ISC Class 12 Chemistry Sample Paper 2022 with Solutions

ISC SEMESTER 2 EXAMINATION
SPECIMEN QUESTION PAPER
CHEMISTRY PAPER 1 (THEORY)

Maximum Marks: 35
Time allowed: One and a half hour

SECTION A - 7 MARKS

Question 1

Fill in the blanks by choosing the appropriate word(s) from those given in the brackets:
(two, four, sec^-1, diamagnetic, acetaldehyde, mol^-1L sec^-1, paramagnetic, formaldehyde, acetone, ethanol)

(i) When the concentration of a reactant of first order reaction is doubled, the rate of reaction becomes ___________ times. The unit of rate constant (k) for the first order reaction is __________. [1 Mark]

Answer: two, sec^-1

Teacher's Note:
a) For a first order reaction, rate = k[A]. Doubling the concentration doubles the rate linearly.
b) The unit of rate constant depends on the overall order of the reaction given by the formula L^(n-1) mol^(1-n) sec^-1.

(ii) The transition metals show ___________ character because of the presence of unpaired electrons while Cu⁺ is ___________ because its electronic configuration is [Ar]3d¹⁰. [1 Mark]

Answer: paramagnetic, diamagnetic

Teacher's Note:
a) Unpaired electrons in d-orbitals impart paramagnetic character to transition metal species.
b) Fully filled orbitals as in Cu⁺ ([Ar]3d¹⁰) contain only paired electrons, resulting in diamagnetism.

(iii) Calcium formate on distillation gives ___________ but the distillation of calcium formate and calcium acetate gives ___________. [1 Mark]

Answer: formaldehyde, acetaldehyde

Teacher's Note:
a) Dry distillation of calcium formate yields methanal (formaldehyde) and calcium carbonate.
b) Dry distillation of a mixture of calcium formate and calcium acetate yields ethanal (acetaldehyde).

Question 2

Select and write the correct alternative from the choices given below.

(i) The type of hybridization involved in Octahedral complexes is: [1 Mark]
(A) sp³
(B) dsp²
(C) sp³d
(D) d²sp³

Answer: (D) d²sp³

Octahedral complexes possess a coordination number of six, which involves either d²sp³ or sp³d² hybridization.

Teacher's Note:
a) Coordination number six dictates an octahedral geometry.
b) Inner orbital octahedral complexes involve d²sp³ hybridization utilizing (n-1)d orbitals.

(ii) One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular mass of 44 amu. The alkene is: [1 Mark]
(A) ethene
(B) propene
(C) 1-butene
(D) 2-butene

Answer: (D) 2-butene

Ozonolysis of 2-butene breaks the double bond symmetrically to yield two moles of ethanal (acetaldehyde), whose molecular mass is 44 amu.

Teacher's Note:
a) The molecular mass of ethanal (CH₃CHO) is 24 + 4 + 16 = 44 amu.
b) Symmetrical alkenes cleave into identical carbonyl fragments during reductive ozonolysis.

(iii) Primary amine when warmed with chloroform and alc. KOH yields: [1 Mark]
(A) cyanides
(B) isocyanides
(C) benzene diazonium chloride
(D) secondary amines

Answer: (B) isocyanides

This is the carbylamine test, specific for primary amines, producing foul-smelling isocyanides (carbylamines).

Teacher's Note:
a) Aliphatic and aromatic primary amines react with chloroform and alcoholic KOH to form foul-smelling isocyanides.
b) Secondary and tertiary amines do not give this carbylamine reaction.

(iv) Assertion: The conversion of fresh precipitate to colloidal state is called peptization.
Reason: It is caused by addition of common ions. [1 Mark]
(A) Both assertion and reason are true and reason is the correct explanation of assertion.
(B) Both assertion and reason are true but reason is not the correct explanation for assertion.
(C) Assertion is true but reason is false.
(D) Assertion is false but reason is true.

Answer: (C) Assertion is true but reason is false.

Peptization is caused by the addition of a suitable electrolyte (peptizing agent) providing specific ions adsorbed by the precipitate, not necessarily common ions.

Teacher's Note:
a) Peptization breaks down a fresh precipitate into colloidal particles through preferential adsorption of ions.
b) The reason is incorrect because the process requires an electrolyte furnishing specific ions, not common ions.

SECTION B - 16 MARKS

Question 3

Name the type of isomerism shown by each of the following pairs of compounds: [2 Marks]

(i) [CoCl₂(NH₃)₄]Cl.H₂O and [CoCl(H₂O)(NH3)₄]Cl₂ [1 Mark]

Answer: Hydrate isomerism

Teacher's Note:
a) Hydrate isomers have the same empirical formula but differ in whether a water molecule acts as a ligand or as water of crystallization.
b) Check the number of chloride ions outside the coordination sphere to confirm hydrate isomerism.

(ii) [Cr(NH₃)₅Br]SO₄ and [Cr(NH₃)₅SO₄]Br [1 Mark]

Answer: Ionisation isomerism

Teacher's Note:
a) Ionisation isomers yield different ions in solution despite having the same chemical composition.
b) Bromide and sulphate ions interchange positions between the coordination sphere and the counter-ion sphere.

Question 4

(i) Write chemical equations to illustrate each of the following name reactions: [2 Marks]

(a) Rosenmund's reduction [1 Mark]

Answer:
CH₃COCl + H₂ Pd/BaSO₄, S→ CH₃CHO + HCl
(Ethanoyl chloride / Acetyl chloride)      (Ethanal / Acetaldehyde)

Teacher's Note:
a) Acid chlorides are hydrogenated to aldehydes using poisoned palladium catalyst (Pd/BaSO₄).
b) Barium sulphate deactivates the catalyst to prevent further reduction of the aldehyde to an alcohol.

(b) Clemmensen's reduction [1 Mark]

Answer:
C₆H₅COCH₃ + 4[H] Zn-Hg / conc. HCl→ C₆H₅CH₂CH₃ + H₂O
(Acetophenone)      (Ethyl benzene)

Teacher's Note:
a) Carbonyl groups of aldehydes and ketones are reduced to methylene groups using zinc amalgam and concentrated hydrochloric acid.
b) This method is suitable for acid-sensitive carbonyl compounds.

OR

(ii) How will you bring about the following conversions? (Give equation). [2 Marks]

(a) Acetic acid to acetone [1 Mark]

Answer:
Step 1: 2CH₃COOH + Ca(OH)₂ → (CH₃COO)₂Ca + 2H₂O
Step 2: (CH₃COO)₂Ca Dry distillation→ (CH₃)₂CO + CaCO₃
(Acetone)

Teacher's Note:
a) Neutralizing acetic acid with calcium hydroxide gives calcium acetate.
b) Dry distillation of calcium acetate yields acetone.

(b) Formaldehyde to urotropine [1 Mark]

Answer:
6HCHO + 4NH₃ → (CH₂)₆N₄ + 6H₂O
(Formaldehyde)                      (Urotropine / Hexamethylenetetramine)

Teacher's Note:
a) Reaction of formaldehyde with ammonia produces hexamethylenetetramine (urotropine).
b) Urotropine is used medically as a urinary antiseptic.

Question 5

What is a zwitter ion? Represent the zwitter ion of glycine. [2 Marks]

Answer:
A zwitter ion is a dipolar ion formed by the internal proton transfer from the carboxylic acid group to the amino group within an amino acid molecule, resulting in a net neutral charge.
Structure of zwitter ion of glycine:
H₃N⁺-CH₂-COO^-

Teacher's Note:
a) Zwitter ions possess both positive and negative charges on the same molecule.
b) Amino acids exist predominantly as zwitter ions in aqueous solution near their isoelectric point.

Question 6

(i) Arrange the following in the increasing order of their basic strength: C₂H₅NH₂, C₆H₅NH₂, (C₂H₅)₂NH. [1 Mark]

Answer:
C₆H₅NH₂ < C₂H₅NH₂ < (C₂H₅)₂NH

Teacher's Note:
a) Aniline is least basic because the lone pair on nitrogen is delocalized into the benzene ring through resonance.
b) Secondary aliphatic amines like diethylamine are more basic than primary amines due to greater electron-donating inductive (+I) effects of two alkyl groups.

(ii) What are the products formed when benzene diazonium chloride reacts with phenol in weak alkaline medium? (Give equation). [1 Mark]

Answer:
C₆H₅N₂Cl + C₆H₅OH Weak NaOH, 0-5°C→ p-Hydroxyazobenzene (Orange dye) + HCl

Teacher's Note:
a) This is an electrophilic aromatic substitution coupling reaction.
b) The reaction yields an intensely colored azo dye (p-hydroxyazobenzene).

Question 7

Give reasons for the following: [2 Marks]

(i) Diabetic patients are advised to take artificial sweeteners instead of natural sweeteners. [1 Mark]

Answer:
Natural sweeteners increase blood glucose levels, which is harmful for diabetic patients who cannot metabolize sugar properly due to insufficient insulin. Artificial sweeteners provide sweetness without increasing carbohydrate intake or blood sugar levels.

Teacher's Note:
a) Artificial sweeteners pass through the body without being metabolized for energy.
b) They help maintain stable blood glucose levels in diabetic patients.

(ii) The use of aspartame is limited to cold foods and drinks. [1 Mark]

Answer:
Aspartame is thermally unstable and decomposes at cooking or baking temperatures, losing its sweet taste.

Teacher's Note:
a) Aspartame undergoes thermal degradation at high temperatures.
b) Consequently, its use is restricted to cold foods and beverages.

Question 8

The rate of reaction becomes four times when the temperature changes from 293K to 313K. Calculate the energy of activation (E_a) of the reaction assuming that it does not change with temperature. (R = 8.314 JK^-1mol^-1) [2 Marks]

Answer:
Given: T₁ = 293 K, T₂ = 313 K, k₂ = 4k₁
Using Arrhenius equation:
\(\log \left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303 R} \left(\frac{T_2 - T_1}{T_1 T_2}\right)
\(\log 4 = \frac{E_a}{2.303 \times 8.314} \left(\frac{313 - 293}{293 \times 313}\right)
\(0.6021 = \frac{E_a}{2.303 \times 8.314} \left(\frac{20}{293 \times 313}\right)
\(E_a = \frac{0.6021 \times 2.303 \times 8.314 \times 293 \times 313}{20} = 52,863 \text{ J mol}^{-1} = 52.86 \text{ kJ mol}^{-1}\)

Teacher's Note:
a) Apply the logarithmic form of the Arrhenius equation relating rate constants at two different temperatures.
b) Ensure proper unit conversion from Joules to kiloJoules per mole in the final answer.

Question 9

Give balanced equation for each of the following: [2 Marks]

(i) Ethylamine and nitrous acid [1 Mark]

Answer:
CH₃CH₂NH₂ + HNO₂ 0-5°C→ CH₃CH₂OH + N₂(g) + H₂O
(Ethylamine)                             (Ethanol)

Teacher's Note:
a) Aliphatic primary amines react with nitrous acid to form unstable aliphatic diazonium salts, which immediately decompose to yield alcohols and nitrogen gas.
b) The reaction is accompanied by the brisk evolution of nitrogen gas.

(ii) Aniline and acetyl chloride [1 Mark]

Answer:
C₆H₅NH₂ + CH₃COCl Pyridine→ C₆H₅NHCOCH₃ + HCl
(Aniline)         (Acetyl chloride)           (Acetanilide)

Teacher's Note:
a) This is an acylation reaction where aniline reacts with acetyl chloride in the presence of a base like pyridine.
b) The product formed is N-phenylacetamide (acetanilide).

Question 10

Give one chemical test for each to distinguish between the following pairs of compound: [2 Marks]

(i) Acetaldehyde and benzaldehyde [1 Mark]

Answer:
Fehling's Test: Acetaldehyde gives a red precipitate of cuprous oxide when warmed with Fehling's solution, whereas benzaldehyde does not give this test.
CH₃CHO + 2Cu²⁺ + 5OH^- Heat→ CH₃COO^- + Cu₂O↓ (Red ppt) + 3H₂O
C₆H₅CHO + 2Cu²⁺ + 5OH^- Heat→ No reaction

Teacher's Note:
a) Aliphatic aldehydes reduce Fehling's solution, while aromatic aldehydes do not.
b) Tollens' reagent can also be used, but Fehling's solution clearly distinguishes between aliphatic and aromatic aldehydes.

(ii) Acetone and acetic acid [1 Mark]

Answer:
Sodium Bicarbonate Test: Acetic acid reacts with NaHCO₃ to produce brisk effervescence of carbon dioxide gas, whereas acetone does not react.
CH₃COOH + NaHCO₃ → CH₃COONa + H₂O + CO₂↑ (Brisk effervescence)

Teacher's Note:
a) Carboxylic acids decompose sodium bicarbonate with the evolution of carbon dioxide gas.
b) Ketones lack acidic carboxyl protons and do not react with sodium bicarbonate.

SECTION C - 12 MARKS

Question 11

(i) Answer the following: [3 Marks]

(a) Define molecularity of a reaction. Give one difference between the order of reaction and its molecularity. [1½ Marks]

Answer:
Molecularity is defined as the total number of reactant species (atoms, ions, or molecules) colliding simultaneously in an elementary step to bring about a chemical reaction.
Difference:

Teacher's Note:
a) Molecularity cannot be zero or fractional and is generally limited to 3.
b) Order can be zero, fractional, negative, or integer values.

(b) The rate constant (k) of a first order reaction is 4.5 × 10^-2 sec^-1. What will be the time required for the initial concentration of 0.4 M of the reactant to be reduced to 0.2 M? [1½ Marks]

Answer:
For a first order reaction:
\(t = \frac{2.303}{k} \log \left(\frac{[A]_0}{[A]_t}\right)
\(t = \frac{2.303}{4.5 \times 10^{-2}} \log \left(\frac{0.4}{0.2}\right)
\(t = \frac{2.303}{4.5 \times 10^{-2}} \log 2
\(t = \frac{2.303}{4.5 \times 10^{-2}} \times 0.3010 = 15.4 \text{ sec}\)

Teacher's Note:
a) Use the integrated rate equation for a first order reaction.
b) Notice that reducing a concentration to half corresponds to the half-life period (\(t_{1/2} = 0.693 / k\)).

OR

(ii) Answer the following: [3 Marks]

(a) For a first order reaction, show that the time required for the completion of 99% reaction is twice the time required for the completion of 90% of the reaction. [1½ Marks]

Answer:
For 99% completion, \([A]_0 = 100\), \([A]_t = 100 - 99 = 1\):
\(t_{99\%} = \frac{2.303}{k} \log \left(\frac{100}{1}\right) = \frac{2.303}{k} \times 2\)      ...(1)
For 90% completion, \([A]_0 = 100\), \([A]_t = 100 - 90 = 10\):
\(t_{90\%} = \frac{2.303}{k} \log \left(\frac{100}{10}\right) = \frac{2.303}{k} \times 1\)      ...(2)
Dividing (1) by (2):
\(t_{99\%} = 2 t_{90\%}\)

Teacher's Note:
a) Substitute percentage completion into the integrated first-order rate expression.
b) Clearly state logarithmic values (\(\log 100 = 2\), \(\log 10 = 1\)) to demonstrate the relation.

(b) For a reaction, rate = k[A]¹[B]^1.5[C]⁰. What is the overall order of reaction? [1½ Marks]

Answer:
Overall order = Sum of powers of concentration terms in the rate law
\(= 1 + 1.5 + 0 = 2.5\)

Teacher's Note:
a) The overall order is the arithmetic sum of the exponents of the concentration terms in the rate law.
b) Zero power indicates that the rate is independent of the concentration of that reactant.

Question 12

(i) What is the basic difference between the electronic configuration of transition and inner transition elements? [1½ Marks]

Answer:
Transition elements involve the filling of d-orbitals in the penultimate shell, with the general outer electronic configuration \((n-1)d^{1-10}ns^{1-2}\).
Inner transition elements involve the filling of f-orbitals in the anti-penultimate shell, with the general electronic configuration \((n-2)f^{1-14}(n-1)d^{0-1}ns^2\).

Teacher's Note:
a) Transition elements belong to d-block, while inner transition elements (lanthanoids and actinoids) belong to f-block.
b) Mentioning shell numbers (penultimate vs. anti-penultimate) is essential for full credit.

(ii) Why are Zn²⁺ ions colourless while Ni²⁺ ions are green in colour? [1½ Marks]

Answer:
Zn²⁺ ions have a completely filled 3d¹⁰ configuration with no unpaired electrons, making d-d transitions impossible, hence they are colourless.
Ni²⁺ ions have a 3d⁸ configuration containing two unpaired electrons, allowing d-d electronic transitions in the visible region, imparting a green colour.

Teacher's Note:
a) Colour in transition metal ions arises primarily from d-d transitions.
b) Presence of unpaired d-electrons is a prerequisite for colour.

Question 13

(i) Write the formula of each of the following compounds: [2 Marks]

(a) Potassium trioxalatoaluminate (III) [1 Mark]

Answer:
K₃[Al(C₂O₄)₃]

Teacher's Note:
a) Identify the central metal (Al), ligands (oxalate, C₂O₄^2-), and counter-ion (potassium, K⁺).
b) Balance charges: Al is +3, each oxalate is -2, total ligand charge is -6, giving a complex anion charge of -3 balanced by three K⁺ ions.

(b) Triammine triaquachromium (III) chloride [1 Mark]

Answer:
[Cr(NH₃)₃(H₂O)₃]Cl₃

Teacher's Note:
a) Neutral ligands ammine (NH₃) and aqua (H₂O) are listed alphabetically inside the coordination sphere.
b) Chromium has an oxidation state of +3, balanced by three chloride counter-ions.

(ii) For the complex ion [Co(NH₃)₆]³⁺, state the oxidation state of central metal atom and the coordination number of the complex ion. [1 Mark]

Answer:
Oxidation state of Co = +3
Coordination number = 6

Teacher's Note:
a) Calculate oxidation state: x + 6(0) = +3, hence x = +3.
b) Coordination number equals the total number of coordinate bonds formed by ligands (six monodentate NH₃ ligands).

Question 14

Give reason for each of the following: [3 Marks]

(i) For ferric hydroxide sol. the coagulating power of phosphate ion is more than chloride ion. [1 Mark]

Answer:
According to the Hardy-Schulze rule, greater the valency of the flocculating ion, higher is its coagulating power. Ferric hydroxide sol is positively charged, so it is coagulated by anions. Phosphate ion (PO₄^3-) carries a higher negative charge than chloride ion (Cl^-), hence its coagulating power is much greater.

Teacher's Note:
a) State the Hardy-Schulze rule clearly.
b) Link the higher ionic charge of the phosphate ion to its enhanced coagulating efficiency.

(ii) Lyophilic colloidal solutions are more stable than lyophobic colloidal solutions. [1 Mark]

Answer:
Lyophilic colloids have a strong affinity for the dispersion medium and are extensively solvated, forming a protective hydration sheath around the colloidal particles that prevents them from coagulating.

Teacher's Note:
a) Solvation provides thermodynamic stability to lyophilic colloids.
b) Lyophobic colloids lack this solvent affinity and rely solely on electrical charges for stability.

(iii) Gelatin is added to ice cream. [1 Mark]

Answer:
Gelatin acts as a protective lyophilic colloid and emulsifier that prevents the crystallization of ice, maintaining a smooth, soft texture in the ice cream.

Teacher's Note:
a) Gelatin stabilizes food emulsions and inhibits crystal growth.
b) It contributes to the desired creamy mouthfeel of frozen dairy products.