Kerala Plus Two Chemistry Question Paper March 2024 with Answers

Reviewing Kerala Syllabus Plus Two Chemistry Previous Year Question Papers and Answers Pdf March 2024 helps in understanding answer patterns.

Kerala Plus Two Chemistry Previous Year Question Paper March 2024

Time: 2 Hours
Total Scores: 60

Answer any 4 questions from 1 to 5. Each carries 1 score. (4 × 1 = 4)

Question 1.
An electrochemical cell can behave like an electrolytic cell when ______________
(a) E_cell = 0
(b) E_cell > E_ext
(c) E_ext > E_cell
(d) E_cell = E_ext
Answer:
(c) E_ext > E_cell

Question 2.
Which of the following is the unit of rate constant for a first-order reaction?
(a) L mol^-1 s^-1
(b) s^-1
(c) L^-1 mol s^-1
(d) L² mol^-2 s^-1
Answer:
(b) s^-1

Question 3.
Coordination number of copper in [Cu(CN)₄]^3- is ______________
Answer:
Co-ordination number = 4, there are four cyanide ligands.

Question 4.
Write the name of the poisonous gas formed when chloroform is oxidized by air in the presence of light.
Answer:
Phosgene OR, Carbonyl Chloride OR, COCl₂

Question 5.
Name the linkage between two monosaccharide units in a disaccharide.
Answer:
Glycosidic linkage OR, C-O-C linkage.

Answer any 8 questions from 6 to 15. Each carries 2 scores. (8 × 2 = 16)

Question 6.
Write any two applications of Henry’s law.
Answer:
Applications of Henry’s law:

• In the preparation of soda water and Soft drinks.
• A medical condition known as Bends in Scuba divers.
• A medical condition known as Anoxia in people living at high altitudes or climbers.

Question 7.
Calculate the standard emf of the cell in which the following reaction takes place.
Zn(s) + Cu²⁺ (aq) → Zn²⁺ (aq) + Cu(s)
(E°CU²⁺ / Cu = 0.34 V & E°Zn²⁺ / Zn = -0.76V)
Answer:

Question 8.
Write any two differences between order and molecularity.
Answer:

Order	Molecularity
1. It is the sum of the powers of the concentration terms in the rate law expression.	1. It is the total number of reactant species that collide simultaneously in a chemical reaction.
2. It is an experimental quantity.	2. It is a theoretical quantity.
3. It can be zero or fractional.	3. It cannot be zero or fractional.
4. It applies to both elementary and complex reactions.	4. It applies only to elementary reactions.

Question 9.
What is the effect of temperature on the rate constant of a reaction? Write the equation used to determine the effect of temperature on the rate constant.
Answer:
When temperature increases, the rate constant of a reaction also increases; when the temperature is increased by 10°, the rate constant is nearly doubled.
k = \(\mathrm{A} \cdot \mathrm{e}^{-\mathrm{Ea} / \mathrm{RT}}\)
ln k = ln A – \(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}}\)
log k = log A – \(\frac{E_a}{2.303 R T}\)

Question 10.
Identify the products X and Y formed in the following reactions:
(i) CH₃CH₂OH + PCl₅ → X + POCl₃ + HCl (1)
(ii) CH₃ – Br + AgF → Y + AgBr (1)
Answer:
(i) X is CH₃-CH₂-Cl or, Chloroethane or, Ethyl chloride.
(ii) Y is CH₃-F or, Fluoromethane or, Methyl fluoride.

Question 11.
The reaction between free-butyl bromide and hydroxide ion yields tert-butyl alcohol following the SN1 mechanism. Write the mechanism.
Answer:
\(S_N{ }^1\) reaction between tert-butyl bromide and hydroxide ion occurs in two steps:
Step-I: The polarised C-Br bond undergoes slow cleavage to produce a carbocation (tert-butyl carbocation) and a bromide ion.

Step-II: The carbocation (tert-butyl carbocation) is then attacked by the nucleophile to form the product.

Question 12.
Write the name and statement of the law that helps to identify the major product in the β-elimination reactions of haloalkanes.
Answer:
Zaitsev rule [Saytzeff rule]:
The rule states that in dehydrohalogenation reactions, the major product is alkene, which has a greater number of alkyl groups attached to the doubly bonded carbon atoms.

Question 13.
Give a reason for the solubility of alcohols in water.
Answer:
Intermolecular hydrogen bonds, H bonds, or, alcohol can form intermolecular hydrogen bonds with water molecules. So they are soluble in water.

Question 14.
(i) What is Tollens’ reagent? (1)
(ii) Which among CH₃CHO and CH₃COCH₃ form a silver mirror on reaction with Tollens’ reagent? (1)
Answer:
(i) Tollen’s reagent is a freshly prepared ammoniacal silver nitrate (AgNO₃) solution or, [Ag(NH₃)₂]⁺OH^–
(ii) CH₃CHO

Question 15.
Among CH₃NH₂ and C₆H₅NH₂, which is more basic? Give reason.
Answer:
CH₃NH₂
Due to the electron releasing inductive (+1) effect of the -CH₃ group, or, less availability of lone pair of electrons on the nitrogen atom of aniline for protonation (resonance effect) results in a more basic nature of amine.

• due to the electron-withdrawing (-1) inductive effect of the phenyl group.
• due to the +R effect of the NH₂ group in aniline.
• due to the less stability of anilinium ion compared to aniline.

Answer any 8 questions from 16 to 26. Each carries 3 scores. (8 × 3 = 24)

Question 16.
Define ideal solutions by citing a suitable example. What are the values of ∆_mixH and ∆_mixV for such a solution?
Answer:
Ideal solutions are solutions that obey Raoult’s law at all concentrations.
E.g., a Mixture of n-hexane and n-heptane, a mixture of bromoethane and chloroethane, a mixture of benzene and toluene, etc…
For ideal solutions, ∆H_mix = 0 and ∆V_mix = 0.

Question 17.
(i) Define the molar conductivity of a solution. How does it vary with concentration? (2)
(ii) State the law that helps to determine the limiting molar conductivity of electrolytes. (1)
Answer:
(i) Molar conductivity of a solution at a given concentration is the conductance of the ‘V’ volume of a solution containing one mole of electrolyte kept between two electrodes with an area of cross-section A and distance of unit length.
or
Molar conductivity is the conductivity of a solution containing 1 mol of an electrolyte, placed in a conductivity cell having a unit distance between the electrodes.
Molar conductivity, ∧_m = k.V [k is the conductivity]
∧_m = k/c [c is the concentration]
Molar conductivity ∧_m = \(\frac{1000 \mathrm{k}}{\mathrm{M}}\) [M is the molarity of the solution]
The molar conductivity decreases with concentration (increases with dilution) for both strong and weak electrolytes.

(ii) The law that helps to determine the limiting molar conductivity of weak electrolytes is Kohirausch’s law. It states that the limiting molar conductivity of an electrolyte is the sum of the individual contributions of the anion and the cation of the electrolyte.

Question 18.
(i) What is meant by the half-life of a reaction? (1)
(ii) A first-order reaction is found to have a rate constant, k = 5.5 × 10^-14 s^-1. Find the half-life of the reaction. (2)
Answer:
(i) It is the time taken to reduce the concentration of a reactant to one-half of its initial concentration.
(ii) Here k = 5.5 × 10^-14 s^-1
\(t_{1 / 2}=\frac{0.693}{k}\)
k = \(\frac{0.693}{5.5 \times 10^{-14}}\) = 1.26 × 10¹³ s

Question 19.
(i) Some transition metal ions are given in the box below. Choose the ions that are colored. (2)
(Z for Sc, Ti, and Cr are 21, 22, and 24 respectively)
Sc³⁺, Ti⁴⁺, Ti³⁺, Cr³⁺
(ii) Give a reason for the formation of colored ions by transition metals. (1)
Answer:
(i) Coloured ions are Ti³⁺ and Cr³⁺
[Electronic configurations: Sc³⁺ – [Ar] 3d⁰, Ti⁴⁺ – [Ar] 3d⁰, Ti³⁺ – [Ar] 3d¹ and [Cr³⁺] – 3d³]
(ii) Colour is due to the d-d transition of electrons or due to the presence of partially filled d-orbitals. when light falls on a transition metal ion, electrons from lower d-orbitals are excited to higher d-levels. The energy required for this transition is absorbed from visible light, and hence, the metal ion exhibits the complementary color of the light absorbed.

Question 20.
What is lanthanoid contraction? What are the consequences of lanthanoid contraction?
Answer:
The regular decrease in the atomic and ionic radii along the lanthanoid series is known as lanthanoid contraction.
Consequences of lanthanoid contraction are:

• The 2nd and 3rd row transition series elements have similar radii [zirconium (Zr) and hafnium (Hf) have almost similar radii].
• Lanthanoids have similar physical properties/They occur together in nature/Their isolation is difficult.
• The basicity of hydroxides of lanthanoids decreases from lanthanum to lutetium. La(OH)₃ is more basiothan Lu(OH)₃.

Question 21.
Draw the hybridization scheme of [Co(NH₃)₆]³⁺ based on the Valence Bond Theory. Predict the geometry and magnetic behavior of the complex.
Answer:
In [Co(NH₃)₆]³⁺, the central atom cobalt is in a +3 oxidation state.
Co3+ – [Ar] 3d⁶ 4s⁰ 4p⁰
Orbitals of Co³⁺ ion

In the presence of the strong ligands NH₃, the electrons in the 3d level get paired. Now, two 3d orbitals, one 4s orbital, and three 4p orbitals undergo d²sp³ hybridization to form 6 new orbitals. Thus the shape of the molecule is octahedral. Due to the absence of unpaired electrons, the complex is diamagnetic.

Question 22.
(i) Write the formulae of the following coordination compounds:
(a) Pentaamminechloridocobalt(III) chloride (1)
(b) Potassiumhexacyanidoferrate(III) (1)
(ii) Which of the above is a heteroleptic complex? (1)
Answer:
(i) (a) [CO(NH₃)₅Cl] Cl₂
(b) K₃[Fe(CN₆)]
(ii) (a) [Co(NH₃)₅Cl] Cl₂ is the heteroleptic complex

Question 23.
(i) Identify the products P & Q in the following reaction:

(ii) What is the product obtained when phenol is treated with concentrated nitric acid? (1)
Answer:
(i) P is o-nitrophenol or, 2-nitrophenol
Or

Q is p-Nitrophenol or, 4-Nitrophenol,

(ii) 2,4,6-Trinitrophenol or, Picric acid
Or

Question 24.
(i) An organic compound A n reaction with CrO₂Cl₂ in CS₂ followed by acidification gives benzaldehyde as a product. Identify the compound A and also name the reaction. (2)
(ii) What is the product obtained when the above organic compound A undergoes side-chain oxidation with acidic potassium permanganate? (1)
Answer:
(i) Toluene or, Methyl Benzene or, The reaction is Etard reaction.

(ii) Benzoic acid or, Benzenecarboxylic acid or, C₆H₅-COOH OR

Question 25.
Describe the Hinsberg test to distinguish primary, secondary, and tertiary amines.
Answer:
Hinsberg Test: Benzenesulphonyl chloride (C₆H₅-SO₂Cl) is known as the Hinsberg reagent, which is used to distinguish the three types of amines. Primary amines react with the Hinsberg reagent to form a precipitate of N-alkylbenzene sulphonamide, which is soluble in alkali. Secondary amines react with Hinsberg reagent to give a precipitate of N, N-dialkylbenzenesulphonamide, which is insoluble in alkali. Tertiary amines do not react with the Hinsberg reagent or, chemical equations.

Question 26.
Write the classification of proteins based on their molecular shape by giving suitable examples.
Answer:
Based on molecular shape, proteins are classified into two – Fibrous protein and Globular protein.
Fibrous Proteins: They have fiber-like structures/They contain linear polypeptide chains/They are generally insoluble in water.
E.g. Keratin or myosin.
Globular proteins: They have a spherical shape/They contain polypeptide chains that are coiled around to give a spherical shape/ They are soluble in water.
E.g. Insulin or albumin.

Answer any 4 questions from 27 to 31. Each carries 4 scores. (4 × 4 = 16)

Question 27.
(i) What are colligative properties? (1)
(ii) The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute is dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. K_b for benzene is 2.53 K kg mol^-1. (3)
Answer:
(i) These are properties of dilute solutions, which depend only on the relative number of solute particles and not on their nature. OR, these are properties that depend on the number of solute particles irrespective of their nature relative to the total number of particles present in the solution.
(ii) Here w₂ = 1.80 g, w₁ = 90 g
T_b = 354.11 K, \(\mathrm{T}_{\mathrm{b}}^0\) = 353.23 K and K_b = 2.53 K kg/mol.
ΔT_b = T_b – \(\mathrm{T}_{\mathrm{b}}^0\)
= 354.11 – 353.23
= 0.88 K
Molar mass, M₂ = \(\frac{1000 \mathrm{k}_{\mathrm{b}} \mathrm{w}_2}{\mathrm{w}_1 \Delta \mathrm{~T}_{\mathrm{b}}}\)
= \(\frac{1000 \times 2.53 \times 1.80}{90 \times 0.88}\)
= 57.5 g/mol

Question 28.
(i) Sketch the diagram of a H₂ – O₂ fuel cell. (1)
(ii) Write the chemical equations for electrode reactions in it. (2)
(iii) Write any two advantages of a fuel cell. (1)
Answer:
(i)

(ii) Anode reaction:
2H₂ (g) + 4OH^– (aq) → 4H₂O (l) + 4e^–
Cathode reaction:
O₂ (g) + 2H₂O (l) + 4e^– → 4OH^– (aq)
Net Reaction:
2H₂ (g) + O₂ (g) → 2H₂O (l)

(iii) Advantages of fuel cells are:

• It is highly efficient.
• It is eco-friendly (pollution-free).
• The cell works continuously as long as the reactants are supplied.
• Water obtained from the H₂ – O₂ fuel cell can be used for drinking.

Question 29.
Explain the different types of structural isomerism in coordination compounds with the help of suitable examples.
Answer:
Coordination compounds show four types of structural isomerism – Ionisation isomerism, Linkage isomerism, Coordination isomerism, and Solvate isomerism.
(i) Ionisation Isomerism: It arises due to the exchange of ions between the inside and outside of the coordination sphere/This type of isomerism arises when the counter ion in a complex can act as a ligand.
OR
E.g. [Co(NH₃)₅SO₄]Br and [Co(NH₃)₅Br]SO₄.
(ii) Linkage Isomerism: It arises in a coordination compound containing an ambidentate ligand, which can bond to the central atom through more than one donor atom.
OR
E.g. [Co(NH₃)₅(ONO)]Cl₂ and [Co(NH₃)₅(NO₂)]Cl₂
(iii) Coordination Isomerism: It arises due to the interchange of ligands between cationic and anionic coordination entities.
OR
E.g. [Co(NH₃)₆][Cr(CN)₆] and [Cr(NH₃)₆][Co(CN)₆]
(iv) Solvate Isomerism (Hydrate isomerism): It arises due to the difference in the no. of solvent molecules (water molecules) bonded to the metal ion as ligand.
OR
E.g. [Cr(H₂O)₆]Cl₃ and [Cr(H₂O)₅Cl]Cl₂.H₂O

Question 30.
(i) Describe the manufacture of ethanol from molasses. (2)
(ii) What is meant by denaturation of alcohol? (1)
(iii) Identify the product obtained when ethanol is treated with Conc. H₂SO₄ at 443 K. (1)
Answer:
(i) Ethanol is manufactured by the fermentation of molasses. The sugar in molasses is converted to glucose and fructose in the presence of an enzyme, invertase. Glucose and fructose are converted to ethanol and carbon dioxide by another enzyme, zymase. OR, the equations:

(ii) Commercial alcohol is made unfit for drinking by mixing it with copper sulphate and pyridine, This is known as denaturation of alcohol.
(iii) CH₂ = CH₂ OR, Ethene OR, Ethylene.

Question 31.
Describe the following reactions:
(i) Cannizaro reaction. (2)
(ii) Stephen reaction (2)
Answer:
(i) Cannizzaro Reaction: Aldehydes having no α-hydrogen atom, when treated with cone, alkali, undergo self-oxidation and reduction (disproportionation) to form alcohol and carboxylic acid salt. This reaction is called the Cannizzaro reaction.

(ii) Stephen Reaction: Nitriles are reduced to imine with stannous chloride in the presence of HCl, which on hydrolysis gives aldehyde. This reaction is known as Stephen’s reaction.