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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 12th Chemistry Important Questions Chapter 8 The d-and f-Block Elements

The d-and f-Block Elements Important Questions

The d-and f-Block Elements Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
General electronic configuration of transition element is :
(a) (n – 1)d^{1 – 10} ns¹
(b) (n – 1)d¹⁰ns²
(c) (n – 1)d¹d^{1 – 10}ns²
(d) (n – 1)d⁵ns¹.
Answer:
(c) (n – 1)d¹d^{1 – 10}ns²

Question 2.
Reason of Lanthanide contraction is :
(a) Negligible screening effect of f – orbitals
(b) Increasing nuclear charge
(c) Decreasing nuclear charge
(d) Decreasing screening effect.
Answer:
(a) Negligible screening effect of f – orbitals

Question 3.
Chromyl chloride test confirms the presence of :
(a) Cl^–
(b) SO₄^2-
(c) Cr³⁺
(d) Cr³and Cl^–.
Answer:
(a) Cl^–

Question 4.
Formula of Mohr’s salt is :
(a) FeSO₄.7H₂O
(b) FeSO₄.(NH₄)2SO₄.6H₂O
(c) CU(OH)₂.CUCO₃.6H₂O
(d) Fe₂O₃.3H₂O.
Answer:
(b) FeSO₄.(NH₄)2SO₄.6H₂O

Question 5.
The outer electronic configuration of chromium is :
(a) 4s¹, 3d⁵
(b) 4s², 3d⁴
(c) 4s⁰, 3d⁶
(d) 4s²,3d⁵.
Answer:
(a) 4s¹, 3d⁵

Question 6.
The equivalent weight of KMnO₄ is alkaline medium will be :
(a) 31.60
(b) 52.66
(c) 79.00
(d) 158.00.
Answer:
(d) 158.00.

Question 7.
The Lanthanide which is widely used :
(a) Lanthanum
(b) Nobelium
(c) Thorium
(d) Cesium.
Answer:
(d) Cesium.

Question 8.
Electronic configuration of Gadolinium is :
(a) [X_e]4f⁶,5d⁹,6s²
(b) [X_e]4f⁷,5d¹6s²
(c) [X_e]4f³,5d⁵,6s²
(d) [X_e]4f⁶,5d²,6s².
Answer:
(b) [X_e]4f⁷,5d¹6s²

Question 9.
In 3d series which element shows highest oxidation state :
(a) Mn
(b) Fe⁺²
(c) Ni
(d) Cr.
Answer:
(a) Mn

Question 10.
Fe, Co, Ni are magnetic substance of which type : (MP 2018)
(a) Paramagnetic
(b) Ferromagnetic
(c) Diamagnetic
(d) Antiferromagnetic.
Answer:
(b) Ferromagnetic

Question 11.
Number of unpaired electrons in Fe⁺² ion is :
(a) 0
(b) 4
(c) 6
(d) 3.
Answer:
(b) 4

Question 12.
In which of the compounds Mn shows highest oxidation state :
(a) K₂MnO₄
(b) KMnO₄
(c) MnO₂
(d) Mn₃O₄.
Answer:
(b) KMnO₄

Question 13.
The atomic radius and ionic radius of Zr and Hf are similar due to :
(a) Diagonal relationship
(b) Both are present in same group
(c) Lanthanide contraction
(d) Similar chemical properties.
Answer:
(c) Lanthanide contraction

Question 14.
Transition elements are coloured due to :
(a) Paired electron in d – orbital
(b) Paired electron in f – orbital
(c) Unpaired electron in d – orbital
(d) None of these
Answer:
(c) Unpaired electron in d – orbital

Question 15.
Stability of ferric ion is due to :
(a) Half filled d – orbital
(b) Half filled f – orbital
(c) Completely filled d – orbital
(d) Completely filled f – orbital.
Answer:
(a) Half filled d – orbital

Question 2.
Fill in the blanks :

1. Metals Fe, Co, Ni are known as …………………….
2. Ionic size of trivalent cations are ……………………. with increase in atomic numbers.
3. The transition metals having lower oxidation state shows ……………………. nature.
4. K₂Cr₂O₇ is a strong ……………………. agent, which gives nascent oxygen.
5. Zn shows only ……………………. oxidation state.
6. f – block elements are known as ……………………. elements.
7. Transition elements and their compounds act as …………………….
8. General electronic configuration of inner transition element is …………………….
9. Chemical form of Potassium manganate is …………………….
10. d – block elements are also known as …………………….

Answer:

1. Ferrous metals
2. Decreases
3. Basic
4. Oxidising, 3
5. +2
6. Inner transition,
7. Catalyst
8. (n – 2)f^{1 – 14} (n – 1)d^{1 – 2}(n – 1)d^{1 – 2}ns²
9. K₂MnO₄
10. Transitional Elements.

Question 3.
State true or false :

1. Mercury is liquid and its oxidation state is +1 and +2.
2. Higher oxidation state of transition elements are acidic in nature.
3. Lanthanides and Actinides both are transition elements.
4. In all transition elements normal oxidation state is +2.
5. Zn, Cd, Hg represent variable oxidation state.
6. Cu⁺² ion is colourless and diamagnetic.
7. Plutonium used as fuel in nuclear reaction and in formation of atomic bomb.
8. Transition elements easily form interstitial compounds.

Answer:

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

Question 4.
Match the following :

Answer:

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

5. Answer in one word / sentence :

1. Which is colourless Cu²⁺or Cu⁺?
2. In a reaction KMnO₄ is replaced by K₂MnO₄ then what will be change in oxidation state of Mn?
3. Which series shows higher oxidation state lanthanides or actinides?
4. Which oxidation state of lanthanum is most stable?
5. Write the equivalent weight of K₂Cr₂O7 in acidic medium.
6. How many unpaired electrons are present in Fe³⁺?
7. Give the name of oxidising agent used in chromyl chloride test.
8. Out of d – block elements, Zn does not show variable valencies, why?
9. Which is the most important oxidation state of Cu?
10. d – block elements can be divided into how many series?
11. What is Lunar caustic?
12. In d – block elements Zn does not exhibit variable oxidation state. Why?
13. What is the alkaline solution of HgCl₂ and KI known as?

Answer:

1. Cu⁺²
2. 1
3. Actinides
4. +3
5. 49
6. 5
7. K₂Cr₂O₇
8. Completely filled ‘d’ orbitals
9. +2
10. 2
11. AgNO₃ (Silver nitrate)
12. Due to fully filled d – orbitals
13. Nessler’s reagent.

The d-and f-Block Elements Very Short Answer Type Questions

Question 1.
Actinide contraction is greater from element to element than lanthanide contraction. Why? (NCERT)
Answer:
This is due to poor shielding effect by 5f electrons in the actinoids than that of 4f electrons in the lanthanoids.

Question 2.
Explain Cu⁺ is colourless while Cu⁺² is coloured.
Answer:
If a transition metal contain unpaired electron, it shows paramagnetism and forms coloured compound. In Cu⁺d – orbital is partially filled (3d⁹) thus Cu⁺ is colourless and diamagnetic while Cu⁺² is coloured and paramagnetic.

Question 3.
Why are Mn²⁺ compounds more stable than Fe²⁺ towards oxidation to their +3 oxidation state? (NCERT)
Answer:
Mn⁺² has stable electronic configuration [Ar]4r⁰3d⁵ and they do not easily change to Mn⁺³, Fe⁺² [Ar] 4s⁰3d⁶ on oxidation forms Fe⁺³ [Ar] 4s⁰3d⁵ a more stable configuration.

Question 4.
What are interstitial compounds? Why are such compounds well known for transition metals? (NCERT)
Answer:
Most of the transition elements form interstitial compounds at high temperature with atoms of non – metallic elements like H,B,C,N, Si etc. Small atoms of these non – metallic elements fit in the interstitial voids of crystal lattice of transition elements. These are called interstitial compounds.

Question 5.
What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses. (NCERT)
Answer:
An alloy is a homogeneous mixture of two or more metals or metals and non – metals. An important alloy contains lanthanoid metal is mischmetal which contains 50% Cerium and 25 % Lanthanum, with small amounts of Nd (Neodymium) and Pr (Praseodymium). It is used in Mg – based alloy to produce bullets, shell and lighter flints.

Question 6.
Ti²⁺, V²⁺ and Cr²⁺ are strong reducing agents. Why?
Answer:
For Ti²⁺, V²⁺ and Cr²⁺, values of M²⁺/M is negative which justify that they are strongly reducing.

Question 7.
Write the unit of magnetic moment.
Answer:
Bohr Magneton (BM).

The d-and f-Block Elements Short Answer Type Questions

Question 1.
Wnat is lanthanoid contraction? What are the consequents of lanthanoid contraction? (NCERT)
Answer:
Interesting feature of the atomic size of lanthanides is that on moving down the group steady decrease in atomic size is observed. The shape of f – orbital is in such a way that its shielding effect is minimum, there fore on addition of extra electron in f – subshell only attractive force increases. The steady decrease (contraction) in size of fourteen lanthanide elements (La³⁺1.06 Å to Lu³⁺ 0.8 Å) by a value of about 0.2Å is known as lanthanide contraction.

Reason:
1. The new electrons in lanthanides instead of going to outermost shell enters (n – 2)f – suborbital as a result of which force of attraction increases between electron and nucleus due to which atom or ion contracts.

2. Electron entering in (n – 2)f – suborbital have negligible or zero shielding effect over electrons present in the last orbit. In addition the shape of f – suborbital is not favourable for the shielding effect of electrons. Thus, lanthanide contraction occur.

Consequences of lanthanide contraction :
1. Change in the properties of lanthanides : Due to lanthanide contraction, little change occurs in the properties of lanthanides. So it is very difficult to obtain them in pure state.

2. Influence over the properties of other elements : Lanthanide contraction have an important influence over the element present before and after it e.g., there is difference in properties of Ti and Zr while Zr and Hf have similar properties.

Question 2.
What are Transition elements? They show metallic character. Why?
Answer:
Elements whose atoms in their ground state or ions in their common oxidation states have incomplete or partially filled d – orbitals are called transitional elements. They are in group 2 to 13. Example : Fe, Ni, Co, etc.
General formula : (n – 1)d^{1 – 10}ns^{1 – 2}

Metallic character of an element depends on its tendency to form cation by loosing one or more electrons from its atom. All transitional elements are metals because they contain one or two electrons in their outermost shell which can be easily lost due to low ionisation energy. Thus, they are metallic in nature.

Question 3.
Why do transition metals exhibit variable oxidation states?
Answer:
Transition metals exhibit variable valency because the energy subshell (n – 1 )d and ns are very close. Thus, possibility to lose electrons from ns subshell as well as from (n -1 )d subshell is very much if there are unpaired electrons. So oxidation states of these metals may increase. In these elements Mn shows maximum variable valencies.

Question 4.
Transition elements form alloy easily. Explain.
Answer:
It is the homogeneous mixture of two or more metals or metals with non – metals. Alloys are made to confer the property of metals. Transition elements have great tendency to form alloys because these elements have similar atomic size and can mutually substitute their positions in their crystal lattice. Alloys are comparatively hard and have higher m.p. than the elements from which they are made.

Question 5.
The radius of Fe²⁺ ion is smaller than the radius of Mn²⁺ion, why?
Answer:
The atomic number of Fe (26) is more than the atomic number of Mn (25). Due to higher value of atomic number, iron nucleus contains more protons. Hence the force of attraction between the nucleus and the electrons of outermost orbit is more. Due to strong attractive force of the nucleus the electron cloud is pulled inwards which results in smaller size of Fe²⁺ ion as compared to Mn²⁺ ion.

Question 6.

1. Transition metals possess the ability to form complex compounds. Explain.
2. Zn, Cd and Hg do not show the properties of Transition elements.
3. Why is Ti known as a wonder metal?

Answer:
1. Cause of formation of complex compounds by Transition metals :

• Small size of ions of these elements and high nuclear charge due to which these ions attract ligands.
• They possess vacant d – orbitals in order to accomodate the electron pair donated by ligand.

2. Elements in which (n – 1) d – orbital is partially filled are known as Transition elements Whereas in Zn [3d¹⁰4s²], in Cd [4d¹⁰ 5s²] and in Hg [5d¹⁰6s²] state is found. Therefore, these do not show the properties of Transition elements.

3. Titanium is a shining white metal. It is extended strong (harder than steel), has high m.p. Good conductor of electric current resistant to corrosion and light metal. Due to all these qualities, it is called wonder metal.

Question 7.

TiO₂ is white whereas TiCl₃ is violet, why?
In first transitional series paramagnetism increases till Cr then it starts de – creasing. Why?

Answer:
1. In TiO₂, Ti is in +4 oxidation state (3d⁰4s⁰) having a vacant rf-orbital hence there is no d – d transition and it is white. On the other hand, in TiCl₃, Ti is in +3 oxidation state (3d¹4s⁰) having one unpaired electron in its 3d – orbital, hence it is coloured.

2. In first transitional series, the number of unpaired electrons till Cr (3d⁵) increases and then due to pairing the number of unpaired electrons decreases. Thus, due to this at first paramagnetismjacreases till Cr and then it decreases.

Question 8.
Write five differences between Lanthanide and Actinide.
Answer:
Differences between Lanthanides and Actinides Elements :
Lanthanides Elements:

• Lanthanides show oxidation state of + 3 mainly and +2 and +4 in few compounds.
• Tendency to form complex compound is low.
• Lanthanide compounds are less basic than actinide compounds.
• These do not form oxo – ions.
• Except promethium all are non – radioactive elements.
• Last electron enters in 4f – subshell.

Actinides Elements:

• Actinides show + 3 oxidation state together with + 4, + 5 and +6 in all compounds.
• Tendency to form complex compund is more than lanthanides.
• Actinide compounds are more basic.
• Actinides form oxo – ions as UO₂⁺, NpO⁺, PuO₂⁺, etc.
• All actinides are radioactive.
• Last electron enters in 5f – subshell.

Question 9.
Write chromyl chloride test with equation.
Answer:
Chromyl Chloride Test:
1. When a metal chloride is heated with solid potassium dichromate and cone. H₂SO₄ orange coloured vapours of chromyl chloride are formed.
K₂Cr₂O₇ + 6H₂SO₄ + 4KCl → 2CrO₂Cl₂ ↑ + 6KHSO₄ + 3H₂O

2. When these fumes are passed in sodium hydroxide solution, yellow solution of sodium chromate is obtained. When lead acetate is added to it in presence of acetic acid yellow precipitate of lead chromate is obtained.

Question 10.
Write any five main differences between d – and f – Block elements.
Answer:
Differences between d and f – Block Elements :
d – Block Elements:

• Two shells n and (n – 1) are incomplete.
• Last electron enters the d – orbital of penultimate shell.
• d – block elements are normally called Transitional element.
• d – block elements are available in nature.
• These elements exhibit variable oxidation state.
• These elements are stable.

f – Block Elements:

• Three shells n, (n – 1) and (n – 2) are incomplete.
• Last electron enters the orbital of antipenultimate (n – 2) shell.
• f – block elements are normally called Inner Transitional element.
• f – block elements are very rare. Therefore, they are known as Rare Earth elements.
• These elements also exhibit variable oxidation state.
• These elements are less stable and many are radioactive.

Question 11.
Explain giving reasons: (NCERT)

1. Transition metals and many of their compounds show paramagnetic behaviour.
2. The enthalpies of atomisation of the transition metals are high.
3. The transition metals generally form coloured compounds.
4. Transition metals and their many compounds act as good catalyst.

Answer:
1. Paramagnetic substance is one which is attracted by magnetic field. It arises due to presence of unpaired electron in atom, ion or molecule. Most of the transition elements and compounds are paramagnetic in nature. This is due to fact that transition elements involve partially filled d – subshell and their atom and ion contain unpaired electron.

2. Transition elements have high effective nuclear charge and a large number of valence electrons. Therefore, they form very strong metallic bonds. As a result, the enthalpy of atomization of transition metals is high.

3. The colour of transitional metal ions is due to partially filled (n – 1 )d orbitals. In transitional metal ions which contain unpaired d electrons, transition of electrons takes place from one d – orbital to another d – orbital. During this transition it absorbs some radiation of visible light and reflects the remaining radiation in the form of coloured light. Thus, the colour of the ion is complementary to the colour absorbed by it.
For example:
[Cu(H₂O)₆]²⁺ ion appears blue because it absorbs the red colour of the visible light for electron promotion and reflects its complementary blue colour.
Colour of some ions:

4. Transition elements act as good catalysts in chemical reaction in the hydrogenation of Ni metal, in contact process of manufacture of SO₃, Pt and in manufacture of NH₃ by Haber process Fe acts as catalyst. In the method of preparation of O₂ by heating KClO₃, MnO₂ acts as catalyst.

Question 12.
How would you account for the following : (NCERT)

1. Of the d⁴ species, Cr²⁺ is strongly reducing while manganese(III) is strongly oxidising.
2. Cobalt(II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised.
3. The d¹ configuration is very unstable in ions.

Answer:
1. Cr⁺² is reducing in nature as its configuration changes from d⁴ to d³ (A stable configuration having half filled t_2g orbitals). On the other hand, Mn⁺³ is oxidising in nature as the configuration changes from d⁴ to d⁵ (A stable configuration having half filled t_2gto e orbitals)

2. Strong ligands force Cobalt (II) to lose One more electron from 3d – subshell and thereby induce d₂sp₃ – hybridisation.

3. The ions with dl configuration try to lose the only electron on d – subshell in order to acquire stable inert gas configuration.

Question 13.
Compare the chemistry of actinoids with that of the lanthanoids with special reference to: (NCERT)

1. Electronic configuration
2. Atomic and ionic sizes
3. Oxidation state and
4. Chemical roactivity.

Answer:
Differences between Lanthanoids and Actinoids :
Lanthanoids:

• Differentiating or last electrons enter in 4f – sub – shell of (n – 2) orbit.
• These elements come after lanthanum so these are called lanthanoids.
• Common oxidation state is +3, other oxidation states are +2 and +4 also.
• Atomic or ionic radius decreases gradually and this is called lanthanide contraction.
• Lanthanoids have smaller tendency to form complexes.
• Lanthanoids do not form oxo – ions.
• Compounds of lanthanoids exhibit less basic in nature.
• Lanthanoids are not radioactive except Promethium.
• Except Pm, other lanthanoids are present in nature in abundance comparatively more than iodine.

Actinoids:

• Differentiating or last electrons enter in 5f – sub – shell of (n – 2) orbit.
• These elements come after actinium so these are called actinoids.
• Common oxidation state in actinoids is also +3 but other oxidation states are higher, example  +4, +5, +6 and +7.
• Atomic or ionic radius also decreases gradually and steadily and this is qallejj actinoid contraction.
• Actinoids have comparatively higher tendency of complex formation.
• Oxo – ions are formed. example UO₂⁺,PuO₂⁺, UO⁺, etc.
• Compounds of actinoids are more basic in nature.
• All the actinoids are radioactive.
• Most of these are not found in nature and are artificially prepared.

Question 14.
What are Inner Transition elements? (NCERT)
Answer:
These are the elements which contain (n-2)f and (n-1)d incomplete orbitals or in which electron enter in the antipenultimate (two energy levels below the outermost orbital) orbital. These are so called because these are found within the transition elements. There are two types of inner transition elements :
(i) Lanthanides series :
The 14 elements after Lanthanum (La₅₇)
i.e., 58^Ce – 71^Lu are called lanthanides.

(ii) Actinide series : 14 elements after Actinide (AC₈₉) i.e., Th₉₀ to LW₁₀₃.

The d-and f-Block Elements Long Answer Type Questions

Question 1.
Describe the preparation of K₂Cr₂O₇ from chromite ore and explain the reactions of K₂Cr₂O₇ with acidic FeSO₄, KI and H₂S.
Answer:
(A) Preparation:
It is prepared from chromite ore or ferrochrome of chrome iron FeCr₂O₄ (FeO.Cr₂O₃). Different steps involved in the process are as follows :

1. Preparation of sodium chromate:
The ore is finely powdered, mixed with sodium carbonate and quick lime and then roasted (heated to redness) in a reverberatory furnace in presence of excess of air when sodium chromate (yellow in colour) is formed with the evolution of CO₂. Quick lime is added to keep the mass porous and thus facilitates oxidation.

The roasted mass is the extracted with water when sodium chromate dissolves com-pletely leaving behind ferric oxide.

2. Conversion of sodium chromate to sodium dichromate:
Sodium chromate is extracted with water and acidified with sulphuric acid to get sodium dichromate.
2Na₂CrO₄ + H₂SO₄ → Na₂Cr₂O₇ + Na₂SO₄ + H₂O
On concentration the less soluble sodium sulphate Na₂SO₄.10H₂O crystallizes out. This is filtered hot and allowed to cool when sodium dichromate Na₂Cr₂O₇.2H₂O separates on standing.

3. Conversion of sodium dichromatic into potassium dichromate:
Hot concentrated solution of sodium dichromate is treated with requisite amount of potassium chloride when potassium dichromate being less soluble crystallizes out on cooling.
Na₂Cr2O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl

(B) Reaction of K₂Cr₂O₇ with acidic FeSO₄, KI and H₂S :
(i) It oxidizes ferrous sulphate to ferric sulphate.

(ii) It liberates I₂ from KI.

These reactions are used in the estimation of iodine and ferrous ion in volumetric an-alysis.
(iii) It oxidizes SO₂ to sulphuric acid.

(iv) It oxidizes H₂S to sulphur.

Question 2.
Explain the oxidizing property of KMnO₄ in acidic, neutral and alkaline medium giving two examples each.
Answer:
KMnO₄ acts as strong oxidizing agent in acidic, neutral and alkaline medium. In acidic medium : It oxidizes in presence of dilute H₂SO₄ and get reduced.
2KMnO₄ +3H₂SO₄ → K₂SO₄ +2MnSO₄ +3H₂O + 5[O]
Example:
1. It oxidizes ferrous salt into ferric salt.

2. It oxidizes oxalate to CO₂ :
2KMnO₄ + 3H₂SO₄ + 5C₂H₂O₄ → K₂SO₄ + 2MnSO₄ + 8H₂O + 10CO₂

3. It oxidizes iodide ion to iodine :
2KMnO₄+10KI+8H₂SO₄ → 6K₂SO₄ + 2MnSO₄+ 8H₂O + 5I₂

4. It oxidizes nitrites to nitrates :
2KMnO₄ + 3H₂SO₄ + 5NaNO₂ → 2MnSO₄ + K₂SO₄ + 5NaNO₃ + 3H₂O

In neutral medium:
In this medium, the reaction begins with neutral ethylene glycol but this does not give neutral reaction because KOH formed in the reaction makes basic in nature.
2KMnO₄ + H₂O → 2KOH + 2MnO₂ + 3[O]
Example:
1. It oxidizes manganous sulphate to manganese dioxide.
2KMnO₄ + 3MnSO₄ + 2H₂O → 5MnO₂ + K₂SO₄ + 2H₂SO₄

2. It oxidizes hydrogen sulphide to sulphur.
2KMnO₄ + 4H₂S → 2MnS + K₂SO₄ + 4H₂O + S

In alkaline medium:
In alkaline medium, reduces to MnO₂ and gives 3 nascent oxygen.
Example:
1. It oxidizes ethylene to ethylene glycol

2. It oxidizes iodide to iodate.

KMnO₄ gives more number of nascent oxygen in acidic medium than in alkaline medium due to which it acts as stronger oxidizing agent in acidic medium.

Question 3.
Describe the preparation of KMnO₄ from pyrolusite and explain its oxidising properties in acidic, basic and neutral medium by suitable example. (MP 2009 Set B, 17)
Answer:
Preparation:
Potassium permanganate is prepared from manganese dioxide. On a large scale, it is prepared from the mineral pyrolusite. The process involves the following steps:

1. Conversion of MnO₂ into potassium manganate:
The finely powdered pyrolusite mineral is fused with potassium carbonate or potassium hydroxide in presence of atmospheric oxygen or an oxidising agent such as potassium nitrate or potassium chlorate. The fused mass turns green due to the formation of potassium manganate.

The fused mass turns green due to the formation of potassium manganate.
2MnO₂ + 2K₂CO₃ + O₂ → 2K₂MnO₄ + 2CO₂
2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O
MnO₂ + 2KOH + KNO₃ → K₂MnO₄ + KNO₂ + H₂O
3MnO₂ + 6KOH + KClO₃ → 3K₂MnO₄ + KCl + 3H₂O

2. Oxidation of potassium manganate into potassium permanganate :
(i) Chemical oxidation:
The fused mass is extracted with water and the solution is filtered. The green solution is then converted to potassium permanganate by bubbling carbon dioxide, chlorine or oxygen through it.
3K₂MnO₄ + 2CO₂ → 2KMnO₄ + MnO₂ ↓ + 2K₂CO₃
2K₂MnO₄+ Cl₂ → 2KMnO₄ + 2KCl
2K₂MnO₄ + H₂O + O₃ → 2KMnO₄ + 2KOH + O₂
The purple solution of potassium permanganate thus obtained is concentrated when it deposits dark purple, needle like crystals having a metallic lustre.

(ii) Electrolytic oxidation: Nowadays, it is largely manufactured by the electrolytic oxidation of the manganate. The manganate solution is electrolysed between iron electrodes separated by diaphragm. The oxygen evolved at the anode converts manganate to permanganate.
2K₂MnO₄ + H₂O + [O] → 2KMnO₄ + 2KOH
MnO₄^2- + e^– Oxidation (At anode)
2K⁺ + 2e^–→ 2K Reduction (At cathode)
2K + 2H₂O →  2KOH + H₂

After the oxidation is completed, the solution is filtered and evaporated under controlled condition to obtain the crystals of potassium permanganate.
(i) Acidified KMnO₄ solution oxidizes Fe(II) ions to Fe(III) ions i.e. ferrous ions to ferric ions.

(ii) Acidified potassium permanganate oxidizes SO₂ to sulphuric acid.

(iii) Acidified potassium permanganate oxidizes oxalic acid to carbon dioxide.

MP Board Class 12th 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:

1. The alum which contains trivalent ion aluminium is called by name of monovalent ion.
K₂SO₄.Al₂(SO₄)₃.24H₂O (Potash alum)

2. The alum which does not contain aluminium as trivalent ion is called by name of both ion.
K₂SO₄.Fe₂(SO₄)₃.24H₂O (Potash ferric alum)

Preparation method:
The crystal of alum are obtained when hot solution of equimolecular quantities of K₂SO₄ and Al₂(SO₄)₃ are mixed and resulted solution is cooled.
K₂SO₄ + Al₂(SO₄)₃ + 24H₂O → K₂SO₄.Al₂(SO₄)₃.24H₂O

Properties:
Alums are octahedral crystalline compounds. Potash alum is milky in colour. It forms acidic solution in water. It is soluble in water but insoluble in organic solvent.

Effects of heat:
On heating at 365 K it looses water molecule. Its water of crystallization is lost and alum swells up and becomes porous. This type of alum is known as burnt alum.

Uses:

1. As blood coagulates.
2. In purifying water.
3. In dyeing and printing clothes.
4. In sizing of paper.

Question 8.
Write the diagonal relationship between Be and Al?
Answer:
Diagonal Relationship of Beryllium with Aluminium; Beryllium shows diagonal relationship with aluminium which is present in third period and third group (13) of periodic table.
Diagonal relationship between beryllium and aluminium is due to the following reason:

1. Nearly equal atomic and ionic radius.
2. Similar electronegativity.
3. Nearly equal electropositivity.
4. Nearly equal polarization power.

Similarities between Be and Al are as follows:
1. Be and Al forms covalent compounds which are soluble in organic solvents.

2. Be and Al oxides are solids having high melting point. They react both with acid and base to produce salt and water due to their amphoteric nature.

3. Be and Al reacts with carbon on heating to form respective carbides which are covalent in nature. Be₂C and Al₄C₃.

4. Carbides of Be and Al forms methane on hydrolysis.
Be₂C + 4H₂O → CH₄ + 2Be(OH)₂
Al₄C₃ + 12H₂O → 3CH₄ + 4Al(OH)₃

5. Be and A1 forms complex fluoro ion in solution.
[BeF₄]^2- and [AlF₆]_3-

6. Polymeric structure are found in BeCl₂ and AlCl₃

7. Solubility of halide of Be and Al is same.

8. Be and A1 develops a protective oxide layer on the surface when treated with nitric acid. Due to this, they are not easily attacked by acids.

9. Hydroxide of Be and Al are amphoteric in nature. They react with acid and base to produce salt and water.

Question 9.
What do you mean by:

1. Inert pair effect.
2. Allotropes.
3. Catenation.

Answer:
1. Inert pair effect:
When the tendency of 5 – electrons to be with 5 – electrons and they do not take part in reactions then this tendency is called inert pair effect. The reason is that the tendency to show +1 oxidation state increases down the group. (MPBoardSolutions.com) It means the tendency of 5 – electron of valence shell to participate in bond formation. Thus, among the heavier elements of p – block, the electron pair in the outer 5 – orbitals is reluctant to take part in chemical bonding. It is called inert pair effect.

2. Allotropes:
If an element exists in different physical forms having different properties than the various forms are called allotropes and the phenomenon is called allotropy.
Example:
Crystalline allotropes of C are diamond, graphite and fullerene.

3. Catenation:
It is the ability of like atoms to link with one another through cova¬lent bonds. This is due to smaller size and higher electronegativity of carbon atom and unique strength of C – C bond.
C >> Si > Ge > Sn >> Pb

MP Board Class 11 Chemistry Important Questions

MP Board Class 12th Chemistry Important Questions Chapter 7 The p-Block Elements

The p-Block Elements Important Questions

The p-Block Elements Objective Type Questions

Question 1.
(A) Choose the correct answer :

Question 1.
In which compound oxygen exhibits +2 oxidation state :
(a) H₂O
(b) Na₂O
(c) OF₂
(d) MgO.
Answer:
(c) OF₂

Question 2.
Reddish brown gas is formed, when nitric oxide oxidizes by air. This gas is :
(a) Na₂O₂
(b) Na₂O₄
(c) NO₂
(d) N₂O₃.
Answer:
(c) NO₂

Question 3.
H₃PO₃ is :
(a) Dibasic acid
(b) Monobasic acid
(c) Tribasic acid
(d) Tetrabasic acid.
Answer:
(c) Tribasic acid

Question 4.
Oxide which shows paramagnetic character :
(a) N₂O₄
(b) NO₄
(C) P₄O₆
(d) N₂O₅
Answer:
(b) NO₄

Question 5.
Ammonia can be dried by :
(a) H₂SO₄
(b) P₂O₅
(c) Anhydrous CaCl₂
(d) CaO.
Answer:
(d) CaO.

Question 6.
Nitric acid change iodine into :
(a) Iodic acid
(b) Hydroiodic acid
(c) Iodine pentaoxide
(d) Iodine nitrate.
Answer:
(a) Iodic acid

Question 7.
NH₃ is a Lewis base which forms complex salt with cations. Following cation does not form complex salt with NH₃ :
(a) Ag⁺
(b) Cu2⁺
(c) Cd2⁺
(d) Pb2⁺.
Answer:
(d) Pb2⁺.

Question 8.
Ammonia is soluble in :
(a) Hg₂Cl₂
(b) PbCl₂
(c) Agl
(d) Cu(OH)₂.
Answer:
(d) Cu(OH)₂.

Question 9.
Bleaching action of SO₂ is due to :
(a) Reduction
(b) Oxidation
(c) Hydrolysis
(d) Acidic nature.
Answer:
(a) Reduction

Question 10.
What happens when SO₂ is passed through acidic K₂Cr₂O₇ :
(a) Solution turns blue
(b) Solution turns colourless
(c) SO₂ reduced
(d) Green chromic sulphate is formed.
Answer:
(d) Green chromic sulphate is formed.

Question 11.
P₂O₂ forms which of the following acid :
(a) H₄P₂O₇
(b) H₃PO₄
(c) H₃PO₃
(d) HPO₃.
Answer:
(c) H₃PO₃

Question 12.
Most acidic halide is :
(a) PCl₅
(b) SbCl₃
(c) BrCl₃
(d) CCl₄
Answer:
(a) PCl₅

Question 13.
The catalyst used in manufacturing of H₂SO₄ is :
(a) Al₂O₃
(b) CrO₃
(c) V₂O₅
(d) MnO₂
Answer:
(c) V₂O₅

Question 14.
The hydrolysis of phosphorus triha lides give:
(a) One monobasjc acid and one dibasic acid
(b) One monobasjc acid and one tribasic acid
(c) One monobasjc acid and a salt
(d) Two dibasic acids.
Answer:
(a) One monobasjc acid and one dibasic acid

Question 15.
In the following reaction:
P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂ PO₂
(a) P oxidises
(b) P oxidises and reduces
(c) P reduces
(d) Na oxidises.
Answer:
(b) P oxidises and reduces

Question 16.
Laughing gas is:
(a) NO
(b) N₂O
(c) N₂O₃
(d) N₂O₅.
Answer:
(b) N₂O

Question 17.
White phosphorus does not contain:
(a) 6 P – P single bond
(b) 4 P – p single bond
(c) 4 lone pair of electron
(d) Bond angle of P – P – P is 600.
Answer:
(b) 4 P – p single bond

Question 18.
On heating NH₄CI and NaNO₂ solution gives :
(a) N₂O
(b) N₂
(c) NO₂
(d) NH₃.
Answer:
(b) N₂

Question 19.
Formula of metaphosphoric acid is:
(a) H₃PO₃
(b) HPO₃
(c) H₃PO₃
(d) H₂PO₃.
Answer:
(b) HPO₃

Question 20.
Gas which cannot be collected on water:
(a) Na
(b) O₂
(c) SO₃
(d) PH₃.
Answer:
(c) SO₃

(B) Choose the correct answer:

Question 1.
Chlorine shows bleaching property in the presence of :
(a) Dry air
(b) Moisture
(c) Sunlight
(d) Pure oxygen.
Answer:
(b) Moisture

Question 2.
Which of the following element forms least number of compounds :
(a) He
(b) Ar
(c) Kr
(d) Xe.
Answer:
(a) He

Question 3.
Which is used for bright advertisement display :
(a) Xe
(b) Ar
(c) Ne
(d) He.
Answer:
(c) Ne

Question 4.
Monazite is a source of:
(a) Ne
(b) Ar
(c) Kr
(d) He.
Answer:
(d) He.

Question 5.
Which of the following is not obtained by the direct reactions of the constituents :
(a) XeF₂
(b) XeF₄
(c) XeO₃
(d) XeF₆.
Answer:
(d) PbI₄.

Question 6.
Which of the following halides are least stable and whose existence is suspicious :
(a) CI₄
(b) Gel₂
(c) Snl₄
(d) PbI₄.
Answer:
(d) PbI₄.

Question 7.
Which of the following is the strongest acid :
(a) HBr
(b) HCl
(c) HF
(d) HI.
Answer:
(d) HI.

Question 8.
Most electronegative element is :
(a) F
(b) Cl
(c) Br
(d) I.
Answer:
(a) F

Question 9.
Which halogen is solid at room temperature :
(a) Cl₂
(b) I₂
(c) Br₂
(d) F₂.
Answer:
(d) F₂.

Question 10.
Which has maximum electron affinity :
(a) F
(b) Cl
(c) Br
(d) I.
Answer:
(b) Cl

Question 11.
Electronic configuration of valence shell of halogen is :
(a) s²p⁵
(b) s²p³
(c) s²p⁶
(d) s²p⁴.
Answer:
(a) s²p⁵

Question 12.
Which element is most alkaline:
(a) F
(b) Cl
(c) Br
(d) I.
Answer:
(d) I.

Question 13.
The strongest reducing agent is:
(a) F^–
(b) Br^–
(c) I^–
(d) Cl^–
Answer:
(c) I^–

Question 14.
Which of the following is weakest acid:
(a) HF
(b) HCl
(c) HBr
(d) HI.
Answer:
(a) HF

Question 15.
Oxidizing property is highest of:
(a) 12
(b) Br₂
(c) F₂
(d) Cl₂.
Answer:
(c) F₂

Question 16.
Which Noble gas does not have octet complete:
(a) Helium
(b) Neon
(c) Argon
(d) Krypton.
Answer:
(a) Helium

Question 17.
Which Halogen sublimates :
(a) Chlorine
(b) Bromine
(c) Iodine
(d) Fluorine.
Answer:

Question 18.
12 easily dissolve in Kl solution to form :
(a) I
(b) KI₂
(c) KI
(d) KI₃
Answer:
(d) KI₃

Question 19.
Gas mixed in oxygen which is used in respiration for asthma patients :
(a) N₂
(b) Cl₂
(c) He
(d) Ne.
Answer:
(c) He

Question 20.
Deacon’s process is used for manufacture of :
(a) Bleaching powder
(b) Chlorine
(c) Nitric acid
(d) Sulphuric acid.
Answer:
(b) Chlorine

Question 21.
Sea grass is the source of industrial manufacture of :
(a) Chlorine
(b) Bromine
(c) Iodine
(d) Fluorine.
Answer:
(c) Iodine

Question 2.
(A) Fill in the blanks :

1. N₂O is a ……………….. oxide.
2. Formula of Caro’s acid is ………………..
3. Concentrated nitric acid has brown colour due to the dissolution of ……………….. gas.
4. Out of nitrogen oxides ……………….. and ……………….. are paramagnetic.
5. Pyrophosphoric acid is ……………….. basic acid.
6. H₂S gas cannot be dried by cone. H₂SO₄ because H₂S ……………….. it.
7. Fuming sulphuric acid dissolves in SO₃ to form ………………..
8. Oxidation state of S in H₂S₂O₈ (Marshall gas) is ………………..
9. NH₃ gives white fumes of ……………….. when it combines with HCl.
10. Elements of 16^th group are known as ………………..
11. ……………….. is used as a refrigerant.

Answer:

1. Neutral
2. H₂SO₅
3. NO₂
4. NO
5. NO₂
6. Tetra
7. Reduces
8. Oleum
9. +6
10. NH₄Cl
11. Chalcogen
12. Liquid NH₃.

(B) Fill in the blanks :

1. …………………. has the highest electron affinity.
2. In the presence of moisture, chlorine acts as a ………………… agent.
3. Bleaching powder is also called as …………………
4. At ordinaiy temperature bromine is a …………………
5. Inter halogen compound AX₅ has ………………… structure.
6. Chlorine was discovered by …………………
7. Neil Bertlett forms first noble compound which is …………………
8. Element with highest electron affinity is …………………
9. In oxy acid halogen present in ………………… hybride state.
10. Gas which is light due to which it is filled in aircraft tyres is …………………
11. ………………… inert gas is mostly used in advertisements.
12. Elements of Group 17^th are commonly known as …………………
13. ………………… is a radioactive inert gas.

Answer:

1. Chlorine
2. Bleaching
3. Calcium chlorohypochloride
4. Liquid
5. Square pyramidal
6. Scheele
7. Xe [PtF₆]
8. Chlorine,
9. sp³
10. Helium
11. Ne (Neon)
12. Halogen
13. Radon.

Question 3.
(A) Answer in one word / sentence :

1. Who protects earth from ultraviolet rays?
2. Give the chemical name of oil of vitriol or king of chemical.
3. Which gas is used for refrigeration?
4. At which temperature water has maximum density?
5. Give the name of compound formed by the dissolution of S03 in sulphuric acid.
6. Write name of an antichlor.
7. Which gas is used for bleaching of oil and elephant teeth?
8. Ammonium salts react with Nessler’s reagent to give which coloured precipitate?
9. On moving down from N to Bi, +3 oxidation state of Bi becomes more stable than +5. Why?
10. State the percentage of N₂ by volume in nature.

Answer:

1. Ozone layer
2. H²SO⁴
3. NH₃
4. 4°C
5. Oleum
6. SO₃
7. Ozone
8. Brown
9. Due to Inert pair effect
10. 80%.

(B) Answer in one word / sentence :

1. Give the name of radioactive halogen.
2. Which noble gas is used in bulb with nitrogen? (MP 2018)
3. Write the name of noble gas which is used in treatment of cancer.
4. Write the formula of Camalite.
5. Which noble gas is maximum available in the atmosphere?
6. What is the shape of XeF₆?
7. Write one use of fluorine.
8. Which type of hybridization is present in XeO₃?
9. Write the oxidation state of F.
10. Write only equation for the preparation of chlorine in the lab.
11. What is the shape of AX₃ type of Interhalogen compound?
12. Write the correct order of strength of halogen acid.
13. Which noble gases do not form compounds?
14. With which noble gas does F form compounds?
15. Sea weeds are the source of which halogen?

Answer:

1. Astatine
2. Ar
3. Rn
4. KCl.MgCl₂.6H₂O
5. Argon
6. Distorted octahedron
7. In preparing fluorocarbon, which is used in refrigeration
8. sp³
9. -1
10. MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂
11. T – shape
12. HF < HCl < HBr < HI
13. He, Ne and Ar
14. Xe
15. Iodine.

Question 4.
Match the following :
I.

Answer:

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

II.

Answer:

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

III.

Answer:

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

General Principles and Processes of Isolation of Elements Very Short Answer Type Questions

Question 1.
What are clathrate compounds?
Answer:
In the cavity or space of crystal lattice of a compound small sized elements like noble gases get trapped within. Such compounds are called clathrate compounds.
Example : Kr₃ (β – quinol).

Question 2.
Why is BiH₃ the strongest reducing agent amongst all the hydrides of Group 15 elements? (NCERT)
Answer:
Among the hydrides of Group – 15, BiH₃ is the least stable because Bi has largest size in group and has least tendency to form covalent bond with small hydrogen atom. Therefore, it can readily lost H atom and has strongest tendency to act as reducing agent.

Question 3.
Why is N₂ less reactive at room temperature? (NCERT)
Answer:
In N₂ molecule, there is a strong pπ – pπ overlap that results in the formation of triple bond between N atoms (N ≡ N). As a consequence, the bond dissociation energy of N₂ is very high and N₂ is less reactive at room temperature.

Question 4.
Why is Helium used in diving apparatus? (NCERT)
Answer:
Helium is used as a diluent for oxygen is modem diving apparatus because of its very low solubility in blood.

Question 5.
Noble gases are inert. Why?
Answer:
Noble gases are inert due to following reasons :
Noble gases are inert because octet of these gases are complete. It is the most stable state and do not have unpaired electrons. Ionisation energy of noble gases are high and electronegativity and electron affinity are zero. Therefore, these gases neither accept electron nor release electron. Thus, noble gases are inert in nature.

Question 6.
Complete the following reactions: (NCERT)
(i) C₂H₂ + O2 →
(ii) 4AI + 3O₂ →
Answer:

Question 7.
Arrange the following in the order of property indicated for each set: (NCERT)
(i) F₂, Cl₂, Br₂,I₂ – increasing bond dissociation enthalpy.
(ii) HF, HCl, HBr, HI – increasing acid strength.
(iii) NH₃, PH₃, ASH₃, SbH₃, BiH₃ – increasing base strength.
Answer:
(i) I₂ < F₂ < Br₂ < Cl₂

Here F₂ has very less bond dissociation energy inspite of having small size. It is due to repulsion between the lone pair of electrons on small size F – atoms.

(ii) HF < HCl < HBr < HI
It depends upon their bond dissociation enthalpy, which decreases from H – F to H – I as the size of atom increases from F to I.

(iii) BiH₃ < SbH₃ < AsH₃ < PH₃ < NH₃.
They are Lewis bases due to presence of lone pair of electrons on central atom of hydrides. The availability of lone pair of electrons is maximum in nitrogen owing to its small size therefore, NH₃ is maximum basic and as the size of central atom increases availability decreases.

Question 8.
Describe the laboratory preparation of chlorine.
Answer:
Laboratory preparation:
HCl and MnO₂ is taken in a round bottom flask. Cone. HCl is added from thistle funnel. The flask is heated slowly so that green yellow colour of Cl₂ gas evolved.
MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂

Question 9.
Fluorine is strong oxidizing agent as compared to chlorine. Why?
Answer:
Because of following points, fluorine is strong oxidizing agent:

1. The size of F atom is smaller than Cl atom.
2. Electronegativity of fluorine is higher than Cl atom.
3. Dissociation energy of fluorine is less than that of chlorine.
4. E° value of fluorine is more than chlorine.

Question 10.
F₂O is not considered to be the oxide of fluorine. Why?
Answer:
Fluorine is the most electronegative element. Its electronegativity is higher than oxygen. In naming a compound less electronegative elements are first and then more electronegative elements are written, Therefore, F₂O or OF₂ is known as oxygen difluoride and not fluorine oxide.

Question 11.
Interhalogen compounds are more reactive than halogen. Why?
Answer:
X – Y bonds of interhalogen compounds are more polar due to difference in the electronegativities of halogen and are generally weaker than X – Y bonds of pure halogen. Therefore, interhalogen compounds are more reactive than halogens.

Question 12.
Helium and Neon do not form compounds with Fluorine. Why?
Answer:
Due to absence of d – orbitals in the valence shell of He and Ne, their electrons cannot get excited into higher energy d – sub – shells. Therefore, He and Ne do not form compounds with fluorine.

Question 13.
Explain why NH₃ is basic while BiH₃ is only feebly basic. (NCERT)
Answer:
Atomic size of N(70pm) is much smaller than that of Bi (148 pm). Therefore, electron density on the N – atom is much higher than that on Bi – atom. Consequently, the tendency of N in NH₃ to donate the pair of electrons is much higher than that of Bi in BiH₃. Thus, NH₃ is basic while BiH₃ is only feebly basic.

The p – Block Elements Short Answer Type Questions

Question 1.
Explain with reason :

1. HF is liquid while other hydrides of halogens are gases at normal temperature.
2. Fluorine does not form polyhalide.

Answer:
1. The electronegativity of fluorine is maximum than other halogens. Therefore, HF molecule is conjugated by H – bond and the boiling point of HF is more to other halogen acids therefore HF is liquid while halides of other halogens are gases at room temperature.
H – z….H – F….H – F….H – F

2. The electronic configuration of fluorine is 1s², 2s², 2p⁵. Due to absence of orbital in its valency shell it does not show higher oxidation state and so it does not made polyhalide.

Question 2.
How is SO₂ an air pollutant? (NCERT)
Answer:
SO₂ dissolves in rain water and produces acid rain. The acid rain contains sulphuric acid.

In addition to H₂SO₄, acid rain also contains HNO₃.

Question 3.
Why are halogens strong oxidising agents? (MP 2018, NCERT)
Answer:
Halogens are strong oxidising agents due to their low bond dissociation enthalpy, high electronegativity and large negative electron gain enthalpy. Halogens have a strong tendency to accept electrons and thus get reduced.
X₂ + 2e^– → 2X^–
As a result, halogen acts as strong oxidising agents. Their oxidising power however decreases from F₂ to I₂ as is evident from their electrode potentials :
E°_F2/F^– = + 2.87 V
E°_Cl2/Cl^– = + 1.09 V
E°_Br2/Br = +1.09 V
E°_Br2/I = + 0.54 V

Question 4.
Why does the reactivity of nitrogen differ from phosphorus? (NCERT)
Answer:
Molecular nitrogen exist as a diatomic molecule having a triple bond between the two nitrogen atoms, N ≡ N (due to it stability to form pπ – pπ multiple bonding). The bond dissociation energy is very high (941.4 kJ mol^-1). Thus, under ordinary conditions, nitrogen behaves as an inert gas. On the other hand, white and yellow phosphorus exists as a triatomic molecule (P₄) having single bonds. The dissociation energy of P – P bond is low (213 kJ mol^-1). Thus, phosphorus is much more reactive than nitrogen.

Question 5.
Give the reason for bleaching action of Cl₂. (NCERT)
Answer:
Bleaching action of chlorine is due to its oxidation. In the presence of moisture, chlorine gives nascent oxygen.
Cl₂ + H₂O → 2HCl + [O]
Because of nascent oxygen, it bleaches colouring substance as
Colouring substance + [O] → Colourless substance.
It bleaches vegetables or organic matter. The bleaching action of chlorine is permanent.

Question 6.
Explain, why:

1. Noble gases are monoatomic?
2. Atomic radii of noble gases are largest?
3. Ionisation energy of noble gases are highest?

Answer:

1. There is no any unpaired electron in electronic configuration of noble gases. So these do not form chemical bonds and are monoatomic.
2. Outermost shell of noble gases are completely filled so atomic radii of noble gases are maximum.
3. All electrons in outermost orbit of noble gases are paired and energy required to make them unpaired is very high. Thus, noble gases have maximum ionisation energy in respective periods.

Question 7.
Why does nitrogen show catenation properties less than phosphorus? (NCERT)
Answer:
The catenation properties depends upon the strength of the element – element bond. The N – N bond strength is much weaker (due to repulsion of lone pairs on nitrogen because of its small size) than the P – P bond strength, therefore, nitrogen shows catenation less than phosphorus.

Question 8.
Write the formulae and structure of any five oxy – acids of sulphur.
Answer:
Main oxyacids of sulphur and oxidation state of sulphur in them are as follows :

Question 9.
Contact process of manufacturing of H₂SO₄ is supposed to be more advance than lead chamber process. Why?
Answer:

1. The acid obtained by contact process is pure while the acid obtained by lead chamber process is impure.
2. The plant for lead chamber process requires large area while plant used in contact process occupies comparatively less area.
3. The catalyst used in contact process is platinised asbestos while in lead chamber process catalyst is gaseous (NO₂ gas) and to maintain its flow regular is tedious.
4. In contact process concentrated sulphuric acid is obtained while in lead chamber process dilute acid is obtained.
5. In contact process running of the plant is less costly while in lead chamber process running of the plant is costly.

Question 10.
Why does NH₃ form hydrogen bond but PH₃ does not? (NCERT)
Answer:
The N – H bond in ammonia is quite polar as nitrogen is highly electronegative in nature. As a result, NH₃ molecules are linked by intermolecular hydrogen bonding.

On the other hand, P – H bond is non – polar as both P and H have same electronegativity. Hence in phosphine no hydrogen bonding is present.

Question 11.
Bond angle in PH₄⁺ is higher than that in PH₃. Why? (NCERT)
Answer:
PH₄⁺ is a regular tetrahedral structure, thus bond angle is 109°28′.
PH₃ molecule has a pyramidal structure. The bond angle is 93°. Hence, bond angle in PH₄⁺ ion is higher.

Question 12.
What happens when white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO₂? (NCERT)
Answer:
When white phosphorus is heated with cone. NaOH solution in an inert atmosphere of CO₂, phosphine gas is prepared.

Question 13.
What happens when PCl₅ is heated? (NCERT)
Answer:
PCl₅ has three equatorial and two axial bonds. Since axial bonds are weaker than equatorial bonds, therefore, when PCl₅ is heated, the less stable axial bond breaks to form PCl₃.

When heated, it sublimes but decomposes on stronger heated.

Question 14.
How are Xenon fluorides XeF₂, XeF₂ and XeF₆ prepared?
Answer:
Xenon Fluorides:
Three fluorides of xenon are important. These are xenon difluoride (XeF₂), xenon tetrafluoride (XeF₄) and xenon hexafluoride (XeF₆).The fluorides of xenon can be prepared by the direct combination between xenon and fluorine under different conditions.

Question 15.
Write the difference between the bleaching action of SO₂ and Cl₂.
Or,
Bleaching of flowers by Cl₂ is permanent while bleaching by SO₂ is temporary, why?
Answer:
Bleaching action of SO₂:
Bleaching by SO₂ is due to the process of reduction in the presence of moisture.
SO₂ + 2H₂O →H₂SO₄ + 2 [H] (nascent)
The nascent hydrogen liberated in the reaction is responsible for bleaching flower but when the flower comes in contact with air. It gets oxidized and become coloured.
Coloured substance + [H] → Colourless \(\underrightarrow { air }\)Colour

Bleaching action of Cl₂:
Bleaching by Cl₂ is due to the process of oxidation in the presence of moisture.
Cl₂ + H₂O → 2HCl + [O ] (nascent)
Coloured substance + [O] > Colourless
The oxygen liberated in the reaction is responsible for bleaching, colour bleached by Cl₂ is permanent and original colour cannot be restored on exposure to air.

Question 16.
Write main differences between the properties of white phosphorus and red pljosfrtiorus. (NCERT)
Answer:
Comparison of properties of White and Red phosphorus :

Question 17.
Give equations for the reaction of chlorine with the following :

1. NH₃
2. NaOH
3. H₂O
4. Bleaching property.

Answer:
Reaction of chlorine :

1. With NH₃ :
(a) Cl₂ reacts with excess of NH₃ and form NH₄Cl and N₂.

(b) When chlorine is taken in excess, an explosive product nitrogen trichloride is formed.
NH₃ +3C₁₂ → NC₁₃ +3HCl

2. With NaOH :
(a) Cold and dilute NaOH reacts with Cl₂ and form chloride and hypochlorite.
Cl₂ + 2NaOH → NaCl + NaClO + H₂O

(b) Hot and concentrated caustic soda reacts with Cl₂ and form chloride and chlorate.
3Cl₂ + 6NaOH → NaC1O₃ + 5NaCl + 3H₂O

3. With H₂O :
Cl₂ reacts with water and form chlorine water which is called hypochlorous acid which decomposes into HCl.

4. Bleaching action:
In presence of moisture chlorine bleaches coloured object and vegetables. It decolourises the coloured substance. This property is due to oxidation.

Thus, bleaching by chlorine is a permanent process.

Question 18.
Give reason :

1. Iodine exhibits some metallic property.
2. Halogens are strongly oxidizing, Why? (MP 2018)

Answer:
1. Ionization energy of iodine is small so it gives out one electron from its valence shell. In some reaction Iodine gives I⁺ ion so it has some metallic property.
I → I⁺ + e^–

2. Halogens are only one electron short to complete their octet and attaining stable structure. So halogens have strong tendency to accept electron. Thus, halogens are strong oxidising agents because of accepting electrons in reduction, oxidising power of halogens decreases from fluorine to iodine.
F₂ > Cl₂ > Br₂ > I₂

The p-Block Elements Long Answer Type Questions

Question 1.
What are interhalogen compounds? How many types are they? Give one example of each type with structure.
Or,
Explain hybridization in ABS and AB? type of interhalogen compounds and draw their structures.
Answer:
Because of difference in the electronegativity value of halogen member it is possible to combine to form compound. When two different halogen combine together to form binary compound, then it is known as interhalogen compound. There are four types of interhalogen compound, which has general formula XY_n, where n = 1, 3, 5 or 7.

Structure of interhalogen compound:
1. AB Type :
Like ClF, BrCl, IBr, ICl. Their geometry is linear. In the ground state electronic configuration of chlorine it is clearly observed that there is one unpaired electron which forms covalent bond with other halogen atom.

2. AB₃ Type :
It is T – shaped. Like ClF₃molecule, Its central atom X shows sp³ hybridization.

3. AB₅ (IF₅ ,BrF₅ )TyPe :
These type of compounds show sp³ d² hybridization. Then structure is square pyramidal in which there is lone pair of electron at one place. Like in the figure formation of IF₅ molecule is shown.

4. AB₇(IF₇) Type :
Its geometry is pentagonal bipyramidal which is formed by sp³d² hybridization. Hybridization in IF7 molecule is sp³d³.

Question 2.
Write the uses of Noble gases.
Answer:
Uses of Noble Gases :
Uses of Helium :

1. It is used in filling air ships and balloons for weather study as it is light and non – inflammable.
2. Helium is used in research work for maintaining very low temperatures.
3. It is used in producing inert atmosphere, for food preservation and as filler in electric transformer.
4. Sea divers use mixture of helium and oxygen for deep sea diving.

This is due to the fact that under high pressure helium is less soluble in blood as compared to nitrogen. If for breathing in deep sea diving air is used then more nitrogen will dissolve in blood under pressure and when the diver comes back at the surface of the sea, the solubility of nitrogen in blood decreases due to decrease in pressure and changes into nitrogen bubbles. This causes severe pain in diver’s body. It may also cause blisters and vomiting.

Uses of Neon:
Neon lights are used for commercial advertisements. It consists of a long tube fitted with electrodes at both ends. On filling the tube with neon gas and passing electric discharge of about 1000 volt potential, a bright red light is produced. Different colours can be obtained by mixing neon with other gases. For producing blue or green light neon is mixed with mercury vapours.

Uses of Argon:

1. Like helium, it is also used to create inert atmosphere in welding of aluminium and stainless steel.
2. It is filled in electric bulbs along with 25% nitrogen.
3. It is also used in radio valves.
4. Argon alone or its mixture with neon is used in tubes for producing lights of different colours.

Uses of Krypton and Xenon:

1. They are also used in electric bulbs in place of argon, but they are costly.
2. Both are used in flash photography.
3. Krypton is mainly used in producing fluorescence, incandescence and electric lights.
4. Krypton lamp is used as indicator at aeroplane terminals for landing of the aeroplane.

Uses of Radon:

1. In treatment of malignant growths (cancer) and non – healing wounds. Also, used in researches associated with radioactivity.
2. Used as a substitute for X – rays in radiology.
3. Radon is soluble in fatty compounds and recently a radon containing ointment has been manufactured for radiotherapy.

Question 3.
Explain manufacture of chlorine under the following heads :
Answer:

1. Labelled diagram of Nelson cell
2. Principle
3. Deacon’s process.

Answer:
1. Nelson cell:
In Nelson cell, graphite anode is suspended in a perforated cylindrical steel cathode provided with asbestos. Brine solution is taken in steel cathode. On electrolysis, chlorine gas is liberated at anode and is drawn off through the outlet at the top. Na reacts with H₂O at bottom to form NaOH and H₂. Hydrogen escapes through the exit.

2. Principle:
Nelson method or By electrolysis:
Chlorine is manufactured by electrolysis of NaCl solution, chlorine is obtained as by product.
NaCl ⇌ Na⁺ + Cl^–

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

At anode : cl⁺ + e^– → Cl
Cl + Cl → Cl₂

Deacon’s process:
Previously chlorine was manufactured by Deacon’s process. In this process, HCl gas is heated with oxygen at 450°C in presence of CuCl₂ catalyst.

Question 4.
Explain Siemen – Halskey ozonizer processor the manufacture of ozone and draw labelled diagram.
Answer:
For commercial production of ozone Siemen – Halskey ozonizer is used. It consists of an iron box fitted with 6 to 8 glass cylinders. Each cylinder is 3 ft high and 10″ in width. The box is divided into three compartments. Cold water flows through the central compartment for cooling. Each cylinder is fitted with an aluminium rod which are resting on insulating glass slab in the lower compartment.

There is annular space between rod and sides of cylinder. A current of air flows through this annular space. The aluminium rods are subjected to high potential of 8,000 to 10,000 volts. The air which enters at bottom escapes at the top and contains about 5% of ozone. If pure oxygen is used in place of oxygen then ozonized oxygen contains about 15% of ozone.

Question 5.
Explain the structures of XeF₂ and XeF₄.
Answer;
Structure of XeF₂:
In XeF₂ molecule Xe atom is in sp³d hybrid state hence, its structure is trigonal bipyramidal but its geometry is linear.

Structures of XeF₄:
XeF₄ formed under second excited state of Xe where two of sp – orbitals are unpaired and 4 unpaired electron develop. Being sp³d² hybridization involved, the structure is octahedral but due to presence of two lone pair of electron geometry molecules get distorted and becomes square planar.

Question 6.
Dtim labelled diagram of laboratory method of phosphine and give chemical emurfion.
Answer:
Phosphine is prepared in laboratory by heating white phosphorus with NaOH in an inert atmosphere of CO₂ and oil gas. Phosphine produced is highly inflammable due to the presence of phosphorus dehydride as impurity. Vapour of phosphine form vortex noing of smoke in contact with air. Alcoholic KOH can be used in place of NaOH.

Question 7.
Describe Brodie’s Ozonizer with diagram.
Answer:
It is U – shaped glass tube, consists Pt wire with dil. H₂SO₄. The whole apparatus is put into glass utensil, outer utensil also contain H₂SO₄ and Pt electrode. Dry O₂ is passed through U – tube under the influence of high electric discharge oxygen gets converted to ozonised oxygen, the outcoming gas contains 20% of ozone.

Question 8.
Give chemical reaction of ozone with

1. K₂MnO₄
2. I₂
3. Ag₂O
4. CH₂ ≡ CH₂ and (v) PbS.

Answer:
1. With K₂MnO₄ : Potassium permangnate is formed.
2K₂MnO₄ + H₂O + O₃ → 2KMnO₄ + 2KOH + O₂

2. With I₂ : Iodic acid is formed.
I₂ + H₂O + 50₃ → 2HIO₃ + 5O₂

3. With Ag₂O : Silver is formed.
Ag₂O + O₃ → 2Ag + 2O₂

4. With CH₂ = CH₂ : Ethene ozonide is formed.

5. With PbS : PbS04 is formed.
PbS + 4O₃ → PbSO₄ + 4O₂

Question 9.
Explain Ostwald’s process of manufacture of nitric acid drawing labelled diagram.
Answer:

Principle of Ostwald’s process:
The mixture of ammonia and air when passed over platinum gauze catalyst at 750 – 900°C, ammonia gets oxidized to nitric oxide.

The reaction is exothermic and the heat liberated maintains the temperature of the catalyst. The nitric oxide is then oxidised to nitrogen dioxide by air which is cooled to 50°C and absorbed by water to produce nitric acid.

Question 10.
Describe the manufacture of H₂SO₄ by contact process. (NCERT)
Answer:
Manufacture of sulphuric acid:
Sulphuric acid is prepared on large scale by contact process. The basic raw material used is either sulphur or iron pyrites.

Principle of contact process:
When pure and dry SO₂ mixed with air is passed over V₂O₅ catalyst it gets oxidised to SO₃ which is absorbed by H₂SO₄ to form oleum i.e,, H₂S₂O₇.
2SO₂ + O₂ → 2SO₃ + 45.2 cal
SO₃ + H₂SO₄ → H₂S₂O₇

Details of the process are given below :
(a) Sulphur or pyrite burners : SO₂ is produced by burning sulphur or roasting pyrites in pyrite burners.

(b) Purification unit: This unit is an assembly of various parts like.
(i) Dust chamber : SO₂ gas produced is passed through dust chamber through which steam is sprayed from top. Dust particles absorb moisture, become heavy and settle down.

(ii) Cooling pipes : Now, the gaseous mixture is passed through cooling pipes to lower the temperature to 100°C.

(iii) Washing tower or scrubbers : It is filled with quartz and showers of cold water continue from top. Dust particles as well as soluble impurities settle down and are removed.

(iv) Drying tower : It is a high tower full of quartz pieces. Cone. H₂SO₄ is sprayed from the top. Gases enter the tower from bottom and get dried in contact with cone. H₂SO₂.

(v) Arsenic purifier : It is filled with shelves containing freshly prepared ferric hydroxide which absorb the impurities of arsenic purified and dried.

(c) Testing box:
Before sending the gaseous mixture to contact chamber it is tested here. A strong beam of light is thrown into the testing box, if the gaseous mixture contains dust or any other particles they become visible, hence gaseous mixture is repurified. The completely pure gas is then sent to the contact chamber.

(d) Preheater:
Pure gas free from dust particle is heated in preheater to 450°C and then passed into contact tower, where SO₂ gets oxidized to SO₃. This reaction is exothermic due to which temperature of contact tower raises to 450°C. After this the pure gases are pumped directly into the contact tower.

(e) Contact tower:
It consists of a big iron container containing several pipes filled with vanadium pentoxide or platinized asbestos or any other suitable catalyst. Temperature is maintained at 450°C. Pure and dry SO₂ reacts with air in these pipes to form sulphur trioxide.

On account of exothermic nature of reaction, sufficient heat is available.

(f) Absorption tower:
The sulphur trioxide is then passed into a tower in which cone, sulphuric acid flows down in the form of spray. Sulphuric acid absorbs SO₃ and becomes more concentrated. This acid, due to excess of SO₃ produces a fog in the tower. The acid obtained is called fuming sulphuric acid or oleum.

Oleum is mixed with calculated quantity of water to form acid of a particular strength.
H₂S₂O₇ + H₂O → 2H₂SO₄
It may be noted that sulphur trioxide is not directly absorbed in water to form sulphuric acid because it forms a dense fog of sulphuric acid which does not condense easily.

Question 11.
Write formula, structure and oxidation state of any three oxy – acids of chlorine
Answer:

Structure:

Structure of HClO:
The central atom Cl in CIO ion of HCIO, forms four hybrid orbitals by sp3 hybridization. On three orbitals lone pair of electron is present and fourth orbital forms σ – bond by overlapping with d – orbital of oxygen atom. It’s structure is linear.

Structure of HClO₃ :

Structure of HClO₄ :
sp³ hybridization is found in it’s central atom Cl, hence structure of ClO₄ ion is tetrahedral.

MP Board Class 12th Chemistry Important Questions

MP Board Class 12th Chemistry Important Questions Chapter 6 General Principles and Processes of Isolation of Elements

General Principles and Processes of Isolation of Elements Important Questions

General Principles and Processes of Isolation of Elements Objective Type Questions

Question 1.
Choose the correct answer :

Question 1.
In the extraction of Fe from haematite, lime stone acts as :
(a) Reducing agent
(b) Slag
(c) Gange
(d) Flux.
Answer:
(d) Flux.

Question 2.
Cupellation is used in the metallurgy of:
(a) Cu
(b) Ag
(c) Al
(d) Fe.
Answer:
(b) Ag

Question 3.
Which is the essential part of photographic plates and Films :
(a) AgNO₃
(b) Ag₂S₂O₃
(c) AgBr
(d)Ag₂CO₃
Answer:
(c) AgBr

Question 4.
Malachite is :
(a) Cu₂S
(b) CUCO₃.CU(OH)₂
(c) Cu₂O
(d) CuCO₃.
Answer:
(b) CUCO₃.CU(OH)₂

Question 5.
AgBr is soluble in hypo because its forms :
(a) Ag₂SO₃
(b) Ag₂S₂O₃
(c) [Ag(S₂O₃)₃]^–
(d) [Ag(S₂O₃)₂]^3-
Answer:
(c) [Ag(S₂O₃)₃]^–

Question 6.
AgCl is soluble in ammonia due to the formation of:
(a) [Ag(NH₃)₄]⁺
(b) [Ag(NH₃)₃]²⁺
(c) [Ag(NH₃)₄]³⁺
(d) [Ag(NH₃)₂]⁺
Answer:
(d) [Ag(NH₃)₂]⁺

Question 7.
When KI mixed with solution of CuSO₄, it forms :
(a) Cul₂
(b) Cul^2-₂
(c) K₂[CuI₄]
(d) Cu_2& I₂ + I₂
Answer:
(d) Cu_2& I₂ + I₂

Question 8.
When KCN reacts with solution of CuSO₄, it forms :
(a) CU(CN)₂
(b) CuCN
(c) K₂[Cu(CN)₄]
(d) K₃[Cu(CN)₄]
Answer:
(d) K₃[Cu(CN)₄]

Question 9.
Na₂S₂O₃ used in photography as a :
(a) Reducing agent
(b) Developer
(c) Fixer
(d) Toning agent
Answer:
(c) Fixer

Question 10.
Calomel is :
(a) Hg₂ Cl₂
(b) HgCl₂
(c) Hg₂ Cl₂ +Hg
(d) Hg + Hgcl₂
Answer:
(a) Hg₂ Cl₂

Question 11.
German silver is an alloy of :
(a) Cu, Zn and Ni
(b) Cu, Zn and Sn
(c) Ag, Cu and Au
(d) Fe, Cr, Ni
Answer:
(a) Cu, Zn and Ni

Question 12.
Heating pyrites to remove sulphur is called :
(a) Roasting
(b) Liquation
(c) Calcination
(d) Smelting
Answer:
(a) Roasting

Question 13.
Reducing agent in the extraction of Iron from haematite is :
(a) CO
(b) C
(c) CO₂
(d) FeO
Answer:
(a) CO

Question 14.
The Chemical Composition of “Cryolite” mineral is :
(a) Al₂O₃
(b) Al₂O₃.12H₂O
(c) KAl Si₃O₈
(d) Na₃AlF₆
Answer:
(d) Na₃AlF₆

Question 15.
In Photography we use :
(a) Agl
(b) NH₃
(c) AgCl
(d) AgBr
Answer:
(d) AgBr

Question 16.
Blister copper is :
(a) Ore of copper
(b) Alloy of copper
(c) Pure copper
(d) Copper containing 1% empuritis.
Answer:
(d) Copper containing 1% empuritis.

Question 2.
Fill in the blanks :

1. Malachite is an ore of …………………
2. In stainless steel, along with iron ………………… and ………………… metals form alloys.
3.  ………………… is used as a purgative.
4. Colloidal solution of ………………… is used as a medicine of eyes.
5. AgNO₃ is known as …………………
6. Chemical formula of corrosive sublimate is …………………
7. Froath floatation process is generally employed for ………………… ores.
8. The process in which metal oxide is reduced by Al is known as …………………
9. Is used for drying ammonia …………………
10.  ………………… is called lunar caustic.
11. The chemical formula of fluorspar is …………………
12. Alkaline solution of HgCl₂ and Kl is known as …………………
13. Red hot steel is slowly cooled when it gets converted to soft steel, this is known as …………………

Answer:

1. Cu
2. Cr, Ni
3. Calomel
4. Ag
5. Lunar caustic,
6. HgCl₂
7. Sulphide
8. Aluminothermic
9. CuO
10. Silver Nitrate
11. CaF₂
12. Nesseler’s reagent
13. Annealing.

Question 3.
Match the following :
I

Answer:

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

II.

Answer:

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

Question 4.
Answer in one word / sentence :

1. The mixture of Cu₂S and FeS obtained from blast furnace is known as.
2. Which metal is purified by polling?
3. After developing which solution is used for the fixing of photographic films? Chemical formula of philosophers’s wool is.
4. Chemical formula of horn silver.
5. Which compound is normally used for toning in photography?
6. Which are known as coinage metals?
7. Give the name of ore used for extraction of Cu.
8. Give the name of ore used for extraction of Iron.
9. What is the name of graph which is drawn between the absolute temperature and standard free energy change for
10. formation of metal oxide?
11. What is the name of iron obtained after the removal of impurities from cast iron?
12. What is the method of slow cooling of hot hard steel known as?
13. What is the method of heating hard steel known as?
14. What is the method of heating steel in pressure of ammonia known as?
15. What is Lunar Caustic?

Answer:

1. Matte
2. Copper
3. Hypo solution (Na₂S₂O₃)
4. ZnO
5. AgCl
6. Aurric chloride
7. Cu, Ag and Au
8. Copperpyrite
9. Haematite
10. Ellinghum diagram
11. Wrought iron
12. Annealing
13. Softening
14. Nitriding
15. AgNO₃

MP Board Class 12th Chemistry Important Questions

MP Board Class 11th Chemistry Important Questions Chapter 10 s – Block Elements

s – Block Elements Important Questions

s – Block Elements Objective Type Questions

Question 1.
Choose the correct answer:

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

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

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

Question 4.
Hydration energy of Mg²⁺ ion will be more than:
(a) Al³⁺
(b) Na⁺
(c) Be²⁺
(d) Mg³⁺
Answer:
(b) Na⁺

Question 5.
Which magnetic property is present in alkaline earth metals:
(a) Diamagnetic
(b) Paramagnetic
(c) Ferro – magnetic
(d) Anti – magnetic
Answer:
(a) Diamagnetic

Question 6.
Solubility of which sulphate is the least:
(a) BaSO₄
(b) MgSO₄
(c) SrSO₄
(d) CaSO₄
Answer:
(a) BaSO₄

Question 7.
Compounds of which element are mainly covalent:
(a) Ba
(b) Sr
(c) Ca
(d) Be
Answer:
(d) Be

Question 8.
Important ore of magnesium is:
(a) Malachite
(b) Kaesiterite
(c) Camallite
(d) Galena.
Answer:
(c) Camallite

Question 2.
Fill in the blanks:

1. Radius of Ca²⁺ ion is less than K⁺ because of ……………………………
2. On heating Rb(ICl₂) decomposes to form ………………………….. and ………………………….
3. Be(OH)₂ is soluble in both acid and base because it is of ……………………… nature.
4. Lithium resembles ………………………… element of group 2.
5. Sodium metal gives blue colour in liquid ammonial this is because of ………………………………
6. Potassium forms three oxides ………………………. and …………………………..

Answer:

1. High positive charge
2. RbCl, ICI
3. Amphoteric
4. Mg
5. e^–(NH₃)₂
6. K₂O, K₂O₂, KO₂

Question 3.
Answer in one word/sentence:

1. Sodium metals are kept in kerosene oil. Why?
2. What is lithopone?
3. Which element is present in teeth and bones?
4. What is Caustic soda?
5. What is Sorrel cement?
6. Write formula of Plaster of Paris?

Answer:

1. Highly reactive
2. BaSO₄ and ZnS
3. Calcium
4. NaOH
5. Mixture of MgCl₂ and MgO
6. (CaSO₄)2H₂O

s – Block Elements Very Short Answer Type Questions

Question 1.
Arrange the alkali metals in the increasing order of their reactivity?
Answer:
The reactivity of alkali metals increases from top to bottom.
Be < Mg < Ca < Sr < Ba < Ra.

Question 2.
Write the name of two ores of lithium with formula?
Answer:
Ores of lithium are:

1. Spodumene LiAl(SiO₃)₂
2. Lepidolite Li₂Al₂(SiO₃)₃.F(OH)₂.

Question 3.
Write the name of two ores of magnesium?
Answer:

1. Camolite
2. Dolomite.

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

Question 5.
What is the formula of global salt?
Answer:
Na₂SO₄.10H₂O.

Question 6.
What is Lithophone?
Answer:
BaSO₄ + ZnS.

Question 7.
What is the formula of hydrolite?
Answer:
CaH₂.

Question 8.
Carnolite is ore of which metal?
Answer:
Magnesium (Mg).

Question 9.
Which alkali metal show radioactivity?
Answer:
Radium.

Question 10.
Which compound is useful in purification of air in aircrafts?
Answer:
Potassium superoxide.

Question 11.
Name one mineral in which Ca and Mg both are present?
Answer:
Name: Dolomite, Formula: MgCO₃.CaCO₃.

Question 12.
Which flux is used for the removal of acidic impurities in metallic processes?
Answer:

1. Limestone CaCO₃
2. Magnesite MgCO₃.

Question 13.
Formula of Nitrolium is?
Answer:
CaCN₂ and C.

Question 14.
Which forms curtain of smoke?
Answer:
SiCl₄.

Question 15.
What is used to join the fractured bone and make statues?
Answer:
Plaster of paris.

Question 16.
What is the order of stability of carbonates of alkali metals?
Answer:
BeCO₃ < MgCO₃ < CaCO₃ < SrCO₃.

Question 17.
Write the configuration of alkali metals?
Answer:
ns^1-2.

Question 18.
What is the formation of Apsum and Gypsum salt?
Answer:
Magnesium sulphate (MgSO₄.2H₂O) and Calcium sulphate (CaSO₄.2H₂O).

s – Block Elements Short Answer Type Questions – I

Question 1.
Why alkali metal do not found in free state?
Answer:
Alkali metals are highly reactive and electropositive because the value of ionization energy is very low due to larger size thus alkali metals easily donate electron and form positive ion. (MPBoardSolutions.com) These positive ion easily combine with oxygen, moisture and other electronegative element present in nature and form ionic compound.

Question 2.
Explain, why sodium is less reactive than potassium?
Answer:
The ionization enthalpy of sodium is higher than that of potassium. Therefore, sodium lose electron less readily as compared to potassium. Hence, sodium is less reactive than potassium.

Question 3.
Potassium carbonate cannot be prepared by Solvay process. Why?
Answer:
Potassium carbonate cannot be prepared by Solvay process because potassium bicarbonate being more soluble than sodium bicarbonate and does not get precipitated when CO₂ is passed through a concentrated solution of KCl with NH₃.

Question 4.
Why is Li₂CO₃ decomposed at a lower temperature whereas Na₂CO₃ at higher temperature?
Answer:
Lithium is less electropositive than sodium and therefore, carbonates of lithium is less stable than that of sodium. Li₂CO₃ is not so stable to heat and therefore, decomposes at lower temperature. (MPBoardSolutions.com) This is because lithium being veiy small in size polarizes a large. CO₃^2- ion leading to the formation of Li₂O and CO₂. On the other hand, Na₂CO₃ is very stable and decomposes at higher.

Question 5.
I – A and II – A group is called s – block elements. Why?
Answer:
I – A and II – A group elements have their electronic configuration of ns¹ and ns². So the last electron fill in 5 – block so the I – A and II – A group is called 5 – block elements.

Question 6.
What is Soral cement and write its uses?
Answer:
When MgCl₂ solution reacts with MgO and product is formed as MgCl₂.2MgO. nH₂O. It is a white paste and called as Soral cement.
Uses:

1. It is used to fill the teeth cavity.
2. They use in porcelain in point.

Question 7.
Find the oxidation state (O.S.) of Na in Na₂O₂?
Answer:
Let the O.S. of Na in Na₂O₂ = x
One peroxide bond (Na – O – O) is present in Na₂O₂ where the O.S. of O = -1

So, in Na₂O₂ the O.S. of Na = +1.

Question 8.
How calcium sulphate is prepared? Write its uses?
Answer:
In the lab, calcium sulphate is formed by Ca oxide, carbonate, chloride. They react with dil. H₂SO₄ to form calcium sulphate.
CaO + H₂SO₄ → CaSO₄ + H₂
Ca(OH)₂ + H₂SO₄ → CaSO₄ + 2H₂O
CaCl₂ + H₂SO₄ → CaSO₄ + 2HCl
CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂

Question 9.
What is Gypsum? How plaster of Paris is formed by gypsum?
Answer:
Calcium sulphate CaSO₄.2H₂O is called gypsum. Gypsum is heated at 120° – 130°C and 3 parts of water molecule are released and plaster of Paris is formed.

The plaster of Paris again absorb the water molecule and again get converted to calcium sulphate (Gypsum).

Question 10.
Why during the formation of quick lime the temperature of furnance is not kept more than 1000°C? Explain with equation?
Answer:
During the formation of quick, lime the temperature is maintained at 1000°C because at high temperature the clay present as impurity in the limestone combines with lime producing fusible silicate which fill the pores of lime. Due to this reason slaking of lime become very difficult.
1000°C.
CaO + SiO₂ \(\underrightarrow { 1000^{ \circ }C } \) CaSiO_3.

Question 11.
Why sodium kept in kerosene oil?
Answer:
Alkali metals are very reactive specially against electronegative elements like oxygen, moisture and CO₂. These are oxidized quickly from oxide and hydroxide due to their reactivity. Sodium is kept in kerosene oil to prevent oxidation on exposure to air.
4Na + O₂ → 2Na₂O
2Na + H₂O → Na₂O + H₂
Na₂O + H₂O → 2NaOH.

Question 12.
Write the formula of lime water. What happens when CO₂ passes through it?
Answer:
The formula of lime water is Ca(OH)₂. Due to the flow of the CO₂ gas in Ca(OH)₂ it becomes milky due to formation of a white precipitate of calcium carbonate. (MPBoardSolutions.com) On passing excess of CO₂, milkyness disappears and calcium bicarbonate is formed which is soluble in water.
Ca(OH)₂ + CO₂ → CaCO₃ + H₂O
CaCO₃ + CO₂ + H₂O → Ca(HCO₃)₂.

Question 13.
Why K₂CO₃ is not prepared by Solvay method?
Answer:
Potassium carbonate cannot be prepared by Solvay method because the potassium salt analogous to NaHCO₃ is KHCO₃ which is much soluble and hence, cannot be obtained by crystallization.

Question 14.
Which metal is used in photochemical cell and why?
Answer:
In photochemical cell potassium and caesium metals are used, as their ionization energy is very low.

Question 15.
Why the extraction of sodium from sodium chloride cannot be done by general reducing agents?
Answer:
Sodium is a strong reducing agent. In electrochemical series, it is present at top position. Due to new availability of strong reducing agent than sodium, it cannot be reduced by normal reducing agents. It can only be reduced by electrolytes.

Question 16.
Why Li and Be have the tendency to form covalent compounds? Explain?
Answer:
Due to small size of Li and Be atoms and their high ionization energy, the electrons of the valence shell are firmly bound to their nuclei. The polarizing power of their ions is also high due to high charge density, hence Li and Be form covalent compounds.

Question 17.
Write the Lewis structure of O^–₂ ion and write the O.N. of each O atom. In this ion what is the average oxidation state of O?
Answer:
Lewis structure of O^–₂ ion =
Oxygen without charge have 6 electrons. So, its O.S. = 0, but oxygen with -1 oxidation state have 7 electrons so its O.S. = —1.
Oxidation state of each oxygen atom = \(\frac{-1}{2}\)
O₂^– = 2x = -1
x = \(\frac{-1}{2}\)

Question 18.
In Solvay’s process, can be obtain sodium carbonate by direct reaction of ammonium carbonate and sodium chloride?
Answer:
No, because the reaction between ammonium carbonate and NaCl.
(NH₄)₂CO₃ + 2NaCl ⇄ Na₂CO₃ + 2NH₄Cl
As the products obtained are highly soluble the equilibrium will not shift to,yards forward direction. This is the reason why NaCO₃ cannot be prepared by reaction of (NH₄)₂CO₃ and NaCl in Solvay’s process.

Question 19.
AH compounds of alkali metals are easily soluble in water but lithium compounds are more soluble in organic solvents? Explain?
Answer:
The size of LT ion is small and thus, it has high polarizing power. This brings covalent character in lithium compounds. Due to covalent character, Li compounds are soluble in organic solvents.

Question 20.
Why are potassium and caesium rather than lithium used n photoelectric cells?
Answer:
Potassium and caesium have much lower ionization enthalpy than that of lithium. Therefore, these metals on exposure to light, easily emit elec arm but lithium does nN. Therefore, K and Cs rather than Li are used in photoelectric cells.

Question 21.
Beryllium chloride (BeCl₂) produces smoke when kept in air. Why?
Answer:
Normally air contains moisture, thus beryllium halide hydrolyses in water and releases HCl due to which it produces smoke in air.
BeCl₂ + 2H₄O → Be(OH)₂ + 2HCl.

Question 22.
On moving downward in the first group the hardness of the elements increases. Why?
Answer:
In the first group on moving downwards along with the increase in size cf the elements their density also increases and the force of attraction between its atoms increases due to which there is an increase in their hardness.

s – Block Elements Short Answer Type Questions – II

Question 1.
Why s – block and p – block elements are called representative elements?
Answer:
The elements present in the s and p-block are called normal or representative elements.

1. The elements of group 1 and 2 constitute the s – block of the periodic table. These are known as s – block elements because the last electron in them enters the s – orbital of the valence shell.
2. The elements of p group of the periodic table these are known as p – block elements because the last electron in them enters the p-orbital of the valence shell.

Question 2.
Why alkaline metals are strong or reducing agents?
Answer:
Due to less ionization energy of alkali metals. They have tendency to loose electron (Get oxidized) and form positive ion (M⁺) also alkali metals have negative value of standard reduction potential. (MPBoardSolutions.com) Therefore, alkali metals are good reducing agent.

Question 3.
Why are BeSO₄ and MgSO₄ readily soluble in water while CaSO₄, SrSO₂ and BaSO₄ are insoluble?
Answer:
The hydration energy of SO₄ of alkaline earth metals decreases down the group. The high hydration energy of Be²⁺ and Mg²⁺ diminishes the lattice energy due to this their SO₂ are soluble in water. But in case of Ca²⁺, Sr²⁺ and Ba²⁺ the hydration energy is low due to this the SO₄ are insoluble in water.

Question 4.
Why the reducing power of lithium is high in solution?
Answer:
Electrode potential is a measure of the tendency of an element to loose electron in aqueous solution. It may depend on the following three factors:

With the small size of its ion, Li has the highest hydration enthalpy. However, ionization enthalpy of Li is highest among alkali metals but hydration (MPBoardSolutions.com) enthalpy predominates over I.P. Therefore, Li is the strongest reducing agent in aqueous solution.

Question 5.
Explain, why can alkali and alkaline earth metals not be obtained by chemical reduction method?
Answer:
The s – block elements themselves are good reducing agents therefore, reducing agent better than s – block elements are not available. Therefore, the chlorides, oxides etc. of s – block elements cannot be reduced to obtain metal.

Question 6.
Explain why alkaline metals form M⁺ cation not M⁺² type of cation?
Answer:
There is one electron in the valence shell of alkali metals. The ionization energy is very low due to bigger size. So they can easily donate electron and can form M⁺ cation. (MPBoardSolutions.com) In the M⁺ state the electronic configuration becomes similar to nobel gases and stable so in this state they become unreactive and the ionization energy become high. So they did not form M²⁺ ion.

Question 7.
In alkali metals, which metal is strongest reducing agent. Why?
Answer:
The reducing character increases from sodium to caesium. However, lithium is the strongest reducing agent among all the alkali metals. Inspite of its highest I.P. This is because of extensive hydration of Li⁺ ions and large amount of energy released during hydration more than compensates the higher I.P. value of Lithium.

Question 8.
Why alkali metals are not found in free state in nature? Or, Why alkali metals always form ionic compounds?
Answer:
Due to bigger size of alkali metals, the ionization energy is very low. So they can easily donate electron and form positive ion. So due to positively charge and highly reactive nature they combine with the (MPBoardSolutions.com) electronegative elements in the nature like moisture, CO₂ etc. and form ionic compounds. Therefore, they cannot be found in free state in nature.

Question 9.
Why alkali metal give flame test?
Answer:
The ionization energy of alkali metal is very low. When these elements or their compounds are heated in Bunsen flame electron present in valency shell absorb energy and easily goes to higher energy level. When these excited electron comes to ground state they emit energy in the form of radiation in visible region and give characteristic colour to flame.

Question 10.
Why Be and Mg do not give flame test?
Answer:
Beryllium and Magnesium atoms in comparison to other alkaline earth metals are comparatively smaller and their ionization energy are very high. (MPBoardSolutions.com) Hence, the energy of the flame is not sufficient to excite their electrons to higher energy levels. These elements therefore do not give any colour in Bunsen flame temperature.

Question 11.
When an alkali metal dissolves in liquid ammonia, the solution acquires different colours? Explain the reason for this type colour change?
Answer:
All alkali metals dissolve in liquid ammonia giving highly conducting deep blue solutions.
M + (x + y )NH₃ M⁺ (NH₃) → x + e^– (NH₃ )_y
When ordinary light falls on these ammoniated electrons. They get excited and jump to higher energy levels by absorbing energy corresponding to red region of the visible light. (MPBoardSolutions.com) As a result transmitted light is blue which imparts blue colour to the solution. However, when the concentration increases the ammoniated metal ion may get bound by free electrons and colour becomes copper bronze.

Question 12.
Why Be and Mg do not give flame test but other metals give? Why?
Answer:
Except beryllium and magnesium all the alkaline earth metals impart characteristic colours to Bunsen flame. Due to small size of Be and Mg atom, the energy required to excite the valency electrons is very high which is not obtained in Bunsen flame. That is why Be and Mg do not impart any colour to flame.

Question 13.
Give two uses of each:

1. Caustic soda
2. Sodium carbonate
3. Quick lime.

Answer:
1. Uses of Caustic soda:

• In soap and paper industry.
• For mercerization of cotton thread.

2. Uses of Sodium carbonate:

• Washing soda is used for washing clothes.
• Removes permanent hardness of water.

3. Uses of Quick lime:

• For making statues, floor, buildings in the form of marble.
• For manufacture of lime, cement, glass and washing soda.

Question 14.
Alkali metals form blue coloured solution when dissolved in ammonia, which is a strong electrolyte. Give reason with equation?
Answer:
Alkali metals dissolve in liquor ammonia to form deep blue coloured solution of high electrical conductivity.
M + (x+y) NH₃ → [M(NH₃)_x] + [e(NH₃)_y]^–
The blue colour of the solution is due to ammoniated electrons positive ion and electrons are responsible for the conductivity.

Question 15.
Na is alkaline or Na₂O? Clarify the statement?
Answer:
Monoxide of all alkali metals are alkaline and form strongly alkaline solution in water.

Na₂O is alkaline because it reacts with water to form NaOH. Na also react with water to form NaOH but it first forms Na₂O and then NaOH, thus Na₂O us alkaline not Na.
4Na + 2H₂O → 2Na₂O + 2H₂
Na₂O + H₂O → 2NaOH

Question 16.
Compare Alkali metals and Alkaline earth metals on the basis of following points:

1. Reaction of heat on carbonate
2. Reaction with nitrogen
3. Solubility of sulphates in water.

Answer:
Comparison of Alkali metals and Alkaline earth metals:

Question 17.
Hydroxides and Carbonates of (Sodium and Potassium) Alkali metals are completely soluble in water whereas that of (Magnesium and Calcium) Alkaline earth metals are partially soluble?
Answer:
All alkali metal carbonates are soluble in water because the value of their Lattice energy is less than their hydration energy.
Lattice energy < Hydration energy (Compound soluble)
Alkaline earth metal carbonates are insoluble because the value of their Lattice energy is more than their hydration energy.
Lattice energy < Hydration energy (Compound insoluble)
∆H_solution = ∆H_{Hydrogen energy} + ∆H_{Lattice energy}

Question 18.
Why are lithium salts commonly hydrated and those of the other alkali ions usually anhydrous?
Answer:
Because of smallest size among alkali metals, Li⁺ can polarize water molecule more easily than the other alkali metal ions and hence get attached to lithium salts as water of crystallization.

Question 19.
Why is LiF almost soluble in water whereas LiCI soluble not only in water but also in acetone?
Answer:
To make compound water soluble, its lattice energy should be low and hydration energy should be high. Lattice energy of LiCl is less than that of LiF. The difference in these two energies for LiCl and LiF is 31 and 14 kJ/mol respectively. (MPBoardSolutions.com) The difference is larger for LiCl, which is soluble in water. Moreover, lattice energy of LiF is much higher than that of LiCl. It makes LiF sparingly soluble in water. Moreover, LiCl is largely covalent, so it is soluble in organic solvent such as acetone.

Question 20.
If the alkali metals are kept open in the air, after sometime the metallic brightness is lost. Why?
Answer:
On exposure to moist air alkali metals soon get covered with a thick crust of their oxide, hydroxides and carbonate hence, their surface get tarnished. For this reason these metals are stored under kerosene and which prevents them from coming in contact with air and moisture.
4M + O₂ → 2M₂O
M₂O + H₂O → 2MOH
2MOH + CO₂ → M₂CO₃ + H₂O
Here, M is any alkali metal.

Question 21.
Among LiCl and RbCl which will ionize more. Why?
Answer:
Among LiCI and RbCl, RbCl is more reactive because LiCl is slightly covalent in nature due to which it is soluble in organic solvent like pyridine and alcohol.

Question 22.
Among alkali metals and alkaline earth metals whose carbonates are soluble in water? Or, The carbonates of alkali metals are souble in water but that of alkaline metals are insoluble?
Answer:
Carbonates of alkali metals are soluble in water because its hydration energy is greater than lattice energy. Whereas alkaline earth metals are smaller in size and have high density, so its hydration energy is lower than lattice energy. So the carbonates of alkaline earth metals are insoluble in water.

Question 23.
Give the differences between the 8eCl₂ and other chlorides of alkaline earth metals?
Answer:
1. Anhydrous halide are deliquescent and they absorb moisture or water forming hydrated salt.
Example: MgCl₂.6H₂O, CaCl₂.6H₂O, BaCl₂.2H₂O, etc.

2. BeCl₂ fumes on hydrolysis in moist air.
BeCl₂ + 2H₂O → Be(OH)₂ + 2HCl

3. BeCl₂ has different structure in solid and in vapour state. In solid state, it exists as polymeric chain in which each Be atom is surrounded by four chlorine atom. Two of the chlorine atom are covalently bonded while, other two are bonded through co – ordinate bond. In vapour state, at a temperature above 1200 K it has linear monomeric structure with zero dipole moment. Below 1200 K it exist as a dimer.

4. Except BeCl₂ and MgCl₂ other metal chloride give characteristic colour to the flame.

5. Anhydrous CaCl₂ has strong affinity for water therefore, it is used as dehydrating agent.

Question 24.
Why solubility of BaSO₄ is less than CaSO₄?
Answer:
The solubility of sulphates decreases from top to bottom as lattice energy is same. But as we move top to bottom the atomic size increases and hydration energy decreases so solubility decreases.

Question 25.
Why ionization energy of Be is more than B?

These orbitals of Be is full – filled so this element is stable but orbitals of B is not full – filled (Completely), so this element is not stable.

Question 26.
How sodium carbonate is formed by soda process? Write its principle?
Answer:
Solvay method or Ammonia soda process: It has replaced Le – Blanck process and is most commonly employed.
Principle:
In this process, concentrated solution of sodium chloride (Brine) is saturated with NH₃ to form ammoniacal sodium chloride. On passing CO₂ gas to this solution, ammonium bicarbonate is formed which reacts with sodium chloride and forms sodium bicarbonate.

Question 27.
How will you prepare

1. Sodium bicarbonate
2. Sodium hydroxide
3. Sodium silicate from sodium carbonate?

Answer:
1. In sodium carbonate (Aqueous solution) pass CO₂ gas and sodium bicarbonate is formed (White precipitate).
Na₂CO₃ + H₂O + CO₂ → 2NaHCO₃

2. Sodium carbonate is boiled with limewater then sodium hydroxide is formed.
Na₂CO₃ + Ca(OH)₂ → 2Na0H + CaCO₃

3. Sodium carbonate is treated with silica to form sodium silicate.

Question 28.
What is Baking soda? Write its method of preparation, properties and uses?
Answer:
Sodium Bicarbonate (NaHCO₃):
It is also known as sodium hydrogen carbonate or baking soda.

Method of preparation:
Sodium bicarbonate is obtained as an intermediate in the manufacture of sodium carbonate by Solvay’s process. It can be also obtained by passing carbon dioxide through aqueous solution of sodium carbonate.
Na₂CO₃ + H₂O + CO₂ → 2NaHCO₃

Physical property:
It is a white, crystalline solid, slightly soluble in water.

Chemical properties:
1. Effect of heat:
When heated to 373 K it decomposes into Na₂CO₃ and CO₂ is set free.

2. Action of water:
Sodium bicarbonate hydrolyses when dissolved in water. Its aqueous solution is alkaline.

Uses:

1. As a source of carbon dioxide in fire extinguisher.
2. It acts as mild antiseptic for skin infection.
3. As a component of baking powder.
4. As a digestive powder to remove acidity of stomach.

Question 29.
Describe the Le – Blanck method of preparation of sodium carbonate with chemical equation? How Solvay method is better than this?
Answer:
Le – BIanck method:
Sodium chloride when heated with cone. H₂SO₄ produces sulphates and HCl gas. Mixture of sodium sulphate, calcium carbonate and coke on heating provide sodium carbonate and calcium sulphide. This mixture is called blanck ash. (MPBoardSolutions.com) Now in this mixture water is added and filtered. Gas is removed as residue while sodium carbonate being soluble goes into filtrate. Evaporation of filtrate provide solid sodium carbonate.
2NaCl + H₂SO₄ → Na₂SO₄ + 2HCl
Na₂SO₄ + 4C → Na₂S + 4CO
Na₂S + CaCO₃ → Na₂CO₃ + CaS

Advantages of Solvay method:

1. It is cheap
2. Pure Na₂CO₃ is formed
3. No harmful smoke is obtained
4. In the middle of reaction NaHCO₃ is formed which is a useful compound.

s – Block Elements Long Answer Type Questions – I

Question 1.
Describe the manufacture of caustic soda by Nelson cell on the following point: Labelled diagram, Chemical reactions.

Answer:
It consists of perforated steel tube Graphite anode lined inside with asbestos act as cathode. It is fitted with NaCl solution (brine) and is suspended in the steel tank. Here asbestos lining separates cathode from anode.

One passing current electrolysis of NaCl takes place. (MPBoardSolutions.com) Chlorine is liberated at anode and escapes out. Sodium ions passed through asbestos and is liberated at cathode they reacts with steam for caustic soda.


At anode:

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

Question 2.
Ions of an element of group 1 participate in the transmission of nerve signals and transport of sugars and amino acids into ceils. This element imparts yellow colour to the flame in flame test and forms an oxide and a peroxide with oxygen. Identify the element and write chemical reaction to show the formation of its peroxide. Why does the element impart colour to the flame?
Answer:
The element is sodium. This imparts yellow colour to the flame. Na⁺ ions participate in the transmission of nerve signals and transport of sugars and amino acids to cell. It forms a monoxide (Na₂O) and a peroxide (Na₂O₂).

The reason for flame colouration is, ionisation energy of Na is low. (MPBoardSolutions.com) Therefore, when Na metal or its salts is heated in Bunsen flame, its valence shell electron is excited to higher energy levels by absorption of energy. When the excited electron returns to the ground state, it emits extra energy in the yellow region of electromagnetic spectrum. Therefore, Na imparts yellow colour to the flame.

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

1. Both beryllium and aluminium form covalent compounds.
2. Both metals are passive towards reaction with concentrated HN03, because they are covered with a layer of oxide on their surface.
3. Both the metals are of weak electropositive nature.
4. Both do not form hydride quickly.
5. Carbides of both the metals react with water to form methane.

Be₂C + 2H₂O → 2BeO + CH₄
Al₄C₃ + 6H₂O → 2AlCl₃ + 3CH₄

6. Oxides of both are soluble in water and form hydroxides which are amphoteric in nature and react with acid and base to form salt.

BeO + 2HCl → BeCl₂ + H₂O
Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O
BeO + 2NaOH → Na₂BeO₂ + H₂O
Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O

7. Both Be and Al do not give colour to the flame.

Question 4.
When water is added to compound (A) of calcium, solution of compound (B) is formed. When carbon dioxide is passed into the solution, it turns milky due to the formation of compound (C). (MPBoardSolutions.com) If excess of carbon dioxide is passed into the solution milkiness disappears due to the formation of compound (D). Identify the compounds A, B, C and D. Explain why the milkyness disappears in the last step?
Answer:
The compound (A) is quick lime, CaO. This combines with water and forms calcium hydroxides Ca(OH)₂.

When CO₂ is passed through the solution having Ca(OH)₂, the solution turns milky due to formation of calcium carbonate, CaCO₃.

When excess of CO₂ is passed into milky solution the milkiness disappears due to formation of calcium bicarbonate Ca(HCO₃)₂ which is soluble.

Question 5.
How is lithium different from other members of its group?
Answer:
Anomalous behaviour of Lithium:
Properties of lithium are different than other members of the group due to the following reasons:

1. Size of lithium atom and ion is very small.
2. Due to small size, polarizing power of lithium ion is high, due to w hich compounds possess covalent character.
3. In comparison to other alkali metals its electropositive nature is less and ionization energy is high.

Lithium differ from alkali metals of group I due to following properties:

1. 1. Lithium is comparatively harder than the other alkali metals.
   2. Melting and boiling point of lithium is comparatively high.
   3. Lithium reacts with oxygen to form only normal oxide (Li₂G), whereas other metals form peroxide (M₂O₂) and superoxide (MO₂).
   4. Lithium hydride (LiH) is more stable as compared to hydrides of other metals.
   5. Lithium hydroxide (LiOH) is a weak base and is partially soluble in water, but hydroxides of other metals of the group are more soluble in water.
   6. Lithium forms nitride (Li₃N) with nitrogen, whereas other metals of the group do not form nitride.
   7. Lithium nitrate when heated decomposes to nitrogen dioxide and oxygen.

4LiNO₃ \(\underrightarrow { \Delta } \) 2Li₂O + 4NO₂ + O₂

Whereas sodium nitrate and potassium nitrate when heated strongly form corresponding nitrate with the release of oxygen.

2NaNO₃ \(\underrightarrow { \Delta } \) 2NaNO₂ + O₂

Question 6.
What happens when:

1. Magnesium is burnt in air –
2. Quick lime is heated with silica –
3. Chlorine reacts with slaked lime –
4. Calcium nitrate is heated?

Answer:

s – Block Elements Long Answer Type Questions – II

Question 1.
How is sodium carbonate manufactured by Solvay process?
Answer:
Principle:
In this process, first concentrated solution of sodium chloride (Brine) is saturated with NH₃ to form ammanica sodium chloride.(MPBoardSolutions.com) On passing CO₂ gas to this, ammonium bicarbonate is formed which reacts with sodium chloride and forms sodium bicarbonate. Precipitate of sodium bicarbonate is filter ed which on calcination gives sodium carbonate.
NH₃ + CO₂ + H₂O → NH₄HCO₃
NH₄CO₃ + 2NaCl → NaHCO₃ + NH₄Cl
Na₂HCO₃ → Na₂CO₃ + H₂O + CO₂

CO₂ formed is used agiain for carbonation NH₄Cl formed on treatment with slaked lime gives NH₃ which is used for saturation of brine.
2NH₄CL + Ca(OH)₂ → CaCl₂ + 2H₂O + 2NH₃

The various steps and reactions involved are given below:

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

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

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

4. Carbonating tower:
The cold solution of brine saturated with ammonia is introduced into the carbonating tank from top. Carbonating tower is fitted with a number of compound diaphragms each made of a horizontal iron plate. Following reactions take place here:
2NH₃ + CO₂ + H₂O → (NH₄)₂CO₃
(NH₄)₂CO₃ + 2NaCl → Na₂CO₃ + 2NH₄Cl
Na₂CO₃ + H₂O + CO₂ → 2NaHCO₃

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

6. Limekiln:
Here limestone is burnt to produce quick lime and carbon dioxide.
CaCO₃ \({ \underrightarrow { \Delta } }\) CaO + CO₂
Formed CaO reacts with water to form slaked lime.
CaO + H₂O → Ca(OH)₂

7. Ammonia recovery tower:
The mother liquor obtained from rotatory filter pump containing ammonium chloride is introduced into the ammonia recovery tank from the top part while the milk of lime is introduced from the top of lower part and steam is introduced into the tower from the bottom, ammonium chloride reacts with milk of lime to produce ammonia.
2NH₄Cl + Ca(OH)₂ → 2NH₃ + CaCl₂ + 2H₂O
Sodium bicarbonate obtained from rotatory filters is ignited in a specially constructed cylindrical vessels when sodium carbonate is formed and CO₂ evolved is collected and used again.
2NaHCO₃ → Na₂CO₃ + H₂O + CO₂

Question 2.
What happens when:

1. Reaction of NaOH with Zn or A1
2. Reaction of NaOH with S
3. Reaction of NaOH with halogen
4. Reaction of NaOH with metal oxides
5. Reaction of NaOH with metal salts.

Answer:
1. Zn + 2NaOH → Na₂ZnO₂ + H₂
2Al + 2NaOH + 2H₂O → 2NaAlO₂ + 3H₂

2. 4S + 6NaOH → Na₂S₂O₃ + 2Na₂S + 3H₂O Hypo
8S + 2Na₂S → 2Na₂S₅

3. 2NaOH + Cl₂ → NaCl + NaCIO + H₂O
(Cold) Sodium hypochlorate
6NaOH + 3Cl₂ → 5NaCl + NaClO₃ + 3H₂O
(Warm)

4. ZnO + 2NaOH → Na₂ZnO₂ + H₂O

5. CuSO₄ + 2NaOH → Na₂SO₄ + Cu(OH)₂
3NaOH + FeCl₃ → Fe(OH)₃ + 3NaCl
2NaOH + 2 AgNO₃ → Ag₂O + 2NaNO₃ + H₂O.

Question 3.
How Lithium show similarity with Magnesium?
Answer:
In the periodic table some elements differ from elements of the same period but show diagonal similarity with elements of next period.
Example:
Elements of second period show similarities with elements of third period. Like Lithium (second) period with magnesium (third period), Beryllium with A1 and Boron shows similarity with Si.

The relationship between Li and Mg:

1. The Li atomic radius is 1.34 Å and atomic radius of Mg is 1.36 Å .
2. The polar capacity of Li and Mg is equal.
3. Li and Mg is harder elements.
4. The Li and Mg is electronegativity is equal to (1.0 and 1.2).
5. The melting and boiling point of Li and Mg is very high.
6. Li and Mg reacts with N₂ and they form nitrite.
7. Li and Mg reacts with O₂ and they form monoxide.
8. Li and Mg reacts with water to release of hydrogen gas.
9. Li and Mg carbonate is heated and release CO₂ gas.
10. LiOH and Mg(OH)₂ is the weak basicity.

Question 4.
Write the formation of calcium oxide and write its properties and uses?
Answer:
Manufacture:
Commercially, calcium oxide is manufactured by heating limestone.

This process is exothermic and reversible. In order to get good yield of lime, carbon dioxide is to be removed from time to time. (MPBoardSolutions.com) The temperature of the reaction should not exceed 900°C because at higher temperature lime and clay reacts to form fusible silicate.

Furnace used for the manufacture of lime contain two fire boxes, one each on each side. Lime stone is added from the top. It decomposes on reach¬ing down. CO₂ produced is collected and stored in cylinders in liquid state. Lime farmed gets collected at the bottom of the furnace.
Physical properties:

1. It is a white solid.
2. Its melting point is high i.e., 2870K.
3. It is highly stable and it does not decompose in oxy-hydrogen flame but produces a brilliant white light called lime light.

Chemical properties:
1. It reacts with moist air absorbing CO₂ to form Ca(OH)₂ and CaCO₃
CaO + H₂O → Ca(OH)₂ (Slaked lime)
CaO + CO₂ → CaCO₃ (Lime stone)

2. When heated with ammonium salt, ammonia gas evolved.
2NH₄Cl + CaO → CaCl₂ + H₂O + 2NH₃

3. When heated with coke at high temperature, calcium carbide is produced.

4. It forms calcium chloride when chlorine gas is passed over hot and dry calcium oxide.
2CaO + 2Cl₂ → 2CaCl₂ + O₂

5. Calcium oxide is a strong basic oxide. It reacts with acid to form salt and water while, with acidic oxide it forms respective salt.
CaO + 2HCl → CaCl₂ + H₂O
CaO + H₂SO₄ → CaSO₄ + H₂O
CaO + CO₂ → CaCO₃ (Calcium carbonate)
CaO + SiO₂ → CaSiO₃ (Calcium silicate)
CaO + SO₂ → CaSO₃ (Calcium sulphite)
3CaO + P₂O₅ → Ca₃(PO₄)₂ (Calcium phosphate)

6. It forms slaked lime when dissolved in water. It is an exothermic reaction.
CaO + H₂O → Ca(OH)₂ + 15,000 cal.

Uses:

1. For making basic lining in furnaces.
2. For drying alcohol and gases.
3. Lime water is used as laboratory reagent and in medicine.
4. For purification of coal gas and in paper industry.
5. As flux in metallurgical process.
6. For producing lime light.
7. For manufacture of ammonia, sodalime, calcium carbide, cement and glass.

Question 5.
Differentiate between Alkali metals and Alkaline earth metals?
Answer:
Differences between Alkali metals and Alkaline earth metals:
Alkali metals:

1. They show +1 oxidation state.
2. Their hydroxides are strong bases.
3. Their carabonate, sulphate and phosphate compounds of alkaline earth plate are soluble in water.
4. Their ionisation energy is comparatively less.
5. They are melleable, ductile and lustrous.

Alkaline earth metals:

1. These show +2 oxidation state.
2. Their hydroxides are comparatively weaker bases.
3. These compounds of alkaline earth metals are insoluble in water.
4. Their ionization energy is high.
5. These are comparatively less.

Question 6.
Explain the Castner – Kellner cell with diagram to obtained sodium hydroxide?
Answer:
Castner – Kellner Cell:
It consists of rectangular iron tank which is fitted with three slate portion which do not touch the bottom but fit into grooves at bottom. The bottom of cell is covered with mercury and this is brought into circulation with the help of an eccentric wheel. (MPBoardSolutions.com) The outer compartmentA is filled with brine and inner Graphite compartment B with caustic soda solution. Two slot graphite electrode project from the ceiling of the outer compartment of the vessel and act anode. The iron cathode consisting of several rods is filled in middle compartment.

There are exit pipes fitted in compartment and for removal of hydrogen in central compartment. On passing current sodium chloride (brine) solution is electrolyzed in the two outer compartment. Chlorine is liberated at anode. (MPBoardSolutions.com) Sodium amalgam formed is pushed by the help of eccentric wheel into the central compartment. In the central compartment B when caustic soda is electrolyzed OH^– move to the mercury layer which act as anode and after discharge react with the sodium of Na – Hg to form caustic soda with the release of H₂ gas. Simultaneously an equivalent amount of sodium liberated at cathode reacts with water to produce more caustic soda.
2Na – Hg + 2H²O → 2NaOH + 2Hg + H₂↑

The mercury obtained can again be used in the cell. Caustic soda solution is evaporated to obtain solid caustic soda which is fused and cast into flakes or sticks.
Cell reactions are as follows:

1. In outer compartments:
Ionization:
2 NaCl ⇄ 2Na⁺ + 2Cl^–; (Ionization)

At cathode:
2Na⁺ + 2e^– → 2Na (Reduction)
2Na + xHg → Hg_xNa₂ (Sodium amalgam)

At anode:
2Cl^– → 2Cl + 2e^–; (Oxidation)
2Cl → Cl^2-

2. In central compartment:
Ionization:
NaOH ⇄ Na⁺ + OH^–; (Ionization)

At cathode:
2Na – Hg + 2H₂0 2NaOH + 2Hg + H₂
Na⁺ + e^– → Na, (Reduction)
2Na +2H₂O 2NaOH + H_2(g)

At anode:
2OH^– → 20H + 2e^–, (Oxidation)
Hg_xNa₂ + 20H → 2NaOH + xHg

MP Board Class 11 Chemistry Important Questions

MP Board Class 12th Chemistry Important Questions Chapter 5 Surface Chemistry

Surface Chemistry Important Questions

Surface Chemistry Objective Type Questions

Question 1.
Choose the correct answer :

Question 1.
In the adsorption process of acetic acid on activated charcoal, acetic acid is :
(a) Adsorber
(b) Absorber
(c) Adsorbent
(d) Adsorbate.
Answer:
(d) Adsorbate.

Question 2.
Cause of stability of lyophobic sol is :
(a) Brownian movement
(b) Tyndall effect
(c) Electric charge
(d) Brownian movement and Electric charge
Answer:
(d) Brownian movement and Electric charge

Question 3.
In the coagulation of As₂S₃ colloidal solution value of whose coagulating power will be minimum :
(a) NaCl
(b) KCl
(c) BaCl₂
(d) AlCl₃.
Answer:
(d) AlCl₃.

Question 4.
Adsorption process is :
(a) Endothermic
(b) Exothermic
(c) No heat change
(d) None of these.
Answer:
(b) Exothermic

Question 5.
Size of colloidal particles in the range of:
(a) 10^-7 to 10^-9 cm
(b) 10^-9 to 10^-1 cm
(c) 10^-5 to 10^-7 cm
(d) 10^-2 to 10^-8 cm.
Answer:
(c) 10^-5 to 10^-7 cm

Question 6.
Which of the following is not used for the preparation of lyophilic colloid :
(a) Starch
(b) Gum
(c) Gelatin
(d) Metal sulphide.
Answer:
(d) Metal sulphide.

Question 7.
Sol which can act like a protective colloid :
(a) As₂S₃
(b) Gelatin
(c) Au
(d) Fe(OH)₃.
Answer:
(b) Gelatin

Question 8.
Fog is an example of colloidal system of:
(a) Liquid dispersed in gas
(b) Gas dispersed in gas
(c) Solid dispersed in gas
(d) Solid dispersed in liquid.
Answer:
(a) Liquid dispersed in gas

Question 9.
Gelatin is mostly used in making ice creams in order to :
(a) Prevent forming a colloidal sol.
(b) Enrich the fragrance
(c) Prevent crystallization and stabilize the mixture
(d) Modify the taste.
Answer:
(c) Prevent crystallization and stabilize the mixture

(a) Auto catalyst
(b) Poison
(c) Negative catalyst
(d) Positive catalyst.
Answer:
(b) Poison

Question 11.
The migration of colloidal particles under the influence of electric fields is :
(a) Cataphoresis
(b) Electrodialysis
(c) Electrophoresis
(d) Electrical dispersion.
Answer:
(c) Electrophoresis

Question 12.
Hardy – Schulze law is related with :
(a) Solution
(b) Coagulation
(c) Solids
(d) Gases.
Answer:
(b) Coagulation

Question 13.
How many phases are present in a colloidal solution :
(a) 1
(b) 2
(c) 3
(d) 4.
Answer:
(b) 2

Question 14.
Which is an emulsion among the following :
(a) Air
(b)Wood
(c) Butter
Answer:
(d) Milk.

Question 15.
Butter is :
(a) A gel
(b) An emulsion
(c) A sol
(d) Not a form of colloid.
Answer:
(a) A gel

Question 16.
Sol which acts as protective colloid is :
(a) Gelatin
(b) Au
(c) As₂S₃
(d) Fe(OH)₃
Answer:
(a) Gelatin

Question 17.
Which is not correct for physisorption :
(a) A reversible process
(b) Needs low heat of adsorption
(c) Needs activation energy
(d) Needs low temperature.
Answer:
(c) Needs activation energy

Question 18.
In which among the following is Tyndall effect not expected :
(a) Suspension
(b) Emulsion
(c) Sugar solution
(d) Gold sol.
Answer:
(c) Sugar solution

Question 2.
Answer in one word/sentence :

1. Conversion of precipitate into colloidal solution is known as. (MP 2016)
2. Dissociation of emulsion into its constituent liquid is known as.
3. The catalyst which is used for hydrogenation of oils.
4. In dissociation of H₂O₂, phosphoric acts as which type of catalyst?
5. Which catalyst converts glucose into alcohol?
6. Cleansing action of soap based on which principle? (MP 2013)
7. Who used the word catalyst for the first time? (MP 2013)
8. Write the size of colloidal particles.
9. What is the effect of temperature on chemical adsorption?
10. Why does colloidal particle show Tyndall effect? (MP 2013)
11. Movement of particles of colloids is called …………….
12. Give an examples of catalytic poisoning.
13. Colloidal Solution of liquid in liquid is known as …………… (MP 2012 Supply 18)

Answer:

1. Peptization
2. Demulsification
3. Ni
4. Negative
5. Zymase
6. Emulsification
7. Berzellius
8. 10^-5 to 10^-7 cm
9. Increases
10. Scattering of light by colloidal particles
11. As₂O₃ is contect process
12. Brownian movement
13. Emulsion.

Question 3.
Fill in the blanks :

1. Rate of physical adsorption ………………. with the increase in temperature.
2. Particles of As₂O₃ sol are ……………….
3. In the contact process of manufacture of H₂SO₄ for Pt catalyst ………………. act as
4. Oxidation of oxalic acid by KMnO₄ is an example of ……………….
5. Biological catalysts are necessarily ……………….
6. Movement of colloidal particles under the effect of electric field is called ……………….
7. Scattering of light by colloidal particles under the effect of electric field is called ……………….
8. Intermediate compound theory is applicable to ………………. catalyst.
9. The substance on whose surface adsorption takes place is called an ……………….
10. Milk is an example of ……………….
11. Blood is a ………………. charged colloid.
12. Adsorption is an ………………. process.
13. Colloidal solution of solid in liquid is called ……………….
14. Catalytic promoter substance is ……………….
15. On adding electrolyte, precipitation of colloidal particles is known as ……………….
16. According to Hardy Schulze law, coagulating capacity of ions depend on the ………………. of ions.

Answer:

1. Decreases
2. Negatively charged
3. As₂O₃, poison
4. Auto catalysis
5. Enzyme
6. Electrophoresi
7. Tyndall effect
8. Homogeneous
9. Adsorbent
10. Emulsion
11. Negatively
12. Exothermic
13. Gel
14. Molybdenum
15. Coagulation
16. Charge.

Question 4.
Match the following :
I.

Answer:

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

II.

Answer:

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

III.

Answer:

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

Surface Chemistry Very Short Answer Type Questions

Question 1.
Why does physisorption decreases with the increase of temperature? (NCERT)
Answer:
It is an exothermic process
Solid + Gas ⇌ Gas/Solid + Heat
Thus, when temperature is raised, reaction proceeds in backward direction and gas adsorbed gets released. It is accordance to Le – Chatelier’s principle.

Question 2.
Why is it necessary to remove CO when ammonia is obtained by Haber’s process? (NCERT)
Answer:
For the preparation of ammonia in Haber’s process solid catalyst is used. It is necessary to remove the CO produced because the gas reacts with iron and forms Fe(CO)₅ which exist in liquid state at the temperature of the chamber and obstructs in the production of NH₃ because at high temperature CO and H₂ react by which production of ammonia decreases. In short CO acts as poison for the iron catalyst used in the process. It is therefore, necessary to remove CO.

Question 3.
Why is adsorption always exothermic? (NCERT)
Answer:
Adsorption leads to decrease in disorder, thus ∆S = +ve. For the process to be performed value of ∆G should be -ve. Thus, in equation ∆G = AH – T ∆S, ∆G will be negative when ∆H will be negative i.e., exothermic. Thus, adsorption is an exothermic process.

Question 4.
What is adsorbent?
Answer:
Solid substances which adsorb the gases or solution are called adsorbent. Like : animal charcoal, silica etc.

Question 5.
What are adsorbate?
Answer:
Gas molecules, vapours or ions which are adsorbed on the surface of solids are called adsorbate.

Question 6.
Delta is formed when river water meets sea water. Explain.
Answer:
In the river water negatively charged particles of soil and sand are present. When river water meets sea – water, various ions Na⁺, K⁺ or Mg⁺⁺ present in sea water causes coagulation and soil particles etc. are settle down. Thus a delta is formed.

Question 7.
How is rain possible by spraying silver iodide on clouds?
Answer:
Clouds are charged due to colloidal nature. Silver iodide is an electrolyte. Spraying it on clouds lead to coagulation by which it rains.

Question 8.
How is the adsorption of a gas related to its critical temperature?
Answer:
Higher is the critical temperature of a gas, greater is the ease of liquification of gas i.e., larger are the van der Waals’ forces of attraction. Therefore, greater is the adsorption.

Question 9.
What happen when a freshly precipitated Fe(OH)₃ is shaken with little amount of dilute solution of FeCI₃?
Answer:
A reddish brown colloidal solution of Fe(OH)₃ is obtained. This process is called peptization. The Fe⁺³ ions from FeCl₃ are absorbed on the surface of the precipitate and form positively charged colloidal solution.
Fe(OH)₃ + Fe⁺³ → [Fe(OH)₃] Fe⁺³

Question 10.
Why is sky blue in colour?
Answer:
Dust particles present in air forms colloidal solution whose particles shows Tyndall effect and hence sky appears blue in colour.

Question 11.
KMnO₄ decolourises slowly initially and then rapidly when added to oxalic acid. Why?
Answer:
Addition of KMn04 to oxalic acid oxidises it. Mn⁺² ion produced in this reaction acts as autocatalyst and hence decolourisation occur rapidly.
2MnO₄^– + 5C₂O₄^-2 + 16H^– → 2Mn⁺² + 10CO₂ + 8H₂O

Question 12.
How does emulsifier provide stability to emulsion? Write the name of two emulsifiers?
Answer:
For the stability of emulsion, emulsifier is added. Emulsifier forms an interfacial film between the emulsifying medium and the suspended particles. For an oil/water emulsion, protein, gum acts as an emulsifier and for water/oil emulsion salts of fatty acids and heavy metals are the main emulsifier.

Surface Chemistry Short Answer Type Questions

Question 1.
What are emulsions? What are their different types? Give examples of each type. (NCERT)
Answer:
The colloidal system in which dispersed phase and dispersion medium both are liqvid, is called emulsion.

Types of emulsions :
1. Oil in Water type (O/W) : In this type of emulsion small droplets of oil are dispersed in the water or dispersion medium, example Milk, Vanishing cream.

2. Water in Oil (W/O) : In this type of emulsion, small quantity of water is dispersed in the form of droplets in oil (dispersion medium), example Butter, Cod – liver.

Question 2.
What is the difference between multimolecular and macromolecular colloids? Give one example of each. How are associated colloids different from these two types of colloids?
Answer:
1. Multimolecular collodis:
In this type of colloids the colloidal particles consist of aggregates of atoms or small molecules with diameter of less than 1 nm. For example, a gold sol may contain particles of different sizes having several atoms of gold. Similarly sulphur sol contains about one thousand of S₈ molecules. These are held together by van derwaals forces.

2. Macromolecular colloids:
In macromolecular colloids the dispersed particles themselves are big molecules, usually polymers. The molecular masses of these macromolecules range from thousands to millions. Synthetic compounds e.g. polyethylene, polystrene, nylon, rubber etc. are macromolecules. Since these molecules have dimensions comparable to those of colloidal particles, there dispersions are known as macromolecular colloids. Most lyophillic sols are of this category.

3. Associated colloids (Micelles):
There are certain substances which behave as normal, strong electrolytes at low concentration, but exhibit colloidal properties at higher concentrations due to the formation of aggregated particles. These are called Micelles. These substances are also called Associated Colloids. Surface active agents like soaps and detergents are of this class.

Question 3.
Explain, what is observed :

1. when a Deam oi ngnt is passed through a colloidal sol?
2. An electrolyte, NaCI is added to hydrated ferric oxide sol?
3. Electric current is passed through a colloidal sol?

Answer:

1. Scattering of light by colloidal particles takes place and path of the. light becomes visible (Tyndall effect).
2. The positively charged colloidal particles of Fe(OH)₃ get coagulated by the oppositely charged CF ions provided by NaCI.
3. On passing electric current, the colloidal particles move towards the oppositely charged electrode where, they lose their charge and get coagulated. This is electrophoresis process.

Question 4.
What do you understand by protective colloids?
Answer:
Protective colloid:
On adding small quantity of electrolyte, Lyophobic colloids easily coagulated but if some quantity of hydrophillic colloid is added in Lyophobic colloids, it minimize the effect of addition of electrolyte and coagulation is prevented or very slowly coagulated. This process is called protection of colloid. The hydrophillic colloid which is responsible for protection against coagulation is called protective colloid.

Example:
If a small amount of gum, gelatin or starch is added to As₂S₃sol, its coagulation of NaCI solution is prevented. Thus starch, gum or gelatin protects the As₂S₃ sol.

Question 5.
Give the method of preparation of colloidal solution of ferric hydroxide and sulphur in water.
Answer:
1. To prepare sol of ferric hydroxide, FeCl₃ solution is added in the boiling water dropwise with shaking. Excess of FeCl₃ and HCl are separated by electrodialysis and sol is also stablized by this.
FeCl₃ + 3H₃O → Fe(OH)₃ + 3HCl

2. H2S gas is passed in the solution of nitric acid (oxidising agent) and colloidal solution of sulphur is obtained.
2HNO₃ + H₂S → S + 2H₂O + 2NO₂.

Question 6.
What is catalysis? Explain induced catalysis with an example.
Answer:
Catalysis:
A catalyst is a substance which alters the rate of chemical reaction without being used in the reaction and the phenomenon is known as catalysis.

Induced Catalysis:
In induced catalysis, one reaction already taking place, catalyse the other reaction also which does not occur separately.

Example:
Sodium sulphite (Na₂SO₃) oxidise in the atmosphere easily while sodium arsenite (Na₂AsO₃) does not oxidise in atmosphere separately. When (Na₂SO₃) and Na₃AsO₃ both are kept together, both are oxidised.

Question 7.
What is coagulation? Write Hardy – Schulze’s law.
Answer:
Coagulation:
Colloidal particles are either positively charged or negatively charged. It is seen that on adding some oppositely charged ions or electrolyte, which cancels the charge of colloidal particles and these particles aggregate and change into precipitates. Precipitation of colloidal particles by electrolytes is called coagulation.

Hardy – Schulze’s law:
Coagulating power of an ion is governed by this law according to which, “Greater the valency of an ion greater will be its coagulating power.”

Thus, to coagulate negative sol, the coagulating power of different cations are found to be in the order :
Al³⁺ > Ba²⁺ > Na⁺

Similarly, for the coagulation of a positive sol of Fe(OH)3, the coagulating power of different anions is found to be in the order :
PO₄^3- > SO₄^2- > Cl_–

Question 8.
What do you understand by catalytic promoter and catalytic poison? Give an example of each.
Answer:
Catalytic promoter : Chemical substances which are used to increase the activity of the catalyst are known as Catalytic promoter.
Example :

Catalytic poison : Substances which are used to reduce the activity of catalyst are known.

Question 9.
What do you understand by Electro – dialysis?
Answer:
Electro – dialysis:
Particles of true solutions pass through parchment paper or cellophane but sol particles cannot pass through such membranes. In dialysis, the sol filled in a bag of parchment or cellophane is suspended in pure water. This process is very slow and takes a long time for completion. However, it can be quickened under the influence of an electric field and the process is known as Electro – dialysis.

Artificial kidney used in medical science works on the principle of dialysis.

Question 10.
What is homogeneous and heterogeneous catalysis? Explain with example.
Answer:
Homogeneous catalysis:
The chemical reaction in which reactants and cata¬lysts are in same physical state, are example of homogeneous catalysis.
Example : In lead chamber process of manufacturing of sulphuric acid reactant and catalyst NO are in same physical state – gaseous.

Heterogeneous catalysis:
The chemical reactions in which reactants and catalysts are in different physical state, are examples of heterogeneous catalysis.
Example : In Haber process manufacture of Ammonia.

Question 11.
What is ‘Gold number’? Explain with example.
Answer:
Gold Number:
Protective actions of different lyophilic colloids are compared with gold number. Gold number can be defined as:
“Gold number is the number of miligrams of protective colloid (Lyophilic) which on adding prevents the coagulation of 10 ml. standard gold sol by addition of 1 ml. of 10% NaCl solution.” Thus smaller the value of gold number, higher is it’s protecting action. For example, gold number of gelatin and gum arobic are 0.005 and 0.15 respectively. It means gelatin is better protective colloid because it’s only 0 005 mg quantity is required to prevent coagulation if 10 ml Au sol by addition of 1 ml. 10% NaCl.

Question 12.
What is the importance of emulsifying agents in emulsifications?
Answer:
The process of making emulsions is called emulsification. Emulsions are formed when suitable liquids are mixed and shaked, but the emulsions are thus formed are not stable. To make these emulsions stable some other substances are added, which are known as emulsifying agents. Soap, gum, starch etc. act as emulsifying agents. In absence of emulsifying agents, disperse drops of liquid meet each other to distroy emulsion state. It is supposed that emulsifying agents form membrane at interphase of oil and water which checks the union of droplets.

Question 13.
Give reason :

1. Milk turns sour on adding acid to it.
2. Alum is added to purify water.
3. Delta is formed where river water meets sea.

Answer:
1. Milk is an emulsion of fats dispersed in water. Albumin and caesin are emulsifiers. Addition of acid destroys emulsifiers and hence milk gets coagulated.

2. In impure water, soil particles, bacteria and other soluble impurities are dissolved. When alum is added, the Al+3 ion present in alum destroys the negative charge of impure water. Due to neutralization of charge the impurities are coagulated and settle down.

3. In the river water negatively charged particles of soil and sand are present. When river water meets sea – water, various salts present in sea water causes coagulation and soil particles etc. are settle down. Thus a delta is formed.

Question 14.
Explain positive and negative catalysis with example.
Answer:
Positive Catalysis:
When the velocity of any chemical reaction increases with the presence of catalyst, this type of catalysis is called positive catalysis.

Negative Catalysis:
Catalysts when decrease the velocity of chemical reaction are called negative catalysts or retarders and the process is known as negative catalysis.

Question 15.
What is peptization? Explain.
Answer:
Peptization:
Peptization is a good method of preparing colloidal sols from precipitates. A fresh precipitate is taken for the purpose and a suitable reason or peptizing agents are added. These peptizing agents are generally dilute solutions of electrolytes of common ion. Peptization process is opposite of the process of coagulation.

Example:
When fresh precipitate of aluminium hydroxide is boiled with dilute HCl mixed water, colloidal solution of Al(OH)₃ is obtained. When an electrolyte is added to a fresh precipitate then the particles of the electrolyte preferentially adsorbs an ion and due to electrostatic repulsion undergo in colloidal state. On adding electrolyte ferric chloride to ferric hydroxide precipitate, sol of ferric hydroxide is obtained.

Question 16.
Write difference between Lyophilic and Lyophobic colloids. (MP2016)
Answer:
Differences between Lyophilic and Lyophobic colloids :

Question 17.
What is Brownian motion?
Answer:
Continuous and zig – zag motion of colloidal particles is called Brownian motion. Such movement was first observed by Robert Brown in 1827, hence it is called Brownian motion. Brownian movement is due to the unequal collision at between molecules of dispersed phase and colloidal particles. On increase of size of colloidal particles or decrease in temperature the motion becomes slower.
Example:

1. Zig – zag motion of dust particles entering through ventilator is Brownian motion.
2. Lycopodium powder particles move in water due to Brownian motion.

Question 18.
Write difference between Physical adsorption and Chemical adsorption.
Answer:
Differences between Physical adsorption and Chemical adsorption :
Physical adsorption:

• Low value of enthalpy of adsorption (20 – 40 kJ mol^-1)
• This type of adsorption involves weak ‘van der Waals’ forces between adsorbent and adsorbate.
• Usually takes place at low tempera¬ture and decreases with increase in temperature.
• It is reversible in nature.
• The extent of adsorption is approxi-mately related to the case of liquification of the gas.
• Forms multimolecular layers.
• On increasing pressure, adsorption also increases.

Chemical adsorption:

• High value of enthalpy of adsorption (40 – 400kJmol^-1)
• This type of adsorption involves strong forces of attraction due to chemical bond formation.
• Takes place at high temperature.
• It is irreversible.
• No such correlation.
• Forms monomolecular layers.
• No any effect of pressure.

MP Board Class 12th Chemistry Important Questions

MP Board Class 12th Chemistry Important Questions Chapter 4 Chemical Kinetics

Chemical Kinetics Important Questions

Chemical Kinetics Objective Type Questions

Question 1.
Choose the correct answer :

Question 1.
For most of the reactions, the value of temperature coefficient lies in between :
(a) 1 and 3
(b) 2 and 3
(c) 1 and 4
(d) 2 and 4.
Answer:
(b) 2 and 3

Question 2.
For First order reaction value of tm is : (MP 2012 Supp.)
(a) \(\frac { 0.693 }{ { k }_{ 1 } }\)
(b) \(\frac { 2.303 }{ { k }_{ 1 } }\)
(c) \(\frac { 0.303 }{ { k }_{ 1 } }\)
(d) \(\frac {0.693}{t}\)
Answer:
(a) \(\frac { 0.693 }{ { k }_{ 1 } }\)

Question 3.
A first order reaction gets completed to 75% in 32 minutes. How much time would have been required for 50% completion :
(a) 24 minute
(b) 16 minute
(c) 8 minute
(d) 4 minute.
Answer:
(b) 16 minute

Question 4.
Hydrolysis of sucrose to glucose and fructose is :

an example of:
(a) First order reaction
(b) Second order reaction
(c) Third order reaction
(d) Zero order reaction.
Answer:
(a) First order reaction

Question 5.
Plants prepare starch in the process of:
(a) Flash photolysis
(b) Photolysis
(c) Photosynthesis
(d) None of these.
Answer:
(c) Photosynthesis

Question 6.
For first order reaction the specific reaction constant depends upon :
(a) Concentration of reactants
(b) Concentration of products
(c) Time
(d) Temperature.
Answer:
(d) Temperature.

Question 7.
In the reaction between A and B to form C, A represents first order and B represents second order. Rate equation will be written as :
(a) Rate = k [A]²[B]
(b) Rate = k [A] [B]²
(c) Rate = k [A]^1/2[B]
(d) Rate = k [A] [B]^1/2
Answer:
(b) Rate = k [A] [B]²

Question 8.
The rate of chemical reaction depends upon : (MP 2017)
(a) Active mass
(b) Atomic mass
(c) Equivalent weight
(d) Molecular mass.
Answer:
(a) Active mass

Question 9.
Arrhenius equation is :
(a) k = e^-E_aRT
(b) k = \(\frac { { E }_{ a } }{ RT }\)
(c) k = log\(\frac { { E }_{ a } }{ RT }\)
(d) k = Ae^-E_a/RT
Answer:
(d) k = Ae^-E_a/RT

Question 10.
Unit of velocity constant of first order reaction :
(a) mol litre^-1 sec^-1
(b) mol^-1 litre⁺¹ sec^-1
(c) sec^-1
(d) mol litre^-1 sec
Answer:
(c) sec^-1

Question 11.
Unit of velocity constant of zero order reaction :
(a) mol^-1 litre^-1 sec^-1
(b) mol^-1 litre⁺¹ sec^-1
(c) sec^-1
(d) mol litre^-1 sec
Answer:
(a) mol^-1 litre^-1 sec^-1

Question 12.
Rate constant of a reaction increased with the increase of which of the following factor as :
(a) Pressure
(b) Temperature
(c) Concentration of reaction
(d) All of these.
Answer:
(b) Temperature

Question 13.
Factor on which the rate constant of the 1st order reaction does not depend :
(a) Temperature
(b) Catalyst
(c) Activation energy
(d) Concentration of reactant.
Answer:
(d) Concentration of reactant.

Question 14.
Minimum energy required for molecules to react is called :
(a) Potential energy
(b) Kinetic energy
(c) Nuclear energy
(d) Activation energy.
Answer:
(d) Activation energy.

Question 15.
Law of mass action prepared by :
(a) Dalton
(b) Guddberg and Waaje
(c) Hunds and Mulliken
(d) Arrhenius.
Answer:
(b) Guddberg and Waaje

Question 16.
Unit of reaction rate is:
(a) mol litre^-1 sec^-1
(b) mol^-1 litre sec^-1
(c) mol^-1 litre^-1 sec
(d) mol litre sec
Answer:
(a) mol litre^-1 sec^-1

Question 2.
Fill in the blanks :

1. The rate of a reaction does not depends on the concentration of the reacting species, then the reaction is of ……………………
2. Half life period of a radioactive element is 140 days. On taking 1 gm element initially the amount left after 560 days will be ……………………
3. Fast reactions are completed in less than …………………… seconds.
4. Unit of rate constant for third order reaction is ……………………
5. Difference in the minimum and maximum energy state of reactants is called ……………………
6. Reactions which take place by the absorption of radiations are called …………………… (MP 2016)
7. The total number of molecules which participate in a reaction is called ……………………
8. In the mechanism of a reaction, the slowest step is called ……………………

Answer:

1. Zero
2. \(\frac {1}{2}\) gm
3. 10^-9
4. mol^-2 Litre² second^-1
5. Activation energy
6. Photo chemical reaction
7. Molecularity
8. Rate determining step.

Question 3.
Answer in one word / sentence :

1. What is the relation between threshold energy and activation energy?
2. What is the expression of rate constant for first order reaction?
3. Write alternative form of Arrhenius equation.
4. Write an example of zero order reaction.
5. What is Quantum Efficiency?
6. Write the expression of half life period for first order reaction.
7. Explain Threshold energy.

Answer:

7. The Number of photons absorbed minimum amount of energy which the reactant molecules should possess for effective collision.

Chemical Kinetics Very Short Answer Type Questions

Question 1.
What is the effect of rate constant on temperature? How can this effect of temperature be measured quantitatively? (NCERT)
Answer:
Rate of reaction increases with increase in temperature and with 10°C rise in
temperature its value becomes two times. Arrhenius equation expresses the effect of temperature on constant
K = A_e^{-E _/RT}
Where A is frequency factor and Ea is activation energy of the reaction.

Question 2.
The rate of reaction for A + 2B → Product, will be equal to it\(\frac {-d[A]}{dt}\) = k[A] [B]² if B is present in excess amount then what will be the order of reaction?
Answer:
It will be 1^st order of reaction.

Question 3.
Write alternative form of Arrhenius equation.
Answer:

Question 4.
For a zero order reaction t_1/2is proportional to what?
Answer:
Initial concentration of reactant [A].

Question 5.
What is Energy Barrier?
Answer:
The minimum energy achieved by the reactant only after which it can be converted to product is known as energy barrier. Reactant molecules cannot form activated complex till they reach this height (activation energy) and cannot be converted to form product

Question 6.
Explain the rate determining step.
Answer:
Some chemical reactions complete in one or more steps. Rate of reaction is determined by the slowest step which is known as rate determining step.

Question 7.
If unit of rate constant is litre/mole/sec. then what will be the order of reaction?
Answer:
Order of reaction = Second.

Question 8.
Order of a reaction is zero can its molecularity be zero?
Answer:
Molecularity of a reaction is always a whole number. Thus, it cannot be zero.

Question 9.
What is specific reaction rate?
Answer:
Specific reaction rate of a reaction at a given temperature is equal to that rate of a reaction when concentration of each reactant is unity.
Reaction : A_2(g) + B_2(g) → 2AB_g
Rate of reaction ∝ [A₂][B₂]
If [A₂] = [B₂] = 1, then
Rate of reaction = k, where k = Specific reaction rate constant.

Question 10.
When is the average rate of the reaction equal to its instantaneous rate?
Answer:
When value of time interval is nearly zero or when time by infinite form is minute then the average rate of reaction is comparable to its instantaneous rate.

Question 11.
What are pseudo unimolecular reactions?
Answer:
There are some reactions in which molecularity is more than one i.e. two or more molecules are present, but in the chemical reaction concentration of only one reactant molecule is changed and it is only responsible for rate of reaction. Thus, order is one. These reactions are called pseudo unimolecular reactions.
Example:

Above reaction is bimolecular but concentration of water does not affect the rate of reaction, thus, rate of reaction is proportional to the concentration of sucrose only.
Rate = k[C₁₂H₂₂O₁₁]
Thus, inversion of sucrose is a first order reaction. It is known as pseudo unimolecular reaction.

Question 12.
What is temperature coefficient?
Answer:
Generally, rate of any reaction increases due to increase in temperature. On increasing temperature by 10°C, velocity of reaction is increased up to 2 – 3 times. If velocity constant of a reaction is and it is increased to increasing temperature by 10°C, the ratio of these two constant is called temperature coefficient, i.e.,
\(\frac { { k }_{ t+10 } }{ { k }_{ t } }\) ≈ 2 to 3
Thus, at various time the ratio of rate of reaction which differ by 10°C is known as temperature coefficient.

Question 13.
As compared to water, gasoline vaporizes faster. Why?
Answer:
Value of activation energy of vaporization of gasoline is less than the value of activation energy of water.

Chemical Kinetics Short Answer Type Questions

Question 1.
The conversion of molecules x to y follows second order kinetics. If concetration of x is increased to three times how will it affect the rate of formation of y? (NCERT)
Solution:
For the reaction x → y
Rate of reaction (r) = k[x]² …. (1)
If concentration of x is increased three times, now
Rate of reaction (r)¹ = k[3x]² = k[9x² ] …. (2)
Dividing equation (2) by Question (1),
\(\frac { { r }^{ 1 } }{ r }\) = \(\frac { k[9{ x }^{ 2 }] }{ k[{ x }^{ 2 }] }\) = 9
Thus, the rate of reactions will become 9 times.

Question 2.
Write down the expression representing the rate of reaction.
Answer:
Rate of reaction is the rate of change in concentration of reactant or concentration of product in unit time interval.

Unit of rate of reaction depend on the units of concentration and time. If concentration is represented in mole per litre and time in second, then unit of rate of reaction is mol per litre per second. If time is expressed in minute then unit of reaction rate is mole per litre per minute.

Question 3.
What is the meaning of instantaneous rate of reaction?
Answer:
The rate of reaction does not remain constant during the whole time interval because rate of reaction depends upon the concentration of reactants. As the concentration of reactants decreases with time, the rate of reaction also decreases with time.
In order to express the reaction rate as accurately as possible, the instantaneous rate of reaction is expressed. For this the time interval (∆t) is taken as small as possible.
Instantaneous rate of reaction = –[\(\frac {∆A}{∆t}\)_{∆t → o}
= – \(\frac {d[A]}{dt}\)
Where, d [A] is a change in concentration of reactant A.

Question 4.
What do you know about molecularity of any reaction?
Answer:
Molecularity of Reaction:
‘Number of moles of reactant participating in elementary step of chemical reaction is called molecularity of the reaction.’
There are many reactions which proceed through the number of steps and each step being independent is called elementary step. The rate of reaction is determined by slowest step and it is known as rate determining step. Here molecularity can be defined as “Number of molecules/atoms or ions participating in rate determining step is called molecularity.”
Example:
(i) Unimolecular reaction :
O₃ → O₃ + O
(ii) Bimolecular reaction:
NO + O₃ → NO₂ + O₂

Question 5.
Differentiate between molecularity and order of reaction. (MP 2018)
Answer:
Differences between Molecularity and Order of reaction :
Molecularity:

• It is the total number of molecules which participate in the reaction.
• It is a theoretical concept.
• It is always a whole number.
• Its value is never zero.
• It does not provide any information about mechanism of reaction.

Order of reaction:

• It is the number of molecules which participate in reaction and whose concentration is changed.
• Order of reaction is determined by experimentally.
• Fractional values are also possible.
• Zero value is possible.
• It provides information about mechanisms of reaction.

Question 6.
Write any four factors which affects rate of a chemical reaction.
Answer:
The rate of reaction depends upon the following factors:
1. Concentration of reactants:
At constant temperature, the rate of a reaction increases by increasing the concentration of the reactants.

2. Temperature of the system:
If the concentration of the reactants are constant then the rate of reaction increases by increasing temperature. For 10 degree rise of temperature, the reaction rate becomes double or triple.

3. Presence of catalyst:
Positive catalyst increases the rate of reaction and negative catalyst decreases the reaction rate.

4. Nature of the reactants:
Nature of reactants also affect the reaction rate. In any chemical reaction some old bonds are broken and new bonds are formed. Thus, in case of more simple molecules the lesser is the number of bond breaking and the rate of reaction increase whereas in case of complex molecules more bonds are broken and rate decreases.

5. Exposure to radiations:
The rate of some reactions increases due to some special radiations. For example: In the absence of light the reaction between hydrogen and chlorine is slow, but in the presence of light, the reaction proceeds at a faster rate.

Question 7.
How does rate of any reaction depend on temperature? Explain.
Answer:
Generally, rate of any reaction increases due to increase in temperature. On increasing temperature by 10°C, velocity of reaction is increased up to 2 – 3 times. If velocity constant of a reaction is and it is increased to increasing temperature by 10°C, the ratio of these two constant is called temperature coefficient, i.e.,
\(\frac { { k }_{ t+10 } }{ { k }_{ t } }\) ≈ 2 to 3
Arrhenius provide the following relation for showing the effect of temperature on velocity constant.
k = A.e^-E_a

Question 8.
Write a short note on activation energy?
Answer:
According to Arrhenius, any chemical reaction is only possible when reacting molecules are activated with minimum energy which is called threshold energy. Kinetic energy of most of the molecules are less than this minimum energy. The excess energy which is required to activate reactant molecules, is called activation energy.
Activation energy can be determined by the use of Arrhenius equation

Question 9.
Write four differences between Rate of reaction and Rate constant. (MP 2018)
Answer:
Differences between Rate of reaction and Rate constant:
Rate of reaction:

• It is expressed in terms of consumption of reactants or formation of product per unit time.
• It depends on concentration of reactant at particular moment.
• It generally decreases with the progress of reaction.
• Its unit is mol L^-1 cm^-1.

Rate constant:

• It is proportionality constant in differential form in rate law or rate equation.
• It is independent of concentration of reactant.
• It does not depend on the progress of reaction.
• It changes according to order of reaction.

Question 10.
Prove that half – life period is independent of the initial concentration for the firsforder reaction.
Solution:
Half – life period for the reaction is that period in which the initial concentration of reactant is reduced to half.
For first order reaction :

Thus, the half – life period is independent of the initial concentration for the first order reaction.

Question 11.
(i) 2N₂O₅ → 4N02 +02
(ii) H₂ +I₂ → 2HI
Reaction (i) is first order reaction and (ii) is second order reaction. Why?
Answer:
(i) In reaction 2N₂O₅ → 4NO₂ + O₂, when graph is plotted between rate of reaction and then again graph is plotted between rate of reaction and [N₂O₅]², it is shown that in the first graph straight line is obtained i.e.
rate ∝ [N₂O₅]
or rate = k[N₂O₅]
Therefore, 2N₂O₅ → 4NO₂ + O₂ is first order reaction.

(ii) In reaction H₂+I₂ → 2HI when graph is plotted between rate of reaction and (H₂) (I₂), it is seen that straight line is obtained.
Therefore rate ∝ [H₂][I₂]
Hence, this reaction is of 2^nd order reaction.

Question 12.
What do you understand by order of reaction? Give example. (MP 2011,12,16)
Answer:
Order of reaction:
The order of a reaction is defined as the sum of all the powers to which concentration terms in the rate law are raised to express the observecfrate of the reaction. Suppose there is a general reaction,
aA + bB + cC → product
For which the rate law is
Rate = – \(\frac {dx}{dt}\) = k[A]^p[B]^q[C]^r
Then the order of the reaction n = p + q + r, where, p, q and r are the orders with
respect to individual reactants and overall order is the sum of the exponents i.e.,p + q + r.
When n = 1 the reaction is of first order, if n = 2 the reaction is of second order and so on.
For example : Decomposition of ammonium nitrite occurs as follows :
NH₄NO₂ → N₂ +2H₂O
Rate of reaction = – \(\frac {dx}{dt}\) = k [NH₄NO₂]
Thus, order of this reaction will be 1.

Question 13.
Write unit of rate constant k for the zero order, first order and second order reaction (MP 2012)
Answer:

Chemical Kinetics Long Answer Type Questions

Question 1.
Determine the expression for zero order reaction.
Answer:
Reactions in which rate of reaction does not depend on the concentration of the rede tan ts are called zero order reactions. Consider a zero order reaction
A → B
Where, A and B are concentration of reactants and products. Since in this type of reaction, rate of reaction does not depend on the concentration of reactant therefore, rate of change in concentration of reactant remains constant.
Rate of reaction = Constant.
Let initial cone, of reactants be a moles /It and after time‘t’ x moles are converted into product. Then cone, of A after time t will be (a – x) mol /It.
\(\frac {-d[A]}{dt}\) or \(\frac {dx}{dt}\) ∝(a – x)⁰
or \(\frac {dx}{dt}\) = k₀(a – x)(a – x)⁰ … (1)
Where, k₀ is velocity constant of a zero order reaction.
\(\frac {dx}{dt}\) ∝ (a-x)(a – x)⁰ … (2)
dx = k₀dt … (3)
Integrating equation. (2), we get
x = k₀t + c … (4)
Where, c = integration constant
When, t = 0 then x = 0
Substituting it in equation. (4), we get
0 = k₀ x O + C
or C = 0  …(5)
Substituting this in equation. (4), we get
x = k₀t  … (6)
or k₀ = \(\frac {x}{t}\) = … (7)
Thus, equation. (7) is the velocity equation for zero order reaction. Unit of k
k₀ = \(\frac {x}{t}\)
= \(\frac {mole/litre}{second}\) = mole litre^-1 second^-1

Question 2.
Write the Arrhenius equation in the form of equation of straight line. What will be the slope of a graph using this equation? Calculate the activation energy for the decomposition in a decomposition reaction in which the value of slope obtained is – 9920 when log k is plotted against \(\frac {1}{T}\)
Solution:
Arrhenius equation : Arrhenius gave a relation between rate constant of reaction and temperature which is known as Arrhenius equation, i.e.,
k = Ae^-E_a/RT
Where, k = Rate constant of reaction, A = Frequency factor, Ea= Activation energy, R = Gas constant and T = Absolute temperature.

Taking logarithm both sides of above equation, we get
log_e k = – \(\frac { { E }_{ a } }{ RT }\) + log_e A
or log₁₀ k = log₁₀ A – \(\frac { { E }_{ a } }{ 2.303RT }\) .. (i)
This is a straight line equation. If log₁₀ k and \(\frac {1}{T}\) is plotted at different temperatures, a straight line is obtained whose slope = \(\frac { { -E }_{ a } }{ 2.303RT }\) From the graph, value of slope = \(\frac { { E }_{ a } }{ 2.303RT }\) can be calculated and activation energy (E_a ) can be calculated.
Slope = \(\frac { { -E }_{ a } }{ 2.303RT }\) (Activation energy)
Or E_a = Slope x 2.303 x R
Or E_a = – 9920 x 2.303 x (-4.58)
∴ Ea = 104633.5808 calorie per gm molecule

Question 3.
Derive an expression for rate constant of first order reaction.
Answer:
The reaction in which rate of reaction depends upon the concentration of one mole, are called first order reaction.
Let this reaction is
A → product
Suppose intial concentration of A is a gram mole and after t second x mole consumed and remaining concentration is (a – x) gram mole.
So after t time, the rate of reaction will be proportional to (a – x)
\(\frac {dx}{dt}\) ∝(a – x)
or \(\frac {dx}{dt}\) = k (a – x) (k = Velocity constant) … (1)
or \(\frac {dx}{(a – x)}\) = kdt … (2)
on intergating this equation,
∫\(\frac {dx}{(a – x)}\) = ∫ kdt
or -ln(a – x) = kt + c
Where c is intergating constant, …. (3)
If t = 0 then x = 0
Putting the value of c from eqn. (3),
-ln a = c .. (4)
Putting the value of c from eqn. (4), into eqn. (3).
-ln(a – x) = kt – ln a
or ln a – ln (a – x) = kt
ln \(\frac {a}{(a – x)}\) = kt
or k = \(\frac {1}{(t)}\)ln\(\frac {a}{(a – x)}\)
charging the base of log,
k = \(\frac {2.303}{(t)}\) log \(\frac {a}{(a – x)}\)
It is the desired expression for first order of reaction

Chemical Kinetics Numerical Questions

Question 1.
A first order reaction has a rate constant 1.15 x 10^-3 s^-1 How long will 5g of this reactant take to reduce to 3g? (NCERT)
Solution:
According to question,

Applying first order kinetic equation

Question 2.
The rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 298K. Calculate E_a.
Solution: