Maharashtra Board Class 12 Chemistry Sample Paper Set 6 with Solutions

Maharashtra State Board Class 12th Chemistry Sample Paper Set 6 with Solutions Answers Pdf Download.

Maharashtra Board Class 12 Chemistry Model Paper Set 6 with Solutions

Time: 3 Hrs.
Max. Marks : 70

General Instructions:

The question paper is divided into four sections.

1. Section A: Q. No. 1 contains Ten multiple choice type of questions carrying One mark each.
   Q. No. 2 contains Eight very short answer type of questions carrying One mark each.
2. Section B: Q. No. 3 to Q. No. 14 are Twelve short answer type of questions carrying Two marks each. (Attempt any Eight).
3. Section C: Q. No. 15 to Q. No. 26 are Twelve short answer type of questions carrying Three marks each. (Attempt any Eight).
4. Section D: Q. No. 27 to Q. No. 31 are Five long answer type of questions carrying Four marks each. (Attempt any Three).
5. Use of log table is allowed. Use of calculator is not allowed.
6. Figures to the right indicate full marks.
7. For each MCQ, correct answer must be written along with its alphabet.
   E.g., (a) ……./(b)……../(c) ………/(d) ………. Only first attempt will be considered for evaluation.
8. Draw well labeled diagrams and write balanced equations wherever necessary.
9. Given data:
   Atomic mass of C = 12, H = 1, O = 16,
   Atomic number (Z): Mn = 25, Tb = 65, Ce = 58, Ar = 18,
   R = 8.314 JK^-1 mol^-1 or 0.083 L bar K^-1 mol^-1
   N_A = 6.022 × 10²³ mol^-1, F = 96500 C

Section – A

Question 1.
Select and write the correct answer for the following multiple choice type of questions: [10 Marks]

i. Blood in human body is highly buffered at pH of ____.
(A) 7.4
(B) 7.0
(C) 6.9
(D) 8.1
Answer:
(A) 7.4

ii. The units of Henry’s law constant are ___.
(A) bar dm³ mol^-1
(B) mol L^-1 bar^-1
(C) L mol^-1 bar^-1
(D) bar L^-1 mol^-1
Answer:
(B) mol L^-1 bar^-1

iii. Which of the following is a trihydric alcohol?
(A) n-Propyl alcohol
(B) Glycerol
(C) Glycol
(D) Glycine
Answer:
(B) Glycerol

iv. Anisole on heating with concentrated HI gives ____
(A) iodobenzene
(B) phenol + methanol
(C) phenol + iodomethane
(D) iodobenzene + methanol
Answer:
(B) phenol + iodomethane

v. The flux added during extraction of iron from haematite ore is ____.
(A) silica
(B) calcium carbonate
(C) sodium carbonate
(D) alumina
Answer:
(B) calcium carbonate

vi. Which of the following occurs in liquid state at room temperature?
(A) HIO₃
(B) HBr
(C) HCl
(D) HF
Answer:
(D) HF

vii. Which of the following is NOT correct?
(A) Gibbs energy is an extensive property.
(B) Electrode potential or cell potential is an intensive property.
(C) Electrical work = – ∆G
(D) If half reaction is multiplied by a numerical factor, the corresponding E° value is also multiplied by the same factor.
Answer:
(D) If half reaction is multiplied by a numerical factor, the corresponding E° value is also multiplied by the same factor.

viii. Choose the compound from the following that will react fastest by \(\mathrm{S}_{\mathrm{N}}{ }^1\) mechanism.
(A) 1-Iodobutane
(B) 1-Iodopropane
(C) 2-Iodo-2-methylbutane
(D) 2-Iodo-3-methylbutane
Answer:
(C) 2-Iodo-2-methylbutane

ix. Pb has fee structure with edge length of unit cell 495 pm. Radius of Pb atom is ____
(A) 205 pm
(B) 185 pm
(C) 260 pm
(D) 175 pm
Answer:
(D) 175 pm

x. Methyl ketone group is identified by _____
(A) Iodoform test
(B) Fehling’s solution
(C) Tollens’ reagent
(D) Schiff’s reagent
Answer:
(A) Iodoform test

Question 2.
Answer the following:

i. Write IUPAC name of the following compounds.

Answer:
2-Bromo-3-methylpent-3-ene

ii. Give two examples of nonideal solutions.
Answer:
a. Ethanol + acetone
b. Carbon disulphide + acetone
c. Phenol + aniline
d. Chloroform + acetone

iii. What is the coordination number and oxidation state of metal ion in the complex [Pt(NH₃)Cl₅]^–?
Answer:
In complex [Pt(NH₃)Cl₅]^–, the coordination number of metal ion is 6 and oxidation state of metal ion is +4.

iv. For a certain reaction ∆H° = 250 kJ and ∆S° = -27 J K^-1. Determine whether the reaction is spontaneous or nonspontaneous.
Answer:
When ∆H positive and ∆S is negative, then ∆G is positive regardless of temperature. Hence, the reaction is nonspontaneous at all temperatures.

v. What is a rate determining step?
Answer:
When a chemical reaction occurs in a series of steps, one of the steps is slower than all other steps. Such a slowest step in the reaction is called a rate determining step.

vi. What is the ratio of octahedral holes to the number of anions in a hexagonal closed packed structure?
Answer:
The ratio of octahedral holes to the number of anions in hexagonal closed packed structure is 1:1.

vii. Define the term: Vulcanization.
Answer:
The process by which a network of cross links is introduced into an elastomer is called vulcanization.

viii. Define ‘sustainable development’.
Answer:
Sustainable development is development that meets the needs of the present, without compromising the ability of future generations to meet their own need.

Section – B

Attempt any EIGHT of the following questions:

Question 3.
What is pseudo first order reaction? Give one example of it.
Answer:
A reaction which is expected to be of hiqher order but follow the first order kinetics is called pseudo first order reaction.
Example: Hydrolysis of methyl acetate.
CH₃COOCH_3(aq) + H₂O_(l) → CH₃COOH_(aq) + CH₃OH_(aq)
The rate law is rate = k [CH₃COOCH₃]

Question 4.
Write the electrode reactions during electrolysis of molten KCl.
Answer:
Electrode? reactions during electrolysis of molten KCl are as follows:

Question 5.
Write formulae of the following complexes
i. Potassium amminetrichloroplatinate(II).
ii. Dicyanoaurate(I) ion
Answer:
i. K[Pt(NH₃)Cl₃]
ii. [Au(CN)₂]^–

Question 6.
How are primary amines prepared from amides?
Answer:
Primary amines having same number of carbon atoms can be obtained by the reduction of amides by LiAlH₄ in ether or by Na/C₂H₅OH.

Question 7.
Ammonia serves as a Lewis base whereas AlCl₃ is a Lewis acid. Explain.
Answer:

1. According to Lewis theory, an acid is a substance which can accept a share in an electron pair.
   In AlCl₃ molecule, the octet of Al is incomplete. Therefore, it can accept an electron pair to complete its octet. Hence, AlCl₃ acts as a Lewis acid.
2. According to Lewis theory, a base is a substance which can donate an electron pair. In ammonia (NH₃) molecule, nitrogen atom has one lone pair of electrons to donate. Hence, NH₃ acts as a Lewis base.

Question 8.
Calculate the spin only magnetic moment of divalent cation of a transition metal with atomic number 25.
Answer:
For an element with atomic number 25, the condensed electronic configuration for its divalent cation will be

There are 5 unpaired electrons, so n = 5. ∴ μ = \(\sqrt{5(5+2)}\) = 5.92 BM

Ans: Spin only magnetic moment of the given ion is 5.92 BM.

Question 9.
Explain, why boiling points of carboxylic acids are higher than corresponding alcohols.
Answer:

1. In liquid phase, carboxylic acids form dimer in which two molecules are held by two hydrogen bonds. Acidic hydrogen of one molecule forms hydrogen bond with carbonyl oxygen of the other molecule.
2. This doubles the size of the molecule resulting in an increase in intermolecular van der Waals forces, which in turn results in high boiling point.

Hence, carboxylic acids have higher boiling points than those of corresponding alcohols of comparable mass.

Question 10.
Aldehydes are more reactive towards nucleophilic addition reactions than ketones. Explain.
Answer:
Reactivity of aldehydes and ketones is due to the polarity of carbonyl group which results in electrophilicity of carbon. In general, aldehydes are more reactive than ketones towards nucleophilic attack. This can be well explained in terms of both the electronic effects and steric effects.

i.’ Influence of electronic effects:

a. Alkyl qroups have electron donating inductive effect (+I). A ketone has two electron donating alkyl groups bonded to carbonyl carbon which are responsible for decreasing its positive polarity and electrophilicity.

b. In contrast, aldehydes have only one electron donating group bonded to carbonyl carbon. This makes aldehydes more electrophilic than ketones.

ii. Steric effects:

a. Two bulky alkyl groups in ketone come in the way of incoming nucleophile. This is called steric hindrance to nucleophilic attack.

b. On the other hand, nucleophile can easily attack the carbonyl carbon in aldehydes because it has one alkyl group and is less crowded or sterically less hindered. Hence, aldehydes are more easily attacked by nucleophiles.

Question 11.
What is the oxidation state of manganese in:
i. \(\mathrm{MnO}_4^{2-}\)
ii. \(\mathrm{MnO}_4^{-}\)
Answer:
i. Mn\(\mathrm{O}_4^{2-}\)
Let the oxidation state of Mn be x.
∴ x + 4(-2) = -2
∴ x – 8 = -2
∴ x = +6

ii. \(\mathrm{MnO}_4^{-}\) [1 Mark]
Let the oxidation state of Mn be x.
∴ x + 4(-2) = -1
∴ x – 8 = -1
∴ x = +7

Question 12.
The pH of a monoacidic weak base is 11.2. Calculate its percent dissociation in 0.02 M solution.
Answer:
Given: pH of monoacidic weak base = 11.2,
Concentration of solution (c) = 0.02 M

To find: Percent dissociation

Formulae:
i. pH + pOH = 14
ii. pOH = -log₁₀[OH^–]
iii. Percent dissociation = α × 100

Calculation:
From formula (i),
pOH = 14 – pH = 14 – 11.2 = 2.8
From formula (ii),
pOH = -log₁₀[OH^–]
∴ log₁₀[OH^–] = – pOH = – 2.8 = – 2 – 0.8 – 1 + 1 = – 3 + 0.2 = \(\overline{3} .2\)
[OH^–] = Antilog_1o \(\overline{3} .2\) = 1.585 × 10^-3 M
For monoacidic base, BOH_(aq) \(\rightleftharpoons\) \(\mathrm{B}_{(a q)}^{+}+\mathrm{OH}_{(a q)}^{-}\)
[OH^–] = α c
α = \(\frac{\left[\mathrm{OH}^{-}\right]}{\mathrm{c}}\) = \(\frac{1.585 \times 10^{-3}}{0.02}\) = 0.07925
From formula (iii),
Percent dissociation = α × 100
= 0.07925 × 100
= 7.925 %

Ans: Percent dissociation is 7.925 %

Question 13.
The vapour pressure of water at 20 °C is 17 mm Hg. What is the vapour pressure of a solution containing 2.8 g urea in 50 g of water?
Answer:
Given: Vapour pressure of pure water = \(p_1{ }^0\) = 17 mm Hg
Mass of urea (W₂) = 2.8 g
Mass of water (W₁) = 50 g
To find: Vapour pressure of the solution (P₁)
Formula: \(\frac{P_1^0-P_1}{P_1^0}\) = \(\frac{W_2 M_1}{M_2 W_1}\)
Calculation: Molar mass of urea (NH₂CONH₂)
= 14 + 2 + 12 + 16 + 14 + 2 = 60 g mol^-1
Molar mass of water = 18 g mol^-1
Now, using formula,
\(\frac{P_1^0-P_1}{P_1^0}\) = \(\frac{W_2 M_1}{M_2 W_1}\)
\(\frac{17 \mathrm{~mm} \mathrm{Hg}-\mathrm{P}_1}{17 \mathrm{mmHg}}\) = \(\frac{2.8 \mathrm{~g} \times 18 \mathrm{gmol}^{-1}}{50 \mathrm{~g} \times 60 \mathrm{gmol}^{-1}}\)
∴ \(\frac{17 \mathrm{~mm} \mathrm{Hg}-P_1}{17 \mathrm{~mm} \mathrm{Hg}}\) = 0.0168
∴ 17 mm Hg – P₁ = 0.0168 × 17 mm Hg
∴ 17 mm Hg – P₁ = 0.2856 mm Hg
∴ P₁ = 17 mm Hg – 0.2856 mm Hg
= 16.71 mm Hg

Question 14.
Write a note on Haloform reaction.
Answer:
Haloform reaction:

1. This reaction is given by acetaldehyde, all methyl ketones (CH₃-CO-R) and all alcohols containing CH₃(CHOH)- group.
2. When an alcohol or methyl ketone is warmed with sodium hydroxide and iodine, a yellow precipitate is formed. Here the reagent sodium hypoiodite is produced in Situ.
3. During the reaction, sodium salt of carboxylic acid is formed which contains one carbon atom less than the substrate.
4. The methyl group is converted into haloform (CHX₃).
   e.g. Acetone is oxidized by sodium hypoiodite to give sodium salt of acetic acid and yellow precipitate of iodoform.

Section – C

Attempt any EIGHT of the following questions:

Question 15.
What is Kohlrausch law of independent migration of ions? How is it useful in obtaining molar conductivity at zero concentration of a weak electrolyte? Explain with an example.
Answer:
Kohlrausch law of independent migration of ions:

i. Kohlrausch law states that “at infinite dilution each ion migrates independent of coion and contributes to total molar conductivity of an electrolyte irrespective of the nature of other ion to which it is associated.”

ii. Both cation and anion contribute to molar conductivity of the electrolyte at zero concentration and thus ∧o is sum of molar conductivity of cation and that of the anion at zero concentration.
Thus, ∧0 = \(\mathrm{n}_{-} \lambda_{-}^0+\mathrm{n}_{-} \lambda_{-}^0\)
where λ₊ and λ_ are molar conductivities of cation and anion, respectively, and n₊ and n_– are the number of moles of cation and anion specified in the chemical formula of the electrolyte.

iii. Determination of molar conductivity of weak electrolyte at zero concentration:
Kohlrausch law is particularly useful in calculating Ao values of weak electrolytes from those of strong electrolytes.

For example, ∧0 of acetic acid can be calculated by knowing those of HCl, NaCl and CH₃COONa as described below:

Because ∧o values of strong electrolytes, HCl, CH₃COONa and NaCl, can be determined by extrapolation method, the ∧o of acetic acid can be obtained.

Question 16.
Write a note on halogen exchange reactions.
Answer:
i. Alkyl iodides are prepared conveniently by treating alkyl chlorides or bromides with sodium iodide in methanol or acetone solution.

ii. The sodium bromide or sodium chloride precipitates from the solution and can be separated by filtration.
This is known as Finkelstein reaction.

iii. Alkyl fluorides are prepared by heating alkyl chlorides or bromides with metal fluorides such as AgF, Hg₂F₂, ASF₃, SbF₃, etc. This reaction is known as Swartz reaction.
R – Cl + AgF → R – F + AgCl↓

Question 17.
What are the types of impurity defects? Explain the formation of vacancies through aliovalent impurity with example.
Answer:
There are two kinds of impurity defects: Substitutional and interstitial impurity defects.

Formation of vacancy through aliovalent impurity:
Vacancies are created by the addition of impurities of aliovalent ions (that is, ions with oxidation state different from that of host ions) to an ionic solid.

e.g. Consider a small amount of SrCl₂ impurity added to NaCl during its crystallization. The added Sr²⁺ ions (O.S. = +2) occupy some of the regular sites of Na⁺ host ions (O.S. = +1). In order to maintain electrical neutrality, every Sr²⁺ ion removes two Na⁺ ions. One of the vacant lattice sites created by removal of two Na⁺ ions is occupied by one Sr²⁺ ion. The other site of Na⁺ ion remains vacant as shown in the figure.

Question 18.
Derive the expression for the maximum work.
Answer:
i. Consider n moles of an ideal gas enclosed in a cylinder fitted with frictionless movable rigid piston. It expands isothermally and reversibly from the initial volume V₁ to final volume V₂ at temperature T. The expansion takes place in a number of steps.

ii. When the volume of a gas increases by an infinitesimal amount dV in a single step, the small quantity of work done
dW = -P_ext dV …….(1)

iii. As the expansion is reversible, P is greater by a very small quantity dp than P_ext. Thus,
P – P_ext = dP or P_ext = P – dP ……(2)
Combining equations (1) and (2),
dW = – (P – dP)dV = – PdV + dP ⋅ dV
Neglecting the product dP dV which is very small, we get
dW = – PdV …… (3)

iv. The total amount of work done during entire expansion from volume V₁ to V₂ would be the sum of infinitesimal contributions of all the steps. The total work is obtained by integration of Equation (3) between the limits of initial and final states. This is the maximum work, the expansion being reversible.
Thus, \(\int_{\text {initial }}^{\text {final }} d W\) = –\(\int_{V_1}^{V_2} P d V\)
Hence, W_max = \(-\int_{V_1}^{V_2} P d V\) …….(4)

v.

vi.
At constant temperature, P₁V₁ = P₂V₂ or \(\frac{V_2}{V_1}\) = \(\frac{P_1}{P_2}\)
Replacing \(\frac{V_2}{V_1}\) in equation (5) by \(\frac{P_1}{P_2}\), we get,
‘W_max = – 2.303 nRT log\(\frac{P_1}{P_2}\) …….(6)
Equations (5) and (6) are expressions for work done in reversible isothermal process.

Question 19.
Convert the following:

Answer:

Question 20.
The solubility product of AgBr is 5.2 × 10^-13. Calculate its solubility in mol dm-3 and g dm-3.
(Molar mass of AgBr = 187.8 g mol^-1)
Answer:
Given: Solubility product (K_sp) of AgBr = 5.2 × 10^-13.
Molar mass of AgBr = 187.8 g mol^-1
To find: Solubility in mol dm^-3 and g dm^-3

Formulae:
i. K_sp = \(x^x y^y S^{x+y}\)
ii. Solubility (g dm^-3) = Molar solubility (mol dm^-3) × Molar mass (g mol^-1)

Calculation:
The solubility equilibrium of AgBr is:

Solubility (g dm^-3) = Molar solubility (mol dm^-3) × Molar mass (g mol^-1)
S = 7.2 × 10^-7 mol dm³ × 187.8 g mol^-1
= 1.35 × 10^-4 g dm^-3

Ans: Solubility in mol dm^-3 is 7.2 × 10^-7 mol dm^-3 and solubility in g dm^-3 is 1.35 × 10^-4 g dm^-3.

Question 21.
Cerium and terbium behaves as good oxidizing agents in +4 oxidation state. Explain.
Answer:
i. Cerium (Ce) and Terbium (Tb) show +4 oxidation states.
Their electronic configurations are given below:
Ce = [Xe] 4f¹ 5d¹ 6s² ; Ce⁴⁺ = [Xe]
Tb = [Xe] 4f⁹ 6s² ; Tb⁴⁺ = [Xe] 4f⁷

ii. It is clear from the configuration of Ce that Ce⁴⁺ is favoured by its noble gas configuration. But it can be easily converted into stable Ce³⁺ by gain of an electron. Due to this reason, Ce⁴⁺ is a good oxidizing agent.

iii. Similarly, Tb⁴⁺ ion is stabilized due to half filled f-subshell. It can be easily converted into stable Tb³⁺ by gain of an electron. Due to this reason, Tb⁴⁺ is a good oxidizing agent.

Question 22.
Write a note on esterification reaction.
Answer:

1. The reaction between alcohol or phenol with a carboxylic acid to form an ester is called esterification.
2. Alcohols and phenols form esters by reaction with carboxylic acid, acid anhydrides and acid halides.
3. Esterification of alcohol or phenol is carried out in the presence of concentrated sulphuric acid. The reaction is reversible and can be shifted in the forward direction by removing water as soon as it is formed.

a. Reaction of alcohol and phenols with carboxylic acid:

b. Reaction of alcohol and phenol with acid anhydride: Alcohols and phenols react with acid anhydrides in presence of acid catalyst to form ester.

c. Reaction of alcohols and phenol with acid chloride: The reaction of alcohol and phenols with acid chloride is carried out in the presence of pyridine (base), which neutralizes HCl.

Question 23.
Derive the integrated rate law for first order reaction.
Answer:
Consider first order reaction,
A → product
The differential rate law is given by
rate = \(-\frac{\mathrm{d}[\mathrm{~A}]}{\mathrm{dt}}\) = k[A] …. (1)
where, [A] is the concentration of reactant at time t.
Rearranging Eq. (1)
\(\frac{d[A]}{[A]}\) = -k dt …. (2)
Let [A]_o be the initial concentration of the reactant A at time t = 0. Suppose [A]_t is the concentration of A at time = t
The equation (2) is integrated between limits [A] = [A]_o at t = 0 and [A] = [A]_t at t = t
\(\int_{[A]_0}^{[A]^{t}} \frac{d[A]}{[A]}\) = \(-\mathrm{k} \int_0^t \mathrm{~d} t\)

Question 24.
i. Why does glucose give positive Tollen’s test and Fehling test?
ii. Give two evidences for presence of formyl group in glucose.
Answer:
i. Hemiacetal group of glucopyranose structure is a potential aldehyde group. It imparts reducing properties to glucose. Thus, glucose gives positive Tollen’s test and positive Fehling test.

ii. a. Glucose gets oxidised to a six carbon monocarboxylic acid called gluconic acid on reaction with bromine water which is a mild oxidizing agent. Thus, the carbonyl group in glucose is in the form of formyl (-CHO).
b. Hemiacetal group of glucopyranose structure is a potential aldehyde (formyl) group. It imparts reducing properties to glucose. Thus, glucose gives positive Tollen’s test or Fehling test.

Question 25.
What is meant by LDP and HDP? Mention the basic difference between the same with suitable examples.
Answer:
i. LDP means low density polythene and HDP means high density polythene.

ii. LDP is a branched polymer of ethene with polymeric chains loosely held. Hence, even though it is tough, it is extremely flexible. Therefore, LDP is used in producing extruded films, sheets, mainly for packaging and household uses like in preparation of squeeze bottles, attractive containers, etc. where low tensile strength and flexibility is required.

On the other hand, HDP is a linear polymer of ethene with closely packed polymeric chains. Hence, it is much stiffer than LDP and has high tensile strength and hardness. Therefore, HDP is used in the manufacture of toys and other household articles like buckets, dustbins, bottles, pipes, laboratory wares and other objects where high tensile strength and stiffness is required.

Question 26.
i. Define green chemistry.
ii. List the 12 principles of green chemistry.
Answer:
i. Green chemistry is the use of chemistry for pollution prevention by environmentally conscious design of chemical products and processes that reduce or eliminate the use or generation of hazardous substances.

ii. The 12 principles of green chemistry are as follows:

1. Prevention of waste or by products
2. Atom economy
3. Less hazardous chemical synthesis
4. Designing safer chemicals
5. Use of safer solvent and auxiliaries
6. Design for energy efficiency
7. Use of renewable feed stocks
8. Reduce derivatives (Minimization of steps)
9. Use of catalysis
10. Design for degradation
11. Real-time analysis pollution prevention
12. Safer chemistry for accident prevention

Section – D

Attempt any THREE of the following questions: [12 Marks]

Question 27.
i. Write short note on: Gabriel phthalimide synthesis.
ii. Write the chemical reaction for the conversion of benzoyl chloride to acetophenone.
Answer:
i. This method is used for the synthesis of primary amine. It involves the following three stages.

i. Formation of potassium salt of phthalimide from phthalimide on reaction with alcoholic potassium hydroxide.

ii. Formation of N-alkyl phthalimide from the potassium salt by reaction with alkyl halide.

iii. Alkaline hydrolysis of N-alkyl phthalimide to form the corresponding primary amine.

ii. Benzoyl chloride to acetophenone:

Question 28.
i. In a first order reaction x → y, 40% of the given sample of a compound remains unreacted in 45 minutes. Calculate rate constant of the reaction.
ii. Write four salient features of \(\mathrm{S}_{\mathrm{N}}{ }^1\) mechanism.
Answer:
i. Given: For a first order reaction, amount of reactant that remains after 45 minutes = 40%

To find: Rate constant (k)
Formula k = \(\frac{2.303}{t}\)log₁₀\(\frac{[A]_0}{[A]_T}\)

Calculation:

ii. Salient features of \(S_N 1\) mechanism:

i. It is two step mechanism.

ii. Heterolysis of C-X bond in the slow and reversible first step forms planar carbocation intermediate.

iii. Attack of the nucleophile on the carbocation intermediate is fast (second step) to form the product.

iv. When SNI reaction is carried out at chiral carbon in an optically active substrate, the product formed is nearly racemic. This indicates that \(S_N 1\) reaction proceeds mainly with racemization. This means both the enantiomers of product are formed in almost equal amount. Racemization in \(S_N 1\) reaction is the result of formation of planar carbocation intermediate. Nucleophile can attack planar carbocation from either side which results in formation of both the enantiomers of the product.

Question 29.
Write molecular formulae and structures of the following compounds:
i. Pyrosulphuric acid
ii. Peroxymonosulfuric acid
Answer:
i. Pyrosulphuric acid:
Molecular formula: H₂S₂O₇
Structure:

ii. Peroxymonosulfuric acid:
Molecular formula: H₂SO₅
Structure:

Question 30.
What are ligands? Classify the following ligands into monodentate, polydentate and ambidentate:
i. SCN^–
ii. \(\mathrm{C}_2 \mathrm{O}_4^{2-}\)
iii. OH^–
iv. CN^–
v. H₂O
vi. CO
Answer:
In coordination compound, the species surrounding the central metal atom or ion are

Sr. No	Ligad	Type of ligand
i.	SCN	Ambidentate
ii.	\(\mathrm{SCN}^{-}\)	Polydentate
iii.	\(\mathrm{C}_2 \mathrm{O}_4^{2-}\)	Monodentate
iv.	\(\mathrm{OH}^{-}\)	Monodentate
v.	\(\mathrm{H}_2 \mathrm{O}\)	Monodentate
vi.	CO	Monodentate

Question 31.
i. Distinguish between isothermal process and adiabatic process.
ii. Calculate emf of the cell at 25 °C. Zn_(s) | Zn²⁺ (0.08 M) || Cr³⁺(0.1 M) | Cr_(s)
E°_zn = – 0.76 V, E°_Cr = -0.74 V
Answer:
i. Difference between isothermal and adiabatic process:

Isothermal process	Adiabatic process
i. It is a process in which temperature of the system remains constant throughout the transformation.	It is a process in which there is no exchange of heat between system and surroundings.
ii. Heat flows from the system to surroundings and vice versa. (Q ≠ O).	System is completely insulated with respect to heat from the surroundings (Q = O).
iii. Temperature remains constant (∆T = O)	Temperature may increase or decrease (∆T ≠ O).
iv. Change in internal energy is zero (∆U = O).	Internal energy may increase or decrease (∆U ≠ O).

Given: \(\mathrm{E}_{\mathrm{Zn}}^0\) = -0.76, \(E_{c_r}^{\circ}\) = -0.74
To find : Emf of the cell (E_cell)

Formulae:

Calculation:

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