CBSE Sample Papers for Class 11 Chemistry Set 7 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 11 Chemistry with Solutions Set 7 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Chemistry Set 7 with Solutions

Time Allowed: 3 hours
Maximum Marks: 70

General Instructions:

• All questions are compulsory.
• This question paper contains 37 questions.
• Questions 1-20 in Section A are objective type-very short answer type questions carrying 1 mark each.
• Questions 21 – 27 in Section B are short answer type questions carrying 2 marks each.
• Questions 28 – 34 in Section C are long-answer I type questions carrying 3 marks each.
• Questions 35-37 in Section D are long-answer II type questions carrying 5 marks each.
• There is no overall choice. However, an internal choice has been provided in six questions of one mark, two questions of two marks, two questions of three marks and two questions of five marks. You must attempt only one of the choices in such questions.
• Use log tables, if necessary. Use of calculator is not allowed.

Section – A

Question 1.
Total number of orbitals associated with third shell will be ___________ . [1]
(A) 2
(B) 4
(C) 9
(D) 3
OR
Which of the following conclusions could not be derived from Rutherford’s a-particle scattering experiment?
(A) Most of the space in the atom is empty.
(B) The radius of the atom is about 10^-10 m while that of nucleus is 10^-15 m.
(C) Electrons move in a circular path of fixed energy called orbits.
(D) Electrons and the nucleus are held together by electrostatic forces of attraction.
Answer:
(C) 9

Explanation:
Total number of orbitals is given as n². Here n = 3 so, total number of orbitals in third shell = 3² = 9

OR

(C) Electrons move in a circular path of fixed energy called orbits.

Explanation:
The concept of movement of electrons in circular path of fixed energy called orbit was given by Bohr.

Question 2.
The pressure of a 1 : 4 mixture of dihydrogen and dioxygen enclosed in a vessel is one atmosphere. What would be the partial pressure of dioxygen ? [1]
(A) 0.8 × 10⁵atm
(B) 0.008 Nm^-2
(C) 8 × 10⁴ Nm^-2
(D) 0.25 atm
Answer:
(C) 8 × 10⁴ Nm^-2

Explanation:
Let the number of moles of dihydrogen and dioxygen be 1 and 4 respectively.
Mole fraction of O₂ = \(\frac{4}{5}\)
Partial Pressure of dioxygen
= Mole fraction × Total pressure = \(\frac{4}{5}\) × 1
= 0.8 atm = 0.8 × 10⁵ N m^-2 = 8 × 10⁴ N m^-2

Question 3.
Consider the reactions given below. On the basis of these reactions find out which of the algebraic relations given in options (A) to (D) is correct ? [1]
(i) C (g) + 4 H (g) → CH4 (g); Δ_rH₄ = x kJ mol^-1
(ii) C (graphite,s) + 2H₂ (g) → CH₄ (g); Δ_rH = y kJ mol^-1
(A) x = y
(B) x = 2y
(C) x > y
(D) x < y Answer: (C) x > y

Explanation:
Same bonds are formed in reaction (i) and (ii) but bonds between reactant molecules are broken only in reaction (ii).

Question 4.
The electronic configuration of the outer most shell of the most electronegative element is : [1]
(A) 2s²2p⁵
(B) 3s²3p⁵
(C) 4s²4p⁵
(D) 5s²5p⁵
OR
Amongst the following elements whose electronic configurations are given below, the one having the highest ionisation enthalpy is:
(A) [Ne]3s²3p¹
(B) [Ne]3s²3p³
(C) [Ne]3s²3p²
(D) [Ar]3d¹⁰4s²4p³
Answer:
(A) 2s²2p⁵

Explanation:
Electronic configuration in (a) is of fluorine while (b), (c) and (d) are of chlorine, bromine and iodine respectively. Fluorine is the most electronegative element.

OR

(B) [Ne]3s²3p³

Explanation:
Electronic configuration of (b) and (d) have half-filled p-orbital but (b) being smaller in size will have highest ionization enthalpy.

Question 5.
What is the product obtained when phenol is distilled with Zn dust ? [1]
(A) Benzene
(B) Chlorobenzene
(C) Toluene
(D) Phenol
OR
Which of the following options will be correct for the stage of half completion of the reactions :
A ⇌ B.
(A) Δ\(\mathrm{G}^{\boldsymbol{\theta}}\) = 0
(B) Δ\(\mathrm{G}^{\boldsymbol{\theta}}\) > 0
(C) Δ\(\mathrm{G}^{\boldsymbol{\theta}}\) < 0
(D) Δ\(\mathrm{G}^{\boldsymbol{\theta}}\) = – RT In 2
Answer:
(A) Benzene

Explanation:
When Phenol reacts with Zn dust, it gives benzene and zinc oxide.

OR

(A) Δ\(\mathrm{G}^{\boldsymbol{\theta}}\) = 0

Explanation:
ΔG° = -RT ln K
At the stage of half completion of reaction, [A] = [B].
∴ K = 1
Thus, ΔG° = 0

Question 6.
The fragrance of flowers is due to the presence of some steam volatile organic compounds called essential oils. These are generally insoluble in water at room temperature but are miscible with water vapour in vapour phase. A suitable method for the extraction of these oils from the flowers is : [1]
(A) distillation
(B) crystallisation
(C) distillation under reduced pressure
(D) steam distillation.
OR
Aromatic hydrocarbons are known as:
(A) alkanes
(B) alkenes
(C) arenes
(D) alkynes
Answer:
(D) steam distillation.

Explanation:
Essential oils are insoluble in water, soluble in steam and have high vapour pressure. Therefore, they can be separated by steam distillation.

OR

(C) arenes

Explanation:
Aromatic hydrocarbons are known as arenes.

Read the given passage and answer the following:
A large number of orbitals are possible in an atom. Qualitatively these orbitals can be distinguished by their size, shape and orientation. An orbital of smaller size means there is more chance of finding the electron near the nucleus. Similarly, shape and orientation mean that there is more probability of finding the electron along with certain directions than along others. The principal quantum number determines the size and to large extent the energy of the orbital. Azimuthal quantum number, T is also known as orbital angular momentum or subsidiary quantum number. It defines the three-dimensional shape of the orbital. Each shell consists of one or more subshells or sub-levels. The number of sub-shells in a principal shell is equal to the value of n. Magnetic orbital quantum nuifiber. ‘ml’ gives information about the spatial orientation of the orbital with respect to a standard set of co¬ordinate axis. The fourth quantum number is known as the electron spin quantum number (ms). An electron spins around its own axis, much in a similar way as the earth spins around its own axis while revolving around the sun.

Answer the questions (7) to (10) given below:

Question 7.
What is the number of subshell in M shell? [1]
(A) 4
(B) 3
(C) 6
(D) 5
Answer:
Option (A) is correct.

Explanation:
There are 4 subshells in M shell-s, p, d and f.

Question 8.
Spin quantum number is denoted as? [1]
(A) m
(B) 1
(C) n
(D) ms
Answer:
(D) ms

Question 9.
Azimuthal quantum number gives: [1]
(A) shape of orbital
(B) number of orbitals
(C) Spin of electron
(D) none of these
Answer:
(A) shape of orbital

Question 10.
What is the value of n for L shell? [1]
(A) 3
(B) 2
(C) 1
(D) 4
Answer:
(B) 2

Question 11.
Out of alcohols and ethers of comparable mass which one have a higher boiling point? [1]
OR
Give the IUPAC name of:

Answer:
Out of alcohols and ethers of comparable molecular mass, alcohols have a higher boiling point because of the presence of intermolecular hydrogen bonding.

OR

1, 3-Butadiene or Buta-1,3 diene

Question 12.
How will you convert: [1]
Acetylene to chlorobenzene
Answer:

Question 13.
Why is an organic compound fused with sodium in Lassaigne’s test ? [1]
Answer:
It is because compounds can be converted from covalent to ionic form by fusing the compound with sodium metal.

Question 14.
Give the IUPAC name of: [1]

Question 15.
Assertion (A): A solution containing a mixture of acetic acid and sodium acetate maintains a constant value of pH on addition of small amounts of acid or alkali.
Reason (R): A solution containing a mixture of acetic acid and sodium acetate acts as a buffer solution around pH 4.75. [1]
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is not the correct explanation of A.
(C) A is true but R is false.
(D) Both A and R are false.
OR
Assertion (A): For any chemical reaction at a particular temperature, the equilibrium constant is fixed and is a characteristic property.
Reason (R): Equilibrium constant is independent of temperature.
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is not the correct explanation of A.
(C) A is true but R is false.
(D) Both A and R are false.
Answer:
(A) Both A and R are true and R is the correct explanation of A.

OR

(C) A is true but R is false.

Explanation:
As the concentration of reactants or products does not affect equilibrium constant at a particular temperature.
Also, equilibrium constant varies with temperature.

Question 16.
Assertion (A): Combustion of all organic compounds is an exothermic reaction.
Reason (R): The enthalpies of all elements in their standard state are zero. [1]
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is not the correct explanation of A.
(C) A is true but R is false.
(P) Both A and R are false.
Answer:
(B) Both A and R are true but R is not the correct explanation of A.

Explanation:
Combustion reactions breaks the bonds of organic compound molecules, and the resulting water and carbon dioxide bonds always release more energy than was used to break them originally. That’s why burning organic compounds produces energy and is exothermic.

Question 17.
Dipole moment of hydrogen halides decreases from HF to HI. Why ? [1]
Answer:
Since the atomic radius increases and electro negativity decreases from HF to HI, therefore dipole moment of hydrogen halides decreases from HF to HI.

Question 18.
Give one point of differences between inductive effect and resonance effect. [1]
Answer:
Inductive effect operates in saturated compounds and resonance effect operates only in unsaturated conjugated systems.

Question 19.
Why molality is preferred over molarity of a solution ? [1]
Answer:
Because molality of a solution does not change with temperature, while molarity of a solution changes with temperature.

Question 20.
Cs shows maximum photoelectric effect, why ? [1]
Answer:
Since, Cs is the most electropositive element of all so it has the minimum ionisation energy and contains the maximum capacity to loose electrons. That’s why Cs shows maximum photoelectric effect.

Section – B

Question 21.
(a) What is limiting reactant in a reaction ?
(b) What is SI unit of molarity ? [2]
OR
(a) Define molar volume of a gas.
(b) State Gay Lussac’s Law of combining volumes of gases.
Answer:
(a) It is the reactant which gets consumed first or limits the amount of product formed.
(b) SI unit of molarity is mole/litre.

OR

(a) It is defined as volume occupied by one of any substance. mole
Molar volume of a gas = 22.4 L at STP (273 K, 1 atm).
(b) According to this law, “When gases combine or are produced in a chemical reaction they do so in a simple ratio by volume provided all gases are at same temperature and pressure.”

Commonly Made Error:
Students often reframe the statement.

Answering Tip:
Always write the law as stated.

Question 22.
Which of the following species will have the largest and smallest size ? [2]
Mg, Mg²⁺, Al, Al³⁺
Answer:
Atomic radii decrease across a period on moving from left to right. Cations are smaller than their parent atoms.
Among isoelctronic species, the one with the larger positive nuclear charge will have smaller radius. Hence, Mg has largest size and Al3+ has smallest size.

Commonly Made Error:
Students get confuse with how the atomic radii vary and write wrong answers.

Answering Tip:
Understand and learn the periodic properties ! thoroughly to avoid mistakes.

Question 23.
Why HCl is polar whereas Cl₂ molecule is non-polar ? [2]
Answer:
In Cl₂, both atoms have same electronegativity. Hence, the shared pair of electrons is attracted equally by both and remains exactly in the centre. No end acquires negative or positive charge. In HCl, chlorine is more electronegative than H. Hence, shared pair of electrons is more attracted towards chlorine which, therefore, acquires negative charge while H acquires positive charge. Thus HCl is polar while Cl₃ is non-polar.

Question 24.
Balance the following equations by the oxidation number method.
Fe²⁺ + H⁺ + Cr₂O₇^2- → Cr³⁺ + Fe³⁺ + H_O [2]
Answer:
Oxidation number method for balancing ionic equation:

Balance increase and decrease in oxidation number.
6Fe²⁺ + H⁺ + Cr₂O₇^2- → 2Cr³⁺ + 6Fe³⁺ + H₂O
Balance charge by multiplying H⁺ by 14.
6Fe²⁺ + 14H⁺ + Cr₂O₇^2- → 2Cr³⁺ + 6Fe³⁺ + H₂O
Balance H and O-atoms by multiplying H₂O by 7
6Fe²⁺ + 14H⁺ + Cr₂O₇^2- → 2Cr³⁺ + 6Fe³⁺ + 7H₂O
This represents a balanced redox reaction.

Question 25.
(a) Show the polarisation of carbon-magnesium bond in the following structure.
CH₃—CH₂—CH₂—CH₂—Mg—X
(b) Compounds with same molecular formula but differing in their structures are said to be structural isomers. What type of structural isomerism is shown by :

OR
Explain how is the electronegativity of carbon atoms related to their state of hybridisation in an organic compound?
Answer:
(a)

Carbon is more electronegative than magnesium.

(b) They show metamerism. (1)

OR

Electronegativity increases with increasing s-character. This is because s-electrons are more strongly attracted by the nucleus than p-electrons.
sp³ – 25% s-character, 75% p-character
sp² – 33% s-character, 67% p-character
sp – 50% s-character, 50% p-character
Hence, the order of electronegativity is sp³ < sp² < sp

Commonly Made Error:
Students forget and write wrong percentage of s and p character for different hybridisation and thus they also do mistake in the order of electronegativity.

Answering Tip:
Understanding the concept of how electronegativity and s-character are related is very important.

Question 26.
Write IUPAC names of the following:

Answer:
(a) (2-Propyl) cyclohexane.
(b) 2, 3-dimethylbutane

Question 27.
For your agricultural field or garden you have developed a compost producing pit. Discuss the process in the light of bad odour, flies and recycling of wastes for a good produce. [2]
Answer:
It is essential to take proper care of the compost producing pit in order to protect ourselves from bad odour and flies.
It should be kept covered to minimize bad odour and prevent flies from entering it. The recyclable waste should not be dumped in the compost producing pit. It should be sent to the industries through vendors for recycling.

Section – C

Question 28.
Differentiate between the following (with examples):
(i) Open and closed system
Answer:

Open System	Closed System
It can exchange matter as well as energy with its surroundings. e.g., Reaction mixture in open beaker.	It does not exchange matter with its surroundings but can exchange energy in the form of heat. e.g., Hot water in a closed container.

(ii) Adiabatic and Isothermal process
Answer:

Adiabatic Process	Isothermal Process
In this process, no heat can flow in or out of the system. i.e,dq = 0	In this process, temperature remains constant. i.e., dT = 0

(iii) State function and path function [3]
Answer:

State Function	Path Function
The variables which depend upon the initial and final states of a system are called state functions. e.g., Internal energy, Pressure, Volume, Enthalpy, etc.	The variables which depend upon the path followed are called path functions. e.g., Heat, Work, etc.

Question 29.
(a) Find out the value of equilibrium constant for the following reaction at 298 K.
2NH₃(g) + CO₂(g) ⇌ NH₂CONH₂(aq) + H₂O(I)
Standard Gibbs energy change, ∆_r\(\mathrm{G}^{\Theta}\) at the given temperature is – 13.6 kj mol^-1. [ 3]
(b) Define entropy of the reaction.
OR
Expansion of a gas in vacuum is called free expansion. Calculate the work done and the change in internal energy when 1 litre of ideal gas expands isothermally into vacuum until its total volume is 5 litre?
Answer:
(a) Given, T = 298 K
∆_r\(\mathrm{G}^{\Theta}\) = – 13.6 kJ mol^-1
R = 8.314 J K^-1 mol^-1
K = ?
∴ ∆_r\(\mathrm{G}^{\Theta}\) = – 2.303RT log K
or log K = \(\frac{-\Delta_{\mathrm{r}} \mathrm{G}^{\Theta}}{2.303 R T}\)
= \(\frac{-\left(-13.6 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1}\right)}{2.303 \times 8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \times 298 \mathrm{~K}}\)
= 2.38
∴ K = antilog 2.38
= 2.4 × 10²

(b) Entropy: Entropy is defined as a measure of degree of randomness or disorderness in an isolated system.
S = \(\frac{q_{\mathrm{rev}}}{\mathrm{T}}\)

OR

W = ?
∆U = ?
V_i = 1 litre
V_f = 5 litre
External pressure (P_ex) = 0
Work done W = – P_ex(V_f – V_i) = – 0(5 – 1) = 0
For isothermal expansion q = 0
According to first law of thermodynamics
∆U = q + W = 0 + 0 = 0

Question 30.
Calculate the standard Gibbs energy change for the formation of propane C3Fl8(g) at 298 K. Given that
∆_fH° for propane = – 103.85 kj/mol
S°M C₃H₈(g) = 270.2 JK^-1 mol^>-1
S°M H₂(g) = 130.68 JK^-1 mol^-1
S°MC_(graphite) = 5.74 JK^-1 mol^-1
Answer:
The formation of C₃H₈(g) is represented by
3C (graphite) + 4H₂(g) → C₃H8(g)
ΔS⁰ = S0(C3H8) – 4S0(H2) – 3S0 (C)
= 270 – 2 – [4 × (130 – 68)] – [3 × 5.74]
= – 269.74 JK^-1 mol^-1
Δ_fG⁰ = Δ_fH⁰ – TΔS⁰
= – 103.85 × 10³ (J mol^-1) – 298 (K) × (- 269-74 JK^-1 mol^-1)
= – 23467-48 J
= – 23.47 kJ.

Question 31.
(i) State the formula of the conjugate acid of each of the following bases : [3]
(a) OH^–
(b) HPO₄^2-
(c) H₂PO₄^–
(d) CH₃NH₂
(e) CO₂^2-
(f) NH₃
(g) CH₃COO^–
(h) HS^–

(ii) Classify the following species into Lewis acids and Lewis bases and show how these act as such :
(a) HO^–
(b) F^–
(c) H⁺
(d) BCl₃
Answer:
(i) The formula of the conjugate acid of the given
(a) H₂O
(b) H₂PO₄^–
(c) H₃PO₄
(d) CH₃NH₃⁺
(e) HCO₃^–
(f) NH₄⁺
(g) CH₃COOH
(h) H₂S

(ii) (a) HO^– : It acts as Lewis base as it can donate lone pair of electrons.
(b) F^– : It acts as Lewis base as it can donate any of its four lone pair of electrons.
(c) H⁺ : It acts as Lewis acid as it can accept lone pair of electrons from bases like hydroxyl ion and fluoride ion.
(d) BCl₃ : It acts as Lewis acid as it can accept lone pair of electrons from species like ammonia or amine molecules.

Question 32.
(a) What is the significance of the term “isolated gaseous atom” while defining ionisation enthalpy?
(b) Na and Mg⁺ have the same number of electrons but removal of electron from Mg⁺ requires more energy. Give reason.
(c) Elements in the same group have similar chemical properties. Give reason. [3]
Answer:
(a) In the gaseous state, the atoms are widely separated, therefore, interatomic forces are minimum. It is because of this reason the term ‘isolated gaseous atom’ has been included in definition of ionization enthalpy.

(b) The effective nuclear charge increases across a period. Both Na and Mg⁺ has same numbers of electrons in outermost shell, but Mg⁺ has more effective nuclear charge than Na. As the effective nuclear charge increases, the hold of the nucleus on the outermost electrons will also increase. Thus, it will be very difficult to remove the electrons from outermost shell. That’s why, it is difficult to remove electron from Mg⁺ as compared to Na.

(c) Elements in the same group in the periodic table have similar chemical properties because their atoms have the same number of electrons in the highest outermost shell.

Question 33.
(a) Wavelengths of different radiations are given below :
λ(a) = 300 nm
λ(b) = 300 pm
λ(c) = 3 nm
λ(d) = 30 A
Arrange these radiations in the increasing order of their energies.
(b) Draw the shapes of the following orbitals :
(i) 3d_xy
(ii) d_z²
Answer:
(a) First of all, convert values of all wavelengths in metre.
λ(A) = 300 nm = 300 × 10^-9 m = 3 × 10^-7 m
λ(B) = 300 μm = 300 × 10^-6 m = 3 × 10^-4 m
λ(C) = 3 nm = 3 × 10^-9 m
λ(D) = 30Å = 30 × 10^-10 m = 3 × 10^-9 m
∵ Energy (E) = \(\frac{h c}{\lambda}\)
∴ E ∝ \(\frac{1}{\lambda}\)
Greater the wavelength of radiation, lesser will be its energy.
Hence, increasing order of energy of these radiations :
λ(B) < λ(A) < λ(C) = λ(D)

(b)
(i) Shape of 3d_xy orbital:

(ii) Shape of d_z² orbital:

Commonly Made Error:
Students get confused while drawing the orbitals and write the lobes on wrong axis.

Answering Tip:
Understanding the concept of drawing the shapes of orbitals is very important. And needs practice to make drawing perfect.

Question 34.
Calculate :
(a) Mass in gram of 5.8 mol N₂O
(b) Number of moles in 8.0 g of O₂
(c) Molar mass if 11.2 L at STP weigh 8.5 g
OR
16 g of an ideal gas SO_x occupies 5.6 L at STE What is its molecular mass? What is the value of x ? [3]
Answer:
(a) Molecular mass of N₂O = 2(14) + 16 = 44
Mass = Molecular mass × No. of moles
= 44 × 5.8
= 255.2 g

(b) Number of moles of
O₂ = \(\frac{\text { Mass of } \mathrm{O}_2}{\text { Molecular mass of } \mathrm{O}_2}\)
= \(\frac{8.0}{32}\)
= 0.25 mol

(c) ∵ 11.2 L at STP weigh = 8.5 g
∴ 22.4 L at STP will weigh = \(\frac{8.5}{11.2}\) × 22.4
= 17 g mol^-1

OR

∵ 5.6 L at STP weigh = 16 g
∴ 22.4 L at STP will weigh = \(\frac{16}{5.6}\) × 22.4 = 64 g
SO_x = 64
32 + x × 16 = 64
16x = 64 – 32 = 32
x = \(\frac{32}{16}\) = 2

Section – D

Question 35.
Write the structures and IUPAC names of different structural isomers of alkenes corresponding to C₅H₁₀. [5]
OR
(a) For the following compounds, write structural formulas and IUPAC names for all possible isomers having the number of double or triple bond as indicated :
(i) C₄H₈ (one double bond)
(ii) C₅H₈ (one triple bond)
(b) How will you convert ethanoic acid into benzene.
Answer:
The various structural isomers and their IUPAC names are given below:

OR

(a) (i) The following structural isomers are possible for C₄H₈ with one double bond :

(b) (ii) The following structural isomers are possible for C₅C₈ with one triple bond:

(b) Ethanoic acid can be converted to benzene in the following steps :

Question 36.
(i) Why 3° carbocation are more stable than 1° carbocation ?
(ii) Compare inductive and electromeric effects.
(iii) Why CCl₃COOH is a stronger acid than (CH₃)₃CCOOH ?
(iv) Define the term nucleophile.
(v) Draw the resonance structure of aniline, using curved arrow for electronic movements.
OR
(a) Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens?
(b) Will CCl₄ gives white precipitate of AgCl on heating it with silver nitrate? Give reason for your answer.
Answer:
(i) Tertiary carbocation has three electron repelling alkyl groups. This increases +I effect on carbon and reduces the positive charge making it more stable.

(ii) Inductive effect describes the transmission of electrical charges between atoms in a molecule whereas electromeric effect describes the transmission of electron pairs between atoms in a molecule. Inductive effect is affected by electronegativity values of atoms whereas electromeric effect is caused by the number of double bonds and their arrangement.

(iii) CCl₃COOH is a stronger acid than (CH₃)₃CCOOH due to presence of three -I effect causing group (-Cl) which increases the acidic strength. Whereas (CH₃)₃COOH contains three +I effect causing group (-CH₃) which decreases the acidic strength.

(iv) Nucleophile: A reagent that brings an electron pair to the reactive site is called a nucleophile (Nu:), i.e., nucleus seeking.
Negatively charged nucleophile: hydroxide (OH^–), cyanide (CN^–), etc.

Neutral nucleophile:

(v)

OR
(a) While testing the Lassaigne’s extract for the presence of halogens, it is first boiled with dilute nitric acid. This is done to decompose NaCN to HCN and Na₂S to H₂S and to expel these gases. That is, if any nitrogen and sulphur are present in the form of NaCN and Na₂S, then they are removed. These ions would otherwise interfere with silver nitrate test for halogens.
NaCN + HNO₃ → NaNO₃ + HCN
Na₂S + 2HNO₃ → 2NaNO₃ + H₂S

(b) CCl₄ will not give the white precipitate of AgCl on heating it with silver nitrate. This is because the chlorine atoms are covalently bonded to carbon in CCl₄. To obtain the precipitate, it should be present in ionic form and for this, it is necessary to prepare the Lassaigne’s extract of CCl₄.

Question 37.
Who proposed VSEPR theory? What are the main postulates of Valence Shell Electron Pair Repulsion (VSEPR) theory ?
Answer:
Sidgwick and Powell (1940) proposed VSEPR theory on the basis of repulsive interaction of the electron pairs in the valence shell of the atoms. This theory was further developed by Nyholm and Gillespie (1950).

Main Postulates of VSEPR Theory :
1. The shape of a molecule depends upon the number of valence shell electron pairs (bonded or non-bonded) around the central atom.

2. Electron pairs in the valence shell repel one another since their electron clouds are negatively charged.

3. These electron pairs tend to occupy such positions in space that minimize repulsion and so maximize distance between them.

4. The valence shell is taken as a sphere with the electron pairs localizing on the spherical surface at maximum distance from one another.

5. A multiple bond is treated as if it is a single electron pair and the two or three electron pairs of a multiple bond are treated as a single super pair.

6. When two or more resonating structures can represent a molecule, the VSEPR model is applicable to any such structure.