in which graphite anode is immersed. In the middle dilute NaOH solution is filled in which iron cathode is immersed. On passing electric current, mercury in middle compartment acts as anode and in the outer compartments act as cathode by induction. Cl₂ is released in the outer compartment at anode.

Sodium released at cathode reacts with mercury to form sodium amalgam. Sodium amalgam is introduced in the central compartment by wheels. Here sodium amalgam acts as anode and iron rod acts as cathode. In this portion NaOH is filled. On passing electric current OH^– ions are released at anode and reacts with sodium present in amalgam to form NaOH and H₂ gas is released.

Reaction : In outer compartment:
2NaCl ⇌ 2Na⁺ +2Cl
At cathode:
2Na⁺ +2e^– → 2Na
2Na + xHg → Hg_xNa₂
At anode:
2Cl^– – 2e^– →2Cl
Cl + Cl → Cl₂

In middle compartment:
NaOH ⇌ Na⁺ + OH^–
At cathode:
2NaHg + 2H₂O → 2NaOH+2Hg + H₂
Na⁺ +e^– →Na
2Na + 2H₂O →2NaOH + H₂
At anode:
2OH^– → 2OH + 2e^–
Hg_x Na₂ +2OH^– → 2NaOH+xHg.

Question 12.
When water is mixed in a compound of calcium
(A) solution of compound
(B) is formed. On passing CO₂ through this solution, due to the formation of compound
(C) it turns milky. On passing excess of carbon dioxide, compound
(D) is formed and due to this milky colour disappears. Identify the compound (A), (B), (C) and (D) and explain the cause of disappearance of milky colour in the last step.
Answer:
On passing CO₂ in the solution of compound
(B), the appearance of milky colour indicates that compound (B) is slaked lime [Ca(OH)₂] and compound
(C) is calcium carbonate. Since compound (B) is formed on adding H₂O to compound (A), thus compound A is quick lime (CaO).
This reactions are as Follows:

Question 13.
Write method of preparation of calcium oxide, chemical properties and its uses.
Answer:
Calcium oxide is formed by heating lime stone.

The reaction is reversible, thus it is necessary to remove CO₂. Temperature of the reaction should be 900°C because at high temperature by the reaction of sand and lime fusible silicate is formed.
Coal is burnt at the side of the furnace. Lime stone is added slowly from the top, which dissociates on coming down. CO₂ gas releases from the upper part.
It is liquefied and stored in the cylinders.

Chemical properties :
1. It absorbs moist air andCO₂ to form Ca(OH)₂ andCaCO₃.
CaO+ H₂O → Ca(OH)₂
CaO+ H₂O → CaCO,
2. It is a strong basic oxide which reacts with acids to form salts.
CaO + 2HCl → CaCl₂ + H₂O
CaO + SiO₂ → CaSiO₃
3. On heating with ammonium salts NH₃ gas is formed.
2NH₄Cl + CaO → CaCl₂ +H₂O + 2NH₃
4. On heating carbon and calcium oxide CaC₂ is obtained.

Uses :

• As a flux in metallurgy,
• In drying of alcohols and gases,
• For producing lime light,
• In the purification of coal gas,
• In making paper.

Question 14.
What happens when :
(1) On reacting NaOH with Zn or Al.
(2) On reacting NaOH with S.
(3) On reacting NaOH with halogen.
(4) On reacting NaOH with metal oxide.
(5) On reacting NaOH with metal salts.
Answer:
1. On reacting NaOH with Zn or Al, complex salt is obtained and hydrogen gas is released.
Zn + 2NaOH →Na₂ZnO₂ + H₂
2Al + 2 NaOH + 2H₂O → 2NaAlO₂ + 3H₂

2. On reacting NaOH with sulphur, hypo is formed.
4S + 6NaOH → Na₂S₂O₃ + 2Na₂S + 3H₂O
8S + 2Na₂S → 2Na₂S₅

3. Reaction with halogen :
(a) On reacting dilute and cold NaOH with chlorine, sodium hypochloride is formed.
2NaOH +Cl₂ → NaCl + NaClO+ H₂O
(b) On reacting hot and cone. NaOH with chlorine, sodium chlorate is obtained.
6NaOH + 3Cl₂ → 5NaCl+NaClO3 + 3H₂O

4. On reacting NaOH with metal oxide, complex salt is obtained.
ZnO + 2NaOH → Na₂ZnO₂+H₂O

5. On reacting NaOH with metallic salts, precipitate of metal hydroxides are obtained.
CuSO₄ + 2NaOH → Na₂SO₄ + Cu(OH)₂
3NaOH + FeCl₃ → Fe(OH)₃ + 3NaCl
2NaOH + 2 AgNO₃ → Ag₂O + 2NaNO₃ + H₂O

Question 15.
State the importance of:
(a) Lime stone,
(b) Cement,
(c) Plaster of Paris.
Answer:
(a) Lime stone CaCO₃:

• With magnesium carbonate, it is used as a flux in the extraction of metals like iron.
• It is used as an antacid, mild abrasive in toothpaste and a filler in cosmetics.

(b) Cement:

• It is an important compound for the construction of buildings.
• It is used in concrete and reinforced concrete, in plastering and in the construction ofbuildings.

(c) Plaster of Paris :

• It is used in building industry as well as plasters in bone fracture.
• It is also used in dentistry, in ornamental work and for making casts and statues and busts.

Question 16.
Write short notes on :
(i) Plaster of Paris and
(ii) Lime.
Answer:
(i) Plaster of Paris : It is formed from gypsum by heating up to 120-130°C in absence of organic fuel.

On adding water it again change into gypsum and becomes hard.
(CaSO₄)₂.H₂O + 3H₂O → 2CaSO₄.2H₂O

Uses :
(a) It is used in plasters in bone fracture.
(b) In preparation of toys, decorative items and in fireproof bricks.

(ii) Lime : Calcium oxide and hydroxide is known as lime.
(A) Quick lime (CaO): It is prepared by heating limestone. .

Properties : (a) It forms limelight on heating with oxygen.
(b) Reaction with air and moisture.

CaO + CO₂ → CaCO₃.

Uses :
(a) As a flux (basic) in metallurgy.
(b) In producing limelight.

Question 17.
Discuss the various reactions that occur in Solvay process.
Answer:
In Solvay process, CO₂ is passed through brine (cone. NaCl so In.) saturated with ammonia when easily soluble sodium bicarbonate is formed.
NaCl + NH₃ + CO₂ + H₂O → NaHCO₃ ↓ + NH4C1
The sodium bicarbonate formed is filtered, dried and heated to form sodium carbonate.

CO₂ used in the carbonating tower is obtained by heating calcium carbonate. CaO formed in this process is dissolved in water to form slaked lime.

Ammonia is regenerated in the regeneration tower by heating NH₄Cl and Ca(OH)₂.
2NH₄Cl + Ca(OH)₂→ 2NH₃ + CaCl₂ + 2H₂O

Question 18.
What are the common physical and chemical characteristics of Alkali metals ?
Answer:
Physical properties :
(i) Alkali metals are silvery white, soft and tight metals.
(ii) These have low density (due to large size). It increases on moving down the group. Though potassium is lighter than sodium.
(iii) Alkali metals have low melting and boiling points. Because due to presence of only one valence electron, weak metallic bond is present between them.
(iv) Alkali metals and their salts give characteristic colour to the flame, because their valence electrons easily get excited from lower energy level to higher energy level.

Chemical properties : Alkali metals are highly reactive due to their low ionization enthalpy. Their reactivity increases down the group.
(i) Reaction with air : Alkali metals form oxide in presence of air and become dull. They form lithium monoxide, sodium peroxide and others form superoxide.
4Li + O₂ → 2Li₂O (Monoxide)
2Na + O₂ → Na₂O₂ (Peroxide)
M + O₂ → MO₂ (Superoxide) (M = K, Rb, Cs)
Lithium reacts directly with nitrogen of the air to form nitride (Li₃N).

(ii) Reaction with water : Alkali metals react with water to form hydroxides and dihydrogen.
2M + 2H₂O → 2M⁺ + 2OH^– + H₂ (M = Alkali metal)
Reaction of Li with water is less than that of sodium.
Alkali metals also react with proton donor species like alcohol, alkyne and gaseous ammonia.

(iii) Reaction with dihydrogen : Alkali metals react with dihydrogen to form ionic hydride.

(iv) Reaction with halogen : Alkali metals easily react with halogens to form ionic halides.
2M + X₂ → 2MX Metal halide . (M = Alkali metal).
(v) Reducing nature : Alkali metals easily lose their valence electrons, thus they are strongly reducing.
Na → Na⁺ + e^–
Lithium is strongly reducing whereas sodium is least reducing.
(vi) Solubility in liquid ammonia : All alkali metals dissolve in liquid ammonia to form deep blue solution.

(vii) All alkali metals react with liquid ammonia to form alloys. With mercury they form amalgams and the reactions are exothermic.

Question 19.
Discuss the general characteristics and gradation in properties of Alkaline earth metals.
Answer:
Elements of group-2 Be, Mg, Ca, Sr, Ba and Ca and together known as Alkaline earth metals (except Be).
General characteristics :
[A] Atomic properties :
(i) Electronic configuration : Their general electronic configuration is represented as [Noble gas]ns².
(ii) Atomic and ionic radius : Their atomic and ionic radii are smaller than the nearest
alkali metals of the same period. Atomic and ionic radii in the group increases with the increase in atomic number. ‘
(iii) Ionization enthalpy : First ionization energy of alkaline earth metals due to their smaller size is more than that of alkali metals[ionization enthalpy ∝ \(\frac{1}{\text { Atomic size }}\) ionization enthalpy of Alkaline earth metals is less than the corresponding alkali metals.

(iv) Hydration enthalpy of Alkaline earth metals like alkali metal ions decreases gradually on moving down the group.
Be²⁺ > Mg²⁺ > Ca²⁺ > Sr²⁺ > Ba²⁺
Hydration enthalpy of Alkaline earth metals is less than the corresponding Alkali metal ions.

[B] Physical properties : (i) These metals are silvery white, lustrous and soft (but harder than alkali metals).
(ii) The melting and boiling of these metals are higher than the corresponding alkali metals due to their smaller size. The trend is however not systematic.
(iii) Flame colouration : Be and Mg do not impart any colour to the flame. The various colours is due to difference in energy required for the excitation of electrons.

(iv) Density : Their density decreases from Be to Ca, then it increases. These metals due to their small size are dense, heavy and harder than corresponding alkali metals.

[C] Chemical properties : Alkaline earth metals are comparatively less reactive than Alkali metals. Reactivity of these metals decreases on moving down the group.
(i) Reaction with air and water: Be and Mg are inert towards air (oxygen) and water because of the formation of an oxide layer on their surface. Though Beryllium powder easily bums in air.

Similarly magnesium is highly electropositive and bums in air with a bright light to form MgO and Mg₃N₂. Ca, Sr and Ba react readily with air to form oxide and nitride.
(ii) Reaction with hydrogen : Except Be, all other metals react with hydrogen on heating to form MH₂ type of hydrides. (M = Mg, Ca, Sr, Ba)
(iii) Reaction with halogen : These metals react with halogens at high temperature to form MX₂ type of halides.
M + X₂ → MX₂ (X = F, Cl, Br)

(iv) Reaction with acids : These metal react readily with acids and liberate H₂ gas.
M+2HCl → MCl₂ + H₂ ↑
(v) Reducing nature : Like Alkali metals, alkaline earth metals are also of strong reducing nature, but less than alkali metals. On moving down the group their reducing nature decreases.
(vi) Like alkali metals, alkaline earth metals also dissolve in liquid ammonia to form bluish black solution.

Question 20.
Compare the alkali metals and alkaline earth metals with respect to :
(i) Ionization enthalpy,
(ii) Basicity of oxides and
(iii) Solubility of hydroxides.
Answer:
(i) Ionization enthalpy : First ionization enthalpy of Alkaline earth metals is higher than that of corresponding alkali metals. But value of second ionization enthalpy of alkaline earth metals is less than the corresponding alkali metals.

(ii) Basicity of oxides: Oxides of alkali and alkaline earth metals dissolve in water to form basic hydroxides. Basicity of hydroxides of alkaline earth metals is less than that of hydroxides of alkali metals.
(iii) Solubility of hydroxides : Solubility of hydroxides of alkaline earth metals is less than that of corresponding alkali metals.

Question 21.
In what ways Lithium shows similarities to Magnesium in its chemical behaviour ?
Answer:
Due to similar size, Lithium and Magnesium show similar properties (Atomic radius Li = 152 pm, Mg = 160 pm).
Ionic radius : Li⁺ = 76 pm, Mg²⁺ = 72 pm. Their similar properties are as follows :
(i) Both Li and Mg are lighter and harder than other members of their group.
(ii) Both Li and Mg react slowly with water.
The s-Block Elements  Class 11 Important Questions Objective Type

1. Choose the correct answer:

Question 1.
Plaster of Paris is :
(a) (CaSO₄)₂.2H₂O
(b) CaSO₄.2H₂O
(c) CaSO₄.H₂O
(d) CaSO₄.
Answer:
(a) (CaSO₄)₂.2H₂O

Question 2.
Lowest melting point compound :
(a) LiCl
(b) NaCl
(c) KCl
(d) RbCl.
Answer:
(a) LiCl

Question 3.
Active constituent of bleaching powder :
(a) CaOCl₂
(b) Ca(OCl)Cl
(c) Ca(O₂Cl₂)
(d) CaCl₂O₂.
Answer:
(b) Ca(OCl)Cl

Question 4.
Solution of sodium in liquid ammonia is reducing due to following reason:
(a) Na atom
(b) NaH
(c) NaNH₂
(d) e^–(NH₃)x.
Answer:
(d) e^–(NH₃)x.

Question 5.
Reaction of sodium with water is vigorous, not with lithium because lithium :
(a) Atomic mass is more
(b) Is a metal
(c) Is more electropositive
(d) Is more electronegative.
Answer:
(d) Is more electronegative.

Question 6.
Order of stability of various chlorides :
(a) LiCl > KC1 > NaCl > CsCl
(b) CsCl > KC1 > NaCl > LiCl
(c) NaCl > KC1 > LiCl > CsCl
(d) KCl>CsCl > NaCl > LiCl.
Answer:
(a) LiCl > KC1 > NaCl > CsCl

Question 7.
Hydration energy of Mg²⁺ ion will be more than :
(a) Al³⁺
(b) Na⁺
(c) Be²⁺
(d) Mg³⁺.
Answer:
(b) Na⁺

Question 8.
Which property of alkaline earth metals increase with increase in atomic number :
(a) Ionization energy
(b) Solubility of hydroxides
(c) Solubility of sulphates
(d) Electronegativity.
Answer:
(b) Solubility of hydroxides

Question 9.
Which is used for drying air :
(a) CaCO₃
(b) Na₂CO₃
(c) NaHCO₃
(d) CaO.
Answer:
(d) CaO.

Question 10.
Metallic lustre of sodium is due to :
(a) Diffusion of Na⁺ ion
(b) Rotation of electrons of outermost shell
(c) Excitation of free electrons Which sulphate has least solubility
(d) bec lattice.
Answer:
(c) Excitation of free electrons Which sulphate has least solubility

11. Which sulphate has least solubility:
(a) BaSO₄
(b) MgSO₄
(c) SrSO₄
(d) CaSO₄.
Answer:
(a) BaSO₄

Question 12.
Order of thermal stability of carbonates of alkaline earth metals is :
(a) BaCO₃ > SrCO₃ > CaCO₃ > MgCO₃
(b) BaCO₃ > SrCO₃ > MgCO₃ > CaCO₃
(c) CaCO₃ > SrCO3 > MgCO3 > BaCO3
(d) MgCO3 > CaCO3 > SrCO3 > BaCO3.
Answer:
(d) MgCO3 > CaCO3 > SrCO3 > BaCO3.

Question 13.
Compounds of which element are mainly covalent:
(a) Ba
(b) Sr
(c) Ca
(d) Be.
Answer:
(d) Be.

Question 14.
Important ore of magnesium is :
(a) Malachite
(b) Cassiterite
(c) Camallite
(d) Galena.
Answer:
(c) Camallite

Question 15.
Functions as air purifier in air craft:
(a) CaO
(b) Ca(OH)₂
(c) KO₂
(d) Anhydrous CaCl₂.
Answer:
(c) KO₂

Question 16.
Solubility of sulphates of alkaline earth metals decreases in moving down the group. Its reason is :
(a) Increase in melting point
(b) Increase in lattice energy
(c) Increase in coordination number
(d) All of these.
Answer:
(b) Increase in lattice energy

Question 17.
Which element does not give flame test:
(a) Be
(b) Ca
(c) Ba
(d) Sr.
Answer:
(a) Be

Question 18.
Which ore is not of lithium : ’
(a) Petalite
(b) Triphylite
(c) Elbite
(d) Spodumine.
Answer:
(c) Elbite

Question 19.
Magnesium is present in :
(a) Haemoglobin
(b) Chlorophyll
(c) Vitamin B₁₂
(d) Vitamin C.
Answer:
(b) Chlorophyll

Question 20.
Mixture of MgCl₂ and MgO is known as :
(a) Sorrel cement
(b) Portland cement
(c) Fitkari
(d) Mortar.
Answer:
(c) Fitkari

Question 21.
Mixture of CaCN₂ and C is known as :
(a) Barytte
(b) Anhydrite
(c) Nitrolim
(d) Iceland spot.
Answer:
(a) Barytte

Question 22.
Lithopone is :
(a) BaSO₄ + BaS
(b) BaSO₄ + ZnS
(c) BaO + ZnS
(d) BaCO₃ + ZnO.
Answer:
(b) BaSO₄ + ZnS

Question 23.
Micro-cosmic salt is:
(a) Na(NH₄) HPO₄.4H₂O
(b) Na(NH₄)H₂O.
(c) Na(NH₃) HPO₄.4H₂O
(d) K(NH₄)HPO₃.2H₂O.
Answer:
(a) Na(NH₄) HPO₄.4H₂O

Question 24.
Which of the following alkali metals gives hydrated salts :
(a) Li
(b) Na
(c) K
(d) Cs.
Answer:
(a) Li

Question 25.
Which of the following alkali metals has lowest melting point:
(a) Na
(b) K
(c) Rb
(d) Cs.
Answer:
(d) Cs.

Question 26.
Which of the following alkaline earth metal carbonate is more stable towards heat:
(a) MgCO₃
(b) CaCO₃
(c) SrCO₃
(d) BaCO₃.
Answer:
(d) BaCO₃.

Question 27.
Co-product of Solvay ammonia process is :
(a) CO₂
(b) NH₃
(c) CaCl₂
(d) CaCO₃.
Answer:
(c) CaCl₂

Question 28.
Increasing order of ionic product is :
(a) BeCl₂ < MgCl₂ < CaCl₂ < BaCl₂
(b) BeCl₂ < MgCl₂ < BaCl₂ < CaCl₂
(c) BeCl₂ < BaCl₂ < MgCl₂ < CaCl₂
(d) BaCl₂ < CaCl₂ < MgCl₂ < BeCl₂.
Answer:
(a) BeCl₂ < MgCl₂ < CaCl₂ < BaCl₂

Question 29.
Which of the following is paramagnetic :
(a) KO₂
(b) SiO₂
(c) TiO₂
(d) BaO₂.
Answer:
(a) KO₂

2. Fill in the blanks:

1. Radius of Ca²⁺ ion is less than K⁺ because of ……………….. .
Answer:
High positive charge

2. Be(OH)₂ is soluble in both acid and base because it is of ……………….. nature.
Answer:
Amphoteric

3. Lithium resemble ……………….. element of group 2.
Answer:
Mg

4. Sodium metal gives blue colour in liquid ammonia this is because of ……………….. .
Answer:
e^– (NH₃)x

5. Potassium forms three oxides ……………….., ……………….. and ……………….. .
Answer:
K₂O, K₂O, KO₂

6. Formula of hydrolith is ……………….. .
Answer:
CaH₂

7. s-block elements are strong ……………….. .
Answer:
Reducing agent

8. Alkaline earth metal ……………….. represent radio-active property.
Answer:
Radium

9. Alkali metal ……………….. represent radio-active property.
Answer:
Francium (Fr)

10. Alkaline earth metal hydroxide ……………….. is amphoteric.
Answer:
Be(OH)₂

11. Chemical formula of camallite is ……………….. .
Answer:
KCl.MgCl₂.6H₂O.

3. Match the following:

‘A’	‘B’
1. To dry air	(a) Ca(OCI)C1
2. Nitrolim is	(b) MgCl2 and MgO
3. Mixture of Sorrel cement	(c) BaSO4 and ZnS
4. Active component of bleaching powder	(d) CaCN2 and C
5. Mixture of Lithopone	(e) CaO
6. CaCO3	(f) Smoke screen
7. SiCl4	(g) Limestone.

Answer:
1.  (e) CaO
2. (d) CaCN₂ and C
3. (b) MgCl₂ and MgO
4. (a) Ca(OCl)Cl
5. (c) BaSO₄ and ZnS
6.  (g) Limestone
7. (f) Smoke screen.

4. Answer in one word/sentence:

1. Arrange the alkaline earth metals in the order of their reactivity.
Answer:
Chemical reactivity of alkaline earth metals increases on moving down the group (Be-Ra).
Be < Mg < Ca < Sr < Ba < Ra.

2. Write name and formula of two ores of lithium.
Answer:
(i) Spodumene LiAl(SiO₃)₂,
(ii) Lepidolite Li₂Al₂(SiO₃)₃.F(OH)₂.

3. Write names of two ores of magnesium.
Answer:
Two ores of magnesium are :
(i) Camallite
(ii) Dolomite.

4. Alkali metals are kept in kerosene oil. Why?
Answer:
Due to high reactivity.

5. Write the order of basic strength and solubility in water of hydroxides of alkaline earth metals.
Answer:
Basic strength and solubility of hydroxides of alkaline earth metals increases from top to bottom in a group.

6. Molecular formula of Glauber’s salt is.
Answer:
Na₂SO₄.10 H₂O.

5. State true or false :

1. Alkali metals are not found in free state but alkaline earth metals exist in free state in earth.
Answer:
False

2. Basicity of hydroxides of alkali metals increases on moving down in the group.
Answer:
True

3. Soda ash means Baking soda.
Answer:
False

4. Magnesium does not impart any colour to Bunsen flame whereas calcium give flame test.
Answer:
True

5. Lil is less ionic as compared to Nal.
Answer:
False

6. Common salt wets because it has less hydration energy.
Answer:
False

7. Due to diagonal relationship properties of sodium are similar to calcium and that of caesium are similar to strontium.
Answer:
False

8. Francium is a radio-active element.
Answer:
True

9. For similar chemical properties sufficient difference in electronegativity is essential.
Answer:
False

10. Na₂O is formed by heating sodium in excess of O₂.
Answer:
False

11. Concentration of K⁺ and Na⁺ is same in blood.
Answer:
False
12.Concentration of Na₊ is very less as compared to K₊ concentration inside a cell.
True.
Answer:

Students get through the MP Board Class 11th Chemistry Important Questions Chapter 9 Hydrogen which are most likely to be asked in the exam.

MP Board Class 11th Chapter 9 Hydrogen

Hydrogen Class 11 Important Questions Very Short Answer Type

Question 1.
What are the causes of temporary and permanent hardness of water? Explain.
Answer:
Temporary hardness of water is due to presence of bicarbonates of Ca and Mg. Permanent hardness is due to presence of chlorides and sulphates of calcium and magnesium.

Question 2.
Write chemical reactions to show the amphoteric nature of water.
Answer:
Water is amphoteric in nature. It behaves as an acid and base both. With stronger acids than itself, it acts as a base where as with stronger bases than itself it acts as an acid.

Question 3.
Is demineralized or distilled water useful for drinking purposes? If not, how can it be made useful?
Answer:
Demineralized or distilled water is not useful for drinking purposes. It can be made useful by mixing appropriate amount of some useful salts in it.

Question 4.
Knowing the properties of H₂O and D₂O, do you think that D₂O can be used for drinking purposes?
Answer:
Heavy water, D₂O is harmful for living organisms because in heavy water, rate of biochemical reactions decreases.

Question 5.
Beryllium forms covalent hydride whereas calcium forms ionic hydride. Why?
Answer:
Due to high. electronegativity Be (1.5) forms covalent hydride whereas electronegativity of Ca is low.

Question 6.
Explain Uno process of the preparation of hydrogen.
Answer:
Uno’s method : Aluminium is treated with hot caustic soda solution. Reaction proceed briskly. Aluminium dissolve and hydrogen gas is produced.
2 Al + 2NaOH + 2H₂O → 2NaAlO₂ +3H₂ Sodium meta aluminate.

Question 7.
How is impure hydrogen purified?
Answer:
On passing hydrogen gas on Platinum black or Palladium metal, hydrogen is adsorbed by the metal. This way impure hydrogen is purified because only pure hydrogen can be adsorbed by Palladium.

Question 8.
Write the disadvantages of hard water.
Answer:
Disadvantages of hard water:

1. More soap is wasted to produce lather for working clothes.
2. Hard water cannot be used in medicine for injections.
3. It is harmful for people suffering from arthritis.

Question 9.
Describe Lanes process for the preparation of hydrogen.
Answer:
In this process water vapours are passed over red hot iron at 600- 900° C and hydrogen gas is formed. Reaction is reversible, hence H2 should be periodically removed.
3Fe + 4H₂O ⇌ Fe₃O₄+4H₂

Question 10.
Explain the method by which temporary and permanent both types of hardness is removed.
Answer:
Chemical method: In this method calcium hydroxide [Ca(OH)₂ ] or sodium carbonate (Na₂CO₃) or sodium phosphate is mixed in hard water when insoluble salts of Ca and Mg are formed and separated and both temporary and permanent hardness are removed.
Ca(HCO₃)₂ + Na₂CO₃ → CaCO₃↓ +2NaHCO₃
MgCl₂+Na₂CO₃ → MgCO₃ ↓ +2NaCl
CaSO₄ + Na₂CO₃ → CaCO₃ ↓ +Na₂SO₄

Question 11.
How is the strength of H₂O₂ expressed?
Answer:
Strength of Hydrogen peroxide solution: The strength of hydrogen peroxide solution is expressed in terms of the volume of oxygen obtained from it. For example, 10 volume, 20 volume etc. The strength is equal to the volume of oxygen produced at NTP which is obtained by heating one unit of that solution.
The ‘10 volume’ solution of hydrogen peroxide will give 10 ml oxygen at NTP when 1 ml of the sample is heated.

Question 12.
Write the uses of hydrogen.
Answer:
Uses of hydrogen :

1. As a reducing agent,
2. For the preparation of ghee by the hydrogenation of oil,
3. For the preparation of artificial petrol by reacting with coal powder,
4. In glass industry, in cooling the glass slowly.

Question 13.
What is the difference in the steps of hydrolysis and hydration?
Answer:
The process of spontaneous reaction of H⁺ ion and OH^– ions of water with the anion and cation of a salt to form the basic acid and water is called hydrolysis.

On the other hand, hydration means formation of hydrated ions or hydrated salts by the addition of water in ions and molecules.

Question 14.
What do you expect the nature of hydrides is, if formed by elements of atomic numbers 15, 19, 23 and 44 with dihydrogen ? Compare their behaviour towards water.
Answer:

• Element of Z = 15, Phosphorus, is a member of p-block, thus it a covalent hydride.
• Element of Z = 19, Potassium, is a member of 5-block, thus it is an ionic hydride.
• Element of Z = 23, Vanadium, is a member of rf-block and group-5. It forms interstitial hydride VH16. It is a non-stoichiometric hydride.
• Element of Z = 44, Rutheum, is a member of d-block and group-8. It does not form any hydride because elements of group-7, 8 and 9 do not form hydride (Hydrideless). With water, only ionic hydride, KH reacts to liberate dihydrogen gas.

Question 15.
What is oil gas and how is it obtained?
Answer:
This gas is prepared by the distillated of kerosene oil. A thin layer of kerosene oil is poured in red hot iron retort by which bigger molecules break up into methane, ethylene and acetylene. It is used in burners.

Question 16.
What is coal gas? Write its chemical composition.
Answer:
Coal gas is a mixture of various gases in which hydrogen is also one of the constituent. Percentage composition of coal gas is as follows :
H₂ = 43.55%,CH₄ = 25.35%,CO₂ = 0.3%, CO = 4.11% N₂ =2.12%,
O₂ = 0-1.5%
Apart from being used as a fuel, coal gas is also used for the production of light.

Question 17.
A gas is produced on reacting sodium with water. Write its name and formula.
Answer:
By the reaction of water on extremely reactive metals like – sodium, potassium, calcium at ordinary temperature, hydrogen gas is released.
2Na + 2H₂O → 2NaOH+H₂
Ca + 2H₂O → Ca(OH)₂+H₂.

Question 18.
What is the cause of hardness of water and hardness is of how many types?
Answer:
Hardness of water is due to the presence of soluble bicarbonates, sulphates, chlorides, salts of calcium and magnesium.
1. Temporary hardness: When hardness of water is removed by boiling or by simple methods then it is known as temporary hardness. Temporary hardness is due to bicarbonates of Ca and Mg.

2. Permanent hardness: If hardness is not removed by boiling or by simple methods, it is known as permanent hardness. It is due to chlorides and sulphates of Ca and Mg.

Question 19.
What is hard and soft water?
Answer:
Soft water: Water which easily form good lather with water is called soft water. Distilled water, rain water, river water etc. are soft water.

Hard water: Water which does not form good lather with water easily is called hard water. It gives less lather and soap spreads and form clots, due to this lot of soap is wasted.

Question 20.
What is distilled water? Write its one use.
Answer:
Pure water obtained by distillation method is known as distilled water. It is used for the preparation of medicines and solutions in the laboratory.

Question 21.
What is the utility of heavy water?
Answer:

• Heavy water is mainly used as a moderator in nuclear reactors.
• It is used as a tracer compound in the study of mechanism of reactions.
• It is used for the preparation of other deuterium compounds like CD₄, D₂SO₄etc.

Question 22.
Acidic solution of hydrogen peroxide behaves as an oxidizing as well as reducing agent. Explain it with the help of chemical equations.
Answer:

• H₂O₂ oxidizes acidified KI to iodine.
  2KI + 2H₂O₂ + H₂SO₄ → I₂ + K₂SO₄ + 2H₂O
• In alkaline medium H202 reduces KMnO₄ to MnO₂.
  2KMnO₄ + 3H₂O₂ →2MnO₂+2KOH + 3O₂ +2H₂O.

Question 23.
Dilute solution of hydrogen peroxide cannot be concentrated by heating. How can it be concentrated ?
Answer:
Dilute solution of H₂O₂ cannot be concentrated because it decomposes much below its melting point.
2H₂O₂ → 2H₂O + O₂
1% H₂O₂ liberated from water is distilled at low pressure and concentrated to 30% by mass. Again it is distilled at low pressure upto 85%. Remaining water is freezed to obtain pure hydrogen peroxide.

Question 24.
For the manufacture of hydrogen peroxide by peroxides phosphoric acid is more useful than sulphuric acid. Why?
Answer:
H₂SO₄ acts as a catalyst in the decomposition of H₂O₂. Thus, in the manufacture of H₂O₂ by peroxides weak acids like H₃PO₄, H₂CO₃ etc. are more suitable than H₂SO₄.

Question 25.
Ionic hydride of a basic metal exhibits effective covalent behaviour and it is unreactive towards oxygen and chlorine. It is used in the synthesis of other useful hydrides. Write the formula of this hydride. Write its reaction with Al₂Cl₆.
Answer:
It is ionic hydride LiH. Due to smallest size it shows covalent behaviour also. LiH is extremely stable. It is extremely unreactive towards oxygen and chlorine. It reacts with Al₂Cl₆ to form Lithium aluminium hydride.
8LiH +Al₂Cl₆ → 2LiAlH₄ +6LiCl.

Question 26.
Why does hard water do not for lather readily with soap?
Answer:
In hard water bicarbonates, chlorides and sulphates of Ca and Mg are dissolved. Soaps are sodium salt of higher fatty acids e.g., sodium stearate (C₁₇H₃₅COONa). Salts of calcium and magnesium react and form precipitates of calcium and magnesium stearate.
C₁₇H₃₅COONa + M₂₊ → (C₁₇H₃₅COO)₂ M + 2Na⁺
Until all the salts are precipitated, no lather is formed with soap and this soap is wasted.

Question 27.
Write four uses of hydrogen peroxide.
Answer:
Uses of hydrogen peroxide :

• It is an antiseptic, used for washing teeth, ears, wounds, etc.
• As an oxidizing agent in laboratory.
• It is used for bleaching hair, silk, wool, feathers, ivory, etc.
• As a rocket fuel (as propellant) by producing oxygen.

Question 28.
Write industrial method of preparation of hydrogen by water gas.
Answer:
Water gas is produced by passing steam over red hot coke.

Water gas is mixed with steam and heated at a temperature of 450 °C in presence of Fe₂O₃ as catalyst and chromic oxide as promoter when carbon monoxide gets oxidized to carbon dioxide.

Question 29.
What is conductivity water? Write its use.
Answer:
Kohlrausch distilled water 42 times at low pressure in a Quartz container and obtained exteremely pure water. This is known as conductivity water. Some potassium permanganate is mixed in distilled water and it is again distilled in a strong glass retort. It is used in the determination of conductivity.

Question 30.
What is deuterium? Write its two uses.
Answer:
Deuterium is an isotope of hydrogen. It is called heavy hydrogen. Its nucleus consists of one proton and one neutron and one electron rotates around the nucleus. That means its atomic weight is 2 and atomic number is 1. It is represented as H₁² or D. In common H₂ gas it is presented 0.0156%. It was discovered by Unbrickwade and Murphy. Deuterium is obtained by the electrolysis of heavy water at cathode and O₂ is obtained at anode.

Uses: 1. It is used to determine the reaction mechanism of chemical and biochemical reactions.
2. It is used in the artificial transmutation of element as a particle in the form of deuteron.
₇N¹⁴+₁D²→ BCl₂+2He⁴.

Question 31.
Sodium reacts with dihydrogen to form crystalline ionic solid. This compound is non-volatile and non-conducting in nature. It reacts readily with water to form dihydrogen gas. Write the formula of this compound and its reaction with water. What will happen on electrolysing it in molten state?
Answer:
Sodium reacts with dihydrogen to form sodium hydride which is a crystalline ionic solid. ‘
2Na + H₂ → 2Na⁺+H^–
It reacts readily with water to form H₂ gas
2NaH + 2H₂O →2NaOH+2H₂
In solid state, electric current does not flow in NaH but in molten state it releases H₂ at anode and Na at cathode.

Question 32.
Write the redox reaction between fluorine and water.
Answer:
Fluorine is a strong oxidizing agent. It oxidizes H₂O to 0₂ and O₃.
2F_2(g) +2H₂O_(l) → O_2(g) + 4H⁺_(aq) +4F^–_(aq)
3F_2(g) +3H₂O_(l) → O_3(g)+6H⁺_(aq) +6F^–_(aq)

Question 33.
Explain: HCl is a gas and HF liquid. Why?
Answer:
Size of F is smaller than Cl and its electronegativity is also more. Thus, it forms stronger H-bond than Cl. That is why HF is a liquid and HC1 a gas.

Question 34.
Write a chemical equation for the manufacture of D₂O₂.
Answer:
D₂O₂ is obtained by the action of BaO₂ on water-soluble D₂SO₄.
BaO₂ + D₂SO₄ → BaSO₄ + D₂O₂.

Question 35.
Why can H₂O₂ not be stored for a long time?
Answer:
H₂O₂ contain peroxide bond. Due to the presence of this bond it is unstable and starts decomposing.
2H₂O₂ → 2H₂O + O₂
As a result H₂O₂ cannot be stored for a long time. This decomposition reaction is catalyzed by glass, therefore the decomposition becomes faster in glass bottle. To stop it, it is stored in a coloured waxy of layered bottle.

Question 36.
H₂O₂ is known as an antichlor. Why?
Answer:
In coloured substances in which bleaching is done by chlorine, some amount of chlorine is left remaining. This way to remove the remaining chlorine H202 is used. Due to this property, it is known as an antichlor.
Cl₂+H₂O₂ → 2HCl + O₂.

Question 37.
On boiling temporary hard water v .th slaked lime it becomes soft. Why?
Answer:
When temporary hard water is boiled with slaked lime then Ca and Mg bicarbonates present in hard water gets converted to insoluble bicarbonates which are removed by Alteration.
Ca(HCO₃)₂ +Ca(OH)₂ → 2CaCO₃ ↓+2H₂O
Mg(HCO₃)₂ + Ca(OH)₂ → MgCO₃ + CaCO₃ + 2H₂O.

Question 38.
Saline hydrides are known to react with water violently producing fire. Can CO₂, a well-known fire extinguisher, be used in this case. Explain.
Answer:
Saline hydrides (NaH, CaH₂ etc.) react with water violently to produce metal oxide and dihydrogen.

Question 39.
If equal weight of liquid water and ice pieces are taken then density of ice is less than water. Why?
Answer:
In ice, water molecules are not denser as in liquid water. In the lattice structure of ice vacant voids are present due to which its volume becomes more and density less.

Question 40.
Explain one method of preparation of Hydrogen peroxide.
Answer:
Peroxide method: H₂O₂ is formed by the action of sodium peroxide on ice-cooled 20% H₂SO₄.
Na₂O₂ + H₂SO₄ → H₂O₂ + Na₂SO₄
On cooling, sodium sulphate separate in the form of crystals Na₂SO₄.5H₂O which is filtered and 30% H₂O₂ solution is obtained.

Question 41.
In which compound oxidation state of hydrogen is negative?
Answer:
When hydrogen combines with lighter and extremely reactive metals like Na, K, Ca etc. then electrovalent hydrides are formed. In these electrovalent hydrides, oxidation state of hydrogen is negative. -1.

Question 42.
Dihydrogen (H₂)- reacts with dioxygen (O₂) to form water. When an isotope of hydrogen which has one proton and one neutron in the nucleus is reacted with oxygen, then write the name and formula of the product formed. Will the reactivity of both the isotopes with oxygen be same? Discuss your answer.
Answer:
Isotope of hydrogen whose nucleus has one proton and one neutron, is deuterium
(D, _1²H).

Reactivity of H₂ and D₂ is the different towards oxygen. D-D bond is stronger as compared to H-H bond. Thus, H₂ is more reactive than D₂.

Hydrogen Class 11 Important Questions Short Answer Type

Question 1.
Justify the position of hydrogen in the periodic table on the basis of its electronic configuration.
Answer:
Hydrogen is the first element placed in the periodic table. Its electronic configuration is 1s¹. Like alkali metals, it can lose one electron and form H⁺ ion. Like halogens, it can gain one electron and form H^– ion. Like alkali metals it forms oxide, halide and sulphide.

Though alternative to alkali metals its ionization enthalpy is high and under normal conditions it does not exhibit metallic behaviour. Actually in terms of ionization enthalpy it exhibits more similarity towards halogens. Like halogens it forms diatomic molecule, reacts with elements to form hydrides and various covalent compounds. Though, alternative to halogens its reactivity is very less.

Thus, on the basis of the above properties, hydrogen should be placed with alkali metals in Group-1 in the periodic table or with halogens in Group-17 in the periodic table. Although it shows similarity in properties with alkali metals and halogens it also show dis-similarities with them. Thus, it should be given a separate place in the periodic table.

Question 2.
Write the name of isotopes of hydrogen and state what is the mass ratio of these isotopes ?
Answer:
Isotopes of hydrogen are Protium (_1¹H), Deuterium (D or _1²H) and Tritium (T or ₃H )
Mass ratio of Protium, Deuterium and Tritium =1 : 2 : 3.

Question 3.
H₂O₂ is used for restoring the colour of lead paintings. Explain this state-ment.
Ans. The lead painting contain white lead which is basic lead carbonate. The colour of these painting gradually becomes blackish due to the formation of PbS by the action of H₂S of the atmosphere. As by the action of H₂O₂ lead sulphide is oxidized to lead sulphate, colour of lead paintings is restored.

Such lead paintings when washed with H₂O₂ become bright due to conversion of black PbS to white PbSO₄.

Question 4.
Draw the figure of all the isotopes of hydrogen and write their names.
Answer:
There are three isotopes of hydrogen :

1. Normal hydrogen or Protium : _1¹H
2. Heavy hydrogen or Deuterium: _1²H
3. Tritium: _3¹H.

Question 5.
Explain ortho and para hydrogen.
Answer:
Only one proton is present in the nucleus of H atom and one electron rotates around it. Like electron, proton also spin around its one axis. If protons of two atoms of H2 molecule rotates in the same direction, then it is called ortho hydrogen, but when proton of two H atoms rotates in opposite directions, then it is called para hydrogen. Para hydrogen is more stable and general hydrogen is the para form.

Question 6.
Hard water is not used in boilers. Why?
Answer:
Hard water is not used in boilers because compounds of calcium and magnesium are deposited on internal sides of boiler. This coating is not conductor of heat, so wastes of fuel takes place. Due to this coating, boilers have to be heated very much for heating water.

At high-temperature Fe of water surface of boiler react with O₂ of air and form Fe₃O₄. Sometimes hot water directly react with iron surface from the gaps between these coatings and form Fe₃O₄ and H₂.
Hydrogen is volatile gas therefore there is risk for explosion. If MgCl₂ is present in hard water, it react with water and form HCl. This acid is corrosive for boiler.
MgCl₂ +H₂O → Mg(OH)Cl + HCl.

Question 7.
How can production of dihydrogen obtained by ‘coal gasification’ be increased?
Answer:
Production of syngas or synthetic gas by coal is known as ‘coal gasification.

CO₂ produced is separated by passing through sodium arsenite solution.

Question 8.
Describe the bulk preparation of dihydrogen by electrolytic method. What is the role of an electrolyte in this process?
Answer:
Hydrogen is obtained by the electrolysis of acidulated water by platinum electrodes.

Role of electrolyte is the conductivity of water.

Question 9.
What happens when :
(i) Calcium hydride is reacted with water.
(ii) H₂O₂ is mixed in acidic KMnO₄.
(iii) Alkaline solution of Potassium ferricyanide is reacted with H₂.
(iv) Reaction of H₂ with N₂ in presence of Feat high temperature and at high pressure.
Answer:
(i) Calcium hydride reacts with water to form calcium hydroxide and hydrogen.
CaH₂ +2H₂O → Ca(OH)₂ +2H₂
(ii) H₂O₂ reacts with acidic KMnO₄ and decourises it.

(iii) H₂O₂ reduces alkaline solution of potassium ferricyanide to potassium ferrocyanide.
2K₃[Fe(CN)₆]+2KOH+H₂O₂ → 2K₄[Fe(CN)₆]+2H₂O+O₂
(iv) At high temperature and high pressure in the presence of Fe as catalyst and Mo as promoter, H₂ is reacted with N₂, ammonia is formed.

Question 10.
At normal temperature H₂O is liquid while H₂S is gas. Why?
Answer:
Electronegativity of oxygen is comparatively higher than the hydrogen so in H₂O molecule, bonding electrons of O-H bond are attracted towards oxygen. Thus, partial positive charge (δ⁺) develops on hydrogen and partial negative (δ^–) charge on oxygen.

When many water molecules are close to each other positively charged hydrogen form a weak bond with negatively charged oxygen of another atom. This type of bond is called hydrogen bonding. Due to hydrogen bonding many molecules are associated very close to each other. Therefore, water is in liquid state.
In H₂S hydrogen bonding is not present because S is less electronegative than oxygen. Therefore, H₂S is a gas.

Question 11.
Write the following reactions with H₂O₂ (Give only chemical equation):
(a) O₃,
(b) PbS,
(c) KI,
(d) Cl₂,
(e) H₂S.
Answer:
(a) Reaction with O₃: Water is formed and liberates oxygen.
₂O₂ + O₃→ H₂O + 2O₂
(b) Reaction with PbS: PbSO₄ is formed.
4H₂O₂ + PbS → PbSO₄ + 4H₂O
(c) Reaction with KI: Iodine is liberated.
2KI + H₂O₂ → 2KOH +I₂
(d) Reaction with CI2: HCl is formed and oxygen is liberated.
Cl₂ +H₂O₂ → 2HCl + O₂
(e) Reaction with H₂S: Hydrogen peroxide oxidizes H₂S to S.
H₂S +H₂O₂ → 2H₂O +S.

Question 12.
Do you expect the carbon hydrides of the type (C_nH_2n+2) to act as lewis acid or base? Justify your answer.
Answer:
(C_nH_2n+2) type of carbon hydrides are electron-rich hydrides. They contain sufficient electrons to form covalent bond. Therefore, they neither behave as Lewis acid nor as Lewis base.

Question 13.
What do you understand by the term “non-stoichiometric hydrides”? Do you expect this type of hydrides to be formed by alkali metals? Justify your answer.
Answer:
These hydrides are formed by cf-block elements (except metals of group 7, 8 and 9) and f-block elements. Hydrides are always non-stoichiometric i.e. are hydrogen deficient. In these hydrides, hydrogen lie in the interstitial sites.
Example : LaH_2.87, YbH_2.55,TiH_15-1.8,PdH_{0.6- 0.8}etc.

This type of hydrides cannot be obtained by alkali metals. By alkali metals ionic or saline hydrides are obtained. These are stoichiometric. Alkali metals are highly electropositive, thus they transfer electron to H-atom and form H^– ion. These H^– ion occupy the interstitial voids.

Question 14.
Complete the following equations :
(i) PbS_(s) + H₂O_2(aq) →
(ii) CO_(g) + 2H_2(g)
Answer:
(i) PbS_(s) + 4H₂O_2(aq) → PbSO₄ + 4H₂O
(ii)

Question 15.
What is water gas? Write its composition and use.
Answer:
Water Gas: This gas is a mixture of carbon monoxide and hydrogen. On passing water vapour on red hot coke mixture of carbon monoxide and hydrogen (water gas) is obtained.

Composition : H₂ = 49%, N₂ = 4%, CO = 44%, CO₂ = 2.7%, CH₄ = 0.3%.

Use :

1. As a fuel in furnaces,
2. In the presence of carburetted water gas,
3. In the manufacture of ammonia,
4. In providing light with unsaturated hydrocarbons.

Question 16.
Hardness of water is of how many types? How is temporary hardness removed?
Answer:
Water which contains soluble bicarbonates, sulphates and chlorides of Ca and Mg is hard water. Hardness of water is of two types :
(i) Temporary hardness,
(ii) Permanent hardness.
Note:
Hardness of water is due to the presence of soluble bicarbonates, sulphates, chlorides, salts of calcium and magnesium.
1. Temporary hardness: When hardness of water is removed by boiling or by simple methods then it is known as temporary hardness. Temporary hardness is due to bicarbonates of Ca and Mg.

2. Permanent hardness: If hardness is not removed by boiling or by simple methods, it is known as permanent hardness. It is due to chlorides and sulphates of Ca and Mg.
There are two methods for the removal of temporary hardness :
1. By Boiling: On boiling soluble bicarbonate of Ca and Mg gets converted to insoluble carbonates and gets precipitated.
Ca(HCO₃)₂ → CaCO₃ + H₂O + CO₂
Mg(HCO₃)₂ → MgCO₃ + H₂O + CO₂

2. Clark’s method: Temporary hard water is mixed with definite amount of slaked lime when soluble bicarbonate converted to insoluble carbonates and gets precipitated.
Ca(HCO₃)₂+Ca(OH)₂ → 2CaCO₃+2H₂O
Mg(HCO₃)₂ +Ca(OH)₂ → MgCO₃ +CaCO₃ + 2H₂O.

Question 17.
Draw Lewis structure of hydrogen peroxide.
Answer:

Question 18.
How do you expect the metallic hydrides to be useful for hydrogen storage? Explain.
Answer:
In metallic hydrides, hydrogen is absorbed in the form of H-atom. In transition metals, adsorption of hydrogen is used for the storage of hydrogen. Metals like Pd, Pt can accomodate bulk volume of hydrogen. This property has strong possibility of hydrogen storage and is used as an energy source. Hydrides on heating give hydrogen and the metal.

Question 19.
How does atomic hydrogen or oxy-hydrogen torch function for cutting and welding purposes? Explain.
Answer:
Atomic hydrogen and oxy-hydrogen torches find use for cutting and welding purposes. Atomic hydrogen atoms produced by dissociation of dihydrogen with the help of electric arc are allowed to recombine on the surface to be welded to generate the temperature of 4000 K.

Question 20.
Compare the structures of H₂O and H₂O₂.
Answer:
In water, oxygen is sp³ hybridized. Due to stronger lone pair-lone pair repulsion than bond pair-bond pair repulsion, HOH bond angle reduces from 109.5° to 104.5° due to which water has a bent shape. It is strongly polar molecule. Structure of hydrogen peroxide is non-polar.

Dipole moment value of H₂O₂ indicates that all the atoms of H₂O₂ do not lie in a plane. Structure of H₂O₂ can be compared to an open book at 94°C. In it H-O-H bond angle is 91°.

Question 21.
Explain structure of hydrogen peroxide.
Answer:
Dipole moment of H₂O₂ is 2.1 D, it indicates that structure of H₂O₂ is non-planar. By X-rays and other physical methods it is known that structure of H₂O₂ is like open book. In gaseous state planes form an angle of 111.5°. In the axis, there are two oxygen atoms and one hydrogen in each plane.

The H-O and O-O bond lengths are 0.95 Å and 1.47 Å respectively. There is a slight change in the crystalline state due to hydrogen bonding. The angle between the planes is 90.2, angle O-O-H = 101.9°, bond length H-O = 0.99 A and O-O is 1.46Å.

According to modem and latest belief probably both forms occur in equilibrium in aqueous solutions involving ionisation which explains the feeble acidic nature of hydrogen peroxide.

Question 22.
Explain that H₂O₂ acts as an oxidizing as well as reducing agent.
Answer:
H₂O₂ liberates oxygen atom easily, so it is strong oxidising agent, but when it reacts with other oxidising agent, it easily extracts oxygen atom from them hence it possesses reducing property too.
(a) Oxidising nature : 2KI + H₂O₂ → 2KOH +I₂
Na₂SO₃ + H₂O₂ → Na₂SO₄ + H₂O
H₂S + H₂O₂ → 2H₂O + S

(b) Reducing nature:
H₂O₂ + O₃ → H₂O + 2O₂
PbO₂ + H₂O₂ → PbO + H₂O + O₂
Cl₂ + H₂O₂ → 2HCl + O.

Question 23.
What is calgon ? How is hardness of water removed by its help?
Answer:
Calgon process: It is a beneficial and industrial method of softening of water. Calgon is poly hexa metaphosphate. Its sodium salt reacts with calcium and magnesium salts forming soluble complex compound by which calcium and magnesium are not left in the form of ions.
Na₂[Na₄(PO₃)₆] + CaSO₄ → Na₂[CaNa₂(PO₃)₆] + Na₂SO₄
Na₂[CaNa₂(PO₃)₆] + CaSO₄ → Na₂[Ca₂(PO₃)₆] + Na₂SO₄.

Question 24.
What do you understand by the term ‘autoprotolysis’ of water? What is its significance?
Answer:
Autoprotolysis of water means, self ionisation of water.

Due to autoprotolysis, nature of water is amphoteric i.e. it reacts both as an acid as well as a base. ‘
For example:

Question 25.
Consider the reaction of water with F₂ and suggest, in terms of oxidation and reduction, which species are oxidized/reduced.
Answer:

In these reactions, water acts as a reducing agent. Thus, it itself oxidizes to oxygen or ozone. Fluorine acts as an oxidizing agent and itself gets oxidized to F^– ion.

Question 26.
How does H₂O₂ behave as a bleaching agent? Write.
Answer:
Due to the production of nascent oxygen H₂O₂ acts as a bleaching agent.
H₂O₂ → H₂O+O
Coloured substance + [O] → Colourless substance
It oxidizes silk, hairs, clothes, paper pulp, wool, oil, fats etc.

Question 27.
How many types of hydrates are there? Explain with example.
Answer:
Water combines with various metal salts and forms hydrate. It is of three types :
1. Complexion cationic water: In complexion, it is present in cation forming ligand with the central metal.
Example: Hexa aqua chromium (III) chloride [Cr(H₂O)₆]Cl₃

2. Complexion oxyanion water: In complex ion water molecule is present with oxyanion.
CuSO₄ +5H₂O → [CU(H₂O)₄]²⁺ + [SO₄ (H₂O)]^2-

3. Crystalline water: In crystal lattice, water molecule is present in the interstitial voids between the molecules which is known as water of crystallization. Some metal salts lose the crystallized water on standing. This is known as efflorescence. Various substances absorb water from air on keeping in air.

Question 28.
Describe permutit process or zeolite process for the removal of hardness of water.
Answer:
Permutit process or Zeolite process: The water is passed over such substances which readily replace the Ca²⁺ and Mg²⁺ ions of hard water by sodium ions. The water does not become hard in presence of sodium salts. Of course too much of sodium salts also make it hard just as in seawater. The water thus obtained is soft. These products are called as zeolites. Zeolite is a technical name given to certain hydrated silicates of aluminium and sodium. It is also known as permutit. It is obtained by fusing sodium carbonate with alumina and silica.
Na₂CO₃+Al₂O₃+2SiO₂ → Na₂Al₂Si₂O₈+CO₂

Na₂Al₂Si₂O₈ is known as sodium zeolite and Al₂Si₂O₈ is zeolite and can be represented by Z. Sodium zeolite reacts with chlorides, sulphates of Ca and Mg as given below and hardness is removed.
CaCl₂ +Na₂Z → CaZ + 2NaCl
MgSO₄ + Na₂Z → MgZ + Na₂SO₄

Reactions have been given by representing sodium zeolite as Na₂Z.
After sufficient use permutit gets exhausted and cannot be used further. It is regenerated and made usable by passing 10% solution of sodium chloride.
CaZ + 2NaCl → CaCl₂+Na₂Z
MgZ + 2NaCl → MgCl₂ + Na₂ Z.

Question 29.
What is meant by ‘demineralized’ water and how can it be obtained?
Answer:
Water which is free from all soluble mineral salts is called demineralised water. To obtain demineralized water, water is flown through cation exchange resin and anion exchange resin.
In cation exchange Ca²⁺, Mg²⁺, Na⁺ and other cations present in water are exchanged
by H⁺ ions and separated whereas in anion exchange Cl^–, HCO₃, SO₄” etc. anions are exchanged by OH^– ions and separated.

Question 30.
What properties of water make it useful as a solvent? What types of compound can it (i) dissolve and (ii) hydrolyse. ,
Answer:
Due to high dipole moment and high dielectric constant water is useful as a solvent.
(i) It can dissolve ionic compounds and these covalent compounds in which H-bond is found with water. (Like ethanol, sugar, glucose etc.)
(ii) Water can hydrolyse various metallic and non-metallic oxides, hydride, carbide, phosphide and other salts.
For example:
P₄O_10(s) + 6H₂O_(l) → 4H₃PO_4(aq)
CaH_2(s) + 2H₂O_(l) → Ca(OH)_2(aq)+2H_2(g)
Al₄C_3(s) + 12H₂O_(l) →4Al(OH)₃ + 3CH₄.

Question 31.
Describe the usefulness of water in biosphere and biological systems.
Answer:
Major portion of body of all living organisms is made up of water. Human body contains about 65% part and some plants contain about 95% part water. Water is an important compound for the survival of all living organisms. Compared to other liquids, specific heat of water, thermal conductivity, surface tension, dipole moment and dielectric constant of water is very high.

Due to these properties role of water in the biosphere is very important. For maintaining the body temperature and normal climatic conditions, heat of vaporization and high specific heat of water is responsible. It acts as a best solvent for the transport of molecules in the metabolism of plants and animals. Water is also necessary for photosynthesis in plants in which O₂ is released in the environment.

Question 32.
Describe the structure of water molecule and also explain that water is a good solvent.
Answer:
In water molecule, 2H atoms and one oxygen atom are linked with covalent bond. Central atom of water molecule is sp³ hybridised and its shape is tetrahedral. In H₂O molecule two sp³ – s sigma bonds and remaining are two lone pair of electrons. Due to repulsion of lone pair-lone pair electrons, bond angle is 104° 27′ in place of 109° 28′ (angle of regular tetrahedron).

Properties of good solvent: Water is a good solvent because water molecule is polar and between its molecules, hydrogen bonding exist. Due to polarity and high dielectric constant ionic compounds and polar covalent compounds are soluble in water. Due to hydrogen bonding those compounds are also soluble which have H-bonding or form hydrogen
bonds with water molecules.

Question 33.
Explain the preparation of hydrogen peroxide from barium peroxide.
Answer:
Preparation of pure hydrogen peroxide :
1. By dilute sulphuric acid: By the action of dil. H₂SO₄on hydrated barium peroxide H< sub>2O₂ is formed. Precipitate of BaSO₄ is also formed which is filtered out.
BaO₂.8H₂O + H₂SO₄ → BaSO₄ ↓+H₂O₂ +8H₂O

2. By the action of Carbon dioxide on Barium peroxide: A paste of hydrated barium peroxide is made in freezed water and carbon dioxide gas is slowly passed through it. Precipitate of barium carbonate is filtered and separated.
BaO₂ + CO₂ + H₂O → BaCO₃ + H< sub>2O₂.

Question 34.
How is hydrogen peroxide concentrated?
Answer:
Hydrogen peroxide solution obtained by any method is dilute, it is Concentrated at low temperature and low pressure because it decomposes on heating at normal pressure.
(a) Vaporization: By vaporizing the H₂O₂ at 70°C on a water bath, 45% H₂O₂ solution is obtained.
(b) Distillation at low pressure: 45% H₂O₂ solution is distilled at 15 mm pressure and 60°C temperature. By this about 90% of H₂O₂ solution is obtained.
(c) Now, this solution is evaporated at vacuum desicators over concentrated H₂SO₄ and 99% cone. H₂O₂ is obtained.
(d) Crystallization: This solution is crystallized by freezing mixture of solid CO₂, ether and crystals of H₂O₂ are obtained. It is melted and pure H₂O₂ liquid is obtained.

Question 35.
How is heavy water manufactured? Write its properties.
Answer:
By the fractional distillation of ordinary water : About one part of heavy water is present in 6000 parts of ordinary water. Heavy water can be separated by the fractional distillation of ordinary water. Boiling point of heavy water is higher than that of ordinary water. Therefore, on fractional distillation of ordinary water, ordinary water is first distilled and heavy water remains back as residue.

Physical properties: Heavy water is colourless, odourless, tasteless liquids. Its freezing point is 3.8°C and boiling point is 101.4°C.
Chemical properties :
1. Electrolysis: By the electrolysis ‘of heavy water in the presence of an electrolyte, deuterium is obtained at cathode and 02 gas is obtained at anode.
2D₂O → 2D₂ + O₂ ,
2. Reaction with nitrides: By the reaction of heavy water with nitrides, heavy ammonia is formed.
Mg₃N₂ + 6D₂O → 3Mg(OD)₂ + 2ND₃ Heavy ammonia.

Question 36.
Density of water is maximum at 4°C. Why?
Answer:
When ice melts to liquid state, the cage like structure present in ice breaks. Number of hydrogen bond decreases and due to breakage of H-bond the vacant space in the structure of ice leads to arrangement of water molecules, the water molecules come very close, as a result density of water increases, but beyond 4°C temperature kinetic energy of water molecules increases as a result water molecules move away. This results in the in¬crease in volume and decrease in density.

Question 37.
Name the classes of hydrides to which H₂O, B₂H₆ and NaH are related.
Answer:
H₂O: Covalent or molecular hydride (Electron rich hydride)
B₂H₆: Covalent or molecular hydride (Electron deficient hydride)
NaH: Ionic or Saline hydride.

Hydrogen Class 11 Important Questions Long Answer Type

Question 1.
Discuss the theory and method of conversion of hard water to soft water by synthetic ion exchange.
Answer:
Synthetic ion exchange resin method is as follows :
1. Cation-exchange resin: These resins are -SO₃H group giant organic molecules insoluble in water. These are treated with NaCl to convert into R-Na. Resin RNa exchanges the Mg₂₊ and Ca₂₊ ions and converts it to soft water.
2RNa_(s)+M²⁺ _(aq) → R₂M_(s) +2Na⁺_(aq) (M = Ca²⁺ or Mg²⁺)

On passing aqueous NaCl solution, it is regenerated. To obtain pure demineralized and deionized water, the water obtained above is again passed through cation-exchanger and anion-exchanger resin.
In cation-exchanger process H⁺ is exchanged by Na⁺, Ca²⁺, Mg²⁺ etc. ions present in water.
2RH_(s) + M⁽²⁺⁾_(aq) ⇌ MR_2(s) + 2H⁺_(aq)
(Cation exchange resin in the form of H⁺)
In this process protons are produced and water becomes acidic.

2. Anion-exchange resin: In this process OH^– are exchanged by Cl^–, HCO^–₃, SO^-2₄ ions present in water.
is substituted ammonium hydroxide anion resin.

This way OH^– ion released, neutralizes H⁺ ions. At the end, cation and anion exchange resin are treated with dilute acid and dilute alkaline solutions and regenerated.

Question 2.
Discuss the position of hydrogen in the periodic table.
Answer:
Position of hydrogen in the periodic table: Hydrogen is placed in the first group and first period of periodic table. It resembles both the alkali metal of group 1 and the halogens of group 17. It also shows some similarities with carbon which is placed in group 14. But considering all the points, its position in group 1 of the periodic table is most suitable.

Resemblance with alkali metals: Hydrogen is placed along with alkali metals because:
(i) s-block element: Hydrogen is s-block element and its electronic configuration is ₁s¹.
(ii) Electronic configuration: It has same number of electrons in its Outermost orbit like Na and K which are alkali metals of first group.
H = 1;Na = 2,8,1;K = 2,8,8,1.
(iii) Valency: Valency of hydrogen and alkali metals is one.
(iv) Formation of cation : Like alkali metals it loses one electron to form a cation.
H – e^– → H⁺
Na-e^– →Na⁺

(v) Electropositive nature: Like alkali metals it is electropositive in nature. It reacts
with electronegative elements like sulphur, oxygen and chlorine etc. .
H₂ +S →H₂S
2Na + S → Na₂S
(vi) Oxidation number: Like alkali metals its oxidation number in salts is + 1.
(vii) Reducing properties : Like alkali metals hydrogen is also a reducing agent.

Resemblance with halogens :
(i) Electronic configuration : Like halogens it is also in short of one electron to the inert gas of adjacent group.
H = 1 He = 2
Cl =17 Ar = 18

(ii) Ionisation potential: Ionisation potential of hydrogen is quite close to ionisation potential ofhalogens but ionisation of alkali metals is very low.
H → 13.5eV, Cl → 13.0eV,Na → 5.1eV.
Where, e^V = potential in electron volt.

(iii) Atomicity : Hydrogen molecule is diatomic like halogen molecules. The value of \(\frac{\mathrm{C}_{p}}{\mathrm{C}_{v}}\) in both H₂ and halogens is = 1.4.
H – H or H₂ Cl – Cl or Cl₂
(iv) Non-metallic nature : Like halogens (F, Cl, Br and I) hydrogen is a non-metal and is a gas.
(v) Combination with metals: Both hydrogen and halogens react with metals.

(vi) Formation of covalent compounds : Both hydrogen and halogens react with carbon, silicon and nitrogen to form covalent-compounds.

Special features of hydrogen nature: In some properties hydrogen also show differences with alkali metals and halogens both as :
(i) Oxide of hydrogen is neutral while oxide of halogen is acidic and oxides of alkali metals are basic in nature.
(ii) Outermost orbit is half-filled like group four elements.
(iii) Its atomic structure is different with all other elements.
On the basis of these facts it is thought that special position should be provided for hydrogen in periodic table but due to electronic configuration of ₁s¹ and its more resemblance, its place is supposed in group first with alkali metals.

Question 3.
Do you expect different products in solution when aluminium (III) chloride and potassium chloride treated separately with (i) Normal water, (ii) Acidified water and (iii) Alkaline water? Write equations wherever necessary.
Answer:
AlCl₃ is a salt of weak base Al(OH)₃ and strong acid HCl. Thus on being reacted with water, it hydrolyses.
AlCl_3(s) + 3H₂O_(l) → Al(OH)₃ + 3H^–_(aq) +3Cl^–_(aq)
Its aqueous solution is of acidic nature. H⁺ ion present in acidified water reacts with Al(OH)₃ to produce Al³⁺_(aq) ion and water. Thus, in acidified water, AlCl₃ exists in the form
Al³⁺_(aq) and Cl^–_(aq) ions. In alkaline water AlCl₃ produces the following products :

KCl is a salt of strong acid HCl and strong base KOH. It does not hydrolyse in normal water. It hydrolyses, in water to give K⁺_(aq) and Cl^–_(aq) ion.

Aqueous solution of KCl is neutral. Thus, in acidified water or in alkaline water, its ion does not react.

Question 4.
Describe laboratory method of preparation of pure and dry hydrogen gas.
Answer:
Preparation of hydrogen (laboratory method) : Hydrogen is prepared in laboratory by action of dil. H₂SO₄ on granulated Zn. Zn is taken into Woulf’s bottle fitted with Thistle funnel and a outlet tube.
Hydrogen formed by this method is collected over water by displacement method.
Zn + H₂SO₄ → ZnSO₄+H₂ ↑

Purification of hydrogen : Hydrogen prepared by this method contains following impurities:
(a) Arsine (AsH₃) and phosphene (PH₃)
(b) H₂S
(c) SO₂,CO₂ and oxides of nitrogen (NO₂)
(d) Water vapours.

In order to remove these impurities, gas is passed in a series of U-tubes containing different solutions.

• Passed in tube filled with AgNO₃ solution to absorb AsH₃ and PH₃.
• Passed in the tube containing lead nitrate solution to remove H₂S.
• Passed in the tube filled with cone. KOH solution to remove SO₂, CO₂ and NO₂.

Question 5.
What is demineralized water? How is it obtained? Or, Explain ion exchange method of softening of hard water with diagram.
Answer:
Water obtained by the removal of cation and anion present in it, is called demineralized water. It is good like distilled water. For this two types of ion exchange resins are used.
1. Cation exchange resin: It is an acidic resin represented by RSO₃H. On flowing hard water over it, Ca²⁺ and Mg₂₊ all cations present in hard water displace hydrogen and occupy its place.
2RSO₃H + CaCl₂ → (RSO₃)₂Ca +2H⁺ + 2Cl^–
2RSO₃H + MgSO₄ → (RSO₃)₂ Mg+2H⁺ + SO^2-₄
This H⁺ ion makes the water acidic.

2. Anion exchange resin: It is a basic resin which is represented by RNH₂. It reacts with water to form RNH₃OH.
R-NH₂+H₂O → RNH₃OH^–
After cation exchange resin, when water is passed over this resin, then the anions like chloride, sulphates liberates the OH~ ion of the resin.
RNH₃OH + Cl^– →RNH₃Cl + OH^–
2RNH₃OH + SO^2-₄ → (RNH₃)₂SO₄ + 2OH^–

This OH^– ion mixes with water and neutralizes the excess H⁺ present in it.
H⁺ +OH^– →H₂O
This way, water coming out through the other chamber is free from all types of cation, anion, acidity and basicity.

Question 6.
What are hydrides? Explain the different types of hydrides with example.
Answer:
When hydrogen combines with certain metals or non-metals (except inert gases) then the compounds formed are called hydrides. These hydrides are of three types according to the nature of chemical bond :
1. Ionic hydride: This type of hydrides are formed when hydrogen combines with strong electropositi ve elements. These hydrides are of ionic nature.
2Li + H₂ →2LiH
2 Na + H₂ →2NaH
These are also known as Saline hydrides.

Properties :

• These are generally non-volatile solid substances.
• They are white crystalline solids. Their crystal structure is unit ion.
• Their boiling and melting points are high.
• Density of these hydrides is less than their metals.
• They are conductors of electricity.
• Aqueous solution of ionic hydrides are basic.
  NaH + H₂O →NaOH + H₂
• Atmospheric oxygen oxidizes and converts them to oxides.
  CaH₂+O₂ →CaO + H₂O

Uses :
1. As a reducing agent,
2. As a solid fuel.

II. Covalent hydrides: When hydrogen combines with p and s-block elements like Be and Mg then covalent hydrides are formed because of less difference in their electronegativity. These are also known as molecular hydrides.
Properties :

1. Covalent hydrides are volatile compounds of low melting and boiling points.
2. These hydrides are weak or bad conductors of electricity.
3. van der Waals’ force is present between their molecules.
4. According to electronegativity difference, their aqueous solution are acidic or alkaline.
5. These are electron deficient compounds.
6. On moving from top to bottom in .the group, stability of their hydride decreases.

III. Metallic hydrides: d-block elements and Be, Mg of s-block combine with hydrogen to form metallic hydrides. These are also known as interstitial hydrides because hydrogen gets arranged in the interstitial voids between the metal atoms.
Properties :

• These are hard and possess metallic lustre.
• They are good conductors of heat and electricity.
• Their density is less than the metals from which they are formed.
• These are unsymmetrical.
• They adsorb sufficient amount of hydrogen on their surface. This is known as adsorption of hydrogen.

Question 7.
Complete the following equations :
(i) H_2(s)+M_mO_o(s)
(ii) CO_(g) + H_2(g)
(iii) C₃H_8(g) + 3H₂O_(g)
(iv) Zn_(s) + NaOH_(aq)
Answer:

Question 8.
How is D₂O obtained from water? Describe the different physical properties of D₂O and H₂O. Give minimum three reactions of D₂O in which hydrogen is exchanged by deuterium.
Answer:
(a) Heavy water D₂O is produced by the electrolysis of water.
(b) Physical properties:
(i) D₂O is a colourless, odourless, tasteless liquid. At 11.6°C its density is maximum = 1.1071 gm L1 (Water at 4°C).
(ii) As compared to normal water solubility of salts in heavy water is less because it is more viscous than normal water.
(iii) Value of nearly all physical constants of D₂O is comparatively more than that of water. This is due to stronger H-bond in D₂O than D₂O due to higher nuclear mass of D₂O than H₂O.

(c) Reactions of exchange of hydrogen by Deuterium:
NaOH + D₂O → NaOD + HOD
HCl + D₂O → DCl + HOD
NH₄Cl + D₂O → NH₃DCl

Question 9.
Calculate the concentration of five-volume
Sol. 5 volume H< sub>2O₂ solution means at N.T.P. by the decomposition of 1L of 5 volume sub>2O₂,5L^-1 O₂ is obtained.

∵ At N.T.P. 22.4 L, 02 is obtained by = 68g H< sub>2O₂
∴ At N.T.P. 5L, O₂ will be obtained by = \(\frac{68 \times 5}{22 \cdot 4} \) g = 15.17g =
15gH< sub>2O₂
But, at N.TP. 5L O₂ is obtained by = 1L of 5 volume H< sub>2O₂
∴ Concentration of H< sub>2O₂ solution = 15g L^-1
or Percent concentration of H< sub>2O₂ solution =\(\frac{15}{100} \times 100\) = 15%

Question 10.
What weight of hydrogen peroxide will be present in 2L volume of 5 molar solution? Determine the amount of oxygen released by the dissociation of 200 ml of this solution.
Solution:
(i) Molar mass of H< sub>2O₂ = 34g mol^-1
Thus, amount of H< sub>2O₂ present in 1L of 1M H< sub>2O₂ solution = 34g
∵ Amount of H< sub>2O₂ present in 2L of 5M H< sub>2O₂ solution = 34 × 5 × 2 = 340g

(ii) Amount of H< sub>2O₂ in 0.2L (or 200 ml) of 5M H< sub>2O₂ solution
= \(\frac{340 \times 0 \cdot 2}{2}\) = 34 g H< sub>2O₂

∵ Amount of O₂ obtained by decomposition of 68g H< sub>2O₂ = 32g .,
∴ Amount of O₂ obtained by decomposition of 34g H< sub>2O₂ = = 16g.

Question 11.
Rohan heard the instructions of a lab assistant for safe storage of a specific chemical. Keep it in dark, mix a small amount of urea in it, and keep it away from dust. This chemical behaves as an oxidizing as well as reducing agent in both acidic and alkaline medium. This chemical is used as pollution controller both in domestic and industrial field.
(i) Write the name of this chemical.
(ii) What precautions should be taken for its storage?
Answer:
(i) Name of the chemical is hydrogen peroxide, H< sub>2O₂. It acts as an oxidizing as well as reducing agent in both acidic and alkaline medium.
(ii) By the presence of small amount of alkali present in light and dust, dissociation of H< sub>2O₂ is catalyzed. That is why it is stored in glass or plastic container sealed with waxy layer in dark. To control its oxidation, urea is mixed in it as a negative catalyst.

Question 12.
Complete the following reactions :
(i) PbS_(s) + H< sub>2O_2(aq) →
(ii) MnO^–_4(aq) + H< sub>2O_2(aq) →
(iii) CaO_(s)+H₂O_(g) →
(iv) AlCl_3(g) + H₂O_(l) →
(v) Ca₃N_2(s)+H₂O_(l) →
Classify above reactions as :
(a) Hydrolysis,
(b) Redox,
(c) Hydration.
Answer:

Question 13.
What do you understand by the following steps :
(i) Hydrogen economy
(ii) Hydrogenation
(iii) Syngas
(iv) Water-gas shift reaction
(v) Fuel cell.
Answer:
(i) Hydrogen economy: Main use of hydrogen and possibilities in the near future, is its use as pollution-free (clean) fuel. Basic principle of hydrogen economy is the transport and storage of energy as liquid hydrogen or gaseous hydrogen.

(ii) Hydrogenation : Combination of hydrogen in a double or triple bonded organic compound in the presence of a catalyst is called hydrogenation. Hydrogenation of vegetable oil in presence of Nickel catalyst forms edible fats (margarine and vanaspati ghee).

(iii) Syn gas: Mixture of CO and H₂ is called syngas or synthetic gas or water gas. Syngas is obtained by the reaction of hydrocarbon or coke with steam at high temperature in the presence of catalyst.

Now a days, ‘Syn gas’ is produced from sewage, sawdust, scrap wood, newspapers etc. The process of producing syngas from coal is called ‘coal gasification’.
(iv) Water-gas shift reaction: To increase the quantity of dihydrogen in water gas CO is treated with steam at 500°C in the presence of iron chromate as catalyst.

This reaction is known as water gas shift reaction.

(v) Fuel cell: In fuel cell, energy produced by the combustion of fuel is directly converted to electrical energy. Hydrogen-oxygen cell is the example of fuel cell. It does not produce any pollution. Fuel cell produces electricity with 70-85% conversion ability

Hydrogen Class 11 Important Questions Objective Type

1. Choose the correct answer:

Question 1.
Which of the following metal adsorb H₂:
(a) Zn
(b) Pb
(c) Al
(d) K.
Answer:
(b) Pb

Question 2.
Radioactive isotope of hydrogen is known as:
(a) Protium
(b) Deuterium
(c) Heavy hydrogen
(d) Tritium.
Answer:
(d) Tritium.

Question 3.
Freezing point of heavy water is:
(a) 0°C
(b) 38°C
(c) 38°C
(d) -0.38°C.
Answer:
(b) 38°C

Question 4.
1 ml of H₂O₂ at N.T.P. gives 10 ml O₂. This is:
(a) 10 volume H₂O₂
(b) 20 volume H₂O₂
(c) 30 volume H₂O₂
(d) 40 volume H₂O₂.
Answer:
(a) 10 volume H₂O₂

Question 5.
The least abundant isotope of hydrogen:
(a) ₁H
(b) ₂D
(c) ₃ T
(d) (a) and (b).
Answer:
(c) ₃ T

Question 6.
Heavy water is obtained :
(a) By boiling water
(b) By fractional distillation of heavy water
(c) By continuous electrolysis of water
(d) By heating H₂O₂.
Answer:
(c) By continuous electrolysis of water

Question 7.
Ortho and Para hydrogen differ by:
(a) Atomic number
(b) Atomic mass
(c) Electron spin
(d) Nuclear spin.
Answer:
(d) Nuclear spin.

Question 8.
Temporary hardness of watef can be removed :
(a) By boiling
(b) By adding CaCl₂
(c) By adding CaCO₃
(d) By adding Ca(OH)₂.
Answer:
(a) By boiling
(b) By adding CaCl₂

Question 9.
H₂O₂ solution:
(a) Turn red litmus blue
(b) Turn red litmus white
(c) Turn blue litmus red
(d) Turn blue litmus white.
Answer:
(b) Turn red litmus white
(d) Turn blue litmus white.

Question 10.
Which hydride do not exist in simple ratio:
(a) Ionic
(b) Molecular
(c) Interstitial
(d) All.
Answer:
(c) Interstitial

Question 11.
Is nO₂ an example of ionic hydride:
(a) LiH
(b) CaH₂
(c) CsH
(d) GeH₂.
Answer:
(c) CsH

Question 12.
In H₂O₂ molecule H-O-O bond angle is :
(a) 99.5°
(b) 95.9°
(c) 97°
(d) 100.5°.
Answer:

Question 13.
Number of neutrons in heavy hydrogen atom is :
(a) 0
(b) 2
(c) 3
(d) 1.
Answer:
(d) 1.

Question 14.
Acts as a fuel for rocket:
(a) N₂ + O₂
(b) H₂ + O₂
(c) O₂ + Ar
(d) H₂ + N₂.
Answer:
(b) H₂ + O₂

Question 15.
Density of water is more than ice :
(a) Due to hydrogen-bond
(b) Due to dipole attraction
(c) Due to polar-chaige polar
(d) Charged polarity.
Answer:
(a) Due to hydrogen-bond

Question 16.
Which of the following is used in calgon process :
(a) Sodium poly metaphosphate
(b) Hydrated sodium aluminium silicate
(c) Cation exchange resin
(d) Anion exchange resin.
Answer:
(a) Sodium poly metaphosphate

Question 17.
Reagent which is used to determine the hardness of water :
(a) Oxalic acid
(b) Disodium salt EDTA
(c) Sodium citrate
(d) Sodium thiosulphate.
Answer:
(b) Disodium salt EDTA

Question 18.
Hydrogen does not reduce :
(a)CuO
(b) Fe₂O₃
(c) SnO₂
(d) Al₂O₃.
Answer:
(d) Al₂O₃.

Question 19.
Write CO₂, H₂O and H₂O₂ in increasing order of their acidity :
(a) CO₂ < H₂O₂ < H₂O
(b) H₂O < H₂O₂ < CO₂
(c) H₂O < H₂O₂₂ > CO₂
(d) H₂O₂ > CO₂ > H₂O.
Answer:
(b) H₂O < H₂O₂ < CO₂

Question 20.
Acts as a moderator in nuclear reactor :
(a) Hard water
(b) Heavy water
(c) Non-ionic water
(d) Mineralized water.
Answer:
(b) Heavy water

2. Fill in the blanks:

1. Hydrogen is released by the action of cone, solution of ………………….. on aluminium.
Answer:
NaOH

2. By the electrolysis of fiised NaH ………………….. gas is released at the anode.
Answer:
H₂

3. Calgon is the commercial name of ………………….. .
Answer:
Sodium hexametaphosphate

4. In the reaction of F₂ and H₂O, H₂O acts as a ………………….. .
Answer:
Reducing agent

5. By the action of water on calcium phosphide ………………….. is formed.
Answer:
Phosphine

6. Oxidized water is known as ………………….. .
Answer:
H₂O₂

7. ………………….. water give less lather with soap.
Answer:
Hard

8. Heavy water is mainly used in ………………….. .
Answer:
Nuclear reactors

9. Hardness of water is due to salts of ………………….. and ………………….. .
Answer:
Ca, Mg

10. ………………….. acts as tracer in the study of biological processes.
Answer:
Heavy water

11. By the reaction of A1 with concentrated ………………….. solution, Hydrogen is released.
Answer:
NaOH

12. Structure of hydrogen peroxide is like ………………….. .
Answer:
Open book

13. Mixture of CO + H₂ is known as ………………….. .
Answer:
Water-gas

14. Volume ability of 2N H₂O₂ will be ………………….. .
Answer:
11.2 volume

15. Absorption of hydrogen by platinum is called ………………….. .
Answer:
adsorption.

3. Match the following:

‘A’	‘B’
1. Aqueous solution of copper sulphate is	(a) Deuterophosphine
2. Aqueous solution of sodium carbonate is	(b) Oil gas
3. Calcium phosphide react with heavy water	(c) H2O2
4. Distillation of kerosene oil form	(d) Acidic
5. Which substance is used to wash statues to turn white from black	(e) Acidic
6. Aqueous solution of ferric chloride is	(f) Basic
7. Hydrogen peroxide H2O2	(g) Moderator
8. Heavy water D2O	(h) Rocket fuel
9. Liquid H2	(i) Insecticide

Answer:
1. (d) Acidic
2. (f) Basic
3. (a) Deuterophosphine
4. (b) Oil gas
5. (c) H₂O₂
6. (e) Acidic
7. (i) Insecticide
8. (g) Moderator
9. (h) Rocket fuel.

4. Answer in one word/sentence:

1. In which compound oxidation state of hydrogen is negative?
Answer:
CaH₂

2. Bleaching property of H₂O₂ is due to oxidation or reduction.
Answer:
H₂O₂ gives oxygen due to which it acts as a bleaching substance.
Thus, bleaching of H₂O is due to oxidation.
H₂O₂ → H₂O + [O]

3. Who discovered hydrogen?
Answer:
Henry Cavendish

4. Formula of heavy water is :
Answer:
D₂O

5. What is used in the form of a moderator?
Answer:
Heavy water (D₂O)

6. What is the mixture of CO and H₂ gas known as?
Answer:
Water gas

7. H₂O₂ reduces Cl₂ to which compound ?
Answer:
To HCl

8. Bleaching property of H₂O₂depend on :
Answer:
Due to nascent oxygen

9. How many isotopes of hydrogen are there?
Answer:
Three isotopes of hydrogen are: ₁H¹, ₁H²,₁H³

10. Which oxide reacts with dilute acid to form H₂O₂?
Answer:
Na₂O₂, BaC₂

11. Ethanol vaporizes faster than water. Why?
Answer:
Due to weak hydrogen bond

12. Which type of elements form interstitial hydrides?
Answer:
Elements of d and f block

13. What is the use of intersititial hydride?
Answer:
In storage of H₂ and catalysing hydrogenation reactions.

5. State true or false:

1. Pure H₂O₂ is a weak acid.
Answer:
True

2. Temporary hardness is removed by boiling water.
Answer:
True

3. Acetylene is formed by the action of CaH₂ with water.
Answer:
False

4. Boiling point of hard water is less than that of soft water.
Answer:
False

5. ²D is an isomer of ¹H.
Answer:
False

6. H₂O₂ cannot be reducing agent.
Answer:
False

7. NH₃ and PH₃ are saline hydrides.
Answer:
False

8. D₂O is more polar than H₂O.
Answer:
False

9. H₂O is an example of ionic hydride.
Answer:
False.