Category: Class 11

MP Board Class 11th Hindi Swati Solutions पद्य Chapter 1 भक्ति

भक्ति अभ्यास

भक्ति अति लघु उत्तरीय प्रश्न

प्रश्न 1.
सही विकल्प चुनिए
(क) तुलसीदास के पद संकलित हैं (2009)
(अ) कवितावली में
(ब) गीतावली में
(स) विनय पत्रिका में
(द) रामचरितमानस में।
उत्तर:
(स) विनय पत्रिका में।

(ख) तुलसीदास के पदों में रस प्रधान है
(अ) शान्त
(ब) श्रृंगार
(स) वीर
(द) वात्सल्य।
उत्तर:
(अ) शान्त रस।

(ग) मीराबाई भक्त थीं (2009)
(अ) राम की
(ब) कृष्ण की
(स) शिव की
(द) विष्णु की।
उत्तर:
(ब) कृष्ण की।

प्रश्न 2.
तुलसीदास किसके चरण छोड़कर नहीं जाना चाहते? (2015, 17)
उत्तर:
तुलसीदास श्रीराम के चरण छोड़कर नहीं जाना चाहते।

प्रश्न 3.
तुलसीदास प्रण करके कहाँ बसना चाहते हैं?
उत्तर:
तुलसीदास प्रण करके श्रीराम के चरण कमलों में बसना चाहते हैं।

प्रश्न 4.
कृष्ण ने किसका घमण्ड चूर करने के लिये गोवर्धन पर्वत धारण किया था? (2016)
उत्तर:
कृष्ण ने इन्द्र का घमण्ड चूर करने के लिए गोवर्धन पर्वत को धारण किया था।

प्रश्न 5.
किसी रोगी की पीड़ा को सबसे अधिक कौन अनुभव कर सकता है?
उत्तर:
किसी रोगी की पीड़ा को सबसे अधिक श्रीराम ही अनुभव कर सकते हैं।

भक्ति लघु उत्तरीय प्रश्न

प्रश्न 1.
कवि के अनुसार राम के चरणों से किन-किनका उद्धार हुआ है?
उत्तर:
कवि के अनुसार राम के चरणों द्वारा पत्थर (अहिल्या), जटायु (पक्षी), मारीच (हिरण) आदि का उद्धार हुआ है।

प्रश्न 2.
‘करहलाज निजपन’ पंक्ति में निजपन’ से कवि का क्या आशय है?(2014)
उत्तर:
‘करहु लाज निजपन’ पंक्ति से कवि का अभिप्राय है कि मैं अपने इस मन के दोषपूर्ण एवं तुच्छ कार्यों का कहाँ तक वर्णन करूँ। आप तो अन्तर्यामी हैं अतः सेवक के मन की प्रत्येक अच्छी-बुरी बात को जानते हैं। मैं अपनी दोषयुक्त बातों को कहाँ तक बताऊँ?

प्रश्न 3.
मीरा ने अपने प्रियतम से मिलने में क्या कठिनाई बताई है? (2008)
उत्तर:
मीरा के प्रियतम श्रीकृष्ण की सेज गगन मण्डल में है, उस स्थान पर पहुँचना कठिन है। इसी कारण मीरा प्रियतम कृष्ण से मिलने में कठिनाई का अनुभव करती हैं।

प्रश्न 4.
मीरा के हृदय की पीड़ा को कौन-सा वैद्य दूर कर सकेगा? (2015)
उत्तर:
मीरा के हृदय की पीड़ा को दूर करने वाला एकमात्र वैद्य साँवला सलोना कृष्ण है।

भक्ति दीर्घ उत्तरीय प्रश्न

प्रश्न 1.
तुलसीदास के अनुसार मानव मन की मूढ़ता क्या है? (2008, 13)
उत्तर:
तुलसीदास जी के अनुसार मानव मूर्ख है। वह प्रभु राम की भक्ति रूपी गंगा को त्यागकर विषय-वासना रूपी ओस के कणों से प्यास बुझाने की अभिलाषा रखता है। जिस प्रकार से धुएँ के समूह को देखकर प्यासा चातक उसे बादल समझकर अपनी प्यास बुझाना चाहता है, लेकिन उसको वहाँ से न तो शीतलता प्राप्त होती है और न ही उसकी प्यास बुझती है, लेकिन उसके नेत्रों को हानि अवश्य पहुँचती है। इसी प्रकार मनुष्य विषय-वासना में मग्न होकर आनन्द की प्राप्ति करना चाहता है, लेकिन उसको आनन्द के स्थान पर अशान्ति की प्राप्ति होती है। जिस प्रकार से मुर्ख बाज दर्पण में अपनी परछाईं को देखकर अन्य बाज समझकर उस पर झपटता है, लेकिन उसको आहार तो नहीं प्राप्त होता है परन्तु उसका मुख अवश्य क्षतिग्रस्त हो जाता है। इसी प्रकार मनुष्य भी मूर्खतावश सांसारिक विषय-वासनाओं में लिप्त रहना चाहता है। यदि मानव ईश्वर के प्रति सच्ची भक्ति करे तो निश्चय ही उसे सांसारिक कष्टों से मुक्ति मिल सकती है, क्योंकि प्रभु राम ही सब के कष्टों को दूर करके तारने वाले हैं।

प्रश्न 2.
तुलसीदास अपना भावी जीवन किस प्रकार बिताने का संकल्प लेते हैं? (2009, 10)
उत्तर:
तुलसीदास जी ने अपने भावी जीवन को श्रीराम की भक्ति में लगाने का संकल्प लिया है, क्योंकि उनका अनुभव है कि मानव जीवन की सार्थकता व मुक्ति का मार्ग राम भक्ति द्वारा ही सम्भव है। तुलसीदास जी का कथन है कि यह संसार मिथ्या है तथा संसार में अनुरक्त व्यक्ति कभी सफलता प्राप्त नहीं कर सकता है। अतः भगवान श्रीराम के चरणों में अपने मन को पूर्ण रूप से समर्पित करके ही जीवन को सफल बनाया जा सकता है।

जिस प्रकार मीरा ने कृष्ण को अपना आराध्य भाग था, उसी प्रकार तुलसीदास जी ने अपना भावी जीवन श्रीराम के चरणों में बिताने का संकल्प ले लिया।.

प्रश्न 3.
मीरा ने हरि के चरणों की कौन-कौन सी विशेषताएँ बताई हैं? (2009)
उत्तर:
मीरा ने हरि के चरणों की विशेषताओं का उल्लेख करते हुए बताया है कि जिन हरि के चरणों ने राजा बलि की दान स्वरूप प्रदान की हुई धरती को तीन पग में नाप लिया था, जिन चरणों का स्पर्श करके गौतम की पत्नी अहिल्या का उद्धार हो गया।

गोपलीला करने के लिये भगवान श्रीकृष्ण ने कालिया नाग का नाश कर भय मुक्त किया था। इन्द्र के अभिमान को नष्ट करने के लिए गोवर्धन पर्वत को अपनी कनिष्ठा उँगली पर धारण कर लिया था। ऐसे भगवान के चरण कमल ही मीरा दासी का उद्धार करने में सक्षम हैं।

प्रश्न 4.
निम्नलिखित पद्यांशों की सप्रसंग व्याख्या कीजिए
(क) ऐसी मूढ़ता या …………..आपने तन की।
(ख) अबलौं नसानी ………… कंचनहिं कसैहों।
(ग) बढ़त पल-पल …………पुनि डार।
(घ) गगन मंडल पै ………..की जिन लाई होय।
उत्तर:
(क) सन्दर्भ :
पूर्ववत्।

प्रसंग :
इस पद में भक्त तुलसीदास अपने अवगुणों को प्रभु के समक्ष प्रदर्शित करते हुए अवगुणों एवं अज्ञान से मुक्त होने की कामना व्यक्त करते हैं।

व्याख्या :
तुलसीदास जी कहते हैं कि मेरे इस मन की ऐसी मूर्खता है कि यह राम की भक्ति रूपी पवित्र गंगा को त्यागकर सांसारिक सुख रूपी ओस की इच्छा करता है। कहने का आशय है कि जिस भाँति कोई प्यासा चातक पक्षी धुएँ के समूह को देखकर उसे बादल समझ ले और जल पीने के लिए दौड़े परन्तु न उसमें शीतलता होती है न जल, अपितु नेत्रों की हानि होती है। इसी भाँति मेरा मन राम की भक्ति को त्यागकर सांसारिक विषयों में सुख समझकर उनकी ओर आकर्षित होता है।

तुलसीदास का कथन है कि मेरे मन की ऐसी स्थिति है कि जैसे फर्श में जड़े हुए काँच में कोई बाज पक्षी अपने शरीर की परछाईं को निहारे और उसे अन्य पक्षी अर्थात् अपना शिकार समझकर उस पक्षी पर भोजन के लिए झपटे। उसे यह ज्ञान नहीं है कि इस प्रकार फर्श पर टकराने से उसके मुँह की ही हानि होगी।

हे कृपालु प्रभु राम ! मैं अपने मन की कुचाल (वाचालता) का कहाँ तक वर्णन करूँ? आप तो मेरी गति को भली-भाँति समझते हैं। हे प्रभु! आप पतित पावन हैं। पतितों एवं दीन-दुःखियों की रक्षा करना आपका प्रण अथवा स्वभाव है, इसीलिए आप इस तुलसीदास के असहनीय कष्टों को हरकर अपने प्राण की लज्जा रखो।
उपर्युक्त पद्यांशों की सप्रसंग व्याख्या ‘सन्दर्भ-प्रसंग सहित पद्यांशों की व्याख्या’ भाग में देखिये।

(ख) सन्दर्भ :
पूर्ववत्।

प्रसंग :
प्रस्तुत पंक्तियों में तुलसीदास ने विगत जीवन के प्रति पश्चाताप का भाव व्यक्त किया है। अब उन्हें विवेक की प्राप्ति हो गई; अत: उन्होंने सुपथ पर चलने का दृढ़ निश्चय किया है।

व्याख्या :
हे नाथ! मेरी अब तक की आयु व्यर्थ में ही नष्ट हो गई, मैं उसका कोई भी सदुपयोग नहीं कर सका। लेकिन अब जो आयु शेष है, उसे मैं व्यर्थ में नष्ट न होने दूंगा, उसका सदुपयोग करूँगा। राम की कृपा से संसार रूपी रात्रि नष्ट हो गयी है, अब में जाग गया हूँ अर्थात् अब मैं इतना मूर्ख नहीं कि जागने के बाद पुनः सोने के लिये बिस्तर बिछा लूँ। मुझे अज्ञान और मोह के बन्धन का बोध हो गया है। अत: मैं इसकी वास्तविकता से परिचित हो गया हूँ। अब पुनः मैं सांसारिक माया में नहीं फसँगा।

मुझे तो राम नाम रूपी सुन्दर चिन्तामणि प्राप्त हो गई है जिसे मैं हृदयरूपी हाथ से खिसकने नहीं दूंगा और राम के सुन्दर श्याम रूप को कसौटी बनाकर उस पर अपने चित्त रूपी कंचन को करूँगा अर्थात् परीक्षा करूँगा कि मेरा मन राम के स्वरूप में लगता है अथवा नहीं। यदि लगता है तो वह शुद्ध स्वर्ण है और यदि नहीं लगता तो उसमें खोट निहित है।

अभी तक मेरा मन इन्द्रियों के वश में था। अतः इन्द्रियों ने मेरा खूब उपहास किया, लेकिन अब मैंने अपने मन को वश में कर लिया है, इसलिए इन्द्रियों को हँसने का अवसर अब नहीं दूंगा। मैं दृढ़ निश्चय करके अपने इस मन रूपी भ्रमर को राम के चरण कमलों में बसाऊँगा। मुझे पूर्ण विश्वास है कि मेरा मन अपनी चंचलता को त्यागकर राम के चरणों में लग जायेगा।

(ग) सन्दर्भ :
पूर्ववत्।

प्रसंग :
इस पद में मीराबाई ने मानव शरीर को पुण्य स्वरूप ठहराकर ईश्वर के प्रति समर्पित होने की प्रेरणा दी है।

व्याख्या :
मीराबाई कहती हैं कि मानव जीवन बार-बार नहीं मिलता है। जीव को पुण्य कर्मों के कारण ही मानव शरीर प्राप्त होता है। यह शरीर पल-पल एवं प्रतिक्षण क्षीण होता जा रहा है, अर्थात् मृत्यु की ओर बढ़ रहा है। जीवन का अन्त होने में तनिक भी विलम्ब नहीं होगा। जिस प्रकार वृक्ष से झड़े हुए पत्ते पुनः डाल पर नहीं लग सकते; तद्नुकूल मानव का अन्त होने पर यह निश्चय नहीं है कि उसे पुनः मानव शरीर प्राप्त होगा। संसार रूपी सागर में विषय-वासनाओं की अत्यन्त तीक्ष्ण धारा है। यदि मनुष्य सुरत अथवा प्रभु में अपना ध्यान लगाये तो शीघ्र ही संसार रूपी सागर से पार हो जायेगा अर्थात् सांसारिक विषय-वासनाओं से उसे छुटकारा मिल जायेगा।

साधु सन्त एवं महन्तों की मण्डली पुकार-पुकार कर कह रही है कि मानव जीवन चार दिनों का मेला है। अत: हे मानव! तू श्रीकृष्ण का आश्रय ग्रहण कर ले, इसी में तेरा हित है।

(घ) सन्दर्भ :
प्रस्तुत पद्यांश हमारी पाठ्य पुस्तक के पाठ ‘मीराबाई’ द्वारा रचित ‘पदावली’ शीर्षक से अवतरित है।

प्रसंग :
प्रस्तुत पद्यांश में मीराबाई की विरह जनित पीड़ा को बड़ा ही मार्मिक वर्णन है। मीराबाई का श्रीकृष्ण से मिलन नहीं हो पा रहा है। अतः उनको असहनीय पीड़ा हो रही है।

व्याख्या :
मीराबाई कहती हैं कि हे सखि! मैं तो श्रीकृष्ण के प्रेम में दीवानी हूँ। उनके प्रति मेरे प्रेम की पीड़ा को मेरे अतिरिक्त दूसरा नहीं जान सकता है। मेरी सेज (सूली) (शय्या) ऐसे दुर्गम स्थान (प्राण दण्ड देने के स्थल) पर है जहाँ मेरे समान सांसारिक प्राणी पहुँच ही नहीं सकता है। श्रीकृष्ण की शैया तो आकाश में है। अत: उनसे मिलना असम्भव है। व्यथा से व्यथित मानव ही व्यथा की पीड़ा का मूल्यांकन कर सकता है। घायल व्यक्ति की पीड़ा को घायल ही जान सकता है। मेरी वियोग की पीड़ा को केवल वही व्यक्ति जान सकता है जिसने वियोग जनित पीड़ा को भी कभी सहन किया हो।

जौहर की पीड़ा को जौहरी (स्वयं को अग्नि की गोद में समर्पित करने वाला) ही जान सकता है। वियोग की पीड़ा से मैं इतनी दुःखी हूँ कि वन-वन भटकती फिर रही हूँ। लेकिन मेरी इस पीड़ा को दूर करने वाला कोई वैद्य आज तक नहीं मिला। मेरे विरह की पीड़ा तो तभी समाप्त होगी, जब श्रीकृष्ण वैद्य के रूप में प्रस्तुत होकर इसका उपचार करें अर्थात् उनके दर्शन से ही मेरा दुःख दूर हो सकता है।

भक्ति काव्य सौन्दर्य

प्रश्न 1.
निम्नलिखित शब्दों के तत्सम रूप लिखिएनिसा, आस, पषान, चरन, सुचि, पुन्य, गरव, भौसागर।
उत्तर:
तत्सम रूप-निशा, आशा, पाषाण, चरण, शुचि, पुण्य, गर्व, भवसागर।

प्रश्न 2.
निम्नलिखित काव्य-पंक्तियों में अलंकार पहचान कर लिखिए
(अ) मन मधुकर पन कै तुलसी रघुपति पद कमल बसैहौं।
(ब) ज्यों गज काँच बिलोकि सेन जड़ छाँह आपने तन की।
(स) बढ़त पल पल घटत छिन छिन चलत न लागे बार।
(द) जौहरी की गति जौहरी जाने, की जिन जौहर होय।
(इ) स्याम रूप सुचि रूचिर कसौटी, चित कंचनहिं कसैहों।
उत्तर:
(अ) रूपक अलंकार
(ब) दृष्टान्त अलंकार
(स) पुनरुक्ति, अनुप्रास अलंकार
(द) पुनरुक्ति अलंकार
(इ) रूपक अलंकार।

प्रश्न 3.
मीरा के पदों में किन-किन बोली अथवा भाषाओं के शब्दों का प्रयोग हुआ है? संकलित अंश से छाँटकर लिखिए।
उत्तर:
मीरा के पदों में ब्रजभाषा के अतिरिक्त राजस्थानी, गुजराती एवं पंजाबी शब्दों का प्रयोग हुआ है।

उदाहरण के लिए राजस्थानी भाषा के शब्द देखें-दीवाणी, जाणे, सोणा, मिलणा, मिल्या।
ब्रजभाषा के उदाहरण- नहिं ऐसो जनम बारम्बार……….
मन रे परसि हरि के चरन……….
जिन चरन प्रभु परसि लीने तरी गौतम धरन………..

प्रश्न 4.
माधुर्य गुण में कोमलकान्त पदावली का प्रयोग किया जाता है। यह गुण प्रायः श्रृंगार, वात्सल्य और शान्त रस में होता है। संकलित अंश से उदाहरण देकर समझाइए।
उत्तर:
संकलित अंश के आधार पर रस के उदाहरण इस प्रकार हैं-श्रृंगार रस-शृंगार रस के दो भेद होते हैं-
(1) संयोग शृंगार,
(2) वियोग शृंगार। स्थायी भाव रति होता है।

वियोग श्रृंगार का उदाहरण :
हे री मैं तो प्रेम दिवाणी मेरा दरद न जाने कोय। उपर्युक्त पंक्ति में वियोग शृंगार है।
अन्य उदाहरण :
दरद की मारी वन-वन डोलूँ, वैद मिल्या नहिं कोय।
मीरा की प्रभु पीर मिटैगी, जब वैद सँवलिया होय।।

वात्सल्य रस :
सोभित कर नवनीत लिए।
मैया मैं तो चंद खिलौना लैहौं।
शान्त रस का स्थायी भाव निर्वेद है।.

उदाहरण देखें :
ऐसी मूढ़ता या मन की …………
…………. करहु लाज निज पन की।
अबलौं नसानी ………… पद कमल बसैहौं।

प्रश्न 5.
निम्नलिखित काव्य पंक्तियों में कौन-सा रस है? उसका स्थायी भाव लिखिए।
(अ) अबलौं नसानी, अब न नसैहों।
(ब) हेरी मैं तो प्रेम दिवाणी, मेरा दरद न जाने कोय।
उत्तर:
(अ) शान्त रस-स्थायी भाव निर्वेद।
(ब) वियोग शृंगार-स्थायी भाव रति।

प्रश्न 6.
“गगन मंडल पै सेज पिया की किस विध मिलणा होय।” पंक्ति में कौन-सी शब्द शक्ति है ?
उत्तर:
शब्द शक्ति लक्षणा है।

प्रश्न 7.
“ऐसी मूढ़ता या मन की परिहरि राम भक्ति सुर सरिता,
आस करत ओसकन की।”
में प्रयुक्त रस और उसका स्थायी भाव लिखिए।
उत्तर:
उपर्युक्त पंक्ति में शान्त रस है तथा स्थायी भाव निर्वेद है।

विनय के पद भाव सारांश

तुलसीदास अपने इष्टदेव श्रीराम से कहते हैं कि आपके चरणों का आश्रय छोड़कर अन्यत्र कहाँ जाऊँ? आप पतितों का उद्धार करने वाले तथा दीनों से अनुराग करने वाले हैं। देवता, राक्षस, मुनि, नाग तथा मनुष्य सभी माया के वशीभूत हैं। अग्रिम पद में अपने को मूर्ख ठहराया है जो राम भक्ति रूपी गंगाजी को त्यागकर ओस कणों से अपनी प्यास बुझाना चाहता है।

अब तुलसी संसार के माया जनित सम्बन्धों को मिथ्या ठहराकर उनके बन्धनों से मुक्त होकर, अपने जीवन को नष्ट नहीं करना चाहते हैं। वे अपने मन रूपी भौरे को श्रीराम के चरण कमलों में अर्पित करना चाहते हैं।

विनय के पद संदर्भ-प्रसंग सहित व्याख्या

[1] जाऊँ कहाँ तजि चरन तुम्हारे।
काको नाम पतित पावन जग, केहि अति दीन पियारे ।।1।।
कौने देव बराइ बिरद-हित, हठिहठि अधम उधारे।
खग, मृग, ब्याध, पषान, विटप जड़, जवन कवन सुर तारे ।।2।।
देव, दनुज, मुनि, नाग, मनुज, सब माया बिबस विचारे।
तिनके हाथ दास तुलसी प्रभु, कहा अपनपौ हारे ।।3।।

शब्दार्थ :
तजि = त्यागकर, छोड़कर; काको = किसका; पावन = पवित्र; जग = संसार; केहि = किसको; बराइ = चुन-चुन कर; बिरद = भक्ति; हठिहठि = हठपूर्वक; अधम = नीच, पापी; उधारे = उद्धार किया; खग = जटायु पक्षी; मृग = हिरण, मारीचि; ब्याध = बहेलिया; पषान = पत्थर, अहिल्या; विटप = वृक्ष; सुर = देवता; दनुज = राक्षस; मनुज = मनुष्य; बिबस = विवश, लाचार; तिनके = उनके।

सन्दर्भ :
प्रस्तुत पद्यांश हमारी पाठ्य पुस्तक के पाठ भक्ति’ के शीर्षक विनय के पद’ से अवतरित है। इसके रचयिता तुलसीदास हैं।

प्रसंग :
प्रस्तुत पद्य में तुलसीदास ने भगवान् राम के चरणों के प्रति अपनी अनन्य भक्ति का उल्लेख किया है।

व्याख्या :
तुलसीदास जी कहते हैं कि हे प्रभु ! मैं आपके चरणों को त्यागकर अन्यत्र कहाँ जाऊँ ? कहीं भी मुझे आश्रय दृष्टिगोचर नहीं होता। आपके समान पापियों को पवित्र करने वाला संसार में अन्य कोई नहीं है, निर्धनों से स्नेह करने वाला कोई दूसरा नहीं है। ऐसा कौन-सा देवता है जिसने हठपूर्वक अपने पतित भक्तों का उद्धार किया है। पक्षी, हिरन, बहेलिया, पत्थर, वृक्ष, जड़, देवता, राक्षस, ऋषि, नाग, मनुष्य सब माया के वशीभूत हैं। तुलसीदास जी कहते हैं कि ऐसे प्रभु राम के सामने मैं अपना सर्वस्व समर्पण करता हूँ।

काव्य सौन्दर्य :

1. हित, हठिहठि, पतित-पावन में अनुप्रास अलंकार है।
2. भाषा में तद्भव शब्दों का प्रयोग है, जैसे-पषान, बिबस, मनुज।
3. ब्रजभाषा का प्रयोग है।

[2] ऐसी मूढ़ता या मन की।
परिहरि राम भक्ति सुर सरिता आस करत ओसकन की।।1।।
धूम समूह निरखि जातक ज्यों, तृषित जानि मति घन की।
नहि तहँ शीतलता न बारि, पुनि हानि होत लोचन की ।।2।। (2012)
ज्यों गच काँच बिलोकि सेन जड़ छाँह आपने तन की।
टूटत अति आतुर अहार बस, छति बिसारि आनन की ।।3।।
कहँ लौ कहाँ कुचाल कृपानिधि, जानत हौं गति जन की।
तुलसीदास प्रभु हरहु दुसह दुखः, करहु लाज निज पन की ।।4।।

शब्दार्थ :
मूढ़ता = मूर्खता; परिहरि = त्यागकर, छोड़कर; ओसकन = ओस की बूंदें; धूम = धुआँ; निरखि = देखकर; सुर-सरिता = गंगा हो; तृषित = प्यास से व्याकुल; गच= भूमि, दीवार; वारि = जल; लोचन = नेत्र; काँच = दर्पण; बिलोकि = देखकर; सेन = बाज, श्येन; जड़ = मूर्ख; तन = शरीर; छति = क्षति, हानि; आनन = मुँह, चोंच; टूटत = झपटकर गिरता है; आतुर = दुःखी; कुचाल = कुचक्र; लाज = लज्जा; निज = अपना; पन = प्रण, प्रतिज्ञा।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
इस पद में भक्त तुलसीदास अपने अवगुणों को प्रभु के समक्ष प्रदर्शित करते हुए अवगुणों एवं अज्ञान से मुक्त होने की कामना व्यक्त करते हैं।

व्याख्या :
तुलसीदास जी कहते हैं कि मेरे इस मन की ऐसी मूर्खता है कि यह राम की भक्ति रूपी पवित्र गंगा को त्यागकर सांसारिक सुख रूपी ओस की इच्छा करता है। कहने का आशय है कि जिस भाँति कोई प्यासा चातक पक्षी धुएँ के समूह को देखकर उसे बादल समझ ले और जल पीने के लिए दौड़े परन्तु न उसमें शीतलता होती है न जल, अपितु नेत्रों की हानि होती है। इसी भाँति मेरा मन राम की भक्ति को त्यागकर सांसारिक विषयों में सुख समझकर उनकी ओर आकर्षित होता है।

तुलसीदास का कथन है कि मेरे मन की ऐसी स्थिति है कि जैसे फर्श में जड़े हुए काँच में कोई बाज पक्षी अपने शरीर की परछाईं को निहारे और उसे अन्य पक्षी अर्थात् अपना शिकार समझकर उस पक्षी पर भोजन के लिए झपटे। उसे यह ज्ञान नहीं है कि इस प्रकार फर्श पर टकराने से उसके मुँह की ही हानि होगी।

हे कृपालु प्रभु राम ! मैं अपने मन की कुचाल (वाचालता) का कहाँ तक वर्णन करूँ? आप तो मेरी गति को भली-भाँति समझते हैं। हे प्रभु! आप पतित पावन हैं। पतितों एवं दीन-दुःखियों की रक्षा करना आपका प्रण अथवा स्वभाव है, इसीलिए आप इस तुलसीदास के असहनीय कष्टों को हरकर अपने प्राण की लज्जा रखो।

काव्य सौन्दर्य :

1. शान्त रस।
2. भक्त अपने अवगुणों को प्रभु के समक्ष व्यक्त कर अपनी दीनता का प्रदर्शन कर रहा है।
3. ब्रजभाषा तथा गेय मुक्तक शैली का प्रयोग है।
4. भ्रान्तिमान रूपक एवं उपमा अलंकार की छटा दर्शनीय है।

[3] अबलौं नसानी, अब न नसैहों।
राम कृपा भक निसा सिरानी, जागे पुनि न डसैहौं।।1।।
पायो नाम चारु चिन्तामनि, उर कर ते न खसैहौं।
स्यामरूप सुचि रुचिर कसौटी, चित कंचनहिं कसैहों।।2।।
परबस जानि हँस्यो इन इन्द्रिन, निज बसहै न हँसैहों।
मन मधुकर पन कै तुलसी रघुपति पद कमल बसैहौं।।3।। (2008, 09)

शब्दार्थ :
अबलौं = अब तक; नसानी = नष्ट की, बिगाड़ी; भव = संसार; निसा = रात्रि; सिरानी = शान्त हो गई, बीत गई; डसैहों = स्वयं को डसाऊँगा; उर = हृदय; चारु = सुन्दर; खसैहौं = गिराऊँगा; सुचि = पवित्र; चित = मन; कंचनहिं = सोने को; परबस = दूसरे के अधीन; पन कै= प्रण करके, प्रतिज्ञा करके बसैहौं = निवास करूँगा, बसाऊँगा; मन-मधुकर = मन रूपी भौंरा।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
प्रस्तुत पंक्तियों में तुलसीदास ने विगत जीवन के प्रति पश्चाताप का भाव व्यक्त किया है। अब उन्हें विवेक की प्राप्ति हो गई; अत: उन्होंने सुपथ पर चलने का दृढ़ निश्चय किया है।

व्याख्या :
हे नाथ! मेरी अब तक की आयु व्यर्थ में ही नष्ट हो गई, मैं उसका कोई भी सदुपयोग नहीं कर सका। लेकिन अब जो आयु शेष है, उसे मैं व्यर्थ में नष्ट न होने दूंगा, उसका सदुपयोग करूँगा। राम की कृपा से संसार रूपी रात्रि नष्ट हो गयी है, अब में जाग गया हूँ अर्थात् अब मैं इतना मूर्ख नहीं कि जागने के बाद पुनः सोने के लिये बिस्तर बिछा लूँ। मुझे अज्ञान और मोह के बन्धन का बोध हो गया है। अत: मैं इसकी वास्तविकता से परिचित हो गया हूँ। अब पुनः मैं सांसारिक माया में नहीं फसँगा।

मुझे तो राम नाम रूपी सुन्दर चिन्तामणि प्राप्त हो गई है जिसे मैं हृदयरूपी हाथ से खिसकने नहीं दूंगा और राम के सुन्दर श्याम रूप को कसौटी बनाकर उस पर अपने चित्त रूपी कंचन को करूँगा अर्थात् परीक्षा करूँगा कि मेरा मन राम के स्वरूप में लगता है अथवा नहीं। यदि लगता है तो वह शुद्ध स्वर्ण है और यदि नहीं लगता तो उसमें खोट निहित है।

अभी तक मेरा मन इन्द्रियों के वश में था। अतः इन्द्रियों ने मेरा खूब उपहास किया, लेकिन अब मैंने अपने मन को वश में कर लिया है, इसलिए इन्द्रियों को हँसने का अवसर अब नहीं दूंगा। मैं दृढ़ निश्चय करके अपने इस मन रूपी भ्रमर को राम के चरण कमलों में बसाऊँगा। मुझे पूर्ण विश्वास है कि मेरा मन अपनी चंचलता को त्यागकर राम के चरणों में लग जायेगा।

काव्य सौन्दर्य :

1. यहाँ तुलसीदास जी ने मानव को अज्ञानता छोड़कर सुपथ पर चलने की प्रेरणा दी है।
2. शान्त रस का परिपाक है।
3. भाषा अलंकारिक गुणों से युक्त है। रूपक अलंकार है।
4. ब्रजभाषा का प्रयोग है।
5. पद्यांशों में प्रसाद गुण है।

पदावली भाव सारांश

मीरा ने अपनी पदावली में अपनी विरह जनित पीड़ा को व्यक्त किया है। उनका कथन है कि वह श्रीकृष्ण के विरह में अत्यन्त व्याकुल हैं। उनके प्रियतम की सेज आकाश मण्डल में शूली ऊपर है जहाँ पहुँचना अत्यन्त दुर्लभ है। वे यत्र-तत्र भटक रही हैं लेकिन उनकी विरह पीड़ा का शमन करने वाला कोई भी नहीं दिखाई देता। उनकी पीड़ा को तो श्रीकृष्ण ही मिटा सकते हैं।

मीरा मानव जीवन को अमूल्य ठहराती हुई कहती हैं कि संसार से पार होने के लिये प्रभु के चरणों का सहारा ही एकमात्र आश्रय है। इन्हीं चरणों ने अनेक पतितों तथा भक्तों का उद्धार किया है।

पदावली संदर्भ-प्रसंग सहित व्याख्या

[1] हेरी मैं तो प्रेम दिवाणी, मेरा दरद न जाणे कोय।
सूली ऊपर सेज हमारी, किस विधि सोणा होय।।
गगन मण्डल पै सेज पिया की, किस विध मिलणा होय।
घायल की गति घायल जानै, की जिन लाई होय।।
जौहरी की गति जौहरी जानै, की किन जौहर होय।
दरद की मारी वन वन डोलूँ, वैद मिल्या नहिं कोय।।
मीरा की प्रभु पीर मिटैगी, जब वैद सँवलिया होय।

शब्दार्थ :
दरद = पीड़ा; सेज = शैया; गगन = आकाश; पिया = प्रियतम; विध = विधि; जौहरी = जौहर करने वाला; वैद = वैद्य; वन-वन = जंगल-जंगल; डोलूँ = विचरण करूँ; प्रभु = ईश्वर; पीर = पीड़ा; मिटैगी = समाप्त होगी; सँवलिया = श्रीकृष्ण का उपनाम।

सन्दर्भ :
प्रस्तुत पद्यांश हमारी पाठ्य पुस्तक के पाठ ‘मीराबाई’ द्वारा रचित ‘पदावली’ शीर्षक से अवतरित है।

प्रसंग :
प्रस्तुत पद्यांश में मीराबाई की विरह जनित पीड़ा को बड़ा ही मार्मिक वर्णन है। मीराबाई का श्रीकृष्ण से मिलन नहीं हो पा रहा है। अतः उनको असहनीय पीड़ा हो रही है।

व्याख्या :
मीराबाई कहती हैं कि हे सखि! मैं तो श्रीकृष्ण के प्रेम में दीवानी हूँ। उनके प्रति मेरे प्रेम की पीड़ा को मेरे अतिरिक्त दूसरा नहीं जान सकता है। मेरी सेज (सूली) (शय्या) ऐसे दुर्गम स्थान (प्राण दण्ड देने के स्थल) पर है जहाँ मेरे समान सांसारिक प्राणी पहुँच ही नहीं सकता है। श्रीकृष्ण की शैया तो आकाश में है। अत: उनसे मिलना असम्भव है। व्यथा से व्यथित मानव ही व्यथा की पीड़ा का मूल्यांकन कर सकता है। घायल व्यक्ति की पीड़ा को घायल ही जान सकता है। मेरी वियोग की पीड़ा को केवल वही व्यक्ति जान सकता है जिसने वियोग जनित पीड़ा को भी कभी सहन किया हो।

जौहर की पीड़ा को जौहरी (स्वयं को अग्नि की गोद में समर्पित करने वाला) ही जान सकता है। वियोग की पीड़ा से मैं इतनी दुःखी हूँ कि वन-वन भटकती फिर रही हूँ। लेकिन मेरी इस पीड़ा को दूर करने वाला कोई वैद्य आज तक नहीं मिला। मेरे विरह की पीड़ा तो तभी समाप्त होगी, जब श्रीकृष्ण वैद्य के रूप में प्रस्तुत होकर इसका उपचार करें अर्थात् उनके दर्शन से ही मेरा दुःख दूर हो सकता है।

काव्य सौन्दर्य :

1. विरह की वेदना का मार्मिक चित्रण मुहावरों द्वारा किया गया है।
2. भाषा-राजस्थानी एवं ब्रजभाषा है।
3. रस-वियोग शृंगार।
4. अनुप्रास-दृष्टान्त, पुनरुक्तिप्रकाश है।
5. गुण-माधुर्य।
6. शब्द-शक्ति-लक्षणा एवं अभिधा है।’

[2] नहिं ऐसो जन्म बारम्बार।
क्या जानूँ कछु पुन्य प्रकटे, मानुसा अवतार।।
बढ़त पल पल घटत छिन छिन, चलत न लागे बार।
बिरछ के ज्यों पाँत टूटे, लागे नहिं पुनि डार।।
भौ सागर अति जोर कहिये, विषय ओखी धार।
सुरत का नर बाँधे बेंडा, बेगि उतरे पार।।
साधु सन्ता ते महन्ता, चलत करत पुकार।
‘दास मीरा’ लाल गिरिधर, जीवना दिन चार।। (2008)

शब्दार्थ :
बारम्बार = बार-बार; पुन्य = पुण्य; पल-पल = हर क्षण; घटत = कम होना, क्षीण होना; बिरछ = वृक्ष; पात = पत्ते; बार = देरी; पुनि = पुनः; डार = डाल; भौ सागर = भवसागर, संसार रूपी समुद्र; ओखी कठिन; सुरत = ध्यान लगाना, भगवान प्रेम; नर = मनुष्य; बेगि = शीघ्रता; बेंडा = आड़ा-तिरछा, कठिन; महन्ता – साधु मण्डली।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
इस पद में मीराबाई ने मानव शरीर को पुण्य स्वरूप ठहराकर ईश्वर के प्रति समर्पित होने की प्रेरणा दी है।

व्याख्या :
मीराबाई कहती हैं कि मानव जीवन बार-बार नहीं मिलता है। जीव को पुण्य कर्मों के कारण ही मानव शरीर प्राप्त होता है। यह शरीर पल-पल एवं प्रतिक्षण क्षीण होता जा रहा है, अर्थात् मृत्यु की ओर बढ़ रहा है। जीवन का अन्त होने में तनिक भी विलम्ब नहीं होगा। जिस प्रकार वृक्ष से झड़े हुए पत्ते पुनः डाल पर नहीं लग सकते; तद्नुकूल मानव का अन्त होने पर यह निश्चय नहीं है कि उसे पुनः मानव शरीर प्राप्त होगा। संसार रूपी सागर में विषय-वासनाओं की अत्यन्त तीक्ष्ण धारा है। यदि मनुष्य सुरत अथवा प्रभु में अपना ध्यान लगाये तो शीघ्र ही संसार रूपी सागर से पार हो जायेगा अर्थात् सांसारिक विषय-वासनाओं से उसे छुटकारा मिल जायेगा।

साधु सन्त एवं महन्तों की मण्डली पुकार-पुकार कर कह रही है कि मानव जीवन चार दिनों का मेला है। अत: हे मानव! तू श्रीकृष्ण का आश्रय ग्रहण कर ले, इसी में तेरा हित है।

काव्य सौन्दर्य :

1. राजस्थानी एवं ब्रजभाषा का प्रयोग है।
2. उपमा, रूपक एवं पुनरुक्ति अलंकार हैं।
3. शब्द गुण माधुर्य से परिपूर्ण है।
4. शान्त-रस का प्रयोग है।
5. कवयित्री ने जीवन को क्षण-भंगुर बताया है।

[3] मन रे परसि हरि के चरन।
सुभग शीतल कमल कोमल, त्रिविध ज्वाला हरन।।
जे चरन प्रह्वाद परसे, इन्द्र पदवी धरन।
जिन चरन ध्रुव अटल कीन्हों, राखि अपने सरन।।
जिन चरन ब्रह्माण्ड भेट्यो, नख सिखौ श्री भरन।
जिन चरन प्रभु परसि लीने, तरी गौतम धरन।।
जिन चरन कालीहि नाश्यो, गोप लीला करन।
जिन चरन धारयो गोवर्धन, गरब मघवा हरन।।
‘दास मीरा’ लाल गिरधर, अगम तारन तरन।

शब्दार्थ :
परसि = स्पर्श, छूना; हरि = भगवान; चरण = चरन, पग; सुभग = सुन्दर; शीतल = ठण्डा; त्रिविध ज्वाला = तीन प्रकार के ताप दैहिक, दैविक, भौतिक; हरन = नष्ट करना; ध्रुव = एक तारा; सरन = शरण; भेट्यो = भेंट करना; नख सिखौ = सिर से पाँव तक; तरी = तारना, मुक्त करना; धरन = स्त्री, पत्नी; कालीहि = कालिया नाग का; नाश्यो = नष्ट करना; गोप = ग्वाल; धारयो = धारण करना; गरब : गर्व, घमण्ड, अभिमान; मघवा = इन्द्र; हरन = हर लेना, छीन लेना; अगम = कठिन, जहाँ पहुँचा न जा सके।

सन्दर्भ :
पूर्ववत्।

प्रसंग :
प्रस्तुत पद में मीरा उद्धार करने वाले प्रभु के चरणों में आश्रय लेने के लिए अपने मन को सम्बोधित करती हुई कह रही हैं।

व्याख्या :
मीराबाई कहती हैं, हे मन! तू भगवान श्रीकृष्ण के कमल के समान कोमल, सुन्दर एवं शीतल चरणों का स्पर्श कर। ईश्वर के चरणों का स्पर्श तीनों प्रकार के तापों की अग्नि को शान्त करने वाला है (तीन प्रकार के ताप दैहिक, दैविक एवं भौतिक)। जिन भगवान के चरण कमलों के द्वारा प्रह्लाद का उद्धार हुआ, इन्द्र के पद को धारण किया।

जिन चरणों ने अपना आश्रय लेने वाले ध्रुव को अटल एवं अमर पद प्रदान किया। भगवान के चरणों ने समस्त भू-मण्डल को माप लिया तथा सिर से लेकर पाँव तक जिन चरणों का स्पर्श करके गौतम की पत्नी अहिल्या का उद्धार हो गया। ग्वाल लीला करते समय जिन चरणों ने विषधर काली नाग का नाश किया, जिन चरणों ने इन्द्र का गर्व नष्ट करने के लिए गोवर्धन पर्वत को अपनी कनिष्ठा उँगली पर धारण किया। मीराबाई कहती हैं कि मैं तो श्रीकृष्ण की दासी हूँ जो कि कठिन से कठिन कार्य अथवा घोर पापों का भी उद्धार करने वाले हैं।

काव्य सौन्दर्य :

1. ब्रजभाषा है, पद गेय एवं लालित्यपूर्ण हैं।
2. अनुप्रास तथा उपमा अलंकार हैं।
3. गुण-माधुर्य।
4. शान्त रस है।

MP Board Class 11th Hindi Solutions

MP Board Class 11th Maths Important Questions Chapter 3 Trigonometric Functions

Trigonometric Functions Important Questions

Trigonometric Functions Objective Type Questions

(A) Choose the correct option :
Question 1.
The value of 1 + cosθ is :
(a) 2sin² θ
(b) \(\frac { { sin }^{ 2 }\theta }{ 2 }\)
(c) 2cos² θ
(d) cos² θ
Answer:
(c) 2cos² θ

Question 2.
The value of \(\frac { cos 11° + sin 11° }{ cos 11° – sin 11° }\) is :
(a) cot 56°
(b) cot 34°
(c) tan 34°
(d) tan 56°
Answer:
(d) tan 56°

Question 3.
The value of sin 18° is :

Answer:

Question 4.
The value of cos 1° cos 2° cos 3° ………… cos 179° is :
(a) 0
(b) 1
(c) – 1
(d) None of these
Answer:
(a) 0

Question 5.
The value of \(\frac { 3 π }{ 2 }\) radian in degree.
(a) 120°
(b) 170°
(c) 220°
(d) 270°
Answer:
(d) 270°

Question 6.
The value of cos² (60 + α) + cos² (60 – α) + cos² α = ……………
(a) 3
(b) \(\frac { 3 π }{ 2 }\)
(c) \(\frac { 1 }{ 2 }\)
(d) \(\frac { 3 }{ 4 }\)
Answer:
(b) \(\frac { 3 π }{ 2 }\)

Question 7.
Amplitude of the function f(x) = 2 sin x is :
(a) π
(b) 2π
(c) 1
(d) 2
Answer:
(d) 2

Question 8.
Period of the function f(x) = tan x is :
(a) π
(b) 2
(c) \(\frac { π }{ 2 }\)
(d) – π
Answer:
(a) π

Question 9.
Solution of the equation 4 sin²θ = 1 is :
(a) nπ ± \(\frac { π }{ 3 }\), n ∈ I
(b) 2nπ ± \(\frac { π }{ 3 }\), n ∈ I
(c) nπ ± \(\frac { π }{ 6 }\), n ∈ I
(d) 2nπ ± \(\frac { π }{ 6 }\), n ∈ I
Answer:
(c) nπ ± \(\frac { π }{ 6 }\), n ∈ I

Question 10.
Maximum value of 3cosθ + 4sinθ is :
(a) 3
(b) 4
(c) 5
(d) 7
Answer:
(c) 5

Question 11.
If tanθ = – \(\frac { 4 }{ 3 }\) , then the value of sinθ is :
(a) – \(\frac { 4 }{ 5 }\) but not \(\frac { 4 }{ 5 }\)
(b) \(\frac { 4 }{ 5 }\) but not \(\frac – { 4 }{ 5 }\)
(c) – \(\frac { 4 }{ 5 }\) or \(\frac { 4 }{ 5 }\)
(d) None of these
Answer:
(c) – \(\frac { 4 }{ 5 }\) or \(\frac { 4 }{ 5 }\)

Question 12.
The value of tan15° + cot15° is :
(a) 1
(b) 3
(c) 2
(d) 4
Answer:
(d) 4

Question 13.
The value of sin50° + sin70° + sin10° is :
(a) 0
(b) 1
(c) – 1
(d) None of these
Answer:
(a) 0

Question 14.
One value of θ which satisfy the equation cosθ +\/3sinθ = 2 is :
(a) \(\frac { π }{ 2 }\)
(b) \(\frac { π }{ 3 }\)
(c) \(\frac { 2π }{ 3 }\)
(d) \(\frac { π }{ 4 }\)
Answer:
(b) \(\frac { π }{ 3 }\)

Question 15.
The general value of θ satisfying the equation cosθ = , tanθ = 1is :
(a) 2nπ + \(\frac { 5π }{ 4 }\)
(b) 2nπ – \(\frac { 5π }{ 4 }\)
(c) 2nπ + \(\frac { π }{ 4 }\)
(d) 2nπ – \(\frac { π }{ 4 }\)
Answer:
(a) 2nπ + \(\frac { 5π }{ 4 }\)

(B) Match the following :

Answer:

1. (d)
2. (c)
3. (a)
4. (b)
5. (g)
6. (j)
7. (e)
8. (i)
9. (h)
10. (f)

(C) Fill in the blanks :

1. The value of cos 18° = ……………….
2. The value of sin 75° = ……………….
3. If tan A = \(\frac { 5 }{ 6 }\) and tan B = \(\frac { 1 }{ 3}\), then the value of A + B = ……………….
4. 2π radian is equal to ………………. right angle.
5. If sinθ + cosθ = 1, then the value of sinθ.cosθ = ……………….
6. The value of \(\frac { 3tanA-{ tan }^{ 3 }A }{ 1-3{ tan }^{ 2 }A }\) = ……………….
7. Solution of the equation cos2θ = cos²θ is ……………….
8. Amplitude of 3cosx is ……………….
9. The value of cot 22\(\frac { 1 }{ 2 }\)° is ……………….
10. If tan θ tan 2θ = 1, then the value of θ is ……………….
11. The value of sin(A + B).sin(A – B) = ……………….
12. The length of the arc of a circle of radian 6 cm substending an angle of 30° at the centre of the circle is ……………….

Answer:

1. \(\frac { \sqrt { 10+2\sqrt { 5 } } }{ 4 }\)
2. \(\frac { \sqrt { 3 } + 1 }{ 2\sqrt { 2 } }\)
3. 45°
4. 4
5. 0
6. tan 3A
7. nπ
8. 3
9. \(\sqrt {2}\) + 1
10. (n + \(\frac { 1 }{ 3}\))\(\frac { π }{ 3 }\)
11. sin²A – sin²B
12. πcm

(D) Write true / false :

1. The value of tan 105° is \(\frac { \sqrt { 3 } + 1 }{ 2\sqrt { 2 } }\)
2. The value of sin 3A is 4 sin³ A + 3sinA
3. The value of cos² A – sin²B is cos(A + B) cos (A – B)
4. The value of cos² 48°- sin² 12 is \(\frac { \sqrt { 5 } + 1 }{ 4 }\).
5. The value of cos² (\(\frac { π }{ 6 }\) + θ) + sin² (\(\frac { π }{ 6 }\) – θ) is 0.
6. If x is a real, then the equation sinθ = x + \(\frac { 1 }{ x }\) has unique solution.
7. Solution of tan² θ + cot² θ = 2 is 2nπ + \(\frac { π }{ 6 }\)
8. If f(x) = sin² x, then f(- x) = sin² x
9. The value of sin\(\frac { 5π }{ 12 }\).cos\(\frac { π }{ 12 }\) is \(\frac { \sqrt { 3 } – 2 }{ 4 }\).
10. If A + B = \(\frac { π }{ 3 }\) and cosA + cosB = 1, then cos(A – B) = – \(\frac { 1 }{ 3 }\).

Answer:

1. False
2. False
3. True
4. True
5. False
6. False
7. False
8. True
9. False
10. True

(E) Write answer in one word / sentence :

1. Find the solution of equation cos² θ – sin² θ – \(\frac { 1 }{ 4 }\) = 0
2. The value of tan 1° tan 2° tan 3° ……………. tan 18° is :
3. The value of sin(A + B) + sin(A – B) is :
4. If (1 + tan x) (1 + tan y) = 2, then find the value of (x + y):
5. The value of sin( \(\frac { π }{ 3 }\) + x) – sin(\(\frac { π }{ 4 }\) – x)

Answer:

1. θ = nπ + (-1)ⁿ\(\frac { π }{ 6 }\)
2. 1
3. 2sinA.cosB
4. \(\frac { π }{ 4 }\)
5. \(\sqrt { 2 }\) sinx

Trigonometric Functions Short Answer Type Questions

Question 1.
A wheel makes 360 revolutions in 1 min. then how many radians measure of an angle does it turn in 1 second? (NCERT)
Solution:
In 60 seconds number of revolutions of wheel = 360
∴ In 1 second number of revolutions of wheel = \(\frac { 360 }{ 60 }\)
∴ In one revolution angle made = 360° = 2π
∴ In 6 revolutions angle made = 2π x 6 = 12π radian

Question 2.
Find the degree measure of an angle substended at the centre of a circle of radius 100 cm by an arc of length 22 cm.
Solution:
We know that, θ = \(\frac { 1 }{ r }\), length of arc = l = 22 cm, radius = r = 100 cm, angle made at the centre = θ = ?
Applying formula θ = \(\frac { 1 }{ r }\)

Question 3.
In a circle of diameter 40 cm, the length of chord is 20 cm, then find the length of minor arc of the chord.
Solution:
Radius of circle = OA = OB = \(\frac { 40 }{ 2 }\) = 20 cm, AB = 20 cm

Hence OAB is an equilateral triangle.
∴ θ = \(\frac { 1 }{ r }\)
⇒ 60° = \(\frac { AB }{ 20 }\)
⇒ AB = 60° x 20 = 60 x \(\frac { π }{ 180 }\) x 20
= \(\frac { 20π }{ 3 }\) cm.

Question 4.
If in two circles, arc of the same length substend angles of 60° and 75° at the centre, then find the ratio of their radii. (NCERT)
Solution:

Question 5.
Find value : (i) sin75°, (ii) tan15°
Solution:

Question 6.
Prove that:
sin²\(\frac { π }{ 4 }\) + cos²\(\frac { π }{ 3 }\) = – \(\frac { 1 }{ 2 }\)
Solution:

Question 7.
Prove that:
cos 27°. tan 27°. tan 63°. cosec 63° = 1. (NCERT)
Solution:
L.H.S. = cos 27°. tan 27°.tan 63°. cosec 63°
= cos 27°. tan 27° tan(90° – 27°). cosec (90° – 27°)
= cos 27°. tan 27° cot 27°. sec 27°
= \(\frac { 1 }{ sec 27° }\).\(\frac { 1 }{ cot 27° }\)cot 27°. sec 27° sec 27° cot 27°
= 1 = R.H.S.

Question 8.
Find the general solution of the following equations :

1. sinθ = \(\frac { \sqrt { 3 } }{ 2 }\)
2. tanθ = 1
3. \(\sqrt { 3 }\) tanθ + 1 = 0.

Solution:
1. sinθ = \(\frac { \sqrt { 3 } }{ 2 }\)
⇒ sinθ = sin\(\frac { π }{ 3 }\)
∴ θ = nπ + ( – 1)ⁿ\(\frac { π }{ 3 }\) [ ∵ sinθ = sinα ⇒ θ = nπ + (- 1)ⁿα, where n ∈ I]

2. tanθ = 1
⇒ tanθ = tan\(\frac { π }{ 4 }\)
∴ θ = nπ + \(\frac { π }{ 4 }\) [ ∵tanθ = tanα ⇒ θ = nπ + α, where n ∈ I]

3. \(\sqrt { 3 }\) tanθ + 1 = 0.
⇒ tanθ = – \(\frac { 1 }{ \sqrt { 3 } }\) = – tan\(\frac { π }{ 6 }\)
⇒ tanθ = tan (π – \(\frac { π }{ 6 }\))
⇒ tanθ = tan\(\frac { 5π }{ 6 }\)
∴ θ = nπ + \(\frac { 5π }{ 4 }\) [ ∵tanθ = tanα ⇒ θ = nπ + α, where n ∈ I]

Question 9.
Solve the following equations :
(i) 4 sin² θ = 1
(ii) 3 tan² θ = 1.
Solution:

Question 10.
Find the principal value of x :
tan x = \(\sqrt { 3 }\)
Solution:
tan x =\(\sqrt { 3 }\)
tan x = tan\(\frac { π }{ 3 }\) = tan(π + \(\frac { π }{ 3}\))
⇒ tan x = tan\(\frac { π }{ 3 }\) = tan\(\frac { 4π }{ 3 }\)
⇒ x = \(\frac { π }{ 3 }\) or \(\frac { 4π }{ 3 }\)
Principal value of x = \(\frac { π }{ 3 }\)
∴ tan x = \(\sqrt { 3 }\) ⇒ tanx = tan\(\frac { π }{ 3 }\)
Hence, general solution of x = nπ + \(\frac { π }{ 3 }\), [ ∵tanθ = tanα ⇒ θ = nπ + α, where n ∈ I]

Question 11.
Find the principal value of x :
sec x = 2.
Solution:

Trigonometric Functions Long Answer Type Questions

Question 1.
Prove that:
\(\frac { sinx – siny }{ cosx + cosy }\) = tan\(\frac { x – y }{ 2 }\)
Solution:

Question 2.
Prove that:
\(\frac { sin3x + sinx }{ cos3x + cosx }\) = tan2x.
Solution:

Question 3.
Prove that:
\(\frac { sinx – sin3x }{ { sin }^{ 2 }x – { cos }^{ 2 }x }\) = 2 sinx.
Solution:

Question 4.
Prove that:
\(\frac { cos9x – cos5x }{ sin17x – sin3x }\) = – \(\frac { sin2x}{ cos10x }\)
Solution:

Question 5.
Prove that:
\(\frac { sin5x + sin3x }{ cos5x + cos3x }\) = tan4x.
Solution:

Question 6.
Prove that:
2sin²\(\frac { π }{ 6 }\) + cosec²\(\frac { 7π }{ 6 }\)cos²\(\frac { π }{ 3 }\) = \(\frac { 3 }{ 2 }\)
Solution:

Question 7.
Prove that:
cot²\(\frac { π }{ 6 }\) + cosec\(\frac { 5π }{ 6 }\) + 3tan²\(\frac { π }{ 6 }\) = 6
Solution:

Question 8.
Prove that:
2sin²\(\frac { 3π }{ 4 }\) + 2cos²\(\frac { π }{ 4 }\) + 2sec²\(\frac { π }{ 3 }\) = 10
Solution:

Question 9.
Prove that:
\(\frac { cos11° + sin11° }{ cos11° – sin11° }\) = tan56°
Solution:

Question 10.
Prove that:
tan 3A – tan 2A – tan A = tan 3A. tan 2A. tan A.
Solution:
tan3A – tan2A – tanA – tan3A.tan2A. tanA = 0
⇒ tan3A – tan2A – tanA (l + tan3A.tan2A) = 0
⇒ tan 3A – tan 2A = tan A (1 + tan 3 A. tan 2 A)
⇒ \(\frac { tan 3A – tan 2A}{ 1 + tan 2A.tan 3A }\)
⇒ tan(3A – 2 A) = tan A
⇒ tanA = tan A.

Question 11.
Prove that:
cos(\(\frac { π }{ 4 }\) + x) + cos(\(\frac { π }{ 4 }\) – x) = \(\sqrt {2}\)cosx
Solution:

Question 12.
Prove that:

Solution:

Question 13.
If α + β = \(\frac { π }{ 4 }\), then prove that :
(1 + tanα)(1 + tanβ) = 2.
Solution:
Given α + β = \(\frac { π }{ 4 }\)
tan(α + β) = tan \(\frac { π }{ 4 }\)

⇒ tan α + tan β = 1 – tanα.tan β
⇒ tan α + tan β.tanα.tan β = 1
⇒ tan α + tan β(1 + tanα) = 1
⇒ 1 + tan α + tan β(1 + tanα) = 1 + 1
⇒ ( 1 + tan α)(1 + tan β) = 2

Question 14.
Find the general solution of cos4x = cos 2x.
Solution:
Given equation :
cos4x = cos2x
⇒ cos4x – cos2x = 0
⇒ – 2sin\(\frac { 4x + 2x}{ 2 }\).sin\(\frac { 4x – 2x}{ 2 }\) = 0
⇒ – 2sin3x.sinx = 0
⇒ sinx.sin3x = 0
⇒ sin3x = 0 or sinx = 0
⇒ 3x = nπ or x = nπ
⇒ x = \(\frac { nπ}{ 3 }\) or x = nπ, where n∈I.

Question 15.
Solve the equation tan 2x = cot(x + \(\frac { π }{ 3 }\))
Solution:

Question 16.
Solve the equation sin 3θ = sin 2θ
Solution:
Given equation sin3θ = sin2θ
sin 3θ – sin 2θ = 0

Question 17.
Solve the equation tan2θ = tan\(\frac { 2 }{ θ }\)
Solution:
tan 2θ = tan\(\frac { 2 }{ θ }\)
⇒ 2θ = nπ + \(\frac { 2 }{ θ }\)
⇒ 2θ – \(\frac { 2 }{ θ }\) = nπ + \(\frac { 2 }{ θ }\)
⇒ 2θ – \(\frac { 2 }{ θ }\) = nπ
⇒ 2θ² – 2 = nπθ
⇒ 2θ² – nπθ – 2 = 0

Question 18.
Solve the equation tanθ.tan2θ = 1
Solution:

Question 19.
If sec x = \(\frac { 13 }{ 5 }\) and x is in fourth quadrant, then find the other five trigonometrical functions (NCERT)
Solution:
secx = \(\frac { 13 }{ 5 }\) ,
Given : x is in fourth quadrant,

Question 20.
If tan x = – \(\frac { 5 }{ 12 }\) and x is in second quadrant, then find the other five trigonometrical functions (NCERT)
Solution:
Given : tan x = – \(\frac { 5 }{ 12 }\) ,

∴ x is in second quadrant, ∵ secx is negative.

∵ x is in second quadrant, hence sinx is positive.
sinx = \(\frac { 5 }{ 13 }\)
cosecx = \(\frac { 1 }{ sin x }\) = \(\frac { 1 }{ 5/13 }\) = \(\frac { 13 }{ 5 }\)

Question 21.
Prove that:
cos 20°. cos 40°. cos 60°. cos 80°= \(\frac { 1 }{ 16 }\) .
Solution:
L.H.S. = cos 20° .cos 40°. cos 60°. cos 80°
= cos 20°.cos 40°.\(\frac { 1 }{ 2 }\). cos80°
= \(\frac { 1 }{ 4 }\)(2 cos 20°. cos 40°). cos 80°
= \(\frac { 1 }{ 4 }\)(cos 60° + cos 20°). cos 80°
= \(\frac { 1 }{ 4 }\) cos 60°. cos 80° + \(\frac { 1 }{ 4 }\) cos 20°. cos 80°
= \(\frac { 1 }{ 4 }\). \(\frac { 1 }{ 2}\). cos 80° + \(\frac { 1 }{ 8 }\)(2 cos 20°. cos 80°)
= \(\frac { 1 }{ 8 }\)cos 80° + \(\frac { 1 }{ 8 }\) (cos 100° + cos 60°)
= \(\frac { 1 }{ 8 }\)cos 80°+\(\frac { 1 }{ 8 }\)[cos(180° – 80°) + \(\frac { 1 }{ 2 }\)
= \(\frac { 1 }{ 8 }\)cos80° + \(\frac { 1 }{ 8 }\)(- cos80°) + \(\frac { 1 }{ 16 }\) [∵cos(180° – θ]= – cosθ]
= \(\frac { 1 }{ 8 }\)cos 80° – \(\frac { 1 }{ 8 }\)cos 80°+ \(\frac { 1 }{ 16 }\)
= \(\frac { 1 }{ 16 }\) = R.H.S.

Question 22.
Prove that:
sin 20°.sin40°.sin60°.sin80°=\(\frac { 3 }{ 16 }\).
Solution:

Question 23.
prove that:
(cosx + cosy)²+ (sinx – siny)² = 4cos²\(\frac { x + y }{ 2 }\)
Solution:

Question 24.
prove that:
(cosx – cosy)²+ (sinx – siny)² = 4sin²\(\frac { x – y }{ 2 }\)
Solution:
L.H.S. = (cosx – cosy)²+ (sinx – siny)²
= cos²x + cos²y – 2 cosxcosy + sin²x + sin²y – 2sinx siny
= cos²x + sin²x + cos²y + sin²y – 2[cosx cosy + sinx siny]
= 1 + 1 – 2cos(x – y) = 2 – 2cos(x – y)
= 2[1 – cos(x – y)] = 2 x 2sin²
= 4 sin²\(\frac { x – y }{ 2 }\) = R.H.S.

Question 25.
Prove that:
sinx + sin3x + sin5x + sin7x = 4 cosx cos2x sin4x.
Solution:
L.H.S. = sinx + sin3x + sin5x + sin 7x
= sin7x + sinx + sin5x + sin3x
= 2sin\(\frac { 7x + x }{ 2 }\).cos\(\frac { 7x – x }{ 2 }\) + 2sin\(\frac { 5x + 3x }{ 2 }\).cos\(\frac { 5x – 3x }{ 2 }\)
= 2sin4x.cos3x + 2sin4x.cosx
= 2sin4x[cos3x + cosx]
= 2sin4x[ 2 x cos\(\frac { 3x + x }{ 2 }\).cos\(\frac { 3x – x }{ 2 }\) ]
= 2sin4x[2cos2x.cosx]
= 4sin4x.cos2x.cosx
= R.H.S

Question 26.
Prove that:

Solution:

Question 27.
Solve the equation and find general solution of sec²2x = 1 – tan2x. (NCERT)
Solution:
The given equation is :
sec² 2x = 1 – tan 2x
⇒ 1 + tan²2x = 1 – tan 2x
⇒ tan² 2x = – tan2x
⇒ tan² 2x + tan2x = 0 ⇒ tan2x (tan2x + 1) = 0
⇒ tan2x = 0 tan2x(tan2x + 1) = 0 ⇒ 2x = nπ or tan2x = – 1

Question 28.
Solve the equation 2 cos² x + 3 sin x = 0
Solution:
2 cos² x + 3 sin x = 0
⇒ 2 (1 – sin²  x)+3 sinx = 0
⇒ 2 – 2sin² x+ 3 sinx = 0
⇒ 2 sin² x – 3 sinx – 2 = 0
⇒ 2 sin² x + sinx – 4sinx – 2 = 0
⇒ sinx(2 sinx + 1) – 2(2sinx + 1) = 0
⇒ (2sinx + 1)(sinx – 2) = 0
⇒ 2 sin x +1 = 0 or sinx – 2 = 0
⇒ 2 sin x = – 1 = 0 or sinx = 2
⇒ 2 sin x = – \(\frac { 1 }{ 2 }\)
⇒ sin x = sin(π + \(\frac { π }{ 6 }\)
⇒ sin x = sin\(\frac { 7π }{ 6 }\)
∴ x = nπ + (- 1)ⁿ\(\frac { 7π }{ 6 }\)

Question 29.
Solve the equation tan² θ + (1 – \(\sqrt { 3 }\)) tanθ = \(\sqrt { 3 }\)
Solution:
tan² θ + (1 – \(\sqrt { 3 }\)) tanθ = \(\sqrt { 3 }\)
⇒ tan² θ + (1 – \(\sqrt { 3 }\)) tanθ – \(\sqrt { 3 }\) = 0
⇒ tan² θ + tanθ – \(\sqrt { 3 }\) tanθ – \(\sqrt { 3 }\) = 0
⇒ tanθ(tanθ + 1) – \(\sqrt { 3 }\)(tanθ + 1) = 0
⇒ (tanθ + 1)(tan θ – \(\sqrt { 3 }\)) = 0
⇒ tanθ + 1 = 0 or tan θ – \(\sqrt { 3 }\) = 0
⇒ tanθ = – 1 or tan θ = \(\sqrt { 3 }\)
⇒ tanθ = tan \(\frac { – π }{ 4 }\) = tanθ = tan\(\sqrt { 3 }\)
⇒ tanθ = tan \(\frac { – π }{ 4 }\) = θ = nπ + \(\frac { π }{ 3 }\)
∴ θ = nπ – \(\frac { π }{ 4}\)

Question 30.
Solve the equation \(\sqrt { 2 }\) secθ + tanθ = 1.
Solution:

Question 31.
Find the general solution of the equation sinx + sin5x + sin5x = 0
Solution:

Question 32.
Prove that:
2cos\(\frac { π }{ 13 }\)cos\(\frac { 9π }{ 13 }\) + cos\(\frac { 3π }{ 13 }\) + cos\(\frac { 5π }{ 3 }\) = 0
Solution:

Question 33.
Prove that:

Solution:

MP Board Class 11th Maths Important Questions

MP Board Class 11th Maths Important Questions Chapter 2 Relations and Functions

Relations and Functions Important Questions

Relations and Functions Objective Type Questions

(A) Choose the correct option :

Question 1.
If A = {2, 4, 5}, B = {7, 8, 9}, then n(A × B) =
(a) 6
(b) 9
(c) 3
(d) 0.
Answer:
(b) 9

Question 2.
If A = { 1, 2, 3, 4, 5} and 5 = {2, 3, 6, 7}, then the number of element in (A × B)∩(B × A) is:
(a) 4
(b) 5
(c) 10
(d) 20.
Answer:
(a) 4

Question 3.
If A and B are two non – empty sets, then:
(a) A × B = {(a, b) : a ∈ B, b ∈ A}
(b) A × B = {(a, b) : a ∈ A, b ∈ B}
(c) {(a, b) : (a, b) ∈ A, (a, b) ∈ B}
(d) None of these.
Answer:
(b) A × B = {(a, b) : a ∈ A, b ∈ B}

Question 4.
If f(x) = log \(\frac { 1+x }{ 1-x } \), then find f [ \(\frac { { 2x } }{ 1+{ x }^{ 2 } }\) ] =
(a) [f(x)]²
(b) [f(x)]³
(c) 2 f(x)
(d) 3 f(x).
Answer:
(c) 2 f(x)

Question 5.
Let A = {1,2} and B = {3,4}, then the number of relation from A to B will be:
(a) 2
(b) 4
(c) 8
(d) 16
Answer:
(d) 16

Question 6.
The range of the function f(x) = \(\sqrt { x-1 } \) is:
(a) [1, ∞)
(b) [0, ∞)
(c) (0, ∞)
(d) (1, ∞)
Answer:
(b) [0, ∞)

Question 7.
If f(x) = \(\frac { x^{ 2 }-1 }{ x^{ 2 }+1 } \), then f ( \(\frac{1}{x}\) ) is:
(a) f(x)
(b) – f(x)
(c) f(-x)
(d) \(\frac { 1 }{ f(x) } \)
Answer:
(b) – f(x)

Question 8.
Domain of the function f(x) = \(\frac { 1 }{ \sqrt { 2x-3 } } \) is:
(a) R – { \(\frac{3}{2}\) }
(b) ( \(\frac{3}{2}\), ∞)
(c) [ \(\frac{3}{2}\), ∞)
(d) None of these.
Answer:
(b) ( \(\frac{3}{2}\), ∞)

(B) Match the following :

Answer:

1. 1. (d)
   2. (a)
   3. (c)

1. (b)
2. (c)

(C) Fill in the blanks:

1. If A = {1, 2} and B = {3, 4, 5}, then the number of subsets of A × B is ………………………….
2. Range of the function f = {(2, 1), (3, 1), (4, 1), (5, 1)} is …………………………..
3. Range of the function f(x) = 11 – 7 sinx is …………………………..
4. If f(x) = x² and g(x) = x + 1, ∀ x ∈ R , then (f + g)x is …………………………
5. If f(x) = 1 – cosx, then the value of f ( \(\frac { \pi }{ 4 } \) ) …………………………….
6. Domain of the function f(x) = \(\frac { 1 }{ \sqrt { (1 – x)(x – 2) } } \) is ……………………………
7. If relation R = {(1, 3), (3, 3), (4, 5)} then the value of R^-1 is ……………………………

Answer:

1. 64
2. {1}
3. [4, 18]
4. x² + x + 1
5. 1 – \(\frac { 1 }{ \sqrt { 2 } } \)
6. (1, 2)
7. {(3, 1), (3, 3), (5, 4)}.

(D) Write true/false :

1. If A, B, C are three sets, then the value of A × (B∪C) is (A∪B) × (A∪C).
2. If A = {x : x² – 5x + 6 = o}, B = {2,4}, C = {4,5}, then A × (B∩C) = {(2, 4), (3, 4)}.
3. The relation R = {(2, 1), (3, 2), (4, 3), (5, 4)} is a function.
4. If a relation on Z is R = {(x, y) : x, y ∈ Z, x² + y² ≤ 4}, then domain of R is {0, ± 1, ± 2}.
5. Domain of the function f(x) = \(\sqrt { a^{ 2 }-x^{ 2 } } \), a > 0 is [0, a].

Answer:

1. False
2. True
3. True
4. True
5. False

(E) Write answer in one word/sentence:

1. If f(x) = x² and g(x) = x + 3, x ∈ R, then the value of (fog)(2).
2. Range of function f(x) = sin x.
3. If mapping f : R → R is defmd by f(x) = x² + 1 the value of f^-1 (26) is:
4. If A = {1, 2, 3} and B = {5, 7}, then the value of A×B is:
5. Domain of the function f(x) = \(\sqrt { 3-2x } \) is:
6. If function f(x) = \(\frac { { x }^{ 2 } }{ 1-{ x }^{ 2 } }\), then the value of f(sin θ) is :

Answer:

1. 25
2. [-1, 1]
3. {-5, 5}
4. {(1, 5), (1, 7), (2, 5), (2, 7), (3, 5), (3, 7)}
5. (-∞, \(\frac { 3 }{ 2 }\))
6. tan² θ

Straight Lines Very Short Answer Type Questions

Question 1.
If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in ( A x B). (NCERT)
Solution:
Given : n(A) = 3, B = {3, 4, 5}, n (B) = 3
∴ n (A x B) = n (A) x n (B) = 3 x 3
⇒ n (A x B) = 9.

Question 2.
If G = {7, 8} and H = {5, 4, 2}, then find G x H. (NCERT)
Solution:
G x H = { 7, 8} x {5, 4, 2}
G x H= { (7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}.

Question 3.
If A x B = { (a, x), (a, y), (b, x), (b, y)} then find A and 5. (NCERT)
Solution:
Given:
A x B = {(a, x), (a, y), (6, x), (6, y)}
A = {a, b} and S = {x, y}.

Question 4.
If A = {1, 2} and B = {3, 4} then find A x B. (NCERT)
Solution:
A x B = { 1, 2} x {3, 4}
A x B = { ( 1, 3), (1, 4), (2, 3), (2, 4)}

Question 5.
The figure shows the relationship between the set P and Q. Write the relation : (NCERT)

1. In set builder form
2. In roster form
3. Find its domain
4. Find its range.

Solution:

1. Set builder form, R = {(x, y) : y = x – 2, x ∈ P and y ∈ Q}.
2. Roster form, R = {(5, 3), (6, 4), (7, 5)}
3. Domain = {5, 6, 7}
4. Range = {3, 4, 5}.

Question 6.
Let A = (1, 2, 3, 4, 6} and R be the relation on A defined by
{(a, b) : a, b ∈ A, b is exactly divisible by a } :

1. Write R in roster form,
2. Find the domain of R
3. Find the range of R. (NCERT)

Solution:
Given: A = {1, 2, 3, 4, 6}
1. R = {{a, b): a, be A, b is exactly divisible by a }
R= {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)}.

2. Domain = {1, 2, 3, 4, 6}.

3. Range = {1, 2, 3, 4, 6}.

Question 7.
Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B. (NCERT)
Solution: A = {x, y, z}, B= {1, 2}
n(A) = 3, n(B) = 2
No. of relations from A to B = 2^mn = 23^3/2 = 2⁶
= 64.

Question 8.
A function f is defined by f (x) = 2x – 5, write down the following values : (NCERT)

1. f(0)
2. f(7)
3. f(-3).

Solution:
Given: f(x) = 2x – 5
1. Put x = 0,
f(0) = 2(0) – 5 = 0 – 5 = – 5.

2. Put x = 7,
f(7) = 2 x 7 – 5 = 14 – 5 = 9.

3. Put x = – 3,
f(- 3) = 2(- 3) – 5 = – 6 – 5 = – 11.

Question 9.
The function ‘t’ which maps temperature in degree celcius into temperature in degree fehrenheit, it is defined as t(c) = \(\frac { 9c }{ 5 }\) +32. Find the following : (NCERT)

1. t (0)
2. t (28)
3. t (- 10)
4. Find c, when t (c) = 212.

Solution:
Given :
t(c) = \(\frac { 9c }{ 5 }\) + 32
1. Put c = 0,
t(0) = \(\frac { 9 × 0 }{ 5 }\) + 32
⇒ t(0) = 32

2. Put c = 28,
t(28) = \(\frac { 9 × 28 }{ 5 }\) + 32 = \(\frac { 252}{ 5 }\) + \(\frac { 32}{ 1 }\)
⇒ t(28) = \(\frac { 252 + 160}{ 5 }\) = \(\frac { 412}{ 5 }\)

3. Put c = – 10,
t(- 10) = \(\frac { 9 x (- 10)}{ 5 }\) + 32
⇒ t(- 10) = – 18 + 32 = 14

4. Put t(c) = 212,
212 = \(\frac { 9c }{ 5 }\) + 32
⇒ \(\frac { 9c }{ 5 }\) = 212 – 32
⇒ \(\frac { 9c }{ 5 }\) = 180
⇒ c = \(\frac { 180 x 5 }{ 9 }\)

Question 10.
Find the domain and range of the function f(x) = |x|.
Solution:
f(x) = – |x|, f(x) < 0
Domain of f = R.
Range of f = {y : y ∈ R, y ≤ 0} = (- ∞, 0].

Straight Lines Short Answer Type Questions

Question 1.
Find the domain and range of the function f(x) = \(\sqrt { 9 – { x }^{ 2 } }\).
Solution:
Given : f(x) = \(\sqrt { 9 – { x }^{ 2 } }\)
Value of f(x) is real, if f(x) ≥ 0
9 – x² ≥ 0
⇒ – (x² – 9) ≥ 0
⇒ x² – 9 ≤ 0
⇒ (x + 3)(x – 3) ≤ 0
∴ Domain = [- 3, 3].

Question 2.
If f(x) = x², then find \(\frac { f(1.1) – f(1)}{ (1.1 – 1)}\)
Solution:
f(x) = x²
f(1.1) = (1.1)² = 1.21
f(1) = 1² = 1

Question 3.
Find domain of function f(x) = \(\frac { { x }^{ 2 } + 2x + 1 }{ { x }^{ 2 } – 8x + 12 }\)
Solution:
Given function is :

f(x) will be defined if,
x² – 8x + 12 ≠ 0
⇒ x² – 6x – 2x + 12 ≠ 0
⇒ x(x – 6) – 2(x – 6) ≠ 0
⇒ (x – 2)(x – 6) ≠ 0
x ≠ 2 and x ≠ 6
Domain of function = R – {2, 6}.

Question 4.
Let f : g → R → R be defined by f(x) = x + 1 and g(x) = 2x – 3 respectively, then find :

1. f + g
2. \(\frac { f }{ g }\)

Solution:
Given: f(x) = x + 1, g (x) = 2x – 3
1. (f + g)x = f(x) + g(x)
= x + 1 + 2x – 3
∴ (f + g)x = 3x – 2

2. \(\frac { f }{ g }\)(x) = \(\frac { f(x)}{ g(x) }\) = \(\frac { x +1 }{ 2x – 3}\)

Question 5.
If f(x) = x² – \(\frac { 1 }{ { x }^{ 2 } }\), then prove that:
f(x) + f\(\frac { 1 }{ x }\) = 0
Solution:

Question 6.
If f(x) = \(\frac { { x }^{ 2 } }{ 1-{ x }^{ 2 } }\), then prove that:
f(sinθ) = tan² θ.
Solution:

Question 7.
If f(x) = x³ + 3x + tanx, then prove that f(x) is an odd function.
Solution:
Given : f(x) = x³ + 3x + tanx
f(- x) = (- x)³ + 3(- x) + tan(- x)
= – x³ – 3x – tanx
= – (x³ + 3x + tanx)
= – f(x).
Hence, f(x) is an odd function.

Question 8.
If f(x) = x² + 2xsinx + 3, then prove that f(x) is an even function.
Solution:
Given: f(x) = x² + 2x sin x + 3
f(- x) = (- x)² + 2(- x) sin(- x) + 3
= x² + 2xsinx + 3, [∵ sin(- x) = – sinx]
= f(x).
Hence, f(x) is an even function

Question 9.
If f(x) = x², g(x) = x + 2, ∀ x ∈R, then find gof and fog. Is gof = fog.
Solution:
Given : f(x) = x², g(x) = x + 2
fog(x) = f[g(x)]
= f(x + 2) = (x+2)².
gof(x) = g[f(x)]
= g(x²) = x² + 2
Hence, fog(x) ≠ gof(x).

Question 10.
If f(x) = e^2x and g(x) = log \(\sqrt {x}\), x> 0, then find the value of fog(x) and gof(x).
Solution:

MP Board Class 11th Maths Important Questions

MP Board Class 11th Maths Important Questions Chapter 10 Straight Lines

Straight Lines Long Answer Type Questions

Question 1.
The slope of a line is double the slope of another line. If tangent of the angle between them is \(\frac {1}{3}\), then find the slope of the line.
Solution:
Let the angle between two lines be θ.

⇒ 1 + 2m² = 3m
⇒ 2m² – 3m + 1 = 0
⇒ 2m² – 2m – m + 1 = 0
⇒ 2m(m – 1) – 1 (m – 1) = 0
⇒ (m – 1)(2m – 1) = 0
⇒ m = 1,\(\frac {1}{2}\)

Question 2.
If three points (h, 0), (a, b) and (0, k) lies on same line, then prove that \(\frac {a}{h}\) + \(\frac {b}{k}\) = 1. (NCERT)
Solution:
Given points are A(h, 0), B(a, b) and C(0, k).

ab = (b – k)(a – h)
⇒ ab = ab – bh – ak + hk
⇒ bh + ak = hk
⇒ \(\frac {bh}{hk}\) \(\frac {ak}{hk}\) \(\frac {hk}{hk}\), (dividing both sides by hk)
⇒ \(\frac {a}{h}\) + \(\frac {b}{k}\) = 1.

Question 3.
Find the equation of the line passing through (- 3, 5) and perpendicular to the line through the points (2, 5) and (- 3, 6). (NCERT)
Solution:
Gradient of AB = m = \(\frac { { y }_{ 2 } – { y }_{ 1 } }{ { x }_{ 2 } – { x }_{ 1 } }\)
Where x₁ = 2 , y₁ = 5, x₂ = – 3, y₂ = 6
∴ m = \(\frac {6 – 5}{- 3 – 2}\) = \(\frac {1}{- 5}\)

Let the gradient of CD = m₁,
AB ⊥ CD
m x m₁ = – 1
⇒ \(\frac {-1}{5}\) x m₁ = – 1
⇒ m₁ = 5
Equation of line CD will be :
y – y₁ = m(x – x₁)
Here x₁ = – 3, y₁ = 5, m = 5 .
⇒ y – 5 = 5(x + 3)
⇒ y – 5 = 5x + 15
⇒ 5x – y + 20 = 0.

Question 4.
Find the equation of straight line passing through the point (2, 2) and cutting off intercepts the axes whose sum is 9. (NCERT)
Solution:
Let the required equation of line is :
\(\frac {x}{a}\) + \(\frac {y}{b}\) = 1 …. (1)
Given: a + b = 9 …. (2)
Line (1) passes through point (2, 2).
∴ \(\frac {2}{a}\) + \(\frac {2}{b}\) = 1
\(\frac {2a + 2b}{ab}\) = 1
2a + 2b = ab
Put b = 9 – a from equation (2), we get
2a + 2(9 – a) = a(9 – a)
⇒ 2a + 18 – 2a = 9a – a²
⇒ a² – 9a + 18 = 0
⇒ a² – 6a – 3a + 18 = 0
⇒ a (a – 6) – 3(a – 6) = 0
⇒ (a – 3)(a – 6) = 0
∴ a = 3 or a = 6
When a = 3, then b = 9 – a = 9 – 3 = 6
Hence the required equation is :
\(\frac {x}{3}\) + \(\frac {y}{6}\) = 1
⇒ 2x + y = 6.
When a = 6, then b = 9 – 6 = 3
Hence the required equation is :
\(\frac {x}{6}\) + \(\frac {y}{3}\) = 1
⇒ x + 2 y = 6.

Question 5.
Find the equation to that straight line which passes through the point (7,9) and such that the portion between the axes is divided by the point in the ratio 3 : 1.
Solution:
Let the required equation of the line is :
\(\frac {x}{a}\) + \(\frac {y}{b}\) = 1 …. (1)
Let the required line cuts the axes at points A (a, 0) and B(0, b),
and Point P intercept the line AB internally in ratio 3 : 1.

Putting values of a and b in eqn. (1), we get
\(\frac {x}{28}\) + \(\frac {y}{12}\) = 1
\(\frac {3x + 7y}{84}\) = 1
Hence required equation is 3x + 7y = 84.

Question 6.
Find the equation of the line which passes through the point (1, 2) and makes a right angled triangle with axes of area \(\frac {9}{2}\) sq units.
Solution:
Area of (∆AOB) = \(\frac {1}{2}\)a.b
⇒ \(\frac {9}{2}\) = \(\frac {1}{2}\)a.b
⇒ ab = 9

Let the equation of line \(\frac {x}{a}\) + \(\frac {y}{b}\) = 1 which passes through point (1, 2)
∴ \(\frac {1}{a}\) + \(\frac {2}{b}\) = 1
From equation (1), b = \(\frac {9}{a}\)
∴ \(\frac {1}{a}\) + \(\frac {2}{9}\) = 1
⇒ 9 + 2a² = 9a
⇒ 2a² – 9a + 9 = 0
⇒ 2a² – 3a – 6a + 9 = 0
⇒ a(2a – 3) – 3(2.a – 3) = 0
⇒ (2a – 3)(a – 3) = 0
∴ b = 6, 3
Equation of line \(\frac {x}{3/2}\) + \(\frac {y}{6}\) = 1
⇒ \(\frac {2x}{3}\) + \(\frac {y}{6}\) = 1
⇒ 12x + 3y = 18
⇒ 4x + y = 6
or \(\frac {x}{3}\) + \(\frac {y}{3}\) = 1
⇒ x + y = 3

Question 7.
If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then prove that \(\frac { 1 }{ { p }^{ 2 } }\) = \(\frac { 1 }{ { a }^{ 2 } }\) + \(\frac { 1 }{ { b }^{ 2 } }\) (NCERT)
Solution:
Equation of given line be :
\(\frac {x}{a}\) + \(\frac {y}{b}\) = 1
⇒ bx + ay = ab
⇒ bx + ay – ab = 0 …. (1)
Let the form of required equation be :
xcosα + ysinβ – p = 0 …. (2)
Comparing equation (1) and (2),

Question 8.
Find the angle between the lines.
y = (2 – \(\sqrt {3}\))x + 6 and y = (2 + \(\sqrt {3}\))x – 8.
Solution:
y = (2 – \(\sqrt {3}\))x + 6 … (1)
m₁ = 2 – \(\sqrt {3}\)
Second line y = (2 + \(\sqrt {3}\))x – 8
m₂ = 2 + \(\sqrt {3}\)
Let θ be the angle between them,

Question 9.
If the perpendicular drawn from origin on line y = mx + c intersect the line at point (- 1, 2), then find the value of m and c. (NCERT)
Solution:
Let the equation of AB be :
y = mx + c … (1)
Gradient of eqn. (1) is m and OP is perpendicular to AB

Gradient of OP m₁ = \(\frac { { y }_{ 2 } – { y }_{ 1 } }{ { x }_{ 2 } – { x }_{ 1 } }\)
m₁ = \(\frac {2 – 0}{- 1 – 0}\) = – 2
∵ Op ⊥ AB
∴ m × m₁ = – 1
⇒ m(- 2) = – 1
⇒ m = \(\frac {1}{2}\)
put value of m in equation (1),
y = \(\frac {1}{2}\)x + c
Above equation passes through point (-1, 2), hence it will satisfy it.
2 = \(\frac {- 1}{2}\) + c
⇒ c = 2 + \(\frac {1}{2}\) = \(\frac {5}{2}\)
∴ m = \(\frac {1}{2}\), c = \(\frac {5}{2}\)

Question 10.
Find the coordinate of the foot of perpendicular from the point (- 1, 3) to the line 3x – 4y – 16 = 0.
Solution:
Equation of line CD is :
3x – 4y – 16 = 0  … (1)
Equation of line AB is perpendicular to equation (1),
4x + 3y + c = 0  … (2)

Equation (2) passes through point (- 1, 3).
∴ 4(- 1) + 3 x 3 + c = 0
– 4 + 9 + c = 0
c = – 5
Put c = – 5 in equation (2), we get 4x + 3y – 5 = 0
Now the intersection of lines (1) and (3) is the foot of perpendicular B
3x – 4y – 16 = 0
4x + 3y – 5 = 0
By cross multiplication method,

Question 11.
Find the equation of line parallel to y – axis and drawn through the point of intersection of line x – 7y + 5 = 0 and 3x + y = 0. (NCERT)
Solution:
Equation of given lines are :
x – 7y + 5 = 0 … (1)
3x + y = 0 … (2)
Equation of straight line passing through the intersection of equation (1) and (2) is :
x – 7y + 5 + λ(3x + y) = 0 … (3)

Put A = 7 in equation (3), we get
x – 7y + 5 + 7(3x + y) = 0
⇒ x – 7y + 5 + 21x + 7y = 0
⇒ 22x + 5 = 0.

Question 12.
Find the value of p so that the three lines 3x + y – 2 = 0, px + 2y – 3 = 0 and 2x – y – 3 = 0 may intersect at one point. (NCERT)
Solution:
Equation of given lines are :
3x + y – 2 = 0 … (1)
px + 2y – 3 = 0 … (2)
2x – y – 3 = 0 … (3)
From eqns. (1) and (3),
3x + y – 2 = 0
2x – y – 3 = 0
Solving by cross multiplication method,

Lines intersect at the one point, hence point (1, – 1) will satisfy equation (2). p(1) – 2 x 1 – 3 = 0
p = 5.

Question 13.
Find the equation of a line drawn perpendicular to the line \(\frac {x}{4}\) + \(\frac {y}{6}\) = 1 through the point where it meets the y – axis. (NCERT)
Solution:
Given equation is :
\(\frac {x}{4}\) + \(\frac {y}{6}\) = 1
Line (1) meets the F – axis at point (0, 6).
From eqn. (1), \(\frac {6x + 4y}{24}\) = 1
⇒ 6x + 4y = 24
⇒ 6x + 4y – 24 = 0 … (2)
Equation of line perpendicular to eqn. (2),
4x – 6y + c = 0 … (3)
Above equation passes through point (0, 6).
∴ 4 x 0 – 6 x 6 + c = 0
c = 36
Put value of c = 36 in equation (3),
4x – 6y + 36 = 0
2x – 3y + 18 = 0.

Question 14.
For what value of k the equation (k – 3) x (4 – k²)y + k² – 7k + 6 = 0 is :

1. Parallel to X – axis
2. Parallel to y – axis
3. Passes through origin.

Solution:
Equation of given line is :
(k – 3)x – (4 – k²)y + k² – 7k+6 = 0
(4 – k²)y = (k – 3)x + k² – 7k + 6
⇒ y = \(\frac { (k – 3)x }{ 4 – { k }^{ 2 } }\) + \(\frac { { k }^{ 2 } – 7k + 6 }{ 4 – { k }^{ 2 } }\)

1. Line parallel to X – axis :
∴ m = 0
⇒ \(\frac { k – 3 }{ 4 – { k }^{ 2 } }\)
⇒ k – 3 = 0
⇒ k = 3.

2. Line parallel to Y – axis :
∴ m = \(\frac {1}{0}\)
⇒ \(\frac { k – 3 }{ 4 – { k }^{ 2 } }\)
⇒ 4 – k² = 0
⇒ k²= 4
⇒ k = ± 2

3. Line passing through origin, if c = 0.
∴ \(\frac { { k }^{ 2 } – 7k + 6 }{ 4 – { k }^{ 2 } }\)
⇒ k² – 7k+6 = 0
⇒ k² – 6k – k+6 = 0
⇒ k(k – 6) – 1(k – 6) = 0
⇒ (k – 6)(k – 6) = 0
∴ k = 1, 6

Question 15.
The line passing through the points (h, 3) and (4,1) meets the line 7x – 9y – 19 = 0 at right angle. Find the value of h.
Solution:
Equation of given line is :
7x – 9y – 19 = 0
⇒ 9y = 7x – 19
⇒ y = \(\frac {7}{9}\)x – \(\frac {19}{9}\) … (1)
Gradient of eqn. (1) m₁ = \(\frac {7}{9}\)
Equation of line passing through points (h, 3) and (4, 1),
m₂ = \(\frac { { y }_{ 2 } – { y }_{ 1 } }{ { x }_{ 2 } – { x }_{ 1 } }\)
m₂ = \(\frac {1 – 3}{4 – h}\) = \(\frac {- 2}{4 – h}\)
∵ Lines are perpendicular.
∴m₁ x m₂= – 1
\(\frac {7(- 2)}{9(4 – h)}\) = – 1
⇒ – 14 = – 36 + 9h
⇒ 9h = 36 – 14
⇒ 9h = 22
∴ h =\(\frac {22}{9}\).

Question 16.
If the lengths of the perpendiculars drawn from origin on the straight lines xcosθ – y sinθ = kcos2θ and xsecθ + y cosec θ = k are p and q then prove that:
p² + 4q² = k²
Solution:
Equation of given lines :
xsecθ + ycosecθ = k … (1)
and x cosθ – y sinθ = k cos2θ … (2)
Length of perpendicular from origin on equation (1),

Adding equation (3) and (4),
4q² + p² = k² sin² 2θ +k² cos² 2θ
= k²(sin² 20 + cos²2θ)
⇒ p² + 4q² = k² .

Question 17.
Find the equation to the straight line passing through the intersection of lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 and makes an equal intercept on both the axes.
Solution:
Equation of given lines are :
4x + 7y – 3 = 0 … (1)
2x – 3y + 1 = 0 … (2)
Equation of straight line passing through intersection of lines (1) and (2),
4x + 7y – 3 + λ(2x – 3y + 1) = 0 … (3)
⇒ x(4 + 2 λ) + y(7 – 3 λ) – 3 + λ = 0
⇒ x(4 + 2 λ) + y(7 – 3 λ) = 3 – λ

Put λ = \(\frac {3}{5}\) in equation (3),
4x + 7y – 3 + \(\frac {3}{5}\)(2x – 3y + 1) = 0
⇒ 20x + 35y – 15 + 6x – 9y +3 = 0
⇒ 26x + 26y = 12
⇒ 13x + 13y = 6.
⇒ 13x + 13y – 6 = 0.

Question 18.
Find the distance of point (-1, 1) from the line 12 (x + 6) = 5 (y – 2). (NCERT)
Solution:
Equation of given line is :
12(x + 6) = 5(y – 2)
⇒ 12x + 72 = 5y – 10
⇒ 12x – 5y + 82 = 0
Length of perpendicular

Question 19.
Find the points on the X – axis, whose distance from the line \(\frac {x}{3}\) + \(\frac {y}{4}\) = 1 are 4 units. (NCERT)
Solution:
Equation of given line is :
\(\frac {x}{3}\) + \(\frac {y}{4}\) = 1
⇒ 4x + 3y = 12
⇒ 4x + 3y – 12 = 0
Let the point on X – axis is (h, 0).
Length of perpendicular drawn from point (h, 0) is 4.

Taking (+) sign,
4h = 20 + 12 = 32
⇒ h = 8
Taking (-) sign,
4h = – 20 + 12
⇒ 4h = – 8
⇒ h = – 2
Point on X – axis are ( – 2, 0) and (8, 0).

Question 20.
Find the distance between two parallel lines : 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0.
Solution:
Equation of given lines are :
15x + 8y – 34 = 0
15x + 8y + 31 = 0
Put x = 0 in eqn. (1),
15(0) + 8y – 34 = 0
⇒ 8y = 34
y = \(\frac {34}{8}\) = \(\frac {17}{4}\)
Point (0, \(\frac {17}{4}\)) is on line (1),
Length of perpendicular from point (0, \(\frac {17}{4}\)) on equation (2),

= \(\frac {65}{17}\) units
Hence the distance between lines is \(\frac {65}{17}\) units

Question 21.
The length L (in centimeters) of a copper rod is a linear function of its Celsius temperature C. In an experiment. If L = 124.942 when C= 20 and L = 125.134 when C = 110. Express L in terms of C. (NCERT)
Solution:
Given : L = 124.942 when C = 20
L = 125.134 when C= 110
In coordinate form points (20, 124.942) and (110, 125.134) are two points.
Required equation of straight line :

Question 22.
The owner of a milk store finds that he can sell 980 L of milk each week at Rs. 14/L and 1220 L of milk each week at Rs. 16/L. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs. 17 L? (NCERT)
Solution:
Let price and litre be denoted in ordered pair (x, y), two points are (14, 980) and (16,1220).
∴ Required equation of straight line will be :
y – y₁ = \(\frac { { y }_{ 2 } – { y }_{ 1 } }{ { x }_{ 2 } – { x }_{ 1 } }\)(x – x₁)
⇒ y – 980 = \(\frac {1220 – 980}{16 – 14}\)(x – 14)
⇒ y – 980 = \(\frac {240}{2}\)(x – 14)
⇒ y – 980 = 120 (x – 14)
⇒ y – 980 = 120 x – 120 x 14
⇒ y = 120 x – 1680 + 980
⇒ y = 120 x – 700
When x = 17, then
y = 120 x 17 – 700
⇒ y = 2040 – 700
⇒ y = 1340
Hence, he will sell weekly 1340 L milk at the rate of Rs. 17 L.

MP Board Class 11th Maths Important Questions

MP Board Class 11th Maths Important Questions Chapter 1 Sets

Sets Important Questions

Sets Objective Type Questions

(A) Choose the correct option:

Question 1.
A set is defind as:
(a) Collection of an object
(b) Collection of well defind object
(c) Nothing can be said
(d) None of these.
Answer:
(b) Collection of well defind object

Question 2.
If A = {1, 2, 3},B = {2, 3, 4} and U = {1, 2, 3, 4, 5, 6},then A’ ∩ B’ =
(a) {2, 3}
(b) {1, 5,6}
(c) {5, 6}
(d) {1, 4, 5, 6}.
Answer:
(c) {5, 6}

Question 3.
A and B are two sets, then A ∩ (A ∪ B) =
(a) A
(b) B
(c) ϕ
(d) None of these.
Answer:
(a) A

Question 4.
A set A = {x : x ∈ R, x² = 16 and 2x = 6} =
(a) ϕ
(b) {14,3,4}
(c) {3}
(d) {4}.
Answer:
(a) ϕ

Question 5.
For a non – empty set A:
(a) A∪A’ = A
(b) A∪A’ = A’
(c) A∪A’ = U
(d) A∪A’ = ϕ .
Answer:
(c) A∪A’ = U

Question 6.
If two non – empty sets A and B are not equal, then:
(a) A⊂A∩B
(b) B⊂A∩B
(c) A∩B⊂B
(d) A∩B⊂ϕ.
Answer:
(c) A∩B⊂B

Question 7.
A and B are two sets and n(A) = 70, n(B) = 60 and n(A∪B) = 110, then n(A∩B) =
(a) 240
(b) 50
(c) 40
(d) 20.
Answer:
(d) 20.

Question 8.
If A = {1,2,3,4,5}, then the number of proper subsets:
(a) 120
(b) 30
(c) 31
(d) 32.
Answer:
(d) 32.

Question 9.
If A, B and C are three sets, then A – (B∪C) is equal to:
(a) (B – A)∩C
(b) (A – B) ∪C
(c) (A – B) ∩(A – C)
(d) (A – B) ∪(A – C)
Answer:
(c) (A – B) ∩(A – C)

Question 10.
Let S = {1,2,3,4}. The total numbers of unordered pair of disjoint subsets of S is equal to:
(a) 25
(b) 34
(c) 42
(d) 41.
Answer:
(d) 41.

(B) Match the following:

Answer:

1. (e)
2. (c)
3. (a)
4. (b)
5. (f)
6. (d)

(C) Fill in the blanks:

1. If the elements of A mid B are 3 and 6, then minimum number of elements in A∪B is …………………………….
2. If X and Y are two sets such that n(X) = 17, n(Y) = 23 and n(X∪Y) = 38, then n(X∩Y) = …………………………
3. If sets A = {1, 2, 3, 4, 5, 6} and B = {2, 4, 6, 8}, then B – A = ……………………………..
4. A set which does not contain any element is called ………………………….
5. If set A has n elements, then the number of subsets of A are …………………………..
6. If A = {1, 2} and B = {3, 4}, then (A∪B)∩ϕ = ……………………………..

Answer:

1. 6
2. 2
3. {8}
4. Empty set
5. 2ⁿ
6. ϕ

(D) Write true/false:

1. If A ∩ B = B, then B ⊂ A.
2. If A = {1, 2, 3, 4} and B = {2, 3, 4, 5, 6}, then A ∆ B = {1, 5, 6}.
3. If n(A – B) = n(A) – n(A n B).
4. For a non – empty sets A is A∩A’- A’.
5. If A = {1, 2, 3, 4}, then n[P(A)] = 16.

Answer:

1. False
2. True
3. True
4. False
5. True.

Question (E)
Write answer in one word/sentence:

1. If A = {1, 2}, then write all the subsets of set A.
2. If A = {4, 5, 8, 12} and B = {1, 4, 6, 9}, then find the value of A – (B – A).
3. If n(U) = 700, n(A) = 200, n(B) = 300, and n(A∩B) = 100, then find the value of n(A’∩B’). ‘
4. Find the value of n[P{P{P(ϕ}}]
5. If A = {a, b, c, d}, then find all subsets of set A.
6. If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, B = {2, 4, 6, 8} and C = {3, 4, 5, 6}, then find the value of (B – C)’.

Answer:

1. ϕ, {1}, {2}, {1,2}
2. {4, 5, 8, 12}
3. 300
4. 4
5. 16
6. {1, 3, 4, 5, 6, 7, 8, 9}.

Sets Very Short Answer Type Questions

Question 1.
Write the following sets in set builder form: (NCERT)

1. {3, 6, 9, 12}
2. {2, 4, 8, 16, 32}
3. {5, 25, 125, 625}
4. {2, 4, 6,},
5. {1, 4, 9, ………… 100}.

Solution:

1. A = {x : x is a natural number, multiple of 3 and x < 15}
2. B = {x : x = 2ⁿ, x∈N and n < 6}
3. C = {x : x = 5ⁿ, and x∈N and n ≤ 4}
4. D = {x : x is an even natural number}
5. E = {x : x = n², x ∈ N and n < 11}.

Question 2.
Write the following sets in roster form: (NCERT)

1. A = {x : x is an odd natural number}
2. B = {x : x is an integer – \(\frac{1}{2}\) < x < \(\frac{9}{2}\)
3. C = {x : x is an integer x² ≤ 4}
4. D = {x : x is a letter of a word LOYAL}
5. E = {x : x is a month of year which does not have 31 days}
6. F – {x : x is a consonant of English alphabet which comes before k}

Solution:

1. A = {1, 3, 5, 7, 9}
2. B = {0, 1, 2, 3, 4}
3. C = {-2, -1, 0, 1, 2}
4. D = {L, 0, Y, A}
5. E = {FEBRUARY, APRIL, JUNE, SEPTEMBER, NOVEMBER}
6. F = {b, c, d, f, g, h, j}.

Question 3.
From following find whether A = B. (NCERT)

1. A = {a, b, c, d}, B = {d, c, b, a}.
2. A = {4, 8, 12, 16}, B = {8,4, 16, 18}.
3. A = {2, 4, 6, 8, 10}, B = {x : x is a positive even integer x < 10}.
4. A = {x : x is a multiple of 10}, B = {10, 15, 20, 25, 30}.

Solution:
1. A = {a, b, c, d}, B = {d, c, b, a}
All the elements of set A is in set B
A = B.

2. A = {4, 8, 12, 16}, B = {8, 4, 16, 18}
All the elements of A is not in set B.
∴ A≠B.

3. Here A = {2, 4, 6, 8, 10} and B = { x : x is an even number and x ≤ 10}
B = {2, 4, 6, 8, 10}
i.e., The elements of set A and B are same.
∴ A = B

4. Here A = {x : x is a multiple of 10} = {10, 20, 30, 40}
and B = {10, 15, 20, 25, 30, …}
The elements of set A and B are not same.
∴ A≠B.

Question 4.
If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10} then find the following: (NCERT)

1. A∪B
2. A∪C
3. B∪C
4. B∪D

Solution:

1. A ∪ B = {1, 2, 3, 4} ∪ {3, 4, 5, 6} = {1, 2, 3, 4, 5, 6}.
2. A ∪ C = {1, 2, 3, 4} ∪ {5, 6, 7, 8} = {1, 2, 3, 4, 5, 6, 7, 8}.
3. B ∪ C = {3, 4, 5, 6} ∪ {5, 6, 7, 8} = {3, 4, 5, 6, 7, 8}
4. B ∪ D = {3, 4, 5, 6} ∪ {7, 8, 9, 10} = {3, 4, 5, 6, 7, 8, 9, 10}

Question 5.
If A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C= {11, 13, 15} and D = {15, 17} then, find the following:

1. A ∩ B
2. B ∩ C
3. A ∩ C
4. B ∩ D

Solution:

1. A ∩ B = {3, 5, 7, 9, 11} ∩ {7, 9, 11, 13} = {7, 9, 11}
2. B ∩ C = {7, 9, 11, 13} ∩ {11, 13, 15} = {11, 13}
3. A ∩ C = {3, 5, 7, 9, 11} ∩ {11, 13, 15} = {11}
4. B ∩ D = {7, 9, 11, 13} ∩ {15, 17} = ϕ

Question 6.
If A = {3, 6, 9, 12, 15, 18, 21}, B = {4, 8, 12, 16, 20}, C = {2, 4, 6, 8, 10, 12, 14, 16} and D = {5, 10, 15, 20} then, find the following: (NCERT)

1. A – B
2. A – C
3. A – D
4. B – A
5. C – A
6. D – A
7. B – C
8. B – D
9. C – B
10. D – B
11. C – D
12. D – C?

Solution:
Let A and B are two sets and A – B is a set of those elements which one present in A and not in B.

1. A – B = {Those elements present in A and not in B}
= {3, 6, 9, 12, 15, 18, 21} – {4, 8, 12, 16, 20}
= {3, 6, 9, 15, 18, 21}.

2. A – C = {Those elements present in A and not in C}
= {3, 6, 9, 12, 15, 18, 21} – {2, 4, 6, 8, 10, 12, 14, 16}
= {3, 9, 15, 18, 21}.

3. A – D = {The elements which are in A but not in D}
= {3, 6, 9, 12, 15, 18, 21} – {5, 10, 15, 20}
= {3, 6, 9, 12, 18, 21}.

4. B – A = {The elements which are in B but not in A}
= {4, 8, 12, 16, 20} – {3, 6, 9, 12, 15, 18, 21}
= {4, 8, 16, 20}.

5. C – A = {The elements which are in C but not in A}
= {2, 4, 6, 8, 10, 12, 14, 16} – {3, 6, 9, 12, 15, 18, 21}
= {2, 4, 8, 10, 14, 16}.

6. D – A = {The elements which are in D but not in A}
= {5, 10, 15, 20} – {3, 6,9, 12, 15, 18, 21}
= {5, 10, 20}.

7. B – C = {The elements which are in B but not in C}
= {4, 8, 12, 16, 20} – {2, 4,6, 8, 10, 12, 14, 16}
= {20}.

8. B – D = {The elements which are in B but not in D}
= {4, 8, 12, 16, 20} – {5, 10, 15, 20}
= {4, 8, 12, 16}.

9. C – B = {The elements which are in C but not in B}
= {2, 4, 6, 8, 10, 12, 14, 16} – {4, 8, 12, 16, 20}
= {2, 6, 10, 14}.

10. D – B = {The elements which are in D but not in B}
= {5, 10, 15,20} – {4, 8, 12, 16, 20}
= {5, 10, 15}.

11. C – D = {The elements which are in C but not in D}
= {2, 4, 6, 8, 10, 12, 14, 16} – {5, 10, 15, 20}
= {2, 4, 6, 8, 12, 14, 16}.

12. D – C = {The elements which are in D but not in C}
= {5, 10, 15, 20} – {2, 4, 6, 8, 10, 12, 14, 16}
= {5, 15, 20}.

Question 7.
Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4}, B = {2, 4, 6, 8} and C then find the following:

1. A’
2. B’
3. (A ∪ C)’
4. (A ∩ B)’
5. (A’)’
6. (B – C)’
7. (B – C)’

Solution:
1. A’= U – A
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4}
⇒ A’= {5, 6, 7, 8, 9}.

2. B’ = U – B
⇒ {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 4, 6, 8}
⇒ B’ = {1, 3, 5, 7, 9}.

3. (A ∪ C) = {1, 2, 3, 4} ∪ {3, 4, 5, 6}
⇒ A ∪ C = {1, 2, 3, 4, 5, 6}
(A ∪ C)’= U – (A ∪ C)
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4, 5, 6}
⇒ (A ∪ C)’ = {7, 8, 9}

4. A ∪ B = {1, 2, 3, 4} ∪ {2, 4, 6, 8} = {1, 2, 3, 4, 6, 8}
(A ∪ B)’ = ∪ – (A ∪ B)
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4, 6, 8}
⇒ (A ∪ B)’ = {5, 7, 9}

5. (A’)’ = ∪ – A’
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {5, 6, 7, 8, 9}
⇒ (A’)’= {1, 2, 3, 4} = A

6. B – C = {2, 4, 6, 8} – {3, 4, 5, 6} = {2, 8}.
(B ∪ C)’ = U – (B – C)
= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {2, 8}
⇒ (B – C)’= {1, 3, 4, 5, 6, 7, 9}.

Sets Short Answer Type Questions

Question 1.
If X and Fare two sets, n(X) = 17, n(Y) = 23 and n(X ∪ F) = 38, then find n(X ∩ F)? (NCERT)
Solution:
We know that n(X ∪ Y) = n(X) + n(Y) – n(X∩Y)
Given: n(X) = 17, n(Y) = 23, n(X ∪ Y) = 38, n(X ∩ Y) = ?
∴ 38 = 17 + 23 – n(X ∩ Y)
⇒ 38 = 40 – n(X ∩ Y)
⇒ n(X ∩ Y) = 40 – 38
⇒ n(X ∩ Y) = 2.

Question 2.
If X and Y are two sets, such that X∪Y has 18 elements, X has 8 elements and F has 15 elements, then how many elements does X ∩ Y have? (NCERT)
Solution:
Given: n(X ∪ Y) = 18, n(X) = 8, n(Y) = 15.
Applying formula n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)
18 = 8 + 15 – n(X ∩ Y)
⇒ n(X ∩ Y) = 23 – 18 = 5.

Question 3.
If S and Tare two sets, such that S has 21 elements, T has 32 elements and S∩T has 11 elements, how many elements does S∪T have? (NCERT)
Solution:
Given: n(S) = 21, n(T) = 32, n(S ∩ T) = 11, n(S ∪ T) = ?
∵ n(S ∪ T) = n(S) + n(T) – n(S ∩ T)
⇒ n(S ∪ T) = 21 + 32 – 11 = 42.

Question 4.
If X and Y are two sets such that X has 40 elements, X∪Y has 60 elements and X∩Y has 10 elements, then how many elements does Y have? (NCERT)
Solution:
Given: n(X) = 40, n(X ∪ Y) = 60, n(X ∩ Y) = 10, n(Y) = ?
Applying formula n(X ∪ Y) = n(X) + n(Y) – n{X ∩ Y)
⇒ 60 = 40 + n(Y) – 10
⇒ 60 = 30 + n(Y)
⇒ n(Y) = 30.

Question 5.
If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10}, then find the following: (NCERT)

1. A ∪ B ∪ C
2. A ∪ B ∪ D
3. B ∪ C ∪ D.

Solution:
1. A ∪ B ∪ C = {1, 2, 3, 4} ∪ {3, 4, 5, 6} ∪ {5, 6, 7, 8}
= {1, 2, 3, 4, 5, 6} ∪ {5, 6, 7, 8}
= {1, 2, 3, 4, 5, 6, 7, 8}.

2. A ∪ B ∪ D = { 1, 2, 3, 4} ∪ {3, 4, 5, 6} ∪{7, 8, 9, 10}
= {1, 2, 3, 4, 5, 6} ∪ {7, 8, 9, 10}
= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.

3. B ∪ C ∪ D = {3, 4, 5, 6} ∪ {5, 6, 7, 8} ∪ {7, 8, 9, 10}
= {3, 4, 5, 6} ∪ {5, 6, 7, 8, 9, 10}
= {3, 4, 5, 6, 7, 8, 9, 10}.

Sets Long Answer Type Questions

Question 1.
In a survey of 600 students in a school, 150 students were found to be taking tea and 225 taking coffee, 100 were taking both tea and coffee. Find how many students were taking neither tea nor coffee? (NCERT)
Solution:
Let C and T be the students taking coffee and tea.
Here, n(T) = 150, n(C) = 225, n(C ∩ T) = 100.
Applying formula n(C ∪ T) = n(T) + n(C) – n(C ∩ T)
n(C ∪ T) = 150 + 225 – 100 = 375 – 100
⇒ n(C ∪ T) = 275
Total number of students = 600 = n(U).
Number of students taking neither tea nor coffee = n(C ∪ T)’
n(C ∪ T)’ = n(U) – n(C ∪ T)
= 600 – 275 = 325.

Question 2.
In a group of 70 people, 37 likes coffee, 52 likes tea and each person likes at least one of these two drinks. Find the number of persons who likes both coffee and tea? (NCERT)
Solution:
Let C and T be the persons who likes coffee and tea.
Given: n(C) = 37, n(T) = 52, n(C ∪ T) = 70, n(C ∩ T) = ?
Applying formula n(C ∪ T) = n(C) + n(T) – n(C ∩ T)
⇒ 70 = 37 + 52 – n(C ∩ T)
⇒ 70 = 89 – n(C ∩ T)
⇒ n(C ∩ T) = 89 – 70
⇒ n(C ∩ T) = 19
∴ Number of people who likes both coffee and tea = 19.

Question 3.
In a group of 65 people, 40 likes cricket, 10 likes both cricket and tennis. How many like tennis only and not cricket? How many like tennis? (NCERT)
Solution:
Let C and T denotes the people who likes cricket and tennis respectively.
Given: n(C ∪ T) = 65, n(C) = 40, n(C ∩ T) = 10
We know that
n(C ∪ T) = n(C) + n(T) – (C ∩ T)
65 = 40 + n(T) – 10
⇒ n(T) = 65 – 30 = 35.
∴ Number of people who likes only tennis and not cricket
= n(T ∩ C)’ = n(T) – (C ∩ T)
= 35 – 10 = 25.

Question 4.
In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper T, 26 read newspaper I, 9 read both H and I. 11 read both H and T, 8 read both T and 1.3 read all three newspapers. Find

1. The number of people who read at least one of the newspapers.
2. The number of people who read exactly one newspaper. (NCERT)

Solution:
Given: n (H) = 25, n (T) = 26, n(I) = 26, n(H ∩ I) = 9, n(H ∩ T) = 11, n(T ∩ I) = 8, n(H ∩ T ∩ I) = 3

1. The number of people who reads at least one of the newspaper = n(H∪T∪I)
= n(H) + n(T) + n(I) – n(H ∩ T) – n(T ∩ I) – n(H ∩ I) + n(H ∩ T ∩ I)
= 25 + 26 + 26 – 11 – 8 – 9 + 3
= 77 – 28 + 3 = 80 – 28 = 52.

2. The number of people who reads exactly one newspaper.
= n(H) + n(T) + n(I) – 2n(H ∩ I) – 2n(H ∩ T) – 2n(T ∩ I) + 3n(H ∩ T ∩ I)
= 25 + 26 + 26 – 2 × 9 – 2 × 11 – 2 × 8 + 3 × 3 = 77 – 18 – 22 – 16 + 9
= 86 – 56 = 30.

Question 5.
In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked product C. If 14 people liked products A and B, 12 people liked products C and A, 14 people liked products B and C and 8 liked all the three products. Find how many liked product C only? (NCERT)
Solution:
Let A, B and C denotes the people liked the products A, B and C respectively.
Given: n(A) = 21, n(B) = 26, n(C) = 29, n(A ∩ B) = 14, n(C ∩ A) = 12, n(B ∩ C) = 14, n(A ∩ B ∩ C) = 8
n(only C) = n(C) – n(C ∩ A) – n(B ∩ C) + n(A ∩ B ∩ C)
= 29 – 12 – 14 + 8
= 37 – 26 = 11.

MP Board Class 11 Maths Important Questions

MP Board Class 11th Maths Important Questions Chapter 9 Sequences and Series Important Questions

Sequences and Series Important Questions 

Sequences and Series Objective Type Questions

(A) Choose the correct answer of the following:

Question 1.
The sum of the cube of first n positive integer is :
(a) \(\frac {n(n + 1)}{2}\)
(b) \(\frac {n(n + 1)(2n + 1)}{6}\)
(c) \(\frac {n(n + 1)(n + 2)}{6}\)
(d) {\(\frac {n(n + 1)}{2}\)}²
Answer:
(d) {\(\frac {n(n + 1)}{2}\)}²

Question 2.
The sum of n term of arithmetic progression is 2n + 3n², its second term will be :
(a) 10
(b) 12
(c) 16
(d) 11
Answer:
(c) 16

Question 3.
If arithmetic mean of a and b is \(\frac { { a }^{ n }+{ n }^{ n } }{ { a }^{ n – 1 }+{ b }^{ n – 1 } }\), then the value of n will be :
(a) 1
(b) 0
(c) – 1
(d) \(\frac {1}{2}\)
Answer:
(a) 1

Question 4.
Which term of the series 8, 4,0, ………….. is – 24 :
(a) 7^th
(b) 28^th
(c) 8^th
(d) 9^th
Answer:
(d) 9^th

Question 5.
If the first term of arithmetic progression be a and last term is l, then the sum of n terms will be :
(a) \(\frac {n}{2}\)[2a – (n – 1)d]
(b) \(\frac {n}{2}\)[2a + (n – 1)d]
(c) \(\frac {n}{2}\)(a + l)
(d ) \(\frac {n}{2}\)(a – l).
Answer:
(c) \(\frac {n}{2}\)(a + l)

Question 6.
The next term of the sequence 2\(\sqrt { 2 }\), \(\sqrt { 2 }\), 0, …………… is :
(a) – \(\sqrt { 3 }\)
(b) \(\frac { 1 }{ \sqrt { 2 } }\)
(c) – \(\sqrt { 2 }\)
(d) \(\sqrt { 2 }\)
Answer:
(c) – \(\sqrt { 2 }\)

Question 7.
If 2x, x + 8, 3x + 1 are in A.P, then value of x will be :
(a) 3
(b) 7
(c) 5
(d) 2
Answer:
(c) 5

Question 8.
The 15^th term from the end of the A.P. 2, 6,10, …………., 86 is :
(a) 30
(b) 32
(c) 46
(d) 48
Answer:
(a) 30

Question 9.
If first 15^th term of an A.P. is a, second term is b and n^th term is 2 a, then sum of the n terms is :
(a) \(\frac {ab}{2(b – a)}\)
(b) \(\frac {2ab}{3(b – a)}\)
(c) \(\frac {3ab}{2(b – a)}\)
(d) \(\frac {3ab}{(b – a)}\)
Answer:
(c) \(\frac {3ab}{2(b – a)}\)

Question 10.
In an A.P. S_n = 3n² + 5n and T_m = 164, then m equals to :
(a) 26
(b) 27
(c) 28
(d) None of these.
Answer:
(b) 27

Question 11.
A.M. of two number is 10 and GM. is 8, the numbers are
(a) a = 4, b = 16
(b) a = 2, b = 8
(c) a = 4, b = 9
(d) a = 2, b = 18.
Answer:
(a) a = 4, b = 16

Question 12.
The A.M. of two numbers is A and G M is G, then relation between them is :
(a) A < G (b) A = G (c) A > G
(d) None of these.
Answer:
(c) A > G

Question 13.
2^1/4.4^1/8.8^1/16 ……………. ∞ =
(a) 1
(b) 2
(c) 3/2
(d) 4
Answer:
(b) 2

Question 14.
If y = x – x² + x³ – x⁴ + ………. ∞, then value of x be (- 1 < x < 1) :
Answer:
(a) y + \(\frac {1}{y}\)
(b) \(\frac {y}{1 + y}\)
(c) y – \(\frac {1}{y}\)
(d) \(\frac {y}{1 – y}\)
Answer:
(d) \(\frac {y}{1 – y}\)

Question 15.
If the second, third and sixth terms of an A.P. are in GP, then the common ratio of the GP. is :
(a) 2
(b) 5
(c) 4
(d) 3
Answer:
(d) 3

Question 16.
Sum up to infinity of 1 + \(\frac {4}{5}\) + \(\frac { 7 }{ { 5 }^{ 2 } }\) + \(\frac { 10 }{ { 5 }^{ 2 } }\) + …………. :
(a) \(\frac {35}{16}\)
(b) \(\frac {37}{16}\)
(c) \(\frac {39}{16}\)
(d) 3
Answer:
(a) \(\frac {35}{16}\)

(B) Match the following :

Answer:

1. (d)
2. (f)
3. (a)
4. (g)
5. (b)
6. (j)
7. (c)
8. (i)
9. (e)
10. (h)

(C) Fill in the blanks :

1. – 7\(\sqrt { 3 }\) will be term of the series 5\(\sqrt { 3 }\), 3\(\sqrt { 3 }\), \(\sqrt { 3 }\), …………….
2. 116 is sum of the term of the series 25,22, 19 …………….
3. If sum of n terms of series is n² + 4n, then 15^th term of the series is …………….
4. 0 will be the term of the series 27, 24,21, 18 …………….
5. 729 will be the term of the series – \(\frac {1}{27}\), \(\frac {1}{9}\), – \(\frac {1}{3}\) …………….
6. The first term of GP. is a, common ratio r < 1 and its last term l, then its sum will be …………….
7. The ratio of the sum of first three term to the sum of first six term is 125 : 152, then common ratio will be …………….
8. First term 16 and fifth term \(\frac {1}{16}\), then its 4^th term will be …………….
9. Sum of n terms of the series x + 2x² + 4x³ + 8x⁴ + ……………. will be …………….
10. If a = 2, d = 2 and n = 50, then the last term of the series will be …………….

Answer:

1. 7
2. 12
3. 33
4. 10
5. 10
6. \(\frac {a – rl}{1 – r}\)
7. \(\frac {3}{5}\)
8. \(\frac {1}{4}\)
9. \(\frac { 1-({ 2x }^{ n })x }{ 1 – 2x }\)
10. 100

(D) Write true / false :

1. 1, 3, 5, 8, ………….. are inA.P.
2. n^th term of a G.P. is a + (n – 1 )d.
3. If 2x, x + 5 and x + 11 are in A.P., then value of x will be – 1.
4. Arithmetic mean of a and b is \(\sqrt { ab }\)
5. Sum of 9 terms of a sequence 24 + 20 + 16 + will be.
6. Four consicutive term of G.P. are \(\frac { a }{ { r }^{ 3 } }\) \(\frac {a}{r}\), ar, ar³.

Answer:

1. False
2. False
3. True
4. False
5. True
6. True.

(E) Write answer in one word / sentence :

1. If a, b, care inA.P., then find the value of ab + ac?
2. Arithmetic mean of (a + b)²and (a – b)² will be?
3. Sum of n arithmetic mean between x and 3x will be.
4. Find the sum of infinite terms of series : 9^1/3.9^1/3.9^1/27 ………….. up to ∞.
5. If a, b, c are in GP., then find the value \(\frac {1}{b}\) + \(\frac {1}{a – b}\) –\(\frac {1}{b – c}\)

Answer:

1. 2b²
2. a² + b²
3. 3nx
4. 3
5. 0

Sequences and Series Short Answer Type Questions

Question 1.
Which term of the sequence 27,24,21,18, …………. is zero? (NCERT)
Solution:
Here a = 27 and d= T₂ – T₁ = 24 – 27 = – 3.
Let the n^th term of series be 0.
∴ T_n = a + (n – 1)d
⇒ 0 = 27 + (n – 1)(- 3)
⇒ 3n – 3 = 27
⇒ 3n = 30
n = 10 or 10^th term.

Question 2.
The last term of the series 8,4,0, ……………. is – 24. Find the total number of terms. (NCERT)
Solution:
Given series 8,4, 0, … (1)
First term = a = 8 Common difference d = 4 – 8 = -4
d = 0 – 4 = – 4
∵ The common difference is same the series is in A.P.
last term l = – 24,
l = a + (n – 1)d,
= – 24 = 8 + (n – 1)(-4)
⇒ – 24 – 8 = (n – 1)(-4)
⇒ – 32 = (n – 1)(- 4)
⇒ (n – 1) = \(\frac {32}{4}\)
⇒ n – 1 = 8
⇒ n = 8 + 1
⇒ n = 9
∴ Number of terms n = 9

Question 3.
Seven times the 7^th term of a series is equal to eleven times of its 11th term. Find the 18^th term of the series. (NCERT)
Solution:
Let first term = a and common difference = d of A.P.
∴ 7^th term = a + 6d and 11^th term = a + 10 d.
∴ According to question,
7(a + 6d) = 11 + (a + 10d)
⇒ 7a + 42d = 11a + 110d
⇒ 7a – 11a = 110d – 42d
⇒ – 4a = 68 d
⇒ a = – 17d
Hence 18^th term = a + 17d
= – 17d + 17d [∵ a = – 17d]
= 0.

Question 4.
Prove that the sum of (m + n)^th term and (m – n)^th terms of an A.P. is twice of its m^th term.
Solution:
Let the first term = a and common difference = d.
T_n = a+(n – 1)d
T_{m + n}= a + (m + n – 1)d … (1)
T_{m – n} = a + (m – n – 1)d … (2)
T_m = a + (m – 1)d … (3)
T_m+n + T_m-n = a + (m + n – 1)d + a(m – n – 1)d
= 2a + (m + n – 1 + m – n – 1)d
= 2a + (2m – 2)d
= 2a + 2(m – 1)d
= 2[a+(m – 1)d]
T_m+n + T_m-n = 2T_m.

Question 5.
Insert three Arithmetic means between 3 and 19. (NCERT)
Solution:
Let three Arithmetic mean be A₁, A₂, A₃,
then 3, A₁, A₂, A₃, 19 are in A.P.
∴ 3 = a, 19 = T₅, let common difference = d
T₅ = a + 4d
T₅ = a + 4d
⇒ 19 = 3 + 4d
⇒ 16 = 4d
⇒ d = 4
Hence A₁ = 3 + 4 = 7, A₂ = 7 + 4 = 11, A₃ = 11 + 4 = 15.

Question 6.
If – 8, A₁, A₂ are in Arithmetic progression (A.P.), then find the value of A₁, A₂. (NCERT)
Solution:
– 8, A₁, A₂, 9 are in A.P.
∴ a = – 8, T₄ = 9, let common difference = d
T₄ = a + 3d
⇒ 9 = – 8 + 3 d
⇒ 3d = 11
⇒ d = \(\frac {17}{3}\)

Instruction :
Write the first five terms of each of the sequence and obtain the corresponding series.

Question 7.
(a) a₁ = 3, a_n = 3a_{n – 1} + 2, where n > 1. (NCERT)
Solution:
Given : a₁ = 3,
a₂ = 3a_{n – 1} + 2, where n > 1
a₂ = 3a_{2 – 1} + 2
⇒ a₂ = 3a₁ + 2
⇒ a₂ = 3 x 3 + 2 = 9 + 2 = 11
a₃ = 3a₁ + 2
= 3a₂ + 2
⇒ a₃ = 3(11)+ 2 = 33+ 2 = 35
a₄ = 3a_{4 – 1} + 2
= 3a₃ + 2
⇒ a₄ = 3(35) + 2 = 105 + 2 = 107
a₅ = 3a_{5 – 1} + 2
= 3a₄ + 2
= 3(107) + 2 = 321 + 2 = 323
Hence series is 3, 11, 35, 107, 323.

Question 7.
(b) a₁ = -1, a_n = \(\frac { { a }_{ n-1 } }{ n }\) where n ≥ 2.
Solution:

Question 8.
A person pays first instalment of Rs. 100 towards his loan. If he increases his instalment every month by Rs. 5, then what will be his 30^th instalment.
Solution:
Given : a = Rs. 100, d= Rs. 5, n = 30,
T_n = a + (n – 1)d
Amount of 30^th instalment
J₃₀ = 100 + (30 – 1) x 5
= 100 + 29 x 5 = 100 + 145
= Rs. 245.
Hence the 30^th instalment = Rs. 245.

Question 9.
Which term of the sequence \(\sqrt { 3 }\), 3, 3\(\sqrt { 3 }\) ……. is 729?
Solution:

Question 10.
How many terms are required in GP. 3,3²,3³ …………, so that their sum would be 120? (NCERT)
Solution:
Given : a = 3, r = \(\frac {9}{3}\) = 3, S_n = 120,
S_n = \(\frac { a({ r }^{ n }-1) }{ r-1 }\)
⇒ \(\frac { 3({ 3 }^{ n }-1) }{ 3-1 }\) = 120
⇒ 3ⁿ – 1 = \(\frac {120 × 2}{3}\)
⇒ 3ⁿ – 1 = 80
⇒ 3ⁿ = 81
⇒ 3ⁿ = 3⁴
⇒ n = 4

Question 11.
Show that ratio between sum of n terms and sum of (n +1)^th term to (2n)^th terms in a G.P. is \(\frac { 1 }{ { r }^{ n } }\)
Solution:
Let 1^st term of G.P. = a and common ratio = r
∴ n terms of GP. is a, ar, ar², ………….. , ar^{n – 1}
Let the sum to n terms be S₁
S₁ = \(\frac { a({ r }^{ n }-1) }{ r-1 }\)
GP. from (n + 1)^th term to (2n)^th term
arⁿ, ar^{n + 1}, …………. , ar^{2n – 1}
Let the sum of this G.P. up to n terms be S₂

Question 12.
If A.M. and GM. of roots of a quadratic equation are 8 and 5 respectively, then find the quadratic equation. (NCERT)
Solution:
Let the roots of equation be α and β
A.M. of roots = \(\frac {α + β}{2}\) = 8
⇒ α +β = 16
G.M. of roots = \(\sqrt {αβ }\) = 5
⇒ αβ = 25
If α, β are the roots of equation
then, x² – (α + β)x + αβ = 0
⇒ x² -16x + 25 = 0.

Question 13.
If the first and n^th term of GP. are a and b respectively and if p is the product of n terms, then prove that
p² = (ab)ⁿ. (NCERT)
Solution:
Let the common ratio of G.P. = r
Given: ar^{n – 1} = b, …(1)
then a.ar.ar²……. ar^{n – 1} = p
⇒ aⁿr^{1+2+3+……+(n – 1)} = p
⇒ aⁿr^{\(\frac {1}{2}\)(n – 1)n} = p
⇒ a²ⁿ.r^{(n – 1)n} = p²
⇒ p² = (a²r^{n – 1n}
⇒ p² = (a.ar^{n – 1})ⁿ
⇒ p² = (ab)ⁿ, [fromeqn. (1)]

Question 14.
If the 4^th term of a GP. is square of its second term and the 1^st term is -3, then find the 7^th term. (NCERT)
Solution:
Let a and r be the 1st term and common ratio of GP.
∴ T_n = ar^n-1
T₄ = ar^4-1 = ar³ … (1)
T₂ = ar^2-1 = ar … (2)
Given : T₄ = (T₂)²
⇒ ar³ = (ar)²
⇒ ar³ = a²r²
⇒ a = r = – 3 (given, a = -3)
T₇ = ar^7-1 = ar⁶
= (-3)(-3)⁶ = (-3)⁷
∴ T₇ = – 2187
Hence 7_th term of G.P. = – 2187.

Question 15.
Find the value of \(\sum _{ k=1 }^{ 11 }{ (2+{ 3 }^{ k }) }\).
Solution:

Sequences and Series Long Answer Type Questions

Question 1.
If m times of the m^th term of an A.P. is equal to n times the n^th term, then prove that (m + n)^th term of the series is zero.
Solution:
Let the 1^st term = a and common difference = d.
then t_m = a + (m – 1)d
t_n = a + (n – 1)d
Given: m[a + (n – 1)d] = n[a + (n – 1)d]
⇒ ma + m(m – 1 )d = na + n(n – 1 )d
⇒ (m² – m)d + (m – n)a = (n² – n)d
⇒ (m² – m – n² + n) d + (m – n)a = 0
⇒ [m² – n² – (m – n)] d + (m – n)a = 0
⇒ [(m – n)(m + n) – (m – n)] d+(m – n)a = 0
⇒ (m – n)[(m + n) – 1] d + (m – n)a = 0
⇒ a + (m + n – 1 )d = 0
⇒ t_m+n=0.

Question 2.
If the 6^th term of A.P. is 12 and 9^th term is 27, then find its r^th term.
Solution:
Let the 1^st term = a and common difference = d.
Given : t₆ = 12
⇒ a + 5d = 12 … (1)
t₉ = 27
⇒ a + 8d = 27 … (2)

Question 3.
If p^th term of an A.P. is \(\frac {1}{q}\) and q^th term is \(\frac {1}{p}\), then prove that (pq)^th
Solution:

Question 4.
If p^th term of an A.P. is \(\frac {1}{q}\) and q^th term is \(\frac {1}{p}\), then prove that (pq)^th terms is \(\frac {1}{2}\) (pq + 1), where p ≠ q.
Solution:

Question 5.
The sum of the series 25, 22, 19, ………….. of A.P. is 116, then find its last term. (NCERT)
Solution:
Given : a = 25, d = 22 – 25 = – 3, S_n = 116
S_n = \(\frac {n}{2}\) [2a + (n – 1)d]
⇒ 116 = \(\frac {n}{2}\)[2 x 25 + (n -1) x (-3)]
⇒ 232 = n[50 – 3n + 3]
⇒ 232 = n[53 – 3n]
⇒ 232 = 53n – 3n²
⇒ 3n² – 53n +232 = 0
⇒ 3n² – 24n – 29n + 232 = 0
⇒ 3n(n – 8) – 29(n – 8) = 0
⇒ (n – 8)(3n – 29) = 0
n = 8, or n = \(\frac {29}{3}\), (which is impossible)
∴ Last term l = a + (n – 1 )d
= 25 + (8 – 1) x (- 3)
⇒ l = 25 – 21 = 4.

Question 6.
If the A.M. of a and b is \(\frac { { a }^{ n }+{ b }^{ n } }{ { a }^{ n – 1 }+{ b }^{ n – 1 } }\), then find the value of it. (NCERT)
Solution:
A.M. of a and b = \(\frac {a+b}{2}\)
According to question,
\(\frac { { a }^{ n }+{ b }^{ n } }{ { a }^{ n – 1 }+{ b }^{ n – 1 } }\) = \(\frac {a+b}{2}\)
⇒ 2aⁿ + 2bⁿ = (a + b) (a^{n – 1} + b^{n – 1})
⇒ 2aⁿ +2bⁿ = aⁿ + bⁿ + ab^{n – 1} + ba^{n – 1}
⇒ aⁿ + bⁿ = ab^{n – 1} + ba^{n – 1}
⇒ aⁿ – ba^{n – 1} = ab^{n – 1} – bⁿ
⇒ a^{n – 1}(a – b) = b^{n – 1}(a – b)
⇒ a^{n – 1} = b^{n – 1}
⇒ \(\frac {a}{b}\)^{n – 1} = \(\frac {a}{b}\)⁰
⇒ n – 1 = 0
⇒ n = 1.

Question 7.
Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7^th and (m – 1)^th mean is 5 : 9 then, find the value of m.
Solution:
Let A₁, A₂, A₃, ……… ,A_m are m A.M. respectively inserted between 1 and 31.
Then 1, A₁, A₂, A₃, ……… A_m 31 are in A.P. Here, 1^st term = 1 and (m + 2)^th term = 31 and let the common difference of sequence be d.

Question 8.
Show that the sum of (m + n)^th and (m – n)^th term of an A.P. is equal to twice the m^th term. (NCERT)
Solution:
Let the first term = a and common difference = d.
T_n = a + (n – 1)d
T_{m + n} =a +(m + n – 1)d … (1)
T_{m – n} = a + (m – n – 1)d … (2)
T_m = a + (m – 1)d … (3)
Adding equation (1) and (2),
T_{m + n} + T_{m – n} = a + (m + n – 1)d + a + (m – n – 1)d
= 2a + (m + n – 1 + m – n – 1)d
= 2a + (2m – 2 )d
= 2a + 2 (m – 1)d
= 2[a + (m – 1)d]
∴ T_{m + n} + T_{m – n} = 2T_m [From equation (3)]

Question 9.
If the sum of three numbers in A.P. is 24 and their product is 440, then find the numbers? (NCERT)
Solution:
Let the three numbers of A.P. are a – d, a, a + d.
Given : a – d + a + a + d = 24
3a = 24
⇒ a = 8
and (a – d) × a × (a + d) = 440
a(a² – d²) = 440
⇒ 8(64 – d²) = 440
⇒ 64 – d² = \(\frac {440}{8}\)
⇒ 64 – d² = 55
⇒ d² = 64 – 55
⇒ d² = 9
⇒ d = ± 3
When a = 8 and d = 3
Then, a – d = 8 – 3 = 5, a = 8, a + d= 8 + 3 = 11
When a = 8 and d = – 3
Then, a – d = 8 + 3 = 11, a = 8, a + d = 8 – 3 = 5
Hence required numbers are 5, 8, 11 or 11, 8, 5.

Question 10.
If the sum of n, 2n and 3n terms in A.P. are S₁ S₂ and S₃, then show that S₃ = 3(S₂ – S₁).
Solution:

Question 11.
Find the sum of all natural numbers lying between 200 and 400, which are divisible by 7. (NCERT)
Solution:
Numbers divisible by 7 are 203, 210, 217, …………. , 399
Here a = 203, d = 210 – 203 = 7, l = 399
l = a + (n – 1)d
399 = 203 + (n – 1) x 7
⇒ 399 – 203 = (n – 1) x 7
⇒ (n – 1)7 = 196
⇒ (n – 1) = \(\frac {196}{7}\)
⇒ (n – 1) = 28
∴ n = 29.
Hence required sum, S₂₉ = \(\frac {n}{2}\)[a+l]
⇒ S₂₉ = \(\frac {29}{2}\)[203 + 399]
⇒ S₂₉ = \(\frac {29}{2}\)[602] = \(\frac {17458}{2}\) = 8729.

Question 12.
If the 5^th, 8^th and 11^th terms of a GP. are p, q and s respectively, then show
that q² = ps. (NCERT)
Solution:
Let the 1st term of GP. = a and common ratio = r.
Then, T_n = ar^{n – 1}
T₅ = ar^{5 – 1} =p
⇒ ar⁴ = P … (1)
T₈ = ar^8-1 = q
⇒ ar⁷ = q … (2)
T₁₁ = ar^{11 – 1} = s
⇒ ar¹⁰ = s
ps = ar⁴.ar¹⁰, [from equation (1) And (3)]
⇒ ps = a²r¹⁴
⇒ ps = (ar⁷)²
⇒ ps = q², [from equation (2)]
∴ q² = ps.

Question 13.
If the first term of G.P. a = 729 and 7^th term is 64, then find S₇. (NCERT)
Solution:

Question 14.
If the 4^th terms of a GP is square of its 2^nd term and first term is – 3, then find its 7^th (NCERT)
Solution:
Let the 1st term of GP. = a and common ratio = r.
Then, T_n = ar^{n – 1}
T₄ = ar^{4 – 1} = ar³ … (1)
T₂ = ar^2-1 = ar … (2)
Given : T₄ = T₂²
ar³ = (ar)₂
ar³ = a²r²
⇒ ar³ = a²r²
⇒ a = r = – 3 (given a = – 3)
T₇ = ar^{7 – 1} = ar⁶
= (-3)(-3)⁶ = (-3)⁷
∴ T₇ = – 2187
Hence, 7^thterm of G. P. = – 2187.

Question 15.
Find the sum of the sequence 8,88,888,8888 ……….. up to n terms. (NCERT)
Solution:
Let the sum of the n terms be S

Question 16.
Find the sunt of the numbers 7,77,777, 7777 up to n terms. (NCERT)
Solution:
Let the sum of n terms is S.

Question 17.
If the A.M. and GM. between two positive numbers a and b are 10 and 8 respectively, then find the numbers. (NCERT)
Solution:
A.M. A = \(\frac {a+b}{2}\)
a + b = 20 … (1)
G.M. G = \(\sqrt {ab}\)
ab = 64 … (2)
(a – b)² = (a + b)² – 4ab
= (20)² – 4 × 64, [From equation (2)]
= 400 – 256
(a – b)² = 144 = (12)² … (3)

Put a = 16 in equation (1), we get,
16 + b = 20
∴ b = 4
Numbers are 16 and 4.
When a – b = – 12, then
a + b = 20
a – b = – 12
On adding 2a = 8
⇒ a = 4
Put a = 4 in equation (1), we get,
4 + 6 = 20
∴ b = 16
Numbers are 4 and 16.
Hence numbers a and b are 4, 16 or 16,4.

Question 18.
The sum of two numbers is 6 times their geometric mean, show that the . numbers are in the ratio (3 + 2\(\sqrt {2}\)) ; (3 – 2\(\sqrt {2}\)). (NCERT)
Solution:
Let the numbers are a and 6.
Given: a + b = 6 \(\sqrt {ab}\)
\(\frac { a+b }{ 2\sqrt { ab } }\) = \(\frac {3}{1}\)
⇒ a + b = 3K …. (1)
and 2 \(\sqrt {ab}\) = K ⇒ 4ab = K²
(a – b)² = (a + b)² – 4ab
= (3K)² – (K)²
= 9K² – K² = 8K²
⇒ a – b = 2\(\sqrt {2}\)K …. (2)

Sequences and Series Very Long Answer Type Questions

Question 1.
If the ratio of the sum of n terms of two A.P. is 5n + 4 : 9n + 6, then find the ratio of their 18^th term.
Solution:
Let the two A.P. are :
a, a + d, a + 2d, ………….
and A, A + D, A +2D …………..

Question 2.
The ratio of the sum of m and n terms of an A.P. is m² : n². Show that the ratio of and u* term is (2m – 1) : (2n – 1). (NCERT)
Solution:
Let the A.P. are a, a + d, a + 2d, ……………
∴ m^th term of A.P. Tm = a + (m – 1)d
n^th term of A.P. Tn= a + (n – 1)d

Question 3.
If the sum of first three terms of a G.P. is \(\frac {39}{10}\) and their product is 1, then find the common ratio and the terms. (NCERT)
Solution:
Let the three terms of G.P. are \(\frac {a}{r}\), a, ar.
Given: \(\frac {a}{r}\) × a × ar = 1
⇒ a³ = 1 ⇒ a = 1
and \(\frac {a}{r}\) + a + ar = \(\frac {39}{10}\)
⇒ a(\(\frac {a}{r}\) + r + 1) = \(\frac {39}{10}\)
⇒ 1 x \(\frac { 1+{ r }^{ 2 }+r }{ r }\) = \(\frac {39}{10}\)
⇒ 10r² + 10r + 10 = 39r
⇒ 10r² – 29r + 10 = 0
⇒ 10r² – 25r – 4r + 10 = 0
⇒ 5r(2r – 5) – 2(2r – 5) = 0
⇒ (5r – 2)(2r – 5) = 0
∴ r = \(\frac {2}{5}\) and r = \(\frac {5}{2}\)
When a = 1 and r = \(\frac {2}{5}\)

Question 4.
Find four numbers forming a G.P. in which the third term is greater than the lint term by 9 and the second term is greater than 4^th term by 18. (NCERT)
Solution:
Let the four terms of G.P. be a, ar, ar²,ar³.
Given: T₃ = T₁ + 9
⇒ T₃ = T₁ + 9
⇒ T₃ = a + 9
⇒ ar² = a + 9 …. (1)
According to question,
T₂ = T₄ + 18
⇒ ar = a³ + 18
⇒ ar – ar³ = 18 …. (2)

Put r = – 2 in equation (1), we get
a(- 2)² – a = 9
⇒ 4a – a = 9
⇒ 3a = 9
⇒ a = 3
∴ Numbers are 3, 3(- 2), 3(- 2)², 3(-2)³, ……………..
3, – 6, 12, – 24, ………………

Question 5.
If S be the sum of it terms of a GP., P be the product and R be the sum of reciprocal of n terms, then prove that
P²Rⁿ = Sⁿ. (NCERT)
Solution:
Let n terms of GP. be a, ar, ar², …………… ar^{n – 1}.
According to the question,

Question 6.
If x = 1 + a + a² + ………….∞ (\(\left| a \right|\)<1)
y = 1 + b + b² + …………….∞ (\(\left| b \right|\)<1)
then prove that
1 + ab + a²b² + …………….∞ = \(\frac {xy}{x + y – 1}\)
Solution:

Question 7.
The sum of infinite terms of a Geometric Progression is 15 and sum of the square of its terms is 45. Find the Geometric Progression. (NCERT)
Solution:
Let a be the first term and r be the common ratio.
∵ \(\left| r \right|\)<1,
Then, \(\frac {a}{1 – r}\) = 1.5 …. (1)
Squaring the terms of GP. the new G.P. is
a², a² r², a² r⁴, a² r⁶, ……….
Sum of infinity of G.P. = \(\frac { { a }^{ 2 } }{ { 1 – r }^{ 2 } }\).
\(\frac { { a }^{ 2 } }{ { 1 – r }^{ 2 } }\) = 45, (given) … (2)
Squaring both sides of equation (1) and the result is divided by equation (2),
\(\frac { { a }^{ 2 } }{ { 1 – r }^{ 2 } }\)² × \(\frac { 1-{ r }^{ 2 } }{ { a }^{ 2 } }\) = \(\frac {15×15}{45}\)
⇒ \(\frac {1 + r}{1 – r}\) = 5
⇒ 1 + r = 5 – 5r
⇒ 6r = 4
⇒ r = \(\frac {2}{3}\)
Put r =\(\frac {2}{3}\) in equation (1), we get

Hence required progression is 5, 5 × \(\frac {2}{3}\), 5 × (\(\frac {2}{3}\))², …………….
Hence 5, \(\frac {10}{3}\), \(\frac {20}{9}\), ………………

Question 8.
A farmer buys a used tractor of Rs. 12,000. He pays Rs. 6,000 cash and agrees to pay the balance in annual instalment of Rs. 500 plus 12% interest on theunpaid amount How much the tractor cost him?
Solution:
Cost of tractor = Rs. 12, 000, down payment = Rs. 6,000
Balance amount = 12,000 – 6,000 = Rs. 6,000

Actual cost of tractor = 12,000 + 4,680 = Rs. 16,680. Ans.

Question 9.
Shamshad Ali buys a scooter for Rs. 22,000. He pays Rs. 4,000 cash and agrees to pay the balance in annual instalment of Rs. 1,000 plus 10% interest on the unpaid amount How much will scooter cost him? (NCERT)
Solution:
Cost of scooter = Rs. 22,000, Cash down payment = Rs. 4,000.
Remaining amount = 22,000 – 4,000 = Rs. 18,000

Actual cost = 22,000 + 17,100 = Rs. 39,100.
Total amount paid for scooter = Rs. 39,100.

Question 10.
A person writes a letter to four of his friends. He asks each one of them, to copy the letter and mail to four different persons with instructions that they move the chain similarly. Assuming that the chain is not broken and that is costs 50 paisa to mail one letter. Find the amount on postage when 8^th set of letter is mailed. (NCERT)
Solution:
First person sends 4 letters.
2^nd step, he sends 4 x 4= 16 letters
3^rd step, he sends 4 x 4 x 4 = 64 letters
Hence 4, 16, 64, 256, …………… is a Geometric series.
Here a = 4, r = \(\frac {16}{4}\) = 4
S_n = \(\frac { { a(r }^{ n } – 1) }{ r – 1 }\) [∵r>1]
∴ Total number of letters till 8^th set = S₈ = \(\frac { { 4(r }^{ 4 } – 1) }{ 4 – 1 }\)
= \(\frac {4}{3}\) (65536 – 1) = \(\frac {4}{3}\) x 65535
Cost for one letter = Rs. 0.50
∴ Hence total cost = \(\frac {4}{3}\) x 65535 x 0.50 = Rs. 43690.

Question 11.
150 workers were engaged to finish a job in a certain number of days, 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed. (NCERT)
Solution:
150, 146, 142, 138, …………, it is a Geometric series.
Let the number of days required complete the work be n.

If the workers not dropped then the work would have complited in (n – 8) days with 150 workers working on each day.
Hence the total workers worked for n days = 150 (n – 8)
= 150n – 1200 ….. (2)
From equation (1) and (2),
150n – 1200 = 152n – 2n₂
2n₂ + 150n – 152n – 1200 = 0
2n₂ – 2n – 1200 = 0
n₂ – n – 600 =0
n₂ – 25n + 24n – 600 = 0
n(n – 25) + 24(n – 25) = 0
(n – 25)(n + 24) = 0
n =25 and n = – 24 (not possible)
∴ Work is completed in 25 days.

MP Board Class 11th Maths Important Questions

MP Board Class 11th Chemistry Important Questions Chapter 14 Environmental Chemistry

Environmental Chemistry Important Questions

Environmental Chemistry Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
Source of non – pollution energy:
(a) Fossil fuel
(b) Sun
(c) Gasoline
(d) Nuclear energy
Answer:
(b) Sun

Question 2.
Which radiations manufacture O₃:
(a) Ultra – violet
(b) Visible
(c) Infra – red
(d) Radio waves
Answer:
(a) Ultra – violet

Question 3.
Which radiations provide Green house effect:
(a) Infra – red
(b) Visible
(c) Ultra – violet
(d) X – rays
Answer:
(c) Ultra – violet

Question 4.
PAN is responsible for:
(a) Depletion of ozone layer
(b) For smog
(c) For Acid rain
(d) Poisonous food
Answer:
(b) For smog

Question 5.
Which is not an air pollutant:
(a) H₂
(b) H₂S
(c) NO₂
(d) O₃
Answer:
(a) H₂

Question 6.
Air pollutant released from jet aeroplanes in the form of an air pollutant:
(a) Photochemical oxidant
(b) Photochemical reductant
(c) Aerosol
(d) Physical pollutant
Answer:
(c) Aerosol

Question 7.
Is not present in Acid rain:
(a) H₂SO₄
(b) HNO₃
(c) H₂SO₃
(d) CH₃COOH
Answer:
(d) CH₃COOH

Question 8.
Is responsible for disease of lungs:
(a) O₂
(b) N₂
(c) CO₂
(d) SO₂
Answer:
(d) SO₂

Question 9.
O₃ is manufactured in:
(a) Troposphere
(b) Stratosphere
(c) Mesosphere
(d) Thermosphere
Answer:
(b) Stratosphere

Question 10.
For acid rain ‘sink’ is:
(a) Leaves
(b) Reservoir
(c) Lime stone
(d) CO₂
Answer:
(c) Lime stone

Question 11.
Primary pollutant is:
(a) SO₃
(b) NO₂
(c) N₂O
(d) NO
Answer:
(d) NO

Question 12.
Most dangerous is:
(a) Smoke
(b) Dust
(c) Smog
(d) NO
Answer:
(c) Smog

Question 2.
Fill in the blanks:

1. The air pollutant released by jet aeroplanes in the form of fluro carbon is …………………………..
2. D.D.T. is …………………….. poisonous pollutant as compared to B.H.C.
3. O₃ is formed in the ……………………… level of atmosphere.
4. A definite tolerable level of pollutants in the environment is expressed by …………………………….
5. Maximum air pollutants are present in …………………………….. level of the atmosphere.
6. Ozone layer prevent us from …………………………….. rays.
7. Oxides of …………………………. and …………………………….. cause acid rain.
8. …………………………… is the main cause of ozone layer depletion.
9. ………………………….. gas is responsible for Green house effect.
10. Substance produces pollution is known as ……………………………..
11. ……………………………… is responsible for lung diseases.
12. SO₂ pollutant is responsible for the disease of ……………………………..

Answer:

1. Aerosol
2. More
3. Stratosphere
4. T.L.V.
5. Troposphere
6. Ultra – violet
7. Nitrogen, sulphur
8. C.F.C
9. CO₂
10. Pollutant
11. Photochemical smog
12. Asthma

Question 3.
Answer in one word/sentence:

1. Name the person who started Chipko Aandolan for the conservation of forest?
2. Region of the atmosphere where living beings exist is known as?
3. What is the name of compound responsible for hole in ozone layer?
4. What is smoke containing fog known as?
5. What is rain water containing small amount of sulphuric acid and nitric acid known as?
6. Which chemistry produces environment friendly chemicals having minimum contribution in pollution?
7. What is decrease in density of ozone gas due to chlorofluorocarbon compound in atmosphere known as?
8. What is the maximum quantity of pollutant having no effect on receptor known as?
9. When do we celebrate World Environment Day?
10. What is PAN?
11. Name the largest sink of earth?
12. When does Bhopal Gas Tragedy occured?
13. Name the disease due to water pollution?

Answer:

1. Sunderlal Bahuguna
2. Troposphere
3. Chlorofluorocarbon
4. Smog
5. Acid rain
6. Green chemistry
7. Ozone hole
8. Threshold limit value
9. 5th June
10. Peroxyacetyl nitrite/ photochemical smog
11. Ocean
12. Midnight of 2nd and 3rd December 1984
13. Jaundice, Diarrhoea

Question 4.
Match the following:
[I]

Answer:

1. (e)
2. (c)
3. (b)
4. (d)
5. (a)

[II]

Answer:

1. (c)
2. (d)
3. (a)
4. (b)
5. (e)

[III]

Answer:

1. (b)
2. (a)
3. (d)
4. (c)

Environmental Chemistry Very Short Answer Type Questions

Question 1.
Write the definition of pollutants?
Answer:
Such substances whose amount in the environment increases more than required and cause harmful effect on human, animal and plant kingdom are called pollutants. Like: CO, CO₂, NO, NO₂, SO₂ etc.

Question 2.
Write names of two air pollutants?
Answer:
SO₂, SO₃.

Question 3.
Name two pollutants which depletes ozone layer?
Answer:

1. Cycle of nitric oxide (NO) and
2. C.F.C. (Chlorofluoro carbon), in which C.F.C. is main.

Question 4.
Ozone is found in?
Answer:
Stratosphere.

Question 5.
Write the two health problems caused by SO₂?
Answer:

1. SO₂ affects the respiratory canal and lungs due to which various health diseases are caused like cancer.
2. Acid rain caused due to SO₂ produces boils on the skin.

Question 6.
What is Acid rain?
Answer:
Various gaseous pollutant present in the atmosphere like SO₂, SO₃, NO₂, NO dissolve in rain drop.
These drops fall with rain and are called acid rain.
SO₂ + H₂O → ₂HSO₃
SO₃ + H₂O → H₂SO₄.

Question 7.
Name any two green house gases?
Answer:
C.F.C. and CO₂.

Question 8.
What is P.A.N.?
Answer:
It is peroxy acyl nitrate which is a photochemical smog.

Question 9.
What is C.F.C.?
Answer:
It is chlorofluoro carbon which is the main cause of depletion of ozone layer.

Question 10.
What is green chemistry?
Answer:
A technique to check pollution in which such chemical reactions are suggested which do not cause pollution and if pollution spreads then it can be destroyed. This is called green chemistry.

Question 11.
What is TLV?
Answer:
A definite value of pollutants can be tolerated. This is expressed by TLV. TLV means ‘Threshold Limit Value’.

Question 12.
What are particulate pollutants?
Answer:
The pollutants mixed up with air in liquid or solid state in such a manner that the remain suspended for a long time is called particulates.

Question 13.
What is specimasion?
Answer:
Many pollutants can be made from element. The method of determining which of the product is more dangerous is called specimasion.

Question 14.
What is sink?
Answer:
Sink is that in which the substance totally gets consumed and then also there is no effect on sink.

Question 15.
Name the biggest sink of the earth?
Answer:
The biggest sink of the earth is the sea.

Question 16.
What is green house effect?
Answer:
The heating of atmosphere due to absorption of infrared radiation by carbon dioxide and other gases is called green house effect.

Question 17.
Explain the mechanism of acid rain?
Answer:
All pollutant gases spread as pollutant particles in the form of smoke by the burning of fossil fuels and other fuels. Due to high temperature of industries and other engines, oxides of nitrogen spread in the atmosphere by the combination of N₂ and O₂. These gases mix in rain drops form acid and then fall on the earth as acid rain drops. This is called acid rain.

Question 18.
The gas which gets leak in 1984 in Bhopal is?
Answer:
CH₃ – N = C = O Methyl isocyanate.

Question 19.
What is the main component for the depletion of ozone layer?
Answer:
Chlorofluorocarbon.

Question 20.
Name the person who started “Chipko Andolan” for forest conservation?
Answer:
Sunderlal Bahuguna.

Question 21.
Write the name of disease caused by water pollution?
Answer:
Cholera, Typhoid, Joindiss etc.

Question 22.
Name the medicinal plant which is helpful in controlling pollution and useful in skin diseases?
Answer:
Neem (Azaderecta indica).

Question 23.
Which gas is responsible for green house effect?
Answer:
Carbon dioxide (CO₂).

Question 24.
What is the main sink of CO pollutant?
Answer:
Biological molecules present in soil.

Question 25.
Write the name of four methods useful in Green chemistry?
Answer:

1. Use of sunlight
2. Micro oven
3. Micro waves
4. Sound waves
5. Use of enzyme.

Question 26.
Which pollutant is responsible for smog?
Answer:
PAN (Peroxy acyl nitrate).

Question 27.
What is sink for acid rain?
Answer:
Lime stone (Marble).

Question 28.
Which sphere is present near to earth?
Answer:
Troposphere.

Question 29.
Full form of B.H.C. is?
Answer:
Benzene Hexa chloride.

Question 30.
Which light is responsible for skin cancer?
Answer:
Ultra – violet light (UV – light).

Question 31.
What is responsible for lung diseases?
Answer:
Sulphur dioxide (SO₂).

Question 32.
What is the main source of emmision of CO?
Answer:
Vehicles.

Question 33.
Oxides of which elements are responsible for acid rain?
Answer:
Oxides of nitrogen and sulphur.

Question 34.
Maximum pollution occurs in which sphere?
Answers:
Troposphere.

Question 35.
What is the main constituent for ozone depletion?
Answer:
C.F.C. (Chlorofluoro carbon).

Question 36.
Which radiation give green house effect?
Answer:
Infrared radiations (IR).

Question 37.
The formation of ozone takes place where in the atmosphere?
Answer:
Stratosphere.

Environmental Chemistry Short Answer Type Questions – I

Question 1.
Define Green Chemistry?
Answer:
Green chemistry is the branch of science in which study of effects of chemicals (like : Origin, transportation, reactions etc.) on environment is studied.

Question 2.
Explain Tropospheric pollution?
Answer:
Tropospheric pollution occurs due to unwanted solids and gas particles present in the air. The pollution occurs due to following two substances:

1. Gaseous air pollutants:
They are sulphur, nitrogen and CO₂, H₂S, hydrocarbon, ozone and other oxidising agents.

2. Particulates:
They are dust, fog, smoke etc.

Question 3.
Which gases are responsible for green house effect?
Answer:
Main gases are CO₂, methane, water vapour, nitrous oxide, chlorofluoro carbon (CFC) and ozone.

Question 4.
What do you mean by BOD (Biochemical oxygen demand)?
Answer:
The total amount of oxygen consumed by microorganism in decomposing the wastes present in a certain volume of a sample of water.

Question 5.
Due to green house effect temperature of the earth is increasing? Which substances are responsible for it?
Answer:
Green house gases like CO, methane, nitrous oxide, ozone and chlorofluoro carbons are responsible for green house effect.

Question 6.
Ozone is a toxic gas and is a strong oxidizing agent even then its presence in the stratosphere is very important Explain what would happen if ozone from this region completely removed?
Answer:
The ozone layer acts as a protective umbrella and does not allow the harmful UV radiations to reach the earth’s surface.(MPBoardSolutions.com) If ozone is completely removed from the stratosphere, the UV radiations will fall directly on the humans, causing skin cancer and on the plants affecting plant proteins.

Question 7.
What are the sources of dissolved oxygen in water?
Answer:

1. Photosynthesis
2. Natural aeration
3. Artificial aeration.

Question 8.
Dissolved Oxygen in water is very important for aquatic life. What processes are responsible for the reduction of dissolved oxygen in water?
Answer:
Dissolve oxygen is essential for sustaining animal and plant life in any aquatic system. The wastes such as domestic, industrial and biodegradable organic compounds are oxygen demanding wastes. These are decomposed by the bacterial population which in turn decreases the oxygen from water.

Question 9.
What are Biodegradable and non – biodegradable pollutants?
Answer:

• Biodegradable pollutants: They can be degrade by microorganisms.
• Example: Sewage, dungs of animals, fruits and vegetable peels etc.
• Non – biodegradable pollutants: They cannot degrade by microorganisms.
• Example: Mercury, Lead, DDT, glass, plastic etc.

Question 10.
What is pollution?
Answer:
Environmental pollution is the effect of undesirable changes in our surroundings that have harmful effects on plants, animals and human beings.

Question 11.
What is pollutant?
Answer:
A substance present in the environment in greater proportion than its natural abundance and resulting into harmful effects, is called a pollutant.

Question 12.
What are contaminants?
Answer:
Some substances which are not present in the environment, but are released in the environment as a result of chemical activities lead to pollution. Such substances are called contaminants. Example: Methyl isocyanate gas (CH₃NCO).

Question 13.
Write the chemical name of the gases depleting ozone?
Answer:
Nitric oxide, atmospheric oxygen and chlorofluoro carbon are responsible.

Question 14.
Which are green house gases?
Answer:
CO₂, ozone and water vapours are green house gases. They have the tendency to absorb IR radiations.

Question 15.
What is polluted air?
Answer:
If some underisable substances get added in the air which affects the health of the organisms, then such air is called polluted air.

Question 16.
Why there is ozone depletion over Antarctica?
Answer:
In the stratosphere compounds formed are converted back into chlorine free radical which deplete ozone layer.

Question 17.
What is the importance of BOD measurement of any water sample?
Answer:
BOD is the measurement of pollution caused by organic biodegradable substances in water sample. The less value of BOD tells that less amount of organic effluent is present in water.

Question 18.
Oxidation of sulphur dioxide into sulphur trioxide in the absence of a catalyst is a slow process but this oxidation occurs easily in the atmosphere. Explain how does this happen. Give chemical reaction for the conversion of SO₂ into SO₃
Answer:
The oxidation of sulphur dioxide into sulphur trioxide can occur both photochemically or non – photochemically. In the near ultraviolet region, the SO₂ molecules react with ozone photochemically.
SO₂ + O₃ \(\underrightarrow { h\nu } \) SO₃ + O₂
2SO₂ + O₂ \(\underrightarrow { h\nu } \) 2SO₃
Non – photochemically, SO₂ may be oxidised by molecular oxygen in presence of dust and soot particles.
2SO₂ + O₂ \(\underrightarrow { Particulates } \) 2SO₃

Question 19.
How is ozone formed in stratosphere?
Answer:
Ozone in the stratosphere is a product of (UV) radiations acting on dioxygen (O₂) molecules. The (UV) radiations split apart molecular oxygen into free oxygen (O) atoms. These oxygen atoms combine with the molecular oxygen to form ozone.

Question 20.
What is chlorosis?
Answer:
Chlorophyll in plants is formed slowly. This is due to presence of SO. This pollution is called chlorosis.

Question 21.
What is Metathesis?
Answer:
Metathesis is the name of that science, in which the study of application of chemical methods for general persons is studied.

Question 22.
What are primary and secondary air pollutants?
Answer:
Primary air pollutants are those which remain as such after their formulation e.g. NO, while secondary pollutants are formed as a result of chemical interaction between primary air pollutants. Example: PAN.

Question 23.
What is photochemical smog?
Answer:
Photochemical smog:
It is formed by photochemical reactions involving solar radiations. The principal constituents are O₃, NO₂ and some photochemical oxidants. It is also called Los Angeles smog.

Question 24.
What is acid rain?
Answer:
Acid rain:
It is the rain water containing sulphuric acid, nitric acid and small amount of hydrochloric acid which are formed from the oxides of sulphur and nitrogen present in the air as pollutants and has a pH of 4 – 5.
CO₂ + H₂O → H₂CO₃
SO₃ + H₂O → H₂SO₄

Question 25.
When CO₂ is called harmful gas?
Answer:
Normal amount of CO₂ in atmosphere is not harmful. Whereas organisms and plants prepare their food with the help of it. But, when the amount of CO₂ increases due to various process then it alters the environmental balance and become harmful.

Question 26.
What is the role of CO₂ in the “Green house effect”?
Answer:
Water vapours are present only near the earth atmosphere but ozone found very far from the earth and CO₂ is found everywhere in the atmosphere. (MPBoardSolutions.com) So for green house effect, CO₂ is more responsible because CO₂ has the tendency to absorb IR radiations. Due to this green house effect produced.

Question 27.
Why acid rain is harmful for Tajmahai?
Answer:
Tajmahai is made up of marbles (CaCO₃). Acid rain contains H₂SO₄ in dilute state. It reacts with marble of Tajmahai and make it discoloured and lustreless.

Question 28.
What is Fly ash pollution?
Answer:
The smaller ash particles are formed when fossil fuel is burnt. Gases produced during burning take the ash particles and pollute the atmosphere. The pollution is called fly ash pollution.

Question 29.
What is contaminated water? Name of diseases occurs due to it?
Answer:
The water which contains dissolved organic matters and salts and micro – organisms is called contaminated water.
Diseases are:

1. Diarrhoea
2. Typhoid
3. Skin diseases etc.

Question 30.
What is Global warming?
Answer:
This increase in average temperature of global air due to increased green house effect is called ‘global warming’.

Question 31.
What is Ionosphere?
Answer:
It is also 40 km thick layer at an altitude of 50 km from the earth surface. It is also called mesosphere. Various ionic reactions taking place in ionosphere is:
O₂ + \(\overset { \bullet }{ O } \) → O₂⁺ + O
O⁺ + N₂ → NO⁺ + N
N₂⁺ + O₂ → N₂ + O₂⁺

Environmental Chemistry Short Answer Type Questions – II

Question 1.
How is the poisonous effect of CO produced on man and animals?
Answer:
CO has poisonous effect on man and plants.
1. It combines with haemoglobin of the blood more strongly than oxygen.
CO + Hb → CO – Hb (Carboxy haemoglobin).

As a result of this amount of haemoglobin available in the blood for the transport of oxygen to the body cells decreases. The normal metabolism is thus, impaired due to less O₂ level. This will cause suffocation and will ultimately lead to death. (MPBoardSolutions.com) Carbon monoxide if present in air can cause mental impairment, respiratory problems, muscular weakness and dizziness.

2. A high concentration of CO (100 ppm or more) will harmfully affect the plants causing leaf drop; reduction of leaf size and premature aging etc.

Question 2.
Write the harmful effects of SO₂ (Sulphur dioxide)?
Answer:
Harmful effects of SO₂ are:

1. SO, affects respiratory tract producing nose, eye and lung irritation. It has been reported that lower concentration of SO₂ causes respiratory weakness.
If present at a concentration of only 2.5 ppm in the environment, then also it leads to dangerous diseases like bronchitis and lung cancer etc.

2. SO₃ produce harmful effect on buildings made of marble and lime stones (CaCO₃). The gas released from Mathura oil refinery is harmful for the Tajmahal.

3. High concentration of SO₂, leads to stiffness of flower buds which eventually fall off from plants.

4. Air polluted by oxides of sulphur enhances the corrosion of metals like copper, zinc, iron etc.

Question 3.
Write the harmful effects of nitrogen dioxide?
Answer:
Nitrogen dioxide is harmful and poisonous. It produces the following harmful effects:

1. It reacts with the ozone present in the atmosphere and decreases its density.
2. Oxides of nitrogen are responsible for the production of photochemical smog.
3. Oxides of nitrogen cause harmful effects on textile fibres like nylon, rayon, cotton fibres etc. NO2 causes cracks in rubber.
4. Increase in concentration of NO₂ in the atmosphere is harmful for plants. It leads to leaf spotting, retards photosynthetic activity, retards plant growth etc.
5. NO₂ creates problems in human respiration and leads to bronchitis.

Question 4.
A farmer was using pesticides on his farm. He used the product of his farm as food for rearing fishes. He was told that fishes were not fit for human consumption because large amount of pesticides had accumulated in the tissues of Fishes. Explain, how did this happen?
Answer:
Pesticides are organic compounds which are used to protect plants from pests. These are mild poisons. These pesticides stick to the plants and also flow into lakes along with the rain water. (MPBoardSolutions.com) Rearing fishes when consume these plants as their food the poisonous pesticides accumulate in the tissues of fishes. Thus, these fishes are not fit for human consumption.

Question 5.
For dry cleaning, in the place of tetrachloroethene, liquefied carbon dioxide with suitable detergent is an alternative solvent. What type of harm to the environment will be prevented by stopping use of tetrachloroethane? Will use of liquefied carbon dioxide with detergent be completely safe from the point of view of pollution? Explain?
Answer:
1. Tetrachloroethene (Cl₂C = CCl₂) is suspected to be carcinogenic and contaminates the ground water. This harmful effect will be prevented by using liquefied CO₂ along with suitable detergent.

2. Use of liquefied CO₂ along with detergent will not be completely safe because detergents also cause pollution as most of the detergents are non – biodegradable. Also, liquefied CO₂ will ultimately enter into the atmosphere and contribute to the green house effect.

Question 6.
What are the harmful effect of Green house effect?
Answer:
Though green house effect was beneficial in maintaining a livable temperature on earth. But excessive CO₂ in the atmosphere due to deforestation and large scale burning of fossil fuel has disturbed the natural balance in favour of higher green house effect. This has led to an increase in average temperature of the earth from 0.3 to 0.6°C over the past century. This increase in average temperature of global air due to increased green house effect is called ‘Global warming’.

The atmospheric CO₂ level is expected to become double sometimes between 2050 – 2150 with a corresponding increase in global temperature from 1 to 3°C. Besides CO₂, other green house gases are methane, water vapour, nitrous oxides, CFCs (Chlorofluorocarbons) and ozone. Methane is produced naturally when vegetation is burnt, digested or rotted in the absence of oxygen. (MPBoardSolutions.com) Large amount of methane are released in paddy fields, coal mines, from rotting garbage dumps and by fossil fuels.

CFCs are man made industrial chemicals used in air conditioning etc. CFCs are also damaging the ozone layer. Nitrous oxide occurs naturally in the environment. In recent years, their quantities have increased significantly due to use of chemicals, fertilizers and the burning of fossil fuels.

Harmful Effects of Global Warming:

• There will be rise in sea level due to increased rate of melting of glaciers. Sea level may rise by 0.5 to 1.5 m during the next 50 to 100 years if present rise in CO₂ continues. This will result in flood and loss of soil particularly in coastal areas.
• Higher global temperature is likely to effect the whole ecosystem by disturbing the life cycle of certain micro and macro organisms.
• Higher temperature is likely to increase incidence of infectious diseases such as malaria, dengue, yellow fever and sleeping sickness.

Question 7.
Write the reasons of water pollution?
Answer:
The sources of water pollution are as follows:

1. Organic pollutants like manure wastes from food processing, rags, paper discards etc.
2. Industrial wastes.
3. Detergents and Fertilizers: The detergents are best available mode for the growth of bacteria.
4. Pollution of water takes place through acids.

Question 8.
How is artificial green house prepared?
Answer:
Synthetic green house:
In nature, coating of CO₂ is forming green house, but synthetic green house can be synthesized by studying its mechanism.

Actually the transparent glass roof and wall of the glass house allow sun rays to pass through and strike the surface of the house. The reflected radiation is of longer wavelength than the incident radiation. A significant portion of reflected radiation absorbed by glass. (MPBoardSolutions.com) As radiation of longer wavelength (Infrared radiation) generates heat, this causes rise in temperature inside the glass house. An effect similar to one in glass house is responsible for keeping the earth’s surface warmer.

Question 9.
For your agricultural field or garden you have developed a compost producing pit Discuss the process in the light of bad odour, flies and recycling of wastes for a good produce?
Answer:
The compost is very useful for agriculture as a fertilizer. But the compost pro-ducing pits may give bad odour and flies. Therefore, the compost producing pit should be set up at a suitable place or in a bin to protect ourselves from bad odour and flies. It must be kept covered so that flies cannot enter into it and there is not much a bad odour.

Question 10.
How house effluents can be used as manure?
Answer:
Waste Management of Household waste:
All the solid household waste should be put in the household garbage box/bin. This should be then put into the community bins so that the municipal workers can take it in their vehicles to the disposable site. Here, the garbage is separated into biodegradable and non – biodegradable materials. (MPBoardSolutions.com) The biodegradable waste is deposited in the land fills. With the passage of time, it is converted into manure compost. Remember that if the waste is not collected in the garbage bins, it may find its way to sewers and some may to eaten up by the cattle.

The non – biodegradable waste like polythene bags, metal scrap, etc. choke the sewers. The polythene bags, if swallowed by cattle, can result into their death. The best way to manage domestic waste is to keep two garbage bins, one for the biodegradable (Non – recyclable) and the other for non – biodegradable (Recyclable) which can be sold to the vendor/dealer.

Question 11.
What do you mean by green chemistry? How will it help in decrease environmental pollutions?
Answer:
By green chemistry we mean a strategy to design chemical process which neither use toxic chemicals nor release the same to the atmosphere. It also means to develop methods of using raw materials more efficiently and generating less wastes.

The creative and innovative skills of green chemistry has developed many new environmental friendly processes, analytical tools, reaction conditions and catalysts etc. A few of these achievement may be listed as follows:

1. Development of new method to improve the yield of ibuprofen upto 99%.
2. Chlorofluorocarbon used as blowing agents for polystyrene foam (Thermocol) sheets have been replaced by CO2.
3. A new technique of catalytic dehydrogenation of ‘diethanolamine’ produces an environment friendly herbicide. This process has avoided the use of highly toxic cyanide and formaldehyde.
4. Organotin: A common antifouling compound used by sea marines has been replaced by a rapidly degradable compound called ‘sea nine’.

Question 12.
Carbon monoxide gas is more dangerous than carbon dioxide gas. Why?
Answer:
Carbon monoxide is highly poisonous in nature. It combines readily with haemoglobin (It has more affinity than oxygen). Due to the formation of carboxyhaemoglobin, the quantity of oxygen to the body cell get reduced i.e. (MPBoardSolutions.com) CO reduces the oxygen carrying capacity of the blood and this leads to oxygen starvation (Anoxia). The deficiency of oxygen produces headache, dizziness, choking cardiac and pulmonary complications leading to paralysis and death. CO₂ does not combine with haemoglobin. However, it is a green house gas and helps in global warming. Hence, it is less dangerous pollutant.

Question 13.
What would have happened if the green house gases were totally missing in the earth’s atmosphere? Discuss?
Answer:
The solar energy radiate back from earth surface is absorbed by the green house gases (i.e. CO₂, CH₄, O₃, CFC and water vapour) present near the earth surface. Thus, they heat up the atmosphere near the earth’? surface and keep it warm. (MPBoardSolutions.com) As a result, they keep the temperature of the earth constant and help in the growth of plants and existence of life on the earth. If there were no green house gases, there would have no vegetation and life on the earth.

Question 14.
Statues and monuments in India are affected by acid rain. How?
Answer:
Statues and monuments are generally made of marble (Taj Mahal). The acid rain contains H₂SO₄ which attacks the marble.
CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂
As a result, the monuments (Taj mahal) are being slowly corroded and the marble is getting discoloured and lustreless.

Question 15.
What are the harmful effect of water pollution? How they can be controlled?
Answer:
The harmful effects of water pollution are:

1. Due to intake of polluted water many diseases like typhoid, dysentry etc. occur.
2. Due to effluents the amount of dissolved oxygen in water decreases.
3. Due to presence of soap and detergent effluents the water become poisonous for fishes.

Control of water pollution:
We have seen that the two sources of water pollution are: Sewage and industrial wastes. They should be removed from water before it is put to use.

Treatment of sewage:
1. Sewage must be churned by machines so that the large pieces may break into smaller ones and may get mixed thoroughly. The churned sewage is passed into a tank with a gentle slope. Heavier particles settle and the water flowing down is relatively pure.

2. Water must be sterilized with the help of chlorination. It kills microbes of sewage fungus as well as some pathogens, spores or cytes. Chlorination is very essential particularly in rainy season.

3. Treatment of water with alum, lime etc, also helps in its purifications.

Treatment of industrial wastes:
The treatment of industrial waste depends upon the nature of the pollutants present. In order to ascertain it, the pH of the medium is first determined and the wastes is then neutralised with the help of suitable acid or alkalis. (MPBoardSolutions.com) The chemical substances present in the industrial wastes dissolve in water can be precipitated by suitable chemical reaction and removed later on from water quite recently. Photocatalysed and ion – exchanges have been developed for the treatment of industrial wastes.

Question 16.
What is soil pollution? Write down methods of prevention of soil pollution?
Answer:
Soil pollution:
Change in physical and chemical property of soil due to humans and natural cause is known as soil pollution.
Soil pollution can be prevented by the following methods:

1. Solid and unusable substances like iron, copper, glass, polythene, etc. should not be hurried under soil.
2. Banning cutting of forest and uncontrolled grazing. Crop cycle to be adopted. Suitable arrangement for irrigation to be made. Control on flood and appropriate use of chemical fertilizers and insecticides.
3. Minimum use of chemical fertilizers, insecticides and pesticides.
4. Special attention on recycling of solid waste on melting.
5. Emphasis on the use of cow – dung and human excreta as bio – gas.
6. Biological insecticides to be used.
7. Using closed mines for disposal off waste.
8. Methods of soil erosion to be checked.
9. Soil management to be adopted.
10. Encouraging the use of biofertilizers.

Question 17.
Write the effects of depletion of ozone layer?
Answer:
Sunlight contains ultraviolet radiations and ozone layer present in the atmosphere prevents ultraviolet radiation to reach the earth surface. Continuous depletion of ozone layer cannot prevent ultraviolet radiations from reaching the earth surface and following disadvantages will occur.

1. Intensity of sunlight will increase and temperature of the environment will become intolerable.
2. Increase in skin diseases.
3. Skin cancer becomes common.
4. Immune system will turn weak.
5. Germination and development of seed slows down.

MP Board Class 11 Chemistry Important Questions

MP Board Class 11th Chemistry Important Questions Chapter 13 Hydrocarbons

Hydrocarbons Important Questions

Hydrocarbons Very Short Answer Type Questions

Question 1.
Full form of T.E.L?
Answer:
Tetra Ethyl Lead.

Question 2.
What is obtained when ethylene dibromide is heated with Zn dust?
Answer:
Ethylene.

Question 3.
The smelling substance in L.P.G is?
Answer:
Ethyl mercaptan.

Question 4.
What is the name of the method in which by electrolysis of potassium acetate, methane is formed?
Answer:
Kolbe’s reaction.

Question 5.
Write the name of the reaction

Answer:
Sabatier and Senderens reaction

Question 6.
Write the names of the products formed in following reaction:

Answer:
(A) CH E= CH
(B) CH3 – CHO
(C) CH3 – CH₂ – OH

Question 7.
Write the reaction in which alkane is formed by alkyl halide and sodium
Answer:
Wurtz – Fittig reaction

Question 8.
What is number of a and n bonds in ethene?
Answer:
5o and 1 n.

Question 9.
In benzene, carbon has which hybridization?
Answer:
sp² hybridization

Question 10.
In HC = CH, which hybridization is present in C – C?
Answer:
sp – sp² hybridization

Hydrocarbons Short Answer Type Questions – I

Question 1.
Trans – alkene is formed by the reduction of alkyne by liquor ammonia. Is butene show geometrical isomerism obtained by reduction of 2 – butyne?

2 – butene shows geometrical isomerism.

Question 2.
Despite their I – effect, halogens are o – and p – directing in haloarenes. Explain?
Answer:
Halogen have (-1) and (+R) effect, these groups are deactivating due to their (-1) effect and they are ortho and para directing due to (+R) effect.

Question 3.
Alkenes are more reactive than alkanes. Why?
Answer:
In alkanes there is only σ – bond between C – C but in alkenes there is one and one σ – bond between C = C. Due to lateral overlapping the π – bond is weaker than the σ – bond. Hence, alkenes are more reactive than alkanes. (MPBoardSolutions.com) The bond energy of π – bond lower than the bond energy of π – bond. Due to this difference of bond energy alkenes are more reactive as compared to alkanes.

Question 4.
What are asymmetric carbon?
Answer:
The carbon atom in which four different groups are attached is called asymmetric carbon. Due to this property, compound shows optical activity.

Question 5.
What do you mean by Chirality?
Answer:
Those molecules which are not superimposable on their mirror images are chiral molecules and this property is called chirality. They are optically active. The Chirality due to presence of asymmetric carbon in the molecule.

Question 6.
What are Alkanes? Which type of bonds are present in it?
Answer:
Alkanes are saturated hydrocarbons because of their low reactivity, they are also called paraffin. The general formula is C_nH_2n+2, In alkanes each carbon atom is sp³ hybridized. Single σ – bonds are present between C – C and C – H.
Example: Methane CH₄, Ethane C₂H₆.

Question 7.
What are Alkenes? In Alkenes C is present in which hybridized state?
Answer:
A saturated hydrocarbon becomes unsaturated when two hydrogens are less in it. Such hydrocarbons are called olefins. In IUPAC system these olefins are called alkenes. Such alkenes are unsaturated and they contain C = C double bond. General formula is
The hybridization of carbon in C = C is sp².
Example: Ethene CH₂ = CH₂, Propylene CH₃ – CH = CH₂.

Question 8.
What are Alkynes? What type of bonds are present in them?
Answer:
Decrease in four hydrogens in saturated hydrocarbons result in the formation of triple bonds between two carbon atoms. The unsaturated hydrocarbon produced known as alkynes. (MPBoardSolutions.com) Their general formula is C_nH_2n-2. Carbon atom of alkyne is sp³ hybridized. Alkynes are also called acetylenes. They contain carbon – carbon triple bond.
Example: Acetylene CH₂ = CH₂, Propyne CH₃ – C = CH₂.

Question 9.
Why Cyclopropane is more reactive than cyclohexane?
Answer:
In cyclopropane the bond angle in C – C – C is 60° due to which ring feelstrain, and so it is reactive and less stable. Whereas in cyclohexane C – C – C have 109°28 and have less strain in the ring, therefore it is stable and less reactive.

Question 10.
Explain functional isomerism with example?
Answer:
The compound having same molecular formula but different functional groups in the molecule are called functional group isomers.
Example:
1. Alcohols and ethers (C₂H_2n+2 – O)

Question 11.
What is dissociation?
Answer:
The method of separation of dor l image isomers from racemic mixture is called dissociation. This is done through biochemical or chemical methods.

Question 12.
How will you obtain nitrobenzene from acetylene?
Answer:

Question 13.
What is Prototrophy?
Answer:
By hydration of Propyne, enol and keto forms are obtained. It shows tautomerism and this type of isomerism is present in that compounds which have at least one hydrogen. This isomerism is due to the transfer of proton from one place to another. This is called prototrophy.

Question 14.
What is conformational stereioisomerism?
Answer:
The phenomenon of easily interconvertibility at room temperature due to free rotation around the carbon – carbon single bond in alkanes and its derivatives is called conformational stereioisomerism.

Question 15.
Explain the process of polymerisation?
Answer:
In polymerisation many simple molecules of a substance combine together to form a big molecule. The simple molecule is called monomer and the big molecule as polymer or macromolecule. Rubber, nylon, bakelite, P.V.C. are examples of high polymers. (MPBoardSolutions.com) Polymerisation of alkenes takes place in presence of Lewis acid BF₃, AlCl₃ or organic and inorganic peroxides.

Polyethylene is used as electrical insulator and as packing materials.

Hydrocarbons Short Answer Type Questions – II

Question 1.
What is Cracking? Write a note on cracking and its uses?
Answer:
Cracking or Pyrolysis:
Higher hydrocarbons when heated to high temperature decomposes into smaller hydrocarbons (lower carbon atom containing molecule). This type of thermal decomposition is known as cracking or pyrolysis.

Pyrolysis is related to free radical reaction. Manufacture of oil gas or petrol gas is based on the concept of pyrolysis. For example, dodecane when heated to 973 K gives a mixture of heptane and pentene. Platinum, palladium or nickel is used as catalyst.

Steam phase cracking, catalytic cracking and liquid phase cracking are the different methods of cracking.

Uses of alkanes:

1. Alkanes are used for making carbon black which is used for making ink, black paints and polish.
2. As a gaseous fuel in industries and as L.P.G.
3. Higher alkanes like petrol, kerosene, lubricant, oil, paraffin, wax, etc.obtained from petroleum are useful.
4. Some halogen derivatives like chloroform and carbon tetrachloride are useful in laboratories and industries.
5. Catalytic oxidation of alkanes give important compounds like alcohol, aldehyde, acid, etc.

Question 2.
Explain Dehydrohalogenation and Dehalogenation with example?
Answer:
Dehydrohalogenation:
When alkyl halide is heated with alcoholic KOH, molecule of hydrogen halide is eliminated forming alkene. This reaction is known as dehydro – halogenation.
Example:
CH₃ – CH₂ – CH₂ – Cl + KOH CH₃ – CH = CH₂ + KCl + H₂O.

Dehalogenation:
Vicinal dihalogen derivatives of alkanes are compounds in which halogen atoms are present on adjacent carbon atoms. They are also called, 1,2 – dihalogen derivative. They form alkenes when heated with Zn dust.
This reaction is known as dehalogenation:

Question 3.
Write notes on Friedei – Crafts reaction?
Answer:
Alkylation:
When benzene or its higher homologous are heated with alkyl halide in presence of anhydrous AlCl₃ alkyl derivatives of benzene are formed.

Acetylation:
When benzene or its higher homologous are treated with acid chloride in presence of anhydrous AlCl₃, acyl benzene or aromatic ketones are formed.

Question 4.
What is Lindlar’s Catalyst? Write its uses?
Answer:
Lindlar catalyst is a palladium supported mixture of calcium carbonate and poisoned with sulphur or quinoline.

Uses:
Aikynes react with hydrogen in presence of nickel powder or platinum or palladium catalyst to form alkenes. This process is known as catalytic hydrogenation.

Question 5.
Geometrical isomerism is found in which type of compounds? Explain with example?
Answer:
Geometrical isomers and Geometrical isomerism or cis – trans isomerism:
This type of isomerism is shown by those alkene derivatives in which different groups are attached with carbons linked with double bond. (MPBoardSolutions.com) For example: compound like abc = cba. When similar atom or group is on same side of bond, it is called cis – isomer and when they are on opposite side of bond, it is called trans – isomer. This type of isomerism is called cis – trans isomerism.

Following compounds does not show geometrical isomerism.

Question 6.
What are Dienes, write their types? Explain with example?
Answer:
Diene is an unsaturated hydrocarbon. In this, two double bonds are present between carbon – carbon chain. On the basis of position of double bonds dienes are of three types:
1. Isolated dienes:
Dienes in which more than one single bonds are present between two double bonds in C – C chain.
CH₂ = CH – CH₂ – CH = CH₂

2. Conjugated dienes:
In these dienes the double bonds are present in alternate position in carbon – carbon chain.

3. Cumulative dienes:
In these dienes the double bonds are present continuously on two C atoms.
CH₃ = C = CH – CH₃
CH₃ – CH = C = CH₂.

Question 7.
Write Diel’s – Alder reaction with equation?
Answer:
When a conjugated diene is heated with ethene, then a cyclic compound is formed. This is called Diel’s – Alder reaction.

Question 8.
What is Huckel’s Rule?
Answer:
According to it, all those planar cyclic compounds exhibit aromatic character, whose ring contain (4n + 2) n electrons. Where n is an integer. Therefore, planar cyclic compounds in which there are 2 (n = 0), 6(n = 1), 10 (n = 2), 14 (n = 3) π electrons exhibit aromatic property.

Benzene and Naphthalene are aromatic compounds on the basis of Huckel’s rules several heterocyclic compounds should also show aromatic properties.

Question 9.
Write Kolbe’s method for manufacuture of alkanes?
Answer:
When sodium or potassium salt of carboxylic acids are electrolysed, alkanes are obtained at anode.
CH₃COONa ⇄ CH₃COO^– + Na⁺

At anode:
CH₃COO^– – e^– → CH₃COO\(\overset { \bullet }{ C } \)
CH₃ – COO^. → \(\overset { \bullet }{ C } \)H₃ + CO₂
\(\overset { \bullet }{ C } \) H₃ + \(\overset { \bullet }{ C } \) H₃ → C₂H₆

At cathode:
Na⁺ + e^– → Na
2Na + 2H₂O → 2NaOH + H₂

Question 10.
Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitution reactions with difficulty?
Answer:
The orbital structure of benzene shows that the π electrons cloud lying above and below the benzene ring is loosely held and therefore it is likely to be attacked by electrophiles which subsequently bring about substitution. The nucleophiles would be repelled by the n – electron ring and hence benzene ring reacts with nucleophiles with difficulty.

Question 11.
Why do alkenes prefer to undergo electrophilic addition reactions while arenes prefer electrophilic substitution reactions? Explain?
Answer:
Alkenes are source of loosely held π – bonds. Due to which they show electrophilic addition reactions. There occurs a tremendous change in the energy during the electrophilic addition of electrons to alkenes therefore they show electrophilic addition reactions.

In Arenes, during the electrophilic addition reactions the aromatic nature of benzene destroyed, but in electrophilic substitution reactions it remains constant. (MPBoardSolutions.com) So the electrophilic substitution reactions are more stable in order of energy.

Question 12.
What is tautomerism? Explain with example?
Answer:
Tautomerism:
This is a special type of functional isomerism in which the isomers differ in the arrangement of atoms but they exist in dynamic equilibrium with each other. For example, acetaldehyde and vinyl alcohol are tautomers which exist in equilibrium as:

It is of different types but the most common among them is the keto – enol tautomerism. This arises due to 1, 3 – migration of a hydrogen atom from one polyvalent atom to other within the same molecule.

Conditions for the molecule to show tautomerism:
1. Presence of electron withdrawing groups such as:

2. Presence pf α – hydrogen:

Question 13.
What is Newman projection formula?
Answer:
Newman projection:
Named after M.S. Newman, who first proposed this method of representing the three – dimensional structure on paper, this is easier to visualize than the one described before. In this projection, the molecule is viewed at the C – C bond head on. (MPBoardSolutions.com) In this formula, front carbon atom is shown by dot and rear carbon atom by a circle. Three hydrogen atoms bonded to the carbon atoms are shown by lines making an angle of 120° with each other.

Newman’s projection formula of ethane shown in fig. gives a planar representation for two – dimensional (2 – D) representation of the molecule. (MPBoardSolutions.com) Staggered form changes into eclipsed form when rotated through 60°. Similarly, eclipsed form also changes into staggered form.

Question 14.
What is Markownikoff s rule?
Answer:
During the addition across unsymmetrical double bond, the negative part of the adding molecule attaches itself to the carbon atom carrying less number of hydrogen atoms.

Markownikoffs rule can be explained on the basis of the mechanism of addition reaction. Stability of intermediate decides the yield of the product. Consider the attacks of H⁺ (an electrophile) on the propene molecule. The two inter – mediate carbocations are formed.

Since, a 2° carbocation (I) is more stable than 1° carbocation (II), therefore, carbocation (I) is predominantly formed. (MPBoardSolutions.com) This carbocation then rapidly undergoes nucleo – philic attack by the Br^– ion forming 2 – bromopropane as the major product. Thus, MarkownikofFs addition occurs through the more stable carbocation intermediate. .

Question 15.
Draw the cis and trans structures of hex – 2 – ene? Which isomer will have higher boiling point and why?
Answer:

As cis isomer is more polar than trans therefore magnitude of dipole – dipole interaction is more than trans. Hence boiling point of cis isomer is more than trans.

Question 16.
An alkene, ‘A’ on ozonolysis gives a mixture of ethanal and pentan – 3 – one. Write structure and IUPAC name of ‘A’? Write structure and IUPAC name of ‘A’?
Answer:
The product of ozonolysis are –

Remove the oxygen atom (= 0) and join the two ends by a double bond, the structure of the alkene ‘A’ is

Question 17.
Explain Racemic mixture with example?
Answer:
The mixture of equal amount of two optical isomers is called Racemic mixture. The rotation of these two isomers are in opposite directions. So it is represented by dl or ±.
Example:

Optical isomerism or Enantiomerism:
This type of isomerism is shown by unsymmetrical compounds. Structural, physical and chemical properties of optical isomers are nearly similar but optical properties are different.(MPBoardSolutions.com) Isomer which rotates plane of polarised light in clockwise direction is called dextrorotatory and one which rotates in anti – clockwise direction is called laevo – rotatory. These two forms are represented as d – and l – or (+) and (-) respectively.
Optical isomers have unsymmetrical i.e., chiral carbon in the molecule.

Question 18.
What is B.H.C.? What is its uses?
Answer:
It is prepared by the chlorination of benzene in the presence of ultraviolet light.

Benzene hexachloride is an addition compound and its y isomer is called Gammaxene. It is an important pesticide. It is also called Lindane or 666.

Question 19.
How will you do the following conversion?

1. Methane to Ethane
2. Ethane to Methane
3. Acetylene to Benzene.

Answer:
1. Methane to Ethane:

2. Ethane to Methane:

3. Acetylene to Benzene:

Question 20.
Write the name of the organic compounds which show geometrical isomerism in cyclic compounds?
Answer:
Some of the cyclic compounds also show geometrical isomerism.
Example:

Question 21.
What is Saytzeff’s rule? Explain with example?
Answer:
Saytzeff’s rule:
According to this:
“If an alkyl halide can eliminate the hydrogen in two different ways, that alkene will be formed in excess in which carbon atoms joined by double bond are more alkylated.

Question 22.
What are the essential conditions for a compound to show aromaticity?
Answer:
The property which gives extra stability to benzene and benzene like compounds is called aromaticity.
Aromaticity is decided by Huckel’s rule. According to Huckel’s rule, a compound will be aromatic if it fulfills the following four conditions:

1. Compound shall be cyclic.
2. Compound should be planar or nearly planar (sp² hybridisation)
3. Compound should be conjugated.
4. Compound should have (4n + 2) π electrons. Where n is a whole number and it may be n = 0, 1, 2, 3,4, 5, 6, …

Question 23.
What is cis and trans isomerism? Explain with example?
Answer:
cis and trans isomerism is also called geometrical isomerism. This isomerism is shown by that compounds which have carbon atoms attached to two different atoms. When the two same groups or hydrogen atoms are present at one side of double bond then the compound is called cis isomer. (MPBoardSolutions.com) But when the groups or H – atoms are present in opposite side of the C double bond then such isomers are called trans isomers.
Example:

Question 24.
Arrange benzene, n – hexane and ethyne in decreasing order of their acidic behaviour? What is the reason for this behaviour?
Answer:
The hybridised state of C in the given compounds are:

The acidic character increases with increase of s – character. So the decreasing order of acidty is:
Ethyne > Benzene > n – hexane.

Question 25.
How would you convert the following compounds into benzene:

1. Ethyne
2. Ethene
3. Hexane.

Answer:
Ethyne:

Ethene:

Hexane:

Question 26.
Explain why the following systems are not aromatic?

Answer:
(i)
 = CH₂ does not have six electrons i.e., (4n + 2) π – electrons in the ring. Therefore, it is not an aromatic compound.

(ii)

Reason: Due to sp³ hybridisation the molecule is not planar. It contains 4π – electrons, So, molecule is not aromatic.

(iii)

It is a conjugated system but does not have (4n + 2) π – electrons.

Hydrocarbons Long Answer Type Questions – I

Question 1.
Explain the conformation of n – butane?
Answer:
Conformations of n – butane:
n – Butane can be considered as dimethyl derivative ethane which is produced when terminal hydrogen of each carbon is replaced by methyl group. To assign conformations to n – butane is a difficult task because it has three carbon – carbon single bonds (one in the middle and two at the ends) which undergo free rotation.

Various conformations of n – butane can be obtained by rotating C₂ or C₃ through 360° in six steps (60° each time). Staggered and eclipsed forms are obtained alternately on rotating C₂ or C₃ by 60°. These conformations are shown below. (MPBoardSolutions.com) Fully eclipsed form is shown in (I) and other eclipsed forms are shown in II and III. On rotation of C₂ – C₃ bond by 120°. The completely staggered form is shown in (IV). It is called Antiform also. Other staggered forms shown in V and VI are called skew or gauche forms. The staggered form IV and gauche forms V and VI are termed conformational diastereoisomers.

Question 2.
Give any four points in favour of Kekule’s formula to show its resonance structures?
Answer:
Factors in favour of Kekule’s structure:
1. Benzene reacts with three molecules of hydrogen to form cyclohexane which proves presence of three double bonds.

2. Benzene reacts with three molecules of chlorine to form benzene hexachloride.

3. Three molecules of acetylene polymerize in a red hot tube to form benzene.
3CH ≡ CH → C₆H₆

4. Oxidation of benzene with air in presence of V₂O₅ gives maleic acid which loses a molecule of water to form maleic anhydride.

Question 3.
A hydrocarbon ‘A’ vapour density is 14, makes the Baeyer’s reagent colourless and perform following reaction:

Write the name and formula of A, B, C and D?
Answer:

(A) → Ethylene
(B) → 1, 2 – dibromoethane or ethylene bromide
(C) → Acetylene
(D) → Acetaldehyde.

Question 4.
Explain the laboratory method of formation of alkene with figure?
Answer:
Laboratory preparation of alkene (ethylene):
Ethylene is obtained in laboratory by heating ethyl alcohol with excess of cone. H₂SO₄ at 170°C.
1. Requirements:
Ethyl alcohol, cone. H₂SO₄, sand bath. Thistle funnel, potassiumm hydroxide, gas jar, stand etc.

2. Chemical equation:
C₂H₅OH + H₂SO₄ → C₂H₅HSO₄ + H₂O

3. Method:
50 cc of ethyl alcohol and 100 cc of conc.H₂SO₄ is taken in a flask, then 8 gm anhydrous Al₂ (SO₄)₃ and 50 gm of sand is added. The mixture of Al₂ (SO₄)₃ and sand checks up formation of foam and facilitates the reaction to take place at 140° C.

Now the flask is fitted with a thermometer, an exit tube and a dropping funnel. Flask is placed on a sand bath and fixed with a stand. The other end of exit tube dips in wash bottle containing NaOH. Another tube from wash bottle leads to a behive shelf placed in a trough of water. (MPBoardSolutions.com) A water jar is Inverted over behive shelf. Flask is heated at a temperature of 150° C and simultaneously mixture of alcohol and cone. H₂SO₄ is added dropwise into the flask. Along with ethylene, CO₂ (by oxidation of alcohol) and SO₂ (by reduction of H₂SO₄) are also present as impurities in flask. These impurities get adsorbed in NaOH solution and pure ethylene is collected in gas jars by downward displacement of water. Ethylene prepared by this method is pure.

4. Labelled diagram:

Question 5.
Explain the Laboratory method of preparation of acetylene? Or, Explain the Laboratory method of preparation of acetylene on following points:

1. Method and Chemical reaction
2. Labelled figure.

Answer:
Preparation of acetylene in laboratory:
Acetylene is prepared by dropping water on calcium carbide. Acetylene obtained by this method contains impurities of PH₃ and NH₃. These are removed by passing the gas through CuSO₄ solution.

Precautions:
Before experiment, air in flask is replaced by oil gas because acetylene forms explosive mixture with air.

Reaction of acetylene with water:
In presence of 1% HgSO₄ and 42% H₂SO₄, acetaldehyde is formed.

Reaction with ammoniacal silver nitrate solution:
When acetylene is passed through ammoniacal AgNO₃ solution, white precipitate of silver acetylide is obtained.

Question 6.
Out of benzene, m – dinitrobenzene and toluene which will undergo nitration most easily, why?
Answer:
CH₃ group is electron releasing while NO₂ group is electron withdrawing. Therefore, maximum electron density will be in Toluene followed by in benzene and least in m – nitrobenzene. Therefore, the ease of nitration decreases in the order:
Toluene > Benzene > m – dinitrobenzene.

Question 7.
How will you obtain:

1. B.H.C. from benzene
2. Acetophenone from benzene
3. P.V.C. from chloroethene
4. Teflon from tetrafluoroethene.

Answer:
1. B.H.C. from benzene:

2. Acetophenone from benzene:

3. P.V.C from chloroethene:

4. Teflon from tetrafluoroethene:

Question 8.
Write notes on following:

1. Sabatier and Senderens reaction
2. Wurtz reaction
3. Duma’s reaction,
4. Swart reaction.

Answer:
1. Sabatier and Senderens reaction:
The reaction of alkene with hydrogen in presence of Ni or Pt, after hydrogenation alkane is obtained.

2. Wurtz reaction:
Reaction of two molecules of alkyl halide with sodium in presence of dry ether, alkane is obtained.

3. Duma’s reaction:
Reaction of sodium salt of monocarboxylic acid with soda lime, after decarboxylation alkanes are formed.

4. Swart reaction:
Reaction of alkyl halide with mercuric chloride, chloroalkanes are obtained. During this reaction the substitution of halogen of alkyl halide occur by Cl.
2C₂H₅ – I + HgF₂ → 2C₂H₅ – F + HgI₂

Question 9.
An unsaturated hydrocarbon ‘A’ adds two molecules of H₂ and on reductive ozonolysis gives butane – 1, 4 – dial, ethanal and propanone. Give the structure of ‘A’. Write its IUPAC name and explain the reaction involved?
Answer:

Thus, the structure of compound ‘A’ may be written as:

Hydrocarbons Long Answer Type Questions – II

Question 1.
Write the equations for the following reactions:

1. Reaction of calcium carbide with water.
2. Reaction of bromine water on ethylene.
3. Heating of ethylene with alkaline KMnO₄.
4. Heating benzene with cone. HNO₃ and cone. H₂SO₄.
5. Heating benzene with methyl chloride in presence of anhydrous AlCl₃

Answer:
1. Reaction of calcium carbide with water:
CaC₂ + 2H.OH → CH = CH + Ca(OH)₂

2. Reaction of ethylene with bromine water:

3. Heating ethylene with alkaline KMnO₄:

4. Heating benzene with cone. HNO₃ and cone. H₂SO₄:

5. Heating benzene with methyl chloride in presence of anhydrous AlCl₃:

Question 2.
How will you obtain following:

1. Acetaldehyde from Acetylene.
2. Mustard gas from Ethylene.
3. Ethane from Grignard reagent
4. Cuprous acetylide from Acetylene.
5. Methane from Aluminium carbide.

Answer:
1. Acetaldehyde from Acetylene:

2. Mustard gas from Ethylene:

3. Ethane from Grignard reagent:

4. Cuprous Acetylide from Acetylene:
CH ≡ CH + Cu₂Cl₂ +2NH₄OH → Cu – C ≡ C – Cu + 2NH₄Cl + 2H₂O

5. Methane from Aluminium carbide:
Al₄C₃ + 12H₂O → 3CH₄ + 4Al(OH)₃.

Question 3.
What is conformation? Describe conformation found in ethane?
Answer:
In alkane the C – C bond is formed by the axial overlapping of sp³ hybrid orbitals of the adjacent carbon atoms. The electron distribution in molecular orbitals of sp³ – sp³ sigma bond is cylindrically symmetrical around the inter nuclear axis. This symmetry permits the free rotation about the bond axis without rupture of the molecule. (MPBoardSolutions.com) This result in a large number of different spatial arrangement of atom or group attached to the carbon atom. Thus, “The different spatial arrangement obtained by the free rotation around the bond axis of a C – C cr bond are called conformers and the molecular geometry corresponding to a conformer is known as conformation.

Conformation of ethane:
If the position of one carbon atom of ethane is fixed in space and the other carbon atom is rotated around the C – C bond, then various conformations of ethane are possible. Out of these the conformers which has the lowest energy is called staggered and the one having highest energy is called eclipsed conformation.

1. Staggered conformation:
In staggered conformation, the hydrogen atom of the two carbon atoms are oriented in such a way that they lie far apart from one another. In other words, they are staggered away with respect to one another.

2. Eclipsed conformation:
In eclipsed conformation, the hydrogen atoms of one carbon are lying directly behind the hydrogen atoms of the other. In other words, hydrogen atoms of one carbon are eclipsing the hydrogen atoms of the other.(MPBoardSolutions.com) The conformations of ethane do not have same stability. The staggered conformation is relatively more stable than the other conformation. The difference in the energy content of staggered and eclipsed conformation is 12.5 kJ mol^-1.

Question 4.
An alkane C₈H₁₈ is obtained as the only product on subjecting a primary alkyl halide to Wurtz reaction. On monobromination this alkane yields a single isomer of a tertiary bromide. Write the structure of alkane and tertiary bromide?
Answer:
From Wurtz reaction of an alkyl halide gives an alkane with double the number of carbon atoms present in the alkyl halide. Here, Wurtz reaction of a primary alkyl gives an alkene (C₈H₁₆), therefore, the alkyl halide must contain four carbon atoms. Now the two possible primaiy alkyl halides having four carbon atoms each are, I and II.

Since, alkane C₈H₁₈ on monobromination yields a single isomer of tertiary alkyl bromide, therefore, the alkene must contain tertiary hydrogen. This is possible, only if primary alkyl halide (which undergoes Wurtz reaction) has a tertiary hydrogen.

Question 5.
Explain the free radical mechanism of halogenation of alkane?
Answer:
Mechanism of halogenation of alkane:
Halogenation of alkanes proceed through the formation of free radicals. Therefore, it is also called free radical substitution. The reaction proceeds in the following steps:

1. Chain initiation step:
The first step involves the homolytic fission of chlorine molecule to form chlorine free radicals. This fission takes place in the presence of light or heat.

Note that Cl – Cl bond being weaker than C – H bond and the C – C bonds. Therefore, undergo cleavage first.

2. Chain propagation:
Chlorine free radical is produced in the first step, attacks methane molecule forming methyl free radical and HCl. Methyl free radical reacts with other chlorine molecule forming methyl chloride and chlorine free radical.

When sufficient amount of methyl chloride has been formed, the Cl produced in reaction (ii) has a greater chance of colliding with a molecule of CH₃Cl rather than a molecule of CH₄. If such a collision occurs, a new free radical (CH₂Cl) is produced.
CH₃Cl + \(\overset { \bullet }{ C } \) l → CH₂Cl + HCl
This \(\overset { \bullet }{ C } \)H₂ Cl then reacts with Cl₂ to give CH₂Cl₂ and another \(\overset { \bullet }{ C } \) l free radical.
\(\overset { \bullet }{ C } \)H₂Cl + Cl₂ → CH₂Cl₂ + \(\overset { \bullet }{ C } \) l
This process continues till all the hydrogen is removed.

3. Chain termination:
The chain reaction steps if two or different free radicals combine amongst themselves without producing new free radicals. The possible termination steps are as follows:
\(\overset { \bullet }{ C } \) l + \(\overset { \bullet }{ C } \) l → Cl₂
\(\overset { \bullet }{ C } \) H₃ + \(\overset { \bullet }{ C } \) H₃ → CH₃ – CH₃
\(\overset { \bullet }{ C } \) H₃ + \(\overset { \bullet }{ C } \) l → CH₃Cl

Question 6.
Explain the conformational isomerism in cyclohexane?
Answer:
Conformations of Cyclohexane:
Like alkanes cyclohexane (a cycloalkane) also exhibits conformational isomerism. Sachse (1890) suggested that if cyclohexane has a planar cyclic hexagonal structure, then it will be highly stable due to the presence of angular strain of the ring composed of sp² hybrid carbon atoms. (MPBoardSolutions.com) However, cyclohexane is quite stable. Sachse suggested that two non – planar models are possible which are free from angular strain. These are called chair and boat conformations as shown in Fig.

These are in a staggered way and eclipsed conformation of alkanes. The chain form is more stable that the boat form by an energy equal to about 7 kcal/mol. In the boat form, like the eclipsed form of ethane, the hydrogen atoms being very close repel each other and the system becomes unstable. (MPBoardSolutions.com) In the boat form there is considerable non – bonded interaction between the flagpole hydrogens and also between other eclipsed hydrogens. As a consequence, this form of cyclohexane can flex into what is known as twist boat form (flexible form) which is stable by about 1.8 kcal/mol than regular boat.



In the chain form, the hydrogen atoms are situated quite far apart from one another. As a result the force of repulsion between the nearest (v icinal) hydrogen atoms is minimum. Thus, out of two main conformations, chair and boat, the chair form is more stable and cyclohexane exists mainly in chair form.

Question 7.
Explain the electrophilic substitution reactions in aromatic hydrocarbons giving two examples?
Answer:
Mechanism of monosubstitution in benzene:
Study of many monosubstitution reactions of benzene show that these reactions follow mechanism of electrophilic substitution. In thesi 40% H₂SO₄tie reagent is an electrophile (E⁺). It can be understood in the following steps:

Step I.
Generation of electrophile:
Dissociation of attacking reagent results in the formation of electrophile (E⁺).
E – Nu → E⁺ + Nu^–

Step II.
Attack of the electrophile on the ring:

Step III.
Abstraction of a proton by the base:
Nucleophile displaces proton from the hybrid in fast step forming desired substituted products. In this way elimination takes place in this step.

In this way in electrophilic substitution reaction of benzene, electrophile is added in first step and then H⁺ ion is eliminated.

Example 1.
Mechanism of nitration of benzene:
Nitration of benzene is done in the ahead steps by treating with a nitrating mixture of concentrated nitric acid and concentrated sulphuric acid:

Step I.
Generation of electrophile (NO₂⁺):

Step II.
Attack of the electrophile on the ring:
Electrophile attacks on the benzene ring forming carbocation. It attains stability by ion resonance.

Step III.
Abstraction of H⁺ ion by the base:
Nucleophile HSO₄^– ion substitutes H⁺ ion of the ring in fast step forming nitrobenzene.

Example 2.
Mechanism of halogenation of benzene:
Mechanism ofhalogenation of benzene can be explained by the action of chlorine on benzene in presence of Lewis acid FeCl₃.

Step I.
Generation of electrophile (Cl⁺):

Step II.
Attack of the electrophile on the ring:
Electrophile attacks on the ring forming carbocation intermediate which is resonance stabilised.

Step III.
Abstraction of H⁺ ion by the base:
FeCl₄^– ion is a base here and it displaces H⁺ ion from hybrid forming desired products in the fast step.

MP Board Class 11 Chemistry Important Questions

MP Board Class 11th Chemistry Important Questions Chapter 12 Organic Chemistry: Some Basic Principles and Techniques

Organic Chemistry: Some Basic Principles and Techniques

Organic Chemistry: Some Basic Principles and Techniques Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
Aniline is generally purified by:
(a) Steam distillation
(b) Simple distillation
(c) Distillation under reduced presence
(d) Sublimation
Answer:
(a) Steam distillation

Question 2.
Glycerol boils at 290°C with slight decomposition. Impure glycerol is purified by:
(a) Steam distillation
(b) Simple distillation
(c) Vacuum distillation
(d) Extraction with solvent
Answer:
(c) Vacuum distillation

Question 3.
The blue or green colour obtained in Lassaigne’s test is due to the formation of:
(a) NaCN
(b) Na₄[Fe(CN)₆]₄
(c) Fe₃[Fe(CN)₆]₄
(d) Fe₄[Fe(CN)₀]₃
Answer:
(d) Fe₄[Fe(CN)₀]₃

Question 4.
In qualitative analysis of organic compound by Lassaigne’s test the violet colour obtained with sodium nitroprusside indicate the presence of:
(a) Nitrogen
(b) Sulphur
(c) Oxygen
(d) Halogen.
Answer:
(b) Sulphur

Question 5.
The blood red colour compound formed during the qualitative analysis of nitrogen and sulphur together is:
(a) Fe₄[Fe(CN)₆]₂
(b) Fe(SCN)₃
(c) KSCN
(d) Na₂S.NaCN.
Answer:
(b) Fe(SCN)₃

Question 6.
Kjeldahl’s as method is used for estimation of:
(a) Sulphur
(b) Netrogen
(c) Halogen
(d) Oxygen
Answer:
(b) Netrogen

Question 7.
A compound with empirical formula C₂H₅O had molecular mass 90. The formula of compound is:
(a) C₄H₁₀O₂
(b) C₂H₅O
(c) C₃H₆O₃
(d) C₅H₁₄O
Answer:
(a) C₄H₁₀O₂

Question 8.
The amount of sulphur present in an organic compound is estimated by changing into:
(a) H₂S
(b) SO₂
(c) H₂SO₄
(d) H₂SO₄
Answer:
(d) H₂SO₄

Question 9.
The reagent used in Carius method to estimate halogen is:
(a) HNO₃ and HCl
(b) HNO₃ and H₂SO₄
(c) Fuming HNO₃ and BaCl₂
(d) Fuming HNO₃ and AgNO₃
Answer:
(d) Fuming HNO₃ and AgNO₃

Question 10.
The gas collected in Duma’s method to estimate of nitrogen in organic compound is:
(a) N₂
(b) NO
(c) NH₃
(d) None of these
Answer:
(a) N₂

Question 11.
An organic compound contain C = 80% and H = 20%. The compound shall be:
(a) C₆H₆
(b) C₂H₅ – OH
(c) C₂H₆
(d) CHCl₃
Answer:
(c) C₂H₆

Question 12.
An organic compound contain C = 39-9%, H = 6‘7% and O = 53.4%. The graphical formula shall be:
(a) CHO
(b) CHO₂
(c) CH₂O₂
(d) CH₂O
Answer:
(d) CH₂O

Question 13.
In an organic compound the ratio of mass is C:H:O = 4:1:5. Its empirical formula shall be:
(a) C₂HO
(b) C₂H₄O₄
(c) CH₄O₂
(d) CH₃O
Answer:
(d) CH₃O

Question 14.
The main source of organic compound is:
(a) Coaltar
(b) Petroleum
(c) Both
(d) None of these
Answer:
(c) Both

Question 15.
But – 1,2 diene contains:
(a) Only sp – hybridized carbon atom
(b) Only sp² – hybridized carbon atom
(c) sp and sp² hybridized carbon atom
(d) sp, sp² and sp³ hybridized carbon atom
Answer:
(d) sp, sp² and sp³ hybridized carbon atom

Question 16.
I.U.P.A.C. name of

(a) 2 – Ethyibut – 2ene
(b) 3 – Ethylbut – 2 – ane
(c) 2 – Methylpent – 3 – ene
(d) 3 – Methylpent – 2 – ene
Answer:
(d) 3 – Methylpent – 2 – ene

Question 17.
The I.U.P.A.C. name of the compound having the formula of Cl₃ – CCH₂CHO is:
(a) 3,3,3 – trichloropropan – l – al
(b) 1,1,1 – trichloropropan – l – al
(c) 2,2,2 – trichloropropan – 1 – al
(d) Chloral.
Answer:
(a) 3,3,3 – trichloropropan – l – al

Question 2.
Fill in the blanks:

1. Paper chromatography is based on the law of …………………………….
2. Column chromatography is based on the law of …………………………….
3. Aniline is purified by ………………………… method.
4. Hybridisation of central carbon atom of a carbocation is ……………………………..
5. Benzoic acid is purified by …………………………….
6. Before test of halogen, sodium extract is heated with ……………………………
7. Isomerism found in organic compounds of same series is ………………………………..
8. Chemical name of freon organic compound which is used in air conditions and refrigerators is ………………………….. and its chemical formula is ……………………..
9. ……………………….. ratio of elements in a compound is called its empirical formula.
10. The process of fractional crystallization of separation of two substances depending on the difference of ……………………………….
11. In Lassaigne’s test, blue or green colour is due to the formation of ……………………………….
12. On adding FeCl₃ solution to sodium extract ……………………………. colour is obtained. The name of the compound is …………………………..
13. In organic compound, presence of amount of halogen can be detected by converting it into …………………………………….
14. A compound contain 80% carbon and 20% hydrogen, its formula will be ……………………………………
15. …………………………. gas is produced by the action of water on calcium carbide.
16. R – CONH₂ is an ……………………………
17. Marsh gas mainly contain …………………………….. gas.

Answer:

1. Distribution
2. Adsorption
3. Steam distillation
4. Sp²
5. Sublimation
6. Cone. HNO₃,
7. Metamerism
8. Difluorodichloro methane CF₂Cl₂
9. Simplest
10. Solvent
11. Ferri – ferro cyanide
12. Red, ferric sulphocyanite
13. Silver halide
14. C₂H₆
15. Acetylene
16. Amide
17. Methane.

Question 3.
Answer in one word/sentence:

1. Method used for separation of components on the basis of adsorption is known as?
2. What is conversion of solid substance on heating into vapours without changing into liquid known as?
3. Which element is detected by Duma’s method?
4. What is method of obtaining pure substance by vaporisation of impure liquid followed by condensation of vapours known as?
5. What is the charge on carbon in carbanion?
6. Which formula represents ratio of atoms of elements present in a molecule of a substance?
7. What is the nature of nucleophile?
8. What are cations carrying positive charge on carbon known as?
9. What is the nature of electrophile?
10. Which effect is responsible for the displacement of electrons of covalent bond towards or aways from carbon atom in an organic molecule?
11. Mixture of KMnO₄ and KOH is known as?

Answer:

1. Chromatography
2. Sublimation
3. Nitrogen
4. Distillation
5. Negative
6. Empirical formula
7. Electronegative
8. Carbocation
9. Electropositive
10. Inductive effect
11. Baeyer’s reagent.

Question 4.
Match the following:
[I]

Answer:

1. (d)
2. (a)
3. (b)
4. (e)
5. (c)

[II]

Answer:

1. (b)
2. (d)
3. (e)
4. (a)
5. (c)

[III]

Answer:

1. (a)
2. (d)
3. (b)
4. (c)

Organic Chemistry: Some Basic Principles and Techniques Very Short Answer Type Questions

Question 1.
What is the chemical name of CH₃ – CH₂ – CHCl – CH₃?
Answer:
iso – butyl chloride.

Question 2.
What is the mixture of KMnO₄ and KOH called?
Answer:
Bayer’s reagent.

Question 3.
IUPAC name of Vinegar?
Answer:
Ethanoic acid.

Question 4.
What is the IUPAC name of grain alcohol?
Answer:
Ethanol.

Question 5.
For the mixture of CuSO₄ and Camphor, Camphor is separated by?
Answer:
Sublimation.

Question 6.
How purification of naphthalene was done?
Answer:
By sublimation.

Question 7.
The purification by Column chromatography occur because?
Answer:
Different absorption.

Question 8.
Purification of petroleum?
Answer:
Fractional distillation.

Question 9.
Balsentein test performed for?
Answer:
In halogen detection.

Question 10.
Free radicals are formed by?
Answer:
Homolytic fission.

Question 11.
Main source of organic compounds are?
Answer:
Coaltar and petroleum.

Question 12.
General formula of alcohol is?
Answer:
C_nH_2n+1OH.

Question 13.
What is Chiral molecule?
Answer:
Those which are not superimposable on their mirror images.

Question 14.
What is the name of the compound Cl – CH₂ – CH₂ – COOH?
Answer:
3 – Chloro propanoic acid.

Question 15.
Write the structural formula of iso – butyl chloride?
Answer:
CH₃CH₂CHClCH₃.

Question 16.
C_nH_2n-2 is formula of?
Answer:
Alkynes.

Question 17.
What is Bayer’s reagent?
Answer:
Alkaline KMnO₄.

Question 18.
Compounds different in configuration are called?
Answer:
Stereo isomers.

Question 19.
What is the IUPAC name of Cl₃C.CH₂CHO?
Answer:
3,3,3 – trichloro propanol.

Question 20.
Which type of isomerism is found in nitro ethane?
Answer:
Tautomerism.

Question 21.
Write the name and formula of Freon?
Answer:
Difluoro – dichloro methane (CF₂Cl₂).

Question 22.
The mixture of o – nitrophenol and p – nitrophenol is separated by which method?
Answer:
Vapour distillation method.

Question 23.
Which gas is present in Marsh gas?
Answer:
Methane.

Question 24.
The decomposition of glycerine occurs before its b.p., by which method it can be purified?
Answer:
Low pressure distillation.

Question 25.
Formalin is formed by which compound?
Answer:
HCHO.

Question 26.
What is the structural formula of gem – dihalide?
Answer:

Question 27.
What is the mechanism of this reaction:
CH₃CH₂I + KOH_(aq) → CH₃CH₂OH + KI.
Answer:
Nucleophilic substitution.

Question 28.
Kjeldahl’s method is used for estimation of which element?
Answer:
Nitrogen.

Question 29.
What is Elution ?
Answer:
The process of separation of products by different absorption rate is called elution

Question 30.
Which is the latest and better technique for the separation and purification of organic compounds?
Answer:
Chromatography method.

Organic Chemistry: Some Basic Principles and Techniques Short Answer Type Questions

Question 1.
Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test?
Answer:
For testing sulphur, the sodium extract is acidified with acetic acid because lead acetate is soluble and does not interfere with the test. If H₂SO₄ were used, lead acetate itself will react with H₂SO₄ to form white ppt. of lead sulphate which interfere the test.

Question 2.
Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound?
Answer:
Carbon dioxide is acidic and it reacts with strong base KOH to form potassium carbonate
2KOH + CO₂ K₂CO₃ + H₂O

This results in increase in mass of potassium hydroxide from the increase in mass of CO₂ produced, the amount of carbon in the organic compound can be calculated by using the formula:

Question 3.
What is homologous series? Write its characteristics?
Answer:
Homologous series is a series of similarly constituted organic compounds in which the members possess the same functional group, have similar or almost similar chemical characteristics, can be represented by the same general formula and the two consecutive members differ by CH₂ group in their molecular formulae.

The various members of a particular homologous series are called homologues. A few homologues of alcohol series (containing straight chain alcohols) are as follows: General formula C_nH_2n+1OH

Characteristics of homologous series:

1. All the members of a series can be represented by the general formula. For example, general formula of alcohol family is C_nH_2n+1OH.
2. The two successive members of a particular family is differ by – CH₂ group or by 14 atomic mass unit (12 + 2 × 1).
3. Different members in a family have common functional group, for example, alcohol family given above.
4. The members of a particular family have almost identical chemical properties and their physical properties such as melting point, boiling point, density, solubility etc. show a proper gradation with the increase in the molecular mass.
5. The members present in a particular series can be prepared almost by similar methods known as the general methods of preparation.

Question 4.
What are primary, secondary, tertiary and quarternary C of organic compound?
Answer:
Primary carbon atom:
Carbon atom in the organic compound which is linked with only one carbon atom is called primaty (p) or (1°) carbon atom.

Secondary carbon atom:
Secondary carbon atom is that carbon atom which is linked with two more carbon atoms in the compound. It is also represented by (2°) or (s) carbon atom.

Tertiary carbon atom:
The carbon atom which is linked with three more carbon atoms, is called tertiary (3°) or (t) carbon atom.

Quarternary carbon atom:
The carbon atom which is linked with four other carbon atoms, is called quartemary carbon atom. It is denoted by 4° or q.
In the following example primary (p), secondary (s) and tertiary (t) carbon atoms are represented:

Where, p = primary, s = secondary, t = tertiary and q = quartemary.

Question 5.
What is resonance? Write its applications?
Answer:
Sometimes it is found that all the known properties of a compound cannot be explained by one structure and for such compounds we draw two or more structures. (MPBoardSolutions.com) Such structures are called resonating structures or canonical forms or contributing structures and the phenomenon is called resonance or mesomeric effect. This is a permanent effect. This effect is transmitted through the chain. There are two types of resonance or mesomeric effect:

1. + R or + M effect:
A group is said to have +R or +M effect when the displacement of the electron pair is away from it. For example,

2. – R or – M effect:
A group is said to have – R or – M effect when the displacement of the electron patr is towards it. For example,

Uses:

1. For determination of structure of benzene.
2. To explain dipole moment.
3. To explain the strength of acids and bases.

Question 6.
What is the reason, that carbon forms large number of compounds?
Or, Explain the special properties of carbon,
Answer:
Anomalous behaviour of First Element of the Group (Carbon):
Carbon the first member of group – 14 shows an anomalous behaviour i.e., differ from the rest of the members of its family. The main reasons for this difference are:

1. Small atomic and ionic size
2. Higher electronegativity
3. Higher ionisation enthalpy
4. Absence of d – orbital in the valence shell

The main points of difference are:

1. It is an important component of animal kingdom.

2. It is also found in free state in nature.
3. It possesses the property of catenation because C – C bond energy is very high 353.3 kJmol^-1.

4. Carbon exist in various allotropic form. Its three crystalline form are diamond, graphite and fullerene.

5. Carbon atom has tendency to form pπ – pπ bond with other carbon atom and also with oxygen, nitrogen, sulphur etc. Due to this, carbon – carbon, carbon – oxygen, carbon – nitrogen, etc. double and triple bonds are possible.

6. Carbon is the only element which forms highly stable open chain, cyclic hydrocarbon and aromatic hydrocarbon with hydrogen.

It is due to its property called catenation. It is the ability of like atoms to link with one another through covalent bonds. This is due to smaller size and higher electronegativity of carbon atom and unique strength of carbon – carbon bond. (MPBoardSolutions.com) Since the bond energy of C – C bond is very large (348 kJ mor1). Carbon forms long straight or branched C – C chains or rings of different size and shape. However, as we move down the group the element – element bond energies decreases rapidly viz C – C (348 kJ mol^-1), Si – Si (297 kJ mol^-1), Ge – Ge (260 kJ mol^-1), Sn – Sn (240 kJ mol^-1), Pb – Pb (81 kJ mol^-1), and therefore, the tendency for catenation decreases in the order:
C >> Si > Ge = Sn > Pb.

7. Carbon forms three types of oxide, monoxide, dioxide and suboxide. Bond energy of carbon monoxide is highest among diatomic molecules.

8. Carbon atom can link with other metals directly through covalent bonds compound formed are called organometallic compound.

Question 7.
Explain metamerism and tautomerism with example?
Answer:
Metamerism:
The compounds having same molecular formula but different number of carbon atoms (or alkyl group) on either side of the functional group are called metamers and phenomenon is called metamerism.

Tautomerism:
This is a special type of functional isomerism in which the isomers differ in the arrangement of atoms but they exist in dynamic equilibrium with each other. For example, acetaldehyde and vinyl alcohol are tautomers.

Question 8.
What are Nucleophile? Explain with example?
Answer:
Nucleophiles:
The species having an atom, unshared or ione pair of electron and seeking positive sets are called nucleophiles.
Neutral nucleophiles: NH₃, H₂O, R – O – R.
Negative nucleophiles: Cl^–, OH^–, NH₂^–, CN^–.

Question 9.
What are Electrophiles? Explain with example?
Answer:
Electrophiles:
The positively charged or neutral species which are deficient of electron and can accept, lone pair of electron are called electrophiles.

Neutral electrophiles:
BF₃, AlCl₃, FeCl₃.

Negative electrophiles:
H₃O⁺, Cl⁺, NO₂⁺.

Question 10.
How nitrogen is tested in any organic compound by Lassaingen’s method?
Answer:
Take 2 ml of Sodium extract, add a 2 ml of freshly prepared solution of ferrous sulphate along with 1 – 2ml of NaOH. Heat and then cool the solution. (MPBoardSolutions.com) Green precipitate of Fe(OH)₃ is obtained. Add cone. HCl so that green precipitate of ferrous sulphate goes into solution. Then 2 – 3 drops of ferric chloride solution is added. If green or blue colour is obtained then the substance contain nitrogen.

Question 11.
How will you test the presence of sulphur in any organic compounds?
Answer:
Sulphur test:
1. Sodium nitropruside solution is added to the sodium extract if violet colour appears. Confirms the presence of sulphur in given organic compound.

2. Sodium extract is acidified with acetic acid and lead acetate solution is added. If black precipitate is obtained confirms the presence of sulphur in organic compound.

Question 12.
A sample of 0.50g of an organic compound was treated according to Kjeldahl’s method. The ammonia evolved was absorbed in SO ml of 0.5M H₂SO₄. The residual acid required 60 mL of 0.5 M solution of NaOH for neutralization.Find the percentage composition of nitrogen in the compound?
Solution:
Volume of acid taken = 50 ml of 0.5 M H₂SO₄
= 25 ml of 1.0 M H₂SO₄
Volume of base used for neutralization of acid
= 60 ml 0.5 M NaOH
= 30 ml 0.1 M NaOH

∴ 30 ml of 1.0 M NaOH = 15 ml of 1.0 M H₂SO₄
∴Volume of acid used by ammonia = 25 – 15 = 10 ml
% Amount of Nitogen = image 17
% N = \(\frac{1.4 × 2 × 10}{0.5}\) = 5 gmN.

Question 13.
Steam distillation is useful for which organic compounds? Explain with example?
Answer:
Steam distillation:
Steam distillation is used to purify those organic compounds which are practically immiscible with water, volatile in steam and has fairly high vapour pressure (low boiling point). In this method, the impure liquid is taken in a heated flask and steam is passed over it with the help of a steam generator (Fig.). (MPBoardSolutions.com) The mixture of steam and the volatile organic compound is condensed and collected in a receiver. From this mixture, water is removed by using separating funnel.

During steam distillation, a liquid boils only when the sum of vapour pressure of the liquid (P₁) and of water (P₂) becomes equal to the atmospheric pressure than its boiling point e.g., a mixture of water and a steam volatile insoluble substance will vaporise close below 373K. (MPBoardSolutions.com) This above technique is used for separating aniline from aniline water mixture and also for separation of p – nitro phenol from p – nitro phenol (o – nitrophenol is steam volatile).

Question 14.
What is the principle of Adsorption chromatography? Explain?
Answer:
Chromatography:
The process by which different components of a mixture are separated by distributing in stationary or mobile phases on the basis of difference in adsorption abilities on any adsorbent, is called chromatography.

Adsorption chromatography:
It is also known as column chromatography. It is based on the fact that when solution of mixture comes in contact with some adsorbent, different components of a mixture get adsorbed to different extent on account of difference in power of adsorption.

Question 15.
0.2 gm of chlorine containing organic substance gave in Carius method 0.2870 gm of AgCl. Determine the percentage of chlorine in the compound?
Weight of organic substance = 0.2 gm.
Weight of AgCl = 0.2870 gm.
Percentage of Chlorine = \(\frac{35.5}{143.5}\) ×
= \(\frac{35.5}{143.5}\) × \(\frac{0.2870 × 100}{0.2}\)
= 35.5%

Question 16.
What is principle of vaccum distillation or distillation under reduced pressure 7 Explain with figure?
Answer:
Distillation under reduced pressure:
Many substances decompose at their boiling points. Hence, they cannot be purified by simple distillation. These compounds are distilled at low temperatures and low pressures. This is known as reduced pressure distillation. (MPBoardSolutions.com) Boiling point of a liquid is the temperature at which the vapour pressure of the liquid is equal to the atmospheric pressure. This means that by lowering the pressure to which a liquid is subjected, the boiling points of the liquid can be lowered. Similarly, if the pressure is increased, the boiling point also increases. This means that a liquid can be made to boil any temperature by varying the pressure.

As shown in the figure distillation under reduced pressure is carried out in a specially designed flask called Claisen flask. (MPBoardSolutions.com) Pressure in the receiver is reduced by vacuum pump which reduces pressure in distillation flask also and liquid begins to boil at low temperature. For example, glycerol is also distilled under reduced pressure. Its b.p. is 290°C, but using 12 mm pressure it can be distilled at 180°C.

Question 17.
How halogens are detected in organic compound?
Answer:
AgNO₃ Test:
If on adding HNO and AgNO₃ in sodium extract, white ppt. comes, then AgCl is present. If the white ppt. is soluble in excess of NH4C1 than Cl is present. On adding dil. HNO₃ and AgNO₃ in sodium extract, if yellow ppt. appears than bromine and iodine is present.

Question 18.
Write the difference between Inductive effect and Electrometric effect:
Inductive effect:

1. It is a permanent effect.
2. This arises due to displacement or σ – bond.
3. Partial negative or positive charge develops.
4. Displacement reaction occurs.
5. Always present in molecule.

Electrometric effect:

1. It is temperory effect.
2. Arises due to displacement of or π – bond.
3. Complete positive and negative charge develops.
4. Addition reaction occurs.
5. This effect arises due to presence of attacking reagent.

Organic Chemistry: Some Basic Principles and Techniques Long Answer Type Questions:

Question 1.
How halogen is detected in an organic compound?
Answer:
Estimation of Halogens Carius method:
In this method estimation of halogens (Cl, Br and I) is done. A known weight of organic compound containing halogen is heated with AgNO₃ and fuming nitric acid. Carbon, hydrogen and sulphur present in the compound gets oxidized and halogens form AgX (Silver halide).

Caluculations:
Suppose = W gm
Weight of organic substance = m gm
Weight of silver halide = x gm
Molecular mass of silver halide = (108 + x) gm
Molecular mass of silver halide = x gm
∴ m gm of silver halide contain halogen = \(\frac{x}{(108 + x)}\) × m gm
W gm of substance contain halogen = \(\frac{x × m}{(108 + x)}\) gm
∴ 100 gm of organic substance contains = \(\frac{x × m × 100}{(108 + x)}\) × W
Hence, percentage of halogen

Question 2.
Write short notes on carbanion?
Answer:
Carbanion:
Carbanions may be defined as negatively charged ions, in which carbon is having negative charge and it has eight electrons in the valence shell e.g.,

Generation of carbanions:
These are mostly generated in the presence of a base by heterolytic cleavage.

Types of alkyl Carbions:
Depending on the carbon bearing negative charge carbions may be of threee types,

Orbital structure of carbanion:
The negatively charged carbon atom in a carbanion is spl hybridized. It is expected to have a tetrahedral geometry. The three hybridized orbitals with one electron each are involved in the σ – bonds with the orbitals of other atom or groups. (MPBoardSolutions.com) The fourth hybridized orbital has overlapped. It is responsible for the negative charge on carbanion and also for the distortion of its geometry. The H actual shape of the carbanion is pyramidal.

Stability of carbanion:
The stability of carbanion can be discussed with the help of inductive effect.

Question 3.
What is Inductive effect? Write its uses?
Answer:
Inductive effect:
In a covalent band between two disimilar atoms having different electro negativities the electron pair does not remain in the centre but gets attracted towards the more elecronegative atoms. (MPBoardSolutions.com) The bond becomes some what polar due to unequal sharing of the electron pair. For example, in the bond

if X is more electronegative than C, the electron pair gets attracted towards S. This shifting of electrons develops a partial negative charge denoted by on δ^– on X and C attains a partial positive charge doneted by δ⁺. Thus,

Now consider a log chain of carbon atoms with a more electronegative element say chlorine attached at one end.

The electron pair of the bond between C₁ and X gets displaced towards electronegative chlorine atom. This results in developing of partial negative charge on chlorine and partial charge on carbon. This displacement is further transmitted to other carbon atoms of the chain but the magnitude of displacement goes on decreasing with the increases in the distance of the carbon atoms from the chlorine atom as shown below:

Thus, it can be concluded that a polar bond induces polarity in the other covalent bonds in a chain. This type of displacement of electrons is referred to as inductive effect (or I effect) or transmission effect. (MPBoardSolutions.com) Thus, inductive effect may be defined as, the permanent displacement of electrons along the chain of carbon atoms due to presence of polar covalent bond in the chain.

Types of inductive effect:
There are two type of inductive effects:

1. Electron withdrawing inductive effect (- I effect):
If the substantiates attached to the end the carbon chain is electron withdrawing, the effect is called – I effect. The decreasing order or – I effect of some atoms or groups is as follows:

2. Electron releasing inductive effect (+1 effect):
If the substantiates attached to the end of the carbon chain is electron releasing, the effect is called +I effect. Alkyl groups are electron releasing in nature. Thus, the decreasing order of +I effect is as follows:

Question 4.
Write the differences between electrophilic and nucleophilic reagents?
Electrophilic Reagent:

1. Deficient electrons.
2. Generally electrons are present in valence shell.
3. They are positive ions.
4. Neutral molecules with incomplete octet accept electrons.
5. They are Lewis acids.

Nucleophilic Reagents:

1. More electron present.
2. Generally 8 electrons are present in valence shell.
3. Negatively charged ion.
4. They are electron pair donor.
5. They are Lewis base.

Question 5.
Explain the Column chromatography technique for purification of organic compounds?
Answer:
Column chromatography:
It is based on the fact that when solution of mixture comes in contact with some adsorbent, different components of a mixture get adsorbed to different extent on account of difference in power of adsorption.

Column chromatography:
There are three steps of column chromatography:

1. Preparation of adsorbent column:
A long tube like burette is filled with a paste of a suitable adsorbent like activated Solvent Separation Continue elution magnesia, alumina, gypsum, silica gel, kieselguhr, etc. in a suitable organic solvent. The paste is prepared in that solvent in which the solution of the mixture to be separated, is prepared. When adsorbent is set, solvent is allowed to flow down.

2. Process of adsorption:
The substance or the mixture to be adsorbed is dissolved in least quantity of the nonpolar solvent like petroleum ether, benzene, etc. The solution is allowed to flow down the column. The different components of mixture get adsorbed in different parts of column. (MPBoardSolutions.com) The compound which is strongly adsorbed remains in the upper part of the column forming a band. Each component of mixture forms a separate band at a definite place. The coloured band formed are seen clearly. In case the bands are not seen due to being colourless, they are made visible by using a suitable indicator.

3. Elution:
In this process the adsorbed substance is extracted by a suitable solvent. The solvent used for this purpose is called eluent and the process is called elution. The solvents are used in the order of increasing polarity. Solvent in the increasing order of polarity are petroleum ether, petroleum ether containing benzene, alcohol with ether and pure ether.

These solvents are added one after the other. The substance which has been least adsorbed gets extracted with a solvent which is least polar while the component which has been adsorbed more strongly than others, is extracted by more polar solvent like alcohol. (MPBoardSolutions.com) By this way various components of mixture can be separated on several steps.

The various components can be separated from solvent by distillation or by using separating funnel. This technique is employed for separation of complex compounds like vitamins and hormones. The method is also employed for determination of purity of substance.

Question 6.
Write the IUPAC name of following compounds:

Answer:

Question 7.
Difference betweeen Aliphatic compound and Aromatic compound?
Answer:
Aliphatic compound:

1. These are open chain compound.
2. Generally C – C bond present.
3. Highly reactive.
4. Halogenation, nitration, sulphonation easily not occur.
5. Combustion energy high.
6. – OH group is neutral.

Aromatic compound:

1. These are closed chain compound.
2. Conjugated single and double bond present.
3. Less reactive.
4. Halogenation, nitration, sulphonation occur.
5. Combustion energy low.
6. – OH group is acidic in nature.

Question 8.
Explain the Duma’s method of determination of nitrogen in organic compound?
Answer:
Duma’s method:
This method can be used for estimation of nitrogen in all types of nitrogenous compounds. A known amount of nitrogen containing organic compound is heated with cupric oxide (CuO) in an atmosphere of CO₂. C and H₂O are oxidised to CO₂ and H₂O while N₂ gas is set free.
C + 2Cuo → CO₂ + 2Cuo
H₂ + Cuo → H₂O + Cu (in organic substance)
Nitrogen + CuO → N₂ + Some amounts of some oxides of N₂
A general equation for nitrogen containing compound is given below:
C_xH_yN_z + (2x + \(\frac{y}{2}\)) CuO → xCO₂ + \(\frac{y}{2}\) H₂O + \(\frac{z}{2}\) N₂ + (2x + \(\frac{y}{2}\)) Cu
If sulphur is present in the organic compound. It is converted into S02. During the above reaction, some oxides of nitrogen also be formed. (MPBoardSolutions.com) Therefore, the gaseous mixture is passed over heated reduced copper gauze which converts oxides of nitrogen back to nitrogen.
2NO + 2Cu → 2CuO + N₂
2NO₂ + 4Cu → 4CuO + N₂
The gaseous mixture containing CO₂, H₂O, SO₂ and N₂ is collected in a graduated nitrometer containing KOH solution. Water vapours are condensed whereas CO₂ and SO₂ are absorbed by KOH solution. Nitrogen is collected in the upper part of nitrometer. Volume of nitrogen is noted at room temperature and pressure.

Apparatus:
The main part of apparatus is combustion tube. It is a long tube, open at both the ends. The tube is packed with

1. Oxidized copper gauze which prevents backward diffusion of gases produced during combustion
2. CuO containing weighed amount of organic compound
3. Coarse CuO which oxidises the organic compound into CO₂, H₂O, SO₂ etc.,
4. A reduced copper oxide i.e., copper which converts oxides of nitrogen (NO, NO₂ etc.) back to nitrogen.

Oxides of nitrogen when passed over hot reduced copper gauze change to nitrogen by losing their oxygen.
Oxides of nitrogen + Cu → N₂ + CuO

The nitrogen set free is passed through SchifFs nitrometer filled with 40% solution of caustic potash. Caustic potash solution absorbs CO₂ while water formed gets condensed.
After the completion of combustion again CO₂ is passed through combustion tube to drive out all the remaining nitrogen gas to nitrometer.

The apparatus is cooled. Reservoir bulb of nitrometer is raised so that level of KOH becomes the same in reservoir bulb and nitrometer which means pressure becomes equal to atmospheric pressure. (MPBoardSolutions.com) Now, the volume of nitrogen in nitrometer is noted. Temperature of reaction and atmospheric pressure is noted from barometer. Aqueous tension at that temperature is noted from tables.

Observations and calculation:
Let,

1. Weight of organic substance = W gm
2. Volume of moist N = V ml
3. Temperature = t°C
4. Atmospheric pressure = P mm of Hg
5. Aqueous tension at t°C = p mm
6. Pressure of dry nitrogen = (P – p) mm

Calculation of volume of nitrogen at N.T.P.:

Question 9.
In an organic compound C = 40%, O = 53.34 % and H = 6.66%. If the vapour density is 30. Determine the molecular formula?
Or
In an organic compound A, C = 40% and H = 6.66%. The vapour density of A is 30. It turns blue litmus red and can react with ash. When its sodium salt is heated with soda lime, the first member of paraffin series obtained. What is A?
Solution:
C = 40%, H = 6.66%
O = 100 – [40 + 6.66] = 100 – 46.66 = 53.34%
In organic compound A:

Empirical formula of A = H₂O
Empirical formula mass = 12 + 2 + 16 = 30
Molecular mass = 2 × Vapour density
= 2 × 30 = 60
 = \(\frac{60}{30}\) = 2
∴ Molecular formula = (CH₂O)₂
= C₂H₄O₂ or CH₃COOH.
It is CH₃COOH

Question 10.
Explain the following with example:

1. Simple distillation
2. Chromatography
3. Crystallization.

Answer:
1. Simple distillation:
This method is employed for the purification of those liquids which boil without decomposition and are associated with non – volatile impurities. Liquids which have a difference of 30 – 40°C in their boiling points are purified by this method. On heating the mixture, vapours of pure substance are formed which condenses as they pass through the air or water condenser. (MPBoardSolutions.com) The pure liquid collects in the receiver while the non – volatile impurities are left behind in the flask. Some glass beads are also added to the distillation flask to avoid bumping.

2. Chromatography:
The process by which different components of a mixture are separated by distributing in stationary or mobile phases on the basis of difference in adsorption abilities on any adsorbent, is called chromatography.

Adsorption chromatography:
It is also known as column chromatography. It is based on the fact that when solution of mixture comes in contact with some adsorbent, different components of a mixture get adsorbed to different extent on account of difference in power of adsorption.

The mixture to be separated and purified is first dissolved in a suitable nonpolar organic solvent like petroleum ether, benzene, chloroform, alcohol, etc. This solution is allowed to flow down an absorption column.

For using this technique there should be proper adsorption column. Adsorbents like activated magnesia, alumina, calcium carbonate, gypsum, etc. is filled in a hard vertical tube (adsorbent column.).

3. Crystallization:
The method by which crystals of a substance can be made is called crystallization. Solids can be separated and purified by crystallization. e.g., nitre, alum, copper sulphate, etc. Some impure solids which differ in solubility in the same solvent, their separations can be achieved by fractional crystallization. (MPBoardSolutions.com) “If two or more components of a mixture which differ in their solubilities are dissolved in a solvent in which their solubilities slightly differ, they can be separated by fractional crystallization.”

Suppose, two solids A and B are dissolved in a solvent. If solubility of A is less as compared to B, then first saturated solution of the mixture is prepared and is allowed to cool. During crystallization first less soluble substance A will crystallize out. (MPBoardSolutions.com) It is separated by filtration. After this crystals of more soluble substance B will separate out. By this technique, crystals of both can be separated. The substances so separated are further crystallized many times to get pure substances.

Question 11.
Write the IUPAC name of following compounds:

(a) CH₃CH = C(CH₃)₂
(b) CH₂ = CH – C = C – CH₂
(c)
(d)
(e)

Answer:

MP Board Class 11 Chemistry Important Questions

MP Board Class 11th Chemistry Important Questions Chapter 11 p – Block Elements

p – Block Elements Important Questions

p – Block Elements Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
Boric acid is a polymer because:
(a) It is acidic in nature
(b) It contains H – bond
(c) Its basicity is one
(d) It has sheet like geometry
Answer:
(b) It contains H – bond

Question 2.
Which acid is not a protonic acid:
(a) H₃BO₃
(b) H₃PO₃
(c) H₂SO₃
(d) HCl₃
Answer:
(a) H₃BO₃

Question 3.
Al₂O₃ can be converted to anhydrous AlCl₃ by heating:
(a) Al₂O₃ with Cl₂ gas
(b) Al₂O₃ with HCl gas
(c) Al₂O₃ with NaCl in solid state
(d) A mixture of Al₂O₃ and Carbon in dry Cl₂ gas
Answer:
(d) A mixture of Al₂O₃ and Carbon in dry Cl₂ gas

Question 4.
Which one of the following is the correct statement:
(a) B₂H₆.2H₃ is known as Inorganic benzene
(b) Boric acid is a protonic acid
(c) Beryllium exhibits co – ordination number of six
(d) Chlorides of both berylium and aluminium have bridged chloride structure in solid phase
Answer:
(d) Chlorides of both berylium and aluminium have bridged chloride structure in solid phase

Question 5.
Shape and hybridization of BF₃ is:
(a) Linear, sp
(b) Planar, sp²
(c) Tetrahedral, sp³
(d) Pyrimidal, sp³
Answer:
(b) Planar, sp²

Question 6.
Reason for the formation of addition product between NH₃ and BF₃ is:
(a) Formation of H – bond between them
(b) Formation of ionic bond between them
(c) Formation of covalent bond between them
(d) Similar structure
Answer:
(c) Formation of covalent bond between them

Question 7.
Dry ice is:
(a) Solid ice without water
(b) Solid sulphur dioxide
(c) Solid carbon dioxide
(d) Solid benzene
Answer:
(c) Solid carbon dioxide

Question 8.
Which halide does not dissociate by water:
(a) CCl₄
(b) SiCl₄
(c) GeCl₄
(d) SnCl₂
Answer:
(a) CCl₄

Question 9.
Whose important constituent is silicon:
(a) Chlorophyll
(b) Haemoglobin
(c) Rocks
(d) Amalgam.
Answer:
(c) Rocks

Question 10.
Silicones are used:
(a) In the preparation of water proof cloth
(b) In the preparation of insulating material
(c) In preparing high elastic rubber
(d) All the above
Answer:
(d) All the above

Question 11.
Poisonous gas found in the smoke released from car:
(a) CH₄
(b) C₂H₂
(C) CO
(d) CO₂
Answer:
(C) CO

Question 2.
Fill in the blanks:

1. When formic acid is heated with cone. H₂SO₄, ………………………. is formed.
2. Bauxite ore containing ferric oxide as impurity is purified by ……………………… method.
3. Alumina containing both Fe₂O₃ and SiO₂ as impurity is subjected to purification by ………………………… process.
4. To obtain cent percent pure aluminium, it is refined by ………………………….. process.
5. Melting point of pure alumina is very high (about 2050°C), to melt it at low temperature ………………………… and ………………………………. are added due to which melting point reduces to ……………………………..
6. When bauxite contains more of silica as impurity then the ore is purified by ………………………………….. process.
7. Process of catenation is maximum in ……………………………..
8. Fullerene is an aliotrope of ……………………………
9. Lamp black is an ………………………………. aliotrope of carbon.
10. Maximum covalency of carbon is ………………………… whereas of Si is …………………………
11. Graphite is a ……………………… of electricity where as silicon is a ……………………..
12. Silicon carbide is called …………………………….

Answer:

1. Carbon mono – oxide
2. Baeyer’s
3. Hall’s
4. Hoope’s
5. Cryolite, Fluorspar, 870°C
6. Serpeck
7. Carbon
8. Carbon
9. Amorphous
10. 4, 6
11. Conductor, semiconductor
12. Carborundum

Question 3.
Answer in one word/sentence:

1. Which type of oxides are formed by elements of Group 13?
2. +1 oxidation state of T1 is more stable as compared to +3?
3. Aluminium reacts with base to form?
4. What are hydrides of boron known as?
5. Boron is m .inly found in which form?
6. What is “Two electron three centre bond” known as?
7. What will happen if excess of ammonia solution is added to copper sulphate solution?
8. What is formed when diborane react with ammonia?

Answer:

1. M₂O₃
2. Inert pair effect
3. Sodium metaaluminate
4. Borane
5. Borax (Na₂B₄O₇.10H₂O)
6. Diborane
7. Complex compound of cupric ammonium sulphite is formed
8. Borazine

Question 4.
Match the following:
[I]

Answer:

1. (b)
2. (c)
3. (e)
4. (a)
5. (d)

[II]

Answer:

1. (e)
2. (d)
3. (a)
4. (b)
5. (c)
6. (f)

p – Block Elements Very Short Answer Type Questions

Question 1.
The +1 oxidation state of Tl is stable than +3 state?
Answer:
Inert pair effect.

Question 2.
What the hydride of boron is called?
Answer:
Borane.

Question 3.
Who is called “Two electron three centre bond”?
Answer:
Diborane.

Question 4.
What happens when ammonia solution is added in excess in CuSO₄ solution?
Answer:
Complex compound of Cupric ammonium sulphate.

Question 5.
What is the formula of Alum?
Answer:
K₂SO₄.Al₂(SO₄)₃.24H₂O.

Question 6.
What is the name of C₆₀ carbon atom?
Answer:
Buckmister fullerenes.

Question 7.
Which carbide is harder than diamond?
Answer:
Boron carbide.

Question 8.
For which property of carbon, it have large number of compounds?
Answer:
Catenation.

Question 9.
What is the good conductor allotrope of carbon?
Answer:
Graphite.

Question 10.
What is water glass?
Answer:
Sodium silicate.

Question 11.
What is the hybridization of B in diborane?
Answer:
sp².

Question 12.
What is carborundum?
Answer:
SiC (Silicon carbide).

Question 13.
Which is electron deficient halide?
Answer:
BCl₃.

Question 14.
What is the property of B₂O₃?
Answer:
Acidic nature.

Question 15.
What is dry ice?
Answer:
Solid CO₂.

Question 16.
What is the formula of dimer of aluminium chloride?
Answer:
Al₂Cl₆.

Question 17.
What is used in welding as oxy acetylene flame?
Answer:
C₂H₂ (Acetylene).

Question 18.
What is Inorganic benzene?
Answer:
Borazene.

Question 19.
Which type of bond is present in diborane?
Answer:
2 electron 3 centre bond (Banana bond).

Question 20.
Which type of hybridization in carbon atom found in diamond?
Answer:
sp² hybridization.

Question 21.
Which is the most abundant metal found on earth surface?
Answer:
Aluminium.

p – Block Elements Short Answer Type Questions – I

Question 1.
How can you explain the higher stability of BCl₃ as compared to TICI?
Answer:
Boron exhibits +3 oxidation state and can form stable BCl₃. Thallium shows oxidation state of +1 as well as +3 but +1 oxidation state is more stable than +3 because of inert pair effect. Therefore, TlCl₃ is not stable. It can form stable TlCl.

Question 2.
Consider the compounds, BCl₃ and CCl₄. How will they behave with water? Justify?
Answer:
The B atom in BCl₃ has only six electrons in the valence shell and hence is an electron deficient molecule. It easily accepts a pair of electrons donated by water and hence BCl₃ undergoes hydrolysis to form boric acid (H₃BO₃) and HCl.
BCl₃ + 3H₂O → H₃BO₃ + 3HCl
In contrast, C atom in CCl₄ has 8 electrons in the valence shell. Therefore, it is an electron precise molecule. As a result, it neither accepts nor donates a pair of electrons from H₂O molecule and hence CCl₄ does not undergo hydrolysis in water.

Question 3.
Why caustic alkali like NaOH are not stored in aluminium vessel?
Answer:
Aluminium can easily dissolve in alkali and form sodium meta aluminate due to this reason caustic alkali like NaOH not stored in aluminium vessel.
2Al + 2NaOH + 2H₂O → 2NaAlO₂ + 3H₂.

Question 4.
What at normal temperature Al does not react with water?
Answer:
In the presence of air, a transparent protective oxide layer is formed on its surface, due to this at normal temperature it does not react with water.

Question 5.
Write the formula of Double salt or Alum?
Answer:
The general formula of double salt is R₂SO₄M₂(SO₄)₃ where R = monovalent metal like Na, K, Rb, Cs or NH₄⁺ ion and M is a trivalent metal like Fe⁺³, Al⁺³ or Cr⁺³.
Example: K₂SO₄.Al₂(SO₄)₃.24H₂O
(Potash Alum)

Question 6.
Write name of ores of aluminium? Write formulae also?
Answer:
Ores of aluminium are:

1. Oxides: Corundum, ruby, sapphire (Al₂O₃), emerald.
2. Hydrated oxides: Diaspore (Al₂O₃.H₂O), bauxite (Al₂O₃.2H₂O), gibbsite (Al₂O₃ – 3H₂O).
3. Fluoride: Cryolite (Na₃AlF₆).
4. Sulphate: Alunite or alum stone [K₂SO₄.Al₂(SO₄)₃.4Al(OH)₃].
5. Silicates: Felspar (K₂O.Al₂O₃.6SiO₂).
6. Phosphate: Phiroza or turquoise [AlPO₄. Al(OH)₃]H₂O.

Question 7.
Why aluminium is a strong reducing agent?
Answer:
Those elements which gives electron in the chemical reaction forms cations, and called reducing agent. The reducing power depends upon the electrode potential. (MPBoardSolutions.com) More the negative value of electrode potential more will be the reducing power. The electrode potential of A1 is – 1.67, so it behaves as a strong reducing agent.

Question 8.
Why gallium is a liquid at room temperature?
Answer:
Gallium changes into liquid when the room temperature is above 30°C. In solid state, its lattice energy is very less due to which the metallic bond can break easily at low temperature.

Question 9.
Write the resonating structure of CO₃^2- and HCO^–₃?
Answer:
The resonance structure of CO₃^2- is:

Resonating structure of HCO^–₃

Question 10.
Why the melting point and boiling point cf Boron is high?
Answer:
The crystal of Boron is formed by the covalent bond between the atoms. 2 atoms combines to form Icosahedron network which have 20 triangular faces and 12 comers. It makes boron very hard. Due to this it have high melting and boiling point.

Question 11.
What is Inorganic benzene? Why it is called Inorganic benzene?
Answer:
When diborane reacts with ammonia at 120°C forming an addition compound diammoniate of diborane B₂H₆.2NH₃. When this diammoniate of diborane is heated at 200°C, a stable cyclic compound B₃N ₃H₆ is formed.
Borazine B₃N₃H₆ has cyclic structure similar to benzene and thus it is called inorganic benzene.

Question 12.
What is corundum?
Answer:
Al is found in more than one crystalline forms. The most hardest crystal is known as corundum, which is used as abrasive.

Question 13.
Write the ores of Boron and write formula also?
Answer:
The ores of Boron are as follows:

1. Borax – Na₂B₄O₇.10H₂O
2. Kemite – Na₂B₄O₇.2H₂O
3. Colemanite – Ca₂[B₃O₄.(0H)₃]₂.2H₂O
4. Orthoboric acid – H₃BO₃.

Question 14.
Prove that Tl⁺³ is an oxidizing agent but Al⁺³ not?
Answer:
Due to inert pair effect, in boron family the stability of +1 oxidation state increases from top to bottom in a gap but the stability of +3 oxidation state decreases. (MPBoardSolutions.com) That is why in comparison to Tl⁺¹ is more stable than Tl⁺³. Tl⁺³ + 2e^– \(\underrightarrow { \Delta } \) Tl⁺¹
It is clear that reduction of Tl⁺³ is occur. Therefore, Tl⁺³ is an oxidizing agent but the oxidation state of Al⁺³ is not possible.

Question 15.
If B – Cl bond has a dipole moment, explain why BCl₃ molecule has zero dipole moment?
Answer:
BCl₃ has polar B – Cl bond, BCl₃ is planar triangular molecule in which three B – Cl bonds are inclined at an angle of 120°. The resultant dipole moment of two B – Cl bonds is cancelled by the dipole moment of third B – Cl bond. The vector sum of the dipole moments of three B – Cl bond is zero.

Question 16.
Boric acid is not protic acid? Why?
Answer:
Protic acid is the one that gives protons in solutions. Boric acid not a protic acid because it does not ionize in water to give proton but it behaves as Lewis acid by accepting electron pair from hydroxide ion.
B(OH)₃ + H₂O ⇄ [B(OH)₄]^– + H⁺

Question 17.
How borax is obtained from colemanite?
Answer:
Colemanite is boiled with concentrated sodium carbonate solution to get borax.
Ca₂B₆O₁₁ + 2Na₂CO₃ → Na₂B₄O₇ +2NaBO₂ + 2CaCO₃
On concentrating the product, crystals of borax is obtained. On passing CO₂ in mother liquor, borax is obtained.
4NaBO₂ + CO₂ → Na₂B₄O₇ + Na₂CO₃.

Question 18.
Why cryolite is used for the extraction of aluminium from alumina?
Answer:
The melting point of pure alumina is very high 2050°C, but in the presence of cryolite and fluorspar it melts at 870°C. In this way cryolite decreases the melting point of alumina. It also act as an electrolyte.

Question 19.
Write the uses of carbon monoxide?
Answer:

1. It is main constituent of water gas (CO + H₂) and producer gas (CO + N₂).
2. Used for the preparation of some metal carbonyls.
3. It is used as reducing agents.

Question 20.
Why diamond is found rare in nature than graphite?
Answer:
The formation of diamond occurs at very high pressure and in liquid state of carbon converts into crystal. But in nature this state is very rare. That is why diamond is found rare in nature.

Question 21.
What is dry ice? Write its main uses?
Answer:
Solid carbon dioxide is called dry ice because its crystal looks like ice but they did not wet the paper and clothes. (MPBoardSolutions.com) At – 78.5°C it converts into solid without changing into liquid. It is used as coolant for preserving edibles and as anaesthetic agent in surgery.

Question 22.
What is carborundum? Write its main use?
Answer:
The structure of silicon carbide is hard like diamond. It is known as carborundum. It is used as abrasive for cutting tools.

Question 23.
Write the name of the compound used as coolant, anaesthetic and solvent, and write formula also?
Answer:

Question 24.
Write the uses of graphite?
Answer:
The uses of graphite are as follows:

1. It is used in lead pencils in place of lead.
2. It is used as moderate in nuclear reactor.
3. Used as electrode in electrolytic and dry cells.

Question 25.
Write the different types of coal?
Answer:
The following types of coal are:

1. Pit: 60% carbon
2. Lignite: 70% carbon
3. Bituminous: 80% carbon
4. Anthracite: 90% carbon.

Question 26.
Write the uses of diamond?
Answer:
The uses of diamond are:

1. It is used for cutting or grinding of hard rocks.
2. It is used in high precision thermometer because it is a good conductor of heat.
3. For making windows of spaceships as it cuts off harmful radiations.

Question 27.
Why CO₂ is acidic? Explain with equation?
Answer:
The aqueous solution of CO₂ is acidic:
CO₂ + H₂O → H₂CO₃
Carbonic acid
It turns blue litmus red and forms salt with base.
2NaOH + CO₂ → Na₂ CO₃ + H₂O
Ca(OH)₂ + CO₂ → CaCO₃ + H₂ O.

Question 28.
Why should not be sleep in a closed room keeping burning sigri?
Answer:
Sigri should not be bum in closed room, as in the smoke from sigri contains large amount of CO. This CO inhaled through respiration combines with the haemoglobin of blood and forms carboxyhaemoglobin which interrupt the blood flow and so the death occurs.

Question 29.
What are carbides?
Answer:
Carbides are those binary compounds of carbon which are formed by carbon with less electronegative atom.
They are of many types:

1. Ionic carbides
2. Metallic carbides
3. Interstitial carbides
4. Covalent carbides.

Question 30.
Write the use of silica gel?
Answer:
Silica gel is an amorphous solid which have 4% moisture. It is used as catalyst in petroleum industries. It is also used in chromatography.

Question 31.
What is Thixotropy?
Answer:
The viscosity of any liquid decreases by shaking it. This property is called Thixotropy. When SiCl₄ is hydrolyzed at high temperature then a thixotropic silica is obtained.

Question 32.
What are interstitial carbides?
Answer:
When carbon atom get inserted in the interstitial places of crystal lattice of transition elements then the compound formed are called interstitial carbides. They are very hard and have high melting point.

Question 33.
What are methonides and acetanilides?
Answer:
Methonides:
The carbides after hydrolysis give methane are called methonides.
Al₄C₃ + 12H₂O → 4Al(OH)₃ + 3CH₄
Acetanilides:
The carbides after hydrolysis gives acetylene are called acetanildes.
CaC₂ + 2H₂O → Ca(OH)₂ + C₂H₂

Question 34.
Be and Ca are member of same group, but Ca forms CaC₂ but Be forms Be₂ C. Why?
Answer:
After the hydrolysis of CaC₂ acetylene is formed, so its structure is CaC₂ whereas after hydrolysis of Be₂ C, methane is formed therefore it is in the form of Be₂ C.

Question 35.
What is silane and germane?
Answer:
The hydrides of Si and Ge are called silane and germane respectively. They are shown by M_n H_2n+2 where M = Si and Ge. In silane n = 1 to 8 and in germane n = 1 to 5.

Question 36.
What is activated charcoal?
Answer:
Charcoal is soft and porous. It adsorbs coloured compounds and gases. It is heated in vapour up to 1100°C then its adsorption power increases and it is known as activated charcoal.

p – Block Elements Short Answer Type Questions – II

Question 1.
What happens when boric acid is heated?
Answer:
Boric acid releases three molecules of water at different temperature and forms boron trioxide.

Question 2.
Explain the structure of BF₃ and BH₄^–, also show the hybridization of B in them?
Answer:
In BF₃, there are 3 bonded electron pairs are present in B, so it is of sp² hybridized and have trigonal planar structure whereas in [BH₄^–] the number of bonding electron = 4. So, the hybridization is sp³ and structure is tetrahedral.

Question 3.
What is Alum? Write its general formula, method of preparation, properties and uses?
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 like Na, K, Rb, Cs or NH₄⁺ radical.
M = Trivalent metal like Fe, A1 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
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. Its one molecule contains 24 molecules of crystallized water.

2. On heating it melts 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:

1. 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.
2. In dyeing and printing cloth, in sizing of paper.
3. In purifying water.
4. In special foam extinguishers.

Question 4.
Aluminium is a less electrical conductor than copper but aluminium wire is used to make transmission cable. Why?
Answer:
Copper is better conductor, but aluminium is lighter metal, its density is very’ low, so it is better conductor in comparison to copper.

Question 5.
Boron forms only covalent compounds. Why?
Answer:
Due to smaller size and high ionization energy, it has very less tendency to form positive ion. Therefore B cannot release three electrons to form positive ion. (MPBoardSolutions.com) For attaining stable configuration it shares electrons with other atoms and form S covalent bond.

Question 6.
Why does boron trihalides behave as a Lewis acid?
Answer:
In BX₃ molecule 3 electrons of Boron atom shared with 3 electrons one each from halogen atom. Thus, total number of electrons in the outermost shell of boron trihalides are six which is short of 2 electrons than noble gas electronic configuration. (MPBoardSolutions.com) Thus, BX₃ is electron deficient compound and have strong tendency to accept lone pair donated by electron rich compound and form addition compound. Thus, Boron trihalides act as Lewis acid.

Question 7.
Why aluminium cannot be obtained by reduction method from its ore?
Answer:
Aluminium is highly electropositive and behaves as reducing agent. So it can be oxidized easily. On the basis of ionization energy and electron affinity that aluminium behaves as electron donor not electron acceptor. Therefore, it cannot be reduced.

Question 8.
Why is CO gas poisonous?
Answer:
CO combines with haemoglobin of blood and forms a stable compound carbo xyhaemoglobin 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 9.
Explain inert pair effect in Boron family?
Answer:
Reason for Inert Pair effect:
As mentioned earlier that tendency to show +1 oxidation state increases down the group. It means the tendency of r – electrons of valence shell to participate in bond formation decreases as we move down the group. (MPBoardSolutions.com) This reluctance of j – electron is termed as inert pair effect. This is due to poor shielding effect of the ns² electrons by the intervening d – and f – electrons. Another reason for the inert pair effect is that as the size of atom increases from Al to Tl.

The energy required to unpair the ns² electrons is not compensated by the energy released in forming the two additional bonds. The inert pair effect becomes more predominant as we go down the group because of increased nuclear charge which outweighs the effect of the corresponding increase in atomic size. The electrons thus becomes more tightly hold (more penetrating) and therefore, becomes more reluctant to participate in bond formation.

Question 10.
How boric acid is prepared from colemanite ?
Answer:
SO₂ gas passed in concentrated aqueous solution of colemanite (Ca₂B₆O₁₁).
Ca₂B₆O₁₁ + 2SO₂ + 9H₂O → 6H₃BO₃ + 2CaSO₃.

Question 11.
What is Borax glass?
Answer:
The anhydrous sodium tetraborate Na₂B₄O₇ is called Borax glass. This is obtained by heating normal boron above its melting point. It is a colourless glass like substance. On absorbing moisture from air converted into decahydrate.
Na₂B₄O₇ + 2H₂O ⇄ H₂B₄O₇ + 2NaOH.

Question 12.
What is the effect of heat on borax?
Answer:
Borax when heated strongly loses molecules of water of crystallization and changes into transparent bead.

Question 13.
How is excessive amount of CO₂ responsible for global warming?
Answer:
We know that CO₂ is very much essential for plants to carry photosynthesis. The gas is produced during various types of combustion reactions and is released into the atmosphere. It is taken up by plants as pointed above. (MPBoardSolutions.com) Thus, a carbon dioxide cycle works in the atmosphere and its percentage remains nearly constant.

However, over the years, combustion reactions have enormously increased. As a result, CO₂ gas is now present in excess in the atmosphere. Like methane, it also behaves like a green house gas and absorbs heat radiated by the earth. Some of the heat is released into the atmosphere while the rest is radiated back to earth. This has resulted in global warming over the years and has brought about major climatic changes.

Question 14.
What is the test for Borate radical?
Answer:
Test of Borate radical:
For testing acidic radical borate BO^3-3 in laboratory, die given salt is heated with ethanol and concentrated sulphuric acid. Vapours oftriethylborate formed bums with green edged flame. Actually, the salt is first converted into boric acid which then reacts with ethanol forming triethylborate.

Question 15.
Explain the structure of AlCl₃?
Answer:
Aluminium trichloride is obtained as a dimer Al₂Cl₆. There are three electrons in the valence shell of aluminium. These electrons get shared with three electrons of chlorine and form AlCl₃. In the valence shell of Al six electrons are present. To complete its octet it requires two electrons. In this condition the Al of AlCl₃ takes electrons of Cl of other AlCl₃ and octet is completed.

Question 16.
What is borax bead test? Explain?
Answer:
In qualitative analysis borax bead test is used for the detection of some coloured ions like Cu²⁺, Ni²⁺, CO²⁺ etc. On heating strongly borax loses water molecule of crystallization and ultimately melts into a transparent bead.

B₂O₃ reacts with certain metal to form metaborates having specific colours. The transparent bead is touched with the speck of the salt. It is then heated in an oxidizing flame and then in reducing flame from the colour of bead in hot and in cold, the basic radical can be predicted.

Question 17.
Give reason why CCl₄ is immiscible in water, whereas SiCl₄ is easily hydrolyzed?
Answer:
CCl₄ is immiscible in water as it is a covalent compound and is not hydrolysed by water because carbon does not have d – orbitals and hence cannot expand its coordination number beyond 4. However, silicon can expand its coordination number beyond 4 due to availability of d – orbitals.
CCl₄ + H₂O → No reaction
SiCl₄ + 4H₂O → Si(OH)₄ + 4HCl

Question 18.
Explain why CO₂ is a gas whereas SiO₂ is a solid. Or, Explain the structure of CO₂ and SiO₂?
Answer:
The structure of CO₂ is linear. The C is sp hybridized and the molecules of CO₂ are attached with weak vander Waals forces. That is why at normal temperature CO₂ is a gas.


SiO₂ is solid. Its structure is like lattice crystal. Every Si is joined with O atom tetrahedrally. Covalent single bond is present between Si – O. This bond is stronger than vander Waals forces. That is why SiO₂ is solid and have high m.p.

Question 19.
Explain back – bonding with example?
Answer:
There are six electrons in the valence shell of BCl₃. Due to electron acceptor BF₃ behaves as Lewis acid. It should be a strong Lewis acid but it works as weak Lewis acid.

In BF₃ the B is sp² hybridized, so BF₃ is a planar molecule. In this molecule 2p, orbital of B is completely vacant. In the 2p_z orbital of fluorine two electrons are present. The 2p_z orbitals of B and 2p_z orbitals of fluorine overlap and form a bond. This is called back – bonding.

Question 20.
(a) BCl₃ is stable but B₂Cl₆ was not found whereas AlCl₃ is unstable. Why?
(b) AlCl₃ is unstable, Al₂Cl₆ is stable? Give the reason?
Answer:
(a) BCl₃ is stable as in the valence shell of BCl₃ electrons are present but due to back – bonding the resonating structure gives stability to BCl₃. In B₂Cl₆ the d – orbitals of B is not vacant so it cannot accept the electron from chlorine and so it is impossible to form B₂Cl₆

(b) In valence shell of AlCl₃ six electrons are present and due to incomplete octet AlCl₃ is unstable. But Al₂Cl₆ dimer, the vacant d – orbital of Al accept the electrons from chlorine and so octet gets completed. So Al₂Cl₆ is stable.

Question 21.
Explain the orbital structure of carbon monoxide?
Answer:
Orbital structure of carbon monoxide: Both C and O in CO is in sp hybrid state.

One of the sp hybrid orbital of carbon overlaps with sp hybrid orbital of oxygen to form σ bond. One sp – orbital each of carbon and oxygen carries lone pair of electrons which remains non – bonded. Half – filled pz orbital of carbon and oxygen overlaps laterally to form π bond. Now, filled py orbital of oxygen overlaps with vacant py orbital of carbon to form coordinate bond.

Question 22.
What is tetrahedral nature of carbon?
Answer:
The atomic number of C is 6. On the basis of this two electrons are present in first shell and four electrons in second shell. So its valency is 4. (MPBoardSolutions.com) According to Levat and van’t Hoff, if in a centre of tetrahedral it is considered to be C atom then H at four comers covalency of C can be shown. The angle between the two valencies is 109°28′. According to Henry’s experiment the covalencies of C is symmetrical. It is presented as tetrahedral in space not in one plane.

Question 23.
Why is graphite used as lubricant?
Answer:
Graphite has layered structure in which the different layers are held together by weak van der Waals forces and hence can be made to slip over one another. Therefore graphite acts as a lubricant.

Question 24.
Why diamond is used as an abrasive?
Answer:
Every carbon in diamond is sp3 hybridized and each carbon form strong covalent bond with other four carbon atoms. In this way the structure of diamond is tetrahedral three dimensional which is very hard. Due to this, diamond is the most hardest element known, therefore it is used as abrasive.

Question 25.
What is silica garden?
Answer:
In sand, crystals of copper sulphate, ferrous sulphate, nickel sulphate, cadmium nitrate, 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 seen to grow in the solution and is known as silica garden.

Question 26.
Carbon monoxide exist but silicon monoxide does not. Why?
Answer:
Carbon can form a n bond after forming a a (sigma) bond with oxygen. At the same time vacant 2p, orbital of carbon can overlap with 2p: orbital of oxygen containing a lone pair of electron. (MPBoardSolutions.com) This is possible because carbon possess some electronegativity to gain a lone pair of electron of oxygen. On the other hand, Si atom is bigger in size and its electronegativity is also less due to which it cannot form 3p_z – 2p_z π bond. That is why SiO is not possible.

Question 27.
If the initial compound in the formation of silicone is RSiCl₃, then give the structure of the product?
Answer:
After the hydrolysis of alkyl trichlorosilane and after its polymerisation, chain isomers (silicones) are obtained.

Question 28.
What is catenation and in which compound it is found maximum?
Answer:
It is the ability of like atoms to link with one another through covalent bonds. This is found maximum in carbon. This is due to smaller size and higher electronegativity of carbon and unique strength of carbon – carbon bond.

Question 29.
Write the structure of graphite?
Answer:
Structure of graphite:
Carbon atoms in graphite are in sp² hybrid state. Carbon atoms get bonded to each other forming hexagonal structure. (MPBoardSolutions.com) Layer of hexagonal arrangement of carbon is held together by vander Waals forces. Carbon – carbon bond distance in hexagonal ring is 142 pm while between carbon of two layers it is 340 pm. These layers can slide over each other due to the presence of weak bond.

Question 30.
Write the balanced equation for preparation of water gas, carburated water gas and producer gas?
Answer:
Water gas:
This is a mixture of CO and H₂
C + H₂O \(\underrightarrow { 1160^{ \circ }C } \) [CO + H₂] – 29,000 calories.
Carburated water gas:
On passing the water gas into hot bricks present in oil, acetylene and ethylene forms and it mixes with water gas and form carburated water gas. In this, the amount of constituents are: CO = 30%, H₂ = 35%, Saturated hydrocarbon = 15.20%, Hydrocarbons = 10%, N₂ = 2.5-5%, CO₂ = 2%.
Producer gas:
Mixture of CO and N₂.
2C + air (O₂ + N₂) → 2CO + N₂

Question 31.
Why carbon cannot form complex compounds whereas other member of group 14 can form? Account for it?
Answer:
The tendency of an element to form complex is favoured by it:

1. High charge density
2. Small size
3. Availability of vacant d – orbital.

From the valence shell electronic configuration of carbon atom it is cleared that it can accompanied maximum of eight electrons in its valence shell while forming covalent bond since carbon atom does not have vacant d – orbital in these tetravalent compound so it cannot form complexes by accomodating any more electron. (MPBoardSolutions.com) On the other hand, in case of other member of group IV are available to their tetracovalent compounds and consequently they can form complexes by accepting lone pair of electron.

p – Block Elements Long Answer Type Questions – I

Question 1.
What is Borane? Write its characteristics and uses?
Answer:
Boron Hydrides or Hydrides of Boron:
Boron forms two series of hydrides. One of its nidoborane with general formula B_nH_n+4 and other is arachnoborane with general formula B_nH_n+6. Nidoborane series (B_nH_n+6): Its first member BH₂ does not exist. Second member B₂H₆ is very important and called as diborane. Other important members are pentaborane (9) B₅H₆, hexaborane (10) B₆H₁₀, octaborane (12) B₈H₁₂.

Arachnoborane series (B_nH_n+6):
Tetraborane B₄H₁₀, pentaborane (11) B₅H₁₁, hexaborane (12) B₆H₁₂ etc. are important members of this series.
Diborane (B₂H₆):
Diborane is the simplest borane.
Preparation:
(I) By the action of lithium aluminium hydride on boron trichloride in the presence of ether.

(II) By passing the silent electric discharge at low pressure through a mixture of boron trichloride or tribromide and excess of hydrogen.

(III) Lab preparation:
By the action of iodine on sodium borohydride in diethylene glycol dimethyl ether (diglyme) as solvent.
2NaBH₄ + I₂ → B₂H₆ + 2NaI + H₂

(IV) Industrial preparation:
Diborane is obtained on large scale by the reduction of BF₃ with NaH at 175°C.
2BF₃ + 6NaH \(\underrightarrow { 175^{ \circ }C } \) B₂H₆ + 6NaF

Physical properties:
It is a colourless gas with foul smell and highly toxic in nature. Uses:

1. As rocket fuel
2. As catalyst in polymerisation reaction.

Question 2.
Explain the structure of diborane? Or, What are electron deficient compounds? Or, What are 2 electron 3 centre compounds?
Answer:
Diborane is prepared by reduction of BF₃ with LiH at 450 K.
2BF₃ + 6LiH \(\underrightarrow { 400^{ \circ }C } \) B₂H₆ + 6LiF
Diborane is also prepared by treatment of BCl₃ with LiAlH₄.
4BCl₃ + 3LiAlH₄ → 2B₂H₆ + 3LiCl + 3AlCl₃.

Structure of diborane:
The structure of diborane is determined by electron diffraction studies. According to this structure sixteen electrons are required for the formation of conventional covalent bond whereas in diborane there are only twelve valence electron three from each boron and six from hydrogen. In this way it is an electron deficient compound.

Diborane has two coplanar BH₂ groups and the remaining two hydrogen atoms lie centrally between BH₂ group. In the structure four hydrogen atoms are terminal hydrogen and two hydrogens are bridge hydrogen. The two BH₂ groups lie in the same plane while the two bridging hydrogen atoms lie in a plane perpendicular to this plane. (MPBoardSolutions.com) Each bridged hydrogen is bonded to the two atoms only by sharing of two electrons. Such covalent bond is called three centre electron pair bond or multi centre bond.

Question 3.
What happens when:

1. Borax is heated strongly.
2. Boric acid is added to water.

Answer:
1. Borax when heated strongly loses molecules of water of crystallization and changes into a transparent bead.

2. It is weak monobasic acid. Boric acid is not a proton donor (i.e protic acid). It accepts a pair of electron from water and acts as a Lewis acid (electron acceptor).
B(OH)₃ + 2HOH → [B(OH)₄]^– + H₃O⁺

Question 4.
A certain salt X, gives the following results:

1. Its aqueous solution is alkaline to litmus.
2. It swells up to glassy material Y on strong heating.
3. When cone. H₂SO₄ is added to a hot solution of X, white crystal of an acid Z separates oct?

Write equation for all the above reaction and identify X, Y and Z?
Answer:
1. Since the aqueous solution of salt (X) is alkaline to litmus, it must be the salt of strong base and a weak acid.

2. Since the salt (X) swells up to glassy material Y on strong heating, therefore X must be Borax and Y must be a mixture of sodium metaborate and boric anhydride.

3. When cone. H₂SO₄ is added to hot solution of X, white crystals of an acid Z separates out. Therefore Z must be orthoboric acid.
Equation for the all above reactions are as follows:

4. 

Question 5.
Explain the following reaction:

1. Silicon is heated with methyl chloride at high temperature in the presence of copper.
2. Silicon dioxide is treated with hydrogen fluoride.
3. CO is heated with ZnO.
4. Hydrated alumina is treated with aqueous NaOH solution.

Answer:
1. A mixture of m ono, di and trimethylchlorosilances along with a small amount of tetra – methylsilance is formed.

2. SiO₂ + 4HF → SiF₄ + 2H₂O
SiF₄ + 2HF → H₂SiF₆

3. ZnO is reduced to zinc metal.
ZnO + CO → Zn + CO₂

4. Aluminium dissolves to form sodium meta – aluminate.

Question 6.
Compare Boron and Carbon?
Answer:
Similarities:

1. Boron and carbon are non – metals.
2. Both show allotropy.
3. Both form more than one type of hydride
4. Both form covalent compounds
5. Compounds of both the elements are soluble in organic solvents.

2NaOH + CO₂ → Na₂CO₃ + H₂O
2NaOH + B₂O₃ → Na₂B₂O₃ + H₂O

Dissimilarities:

Dissimilarities between Boron and Carbon:
Boron:

1. Electronic configuration is 1s²,2s²,2p¹.
2. Valency is 3.
3. Not formed double and triple bond.
4. Electron deficient compounds

Carbon:

1. Electronic configuration is 1s²,2s²,2p¹.
2. Valency is 4.
3. It forms double and triple bond.
4. Not electron deficient compounds.

Question 7.
Compare the structure of BCl₃ and AlCl₃?
Answer:
BCl₃ is an electron deficient compound which present always as monomer. In BCl₃, the B is sp² hybridized. Therefore, its structure is triagonal and bond angle is 120°. The atomic radius of Boron is small and Chlorine bridge is unstable, therefore, it will not form dimer.

AlCl₃ is always present as dimer. In this structure, the A1 atom accept lone pair of electron from the Cl atom of another Al atom and octate is completed and attain stability.

Question 8.
Explain the structure of diborane and boric acid?
Answer:
Structure of diaborane:
Diborane LS an example of ‘three centre two electron pair bond’ or banana bond in which two boron atom and one hydrogen atom has only 2 electrons available for bonding. These are two banana bonds of this type in diborane.

Two boron atoms and two hydrogen atoms (Total four hydrogen atom) forming covalent bond with each boron atom lies in the same plane. (MPBoardSolutions.com) Among remaining two hydrogen atom one lies above and other below the plane of boron atoms forming banana bond. These hydrogens are called bridging hydrogen.

Bond length of regular B – H covalent bond is 119 pm.
Bond length of banana B – H bond is 134 pm.
Interatomic distance between two boron atom is 178 pm. The reason for the formation of banana bond is that there are only 12 electrons in diborane (6 from two boron and 6 from six hydrogen) while, for the formation of 8 covalent bonds between 2 boron and 6 hydrogen, total 16 electrons are required. Thus, diborane is an electron deficient compound.

Structure:
Crystal of boric acid are soapy in touch and have layer like structure similar to graphite
BO₃^3- is bonded to hydrogen covalentlly and through hydrogen bond.

Question 9.
Describe abnormal behaviour of boron with respect to other elements of group – 13?
Answer:
The first element boron of group-13 differs from the rest elements due to its:

1. Small size
2. High electronegativity
3. Absence of d – orbitals

Abnormal behaviour:

1. Boron is non – metal and others are metals.
2. Boron exhibits allotropy while others do not.
3. Oxides and hydroxides of B are acidic in nature. Oxides of A1 and Ga are amphoteric while articles and hydroxides of In and T1 are basic.
4. Melting point of B is higher in comparison to rest of other elements.
5. Boron combines with metals to form borides. Whereas others do not form borides.
6. B does not form alum, but Al, Ga and In form alums. ,
7. B does not form B⁺³ but others elements of this group form m⁺³ ions.
8. B generally forms covalent compounds whereas other elements of the group -13 form both covalent and electrovalent compounds.

Question 10.
Write a not on fullerene?
Answer:
After 1985, new allotropes of carbon were discovered which contains cluster of C₃₂, C₃₀, C₆₀, C₇₀ etc. they are known as fullerene. C₆₀ cluster is of special importance which is known as Buck – minister fullerene.

The structure of fullerene resembles with that of a soccer ball with six membered as well as five membered rings the carbon atom in fullerene have been found to be equivalent and are connected by both single bond and double bond. These are called Bucky Ball. Physical Properties:

1. It is soluble in organic solvent.
2. Its molecule is sufficiently stable.
3. It forms complex with platinum.

Chemical properties:
1. Combustion:
In limited supply of air it forms CO and in complete combustion it forms CO₂.
2C + O₂ → 2CO + 52 Kcal
C + O₂ → CO₂ + 94 Kcal

2. Action with metal:
It reacts with metal at high temperature to form carbides.
Ca + 2C → CaC₂
4AI + 3C → Al₄C₃

3. Oxidizing property:
Oxides of many metals are reduced to metal at high temperature.
ZnO + C → Zn + CO
Fe₂O₃ + 3C → 2Fe + 3CO.

Uses:

1. As Lubricant
2. As conductor.

Question 11.
Why carbon shows dissimilarity with the other members of its group?
Answer:
Carbon shows dissimilarity with the other members because the reason is:

1. Atomic radius and ionic radius is small.
2. High ionisation energy.
3. High electron affinity.
4. Absence of d – orbital.

Anomalous behaviour:

1. The melting and boiling points of C is high in comparison to other members.
2. Property of catenation is high.
3. Carbon forms multiple bonds, but other members don’t.
4. Mono – oxide of C is known, but of other members are not found.
5. The maximum valency of C is 4, but other members 6.
6. Carbon does not form compounds like other members.
7. Carbon forms more than one type of hydride.
8. CO₂ is a gas, but dioxide of other members are solid.

p – Block Elements Long Answer Type Questions – II

Question 1.
Write short notes on following:

1. Freon
2. Silicone.

Answer:
1. Freon: Dichloro – difluoro methane is known as freon. Freon is formed by the reaction of CCl₄ with HF or SbF₃ in presence of SbCl₅. Freon is used as coolent in refrigerators and A.C.

2. Silicone:
Silicones are polymers containing R₂SiO units. Name silicone is given to it because of the resemblance of formula of its basic unit with that of ketones R₂CO. Silicones are also found in form of polymers while ketones can exist as independent molecule as well as polymer.

Preparation:
Alkyl chloride is heated with silicon in presence of copper powder to produce dialkyl – dichloro – silane.

This compound on hydrolysis produces dihydroxysilane.

Due to the presence of two – OH group on a silicone atom, these molecules condenses together to form polymers.

In this way, due to condensation of large number of units, unite together. Cross – linking occurs if, trihydroxysilane units are involved and a two – dimensional polymer is formed.

Uses:

1. Silicones are insulators, heat resistant, non-reactive and water repellant. Due to this, it is used for making waterproof cloth and paper.
2. As a lubricant.
3. As insulators in electric appliances.
4. Silicone oil is used in high temperature thermostate and vacuum pumps.

Question 2.
Explain Cold – schmidt alumino thermic process with a labelled diagram? Or, Explain the thermite welding process?
Answer:
Thermite process:
It is also called Gold – schmidt 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. (MPBoardSolutions.com) 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 0₂ 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 3.
Write short notes on zeolites?
Answer:
These are micro porous aluminosilicate having general formula M_x/nⁿ⁺[AlO₂]_x[SiO₂]_y^x- mH₂O. Aluminosilicates are obtained by substituting some of the silicon atoms in the three dimensional network of silicon – dioxide by Al atoms. The negative charge carried by aluminosilicate, framework is neutralized by exchangeable cations such as Na⁺, K⁺ or Ca²⁺ of valence n, while m water molecule (mH₂O) fill the voids.

Uses:

1. For the manufacture of cement and bricks.
2. In porcelain and glass industries.
3. In pottery industries.

Question 4.
Write similarities and dissimilarities between Boron and Aluminium?
Answer:
Similarities:
1. Both possess similar outer electronic configuration ns²np¹.

2. B and Al both are trivalent due to presence of 3 electron in outermost shell.

3. Both form trioxide with oxygen.
4B + 3O₂ → 2B₂O₃
4Al + 3O₂ → 2Al₂O₃.

4. Boron and aluminium both react with base and form H₂ gas.
2B + 6NaOH → 2Na₃BO₃ + 3H₂
2Al + 2NaOH + 2H₂O → 2NaAlO₂ + 3H₂.

5. Boron and aluminium both form similar type of compounds with alkyl radicals.
2BCl₃ + 3Zn (CH₃)₂ → 2B(CH₃)₃ + 3ZnCl₂
2Al + 3Hg(CH₃)₂ → 2Al(CH₃)₃ + 3H₂.

6. Nitride of both are hydrolyzed and give ammonia.
BN + 3H₂O → H₃BO₃ + NH₃
AIN + 3H₂O → Al(OH)₃ + NH₃.

Dissimilarities:

Boron:

1. It is a non – metal with three dimensional structure.
2. It exhibits allotropy.
3. It does not form tripositive B⁺³ ion.
4. B₂O₃ is acidic in nature.
5. It forms stable hydride.
6. It reacts with metals and form borides.
7. It is bad conductor of electricity.
8. Maximum covalency of boron is 4.

Aluminium:

1. Aluminium is a metal with metallic structure.
2. It does not exhibit allotropy.
3. It forms tripositive Al⁺³ ion.
4. Al₂O₃ is amphoteric.
5. It forms unstable hydride.
6. It forms alloys.
7. It is good conductor of electricity.
8. Maximum co valency of Aluminium is 6.

Question 5.
Write short notes on:

1. Sodium zeolite
2. Sodium silicate.

Answer:

1. 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. (MPBoardSolutions.com) When hard water is passed through permutit kept in a column then calcium and magnesium ions are displaced by sodium. Sodium 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.

2. Sodium silicate or water glass:
It is bright like glass and soluble in water, therefore, it is known as water glass. It contains sodium meta – silicate and silica in excess and its formula is Na₂SiO₃. SiO₂ or Na₂SiO₃.3SiO₂.
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.

Question 6.
Describe, diagonal relationship between B and Si?
Answer:
B and Si show resemblances in properties. The resemblances arise due to the fact that both B and Si are non-metals and have small difference in size. Their following properties are similar:

1. Both are non – metal, non – conductors unable to form cations and possess high melting point.
2. B and Si both do not exist in free state in nature and only exist in the form of their oxides.
3. Both do not react with dilute acid.
4. Both react with nitrogen at high temperature to form nitrides.
   2B + N₂ → 2BN
   Si + 2N₂ → SiN₄
5. Both form covalent hybride.
6. Both react with metals to form boride and silicates.
7. Both form covalent chlorides which are hydrolyzed in aqueous solution.
   BCl₃ + 3H₂O → H₃BO₃ + 3HCl
   SiCl₄ + 3H₂O → H₂SiO₃ + 4HCl
8. H₃BO₃ and H₄Si₄ are very weak acids.

Question 7.
What is Alum? Write its general formula, method of preparation and properties?
Answer:
Double salts which can be represented by the general formula R₂SO₄.M₂(SO₄)₃. 24H₂O are called alums.
Here, R = Monovalent metal, like: Na, K, NH₄⁺
M = Trivalent metal, like Fe, Cr, Al Nomenclature of alum: