SOLUTIONS
2025 BOARD EXAM
CBSE CLASS 12-PCB CHEMISTRY Board Paper 2025 — Set 5
2025
CHEMISTRY
CLASS 12-PCB
CBSE EXAMINATION PAPER-2025
CHEMISTRY
(Solved)
General Instructions :
Read the following instructions carefully and follow them :
- This question paper contains 24 questions. All questions are compulsory.
- This question paper is divided into 4 sections.
- Section A – questions number 1 to 2 are case based questions
- Section B – questions number 3 to 14 are multiple choice questions
- Section C – questions number 15 to 18 are very short answer
- Section D – questions number 19 to 24 are short answer
- There is no overall choice given in the question paper. However, an internal choice has been provided in few questions.
- Use of calculator is NOT allowed.
Section A
The a-amino acids are the building blocks of proteins. All a-amino acids exist as zwitter ion due to which they show amphoteric behaviour. All amino acids are joined through peptide bond. Proteins are broadly classified as globular proteins and fibrous proteins. Globular proteins are water soluble, whereas fibrous proteins are not. The complete structure of protein is discussed at four different levels i.e. primary, secondary, tertiary and quaternary structures. Protein loses its biological activity in denatured form.
(1) Define the following : (i) Peptide linkage (ii) Denatured protein
(2) Write the names of two different secondary structures of proteins.
(3) Why do amino acids show amphoteric behaviour ?
(4) How can you differentiate between Fibrous protein and Globular protein ?
Section B
According to Werner’s theory of coordination compounds:
The correct option is 'Primary valences are ionisable.' According to Werner’s theory, primary valences denote the ionisable part of the metal ion and are satisfied by negative ions, while secondary valences are non-ionisable.
Which of the following complex ion is not optically active?
A molecule is optically active if it cannot be superimposed on its mirror image. The trans isomer of [Co(en)₂Cl₂]⁺ has a plane of symmetry, meaning its mirror image can be superimposed on the original molecule, making it optically inactive.
Which of the following is the softest metal?
Among the given options (Fe, Zn, Cu, Sc), Zinc (Zn) is the softest metal. The context mentions that transition metals are generally very hard, with exceptions like Zn, Cd, and Hg, which do not display the typical hardness properties of transition elements. This indicates that zinc is comparatively softer than the other metals listed.
The freezing point of one molal KCl solution, assuming KCl to be completely dissociated in water, is: (Kf for water = 1·86 K kg mol⁻¹)
To calculate the depression in freezing point (ΔTf), we use the formula ΔTf = i * Kf * m, where i (van't Hoff factor for KCl) = 2 (since KCl dissociates into K+ and Cl-), Kf for water = 1.86 K kg mol⁻¹, and m = 1 molal. Thus, ΔTf = 2 * 1.86 * 1 = 3.72°C. The freezing point will decrease, so the final freezing point = 0°C - 3.72°C = -3.72°C.
A solution of acetone in ethanol:
The correct answer is 'shows a positive deviation from Raoult’s law.' This is because when acetone is added to ethanol, it disrupts the hydrogen bonding between ethanol molecules, leading to an increase in vapor pressure. This implies that the solution will exhibit positive deviation from Raoult's law, where the mixture's vapor pressure is higher than expected for an ideal solution.
Which of the following cell converts the energy of combustion of fuel into electrical energy?
The correct answer is 'Fuel cell' because, as stated in the context, fuel cells are designed specifically to convert the energy from the combustion of fuels like hydrogen, methane, and methanol directly into electrical energy.
The unit of rate and rate constant are same for a:
The correct option is 'Zero order reaction'. For a zero order reaction, the unit of rate (mol L⁻¹ s⁻¹) is the same as the unit of the rate constant (k), which is also expressed as mol L⁻¹ s⁻¹ according to the provided context.
Pyranose ring of glucose is formed due to the reaction between:
The correct option is C₁ and C₅. The pyranose ring structure of glucose, which is a six-membered ring, is formed when the hydroxyl group (-OH) on carbon atom 5 (C-5) reacts with the aldehyde group (C-1) to create a cyclic structure, leading to the absence of the aldehyde (-CHO) group in the cyclic form.
Assertion (A) : Actinoids show wide range of oxidation states.
Reason (R) : Actinoids are radioactive in nature .
Assertion (A) : Hydrolysis of an ester follows first order kinetics.
Reason (R) : The concentration of water does not get altered much during the reaction.
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). The hydrolysis of an ester like ethyl acetate is indeed a second order reaction generally, but behaves as pseudo first order due to the large excess of water, which means its concentration remains almost constant throughout the reaction. Therefore, while both statements are true, the reason provided does not directly explain why the assertion holds.
Assertion(A): Boiling point of (CH3)3N is higher than that of CH₃CH₂CH₂NH₂.
Reason (R) : Hydrogen bonding is more extensive in CH₃CH₂CH₂NH₂.
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (CH3)3N (trimethylamine) has steric hindrance that reduces the ability to form hydrogen bonds compared to CH₃CH₂CH₂NH₂ (propylamine), which has more hydrogen bonding due to its structure, resulting in a higher boiling point for propylamine despite the assertion.
Assertion (A) : Phenol is strongly acidic as compared to ethanol.
Reason (R) : Phenoxide ion is more stable than ethoxide ion.
The correct answer is 'Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A)'. Phenol is more acidic than ethanol due to the stability of the phenoxide ion, which is a result of the delocalization of the negative charge. In contrast, the negative charge in the ethoxide ion is localized, making it less stable. This greater stability of the phenoxide ion compared to the ethoxide ion explains why phenol is more acidic.
Section C
State Henry’s law. Why are aquatic species more comfortable in cold water as compared to warm water ?
Aquatic species are more comfortable in cold water because solubility of gases like oxygen increases as temperature decreases. So colder water holds more dissolved oxygen than warm water, which aquatic animals need to survive.
Write IUPAC names of the following coordination compounds :
(i) [CoCl₂(en)₂]SO₄
(ii) K₃[Fe(C₂O₄)₃]
(ii) The complex ion is [Fe(C2O4)3]3-, where C2O42- is the oxalato ligand, a bidentate ligand. There are three oxalato ligands attached to Fe. The cation is K+, with three potassium ions balancing the charge. The oxidation state of Fe is +3. The IUPAC name is 'Potassium trioxalatoiron(III)'.
Differentiate between :
(i) Double salt and Complex compound
(ii) Didentate ligand and Ambidentate ligand
Complex compound is formed by a central metal ion bonded to ligands forming a coordinate bond, and it does not dissociate into the original salts in solution.
(ii) Didentate ligand is a ligand that attaches to the central metal atom through two donor atoms.
Ambidentate ligand is a ligand that has two different donor atoms but attaches to the metal through only one donor atom at a time.
How do you explain the following ?
(a) Presence of an aldehydic group in glucose.
(b) Presence of five – OH groups in glucose.
(b) Glucose has five –OH groups, one on each carbon except the aldehyde carbon. This is confirmed by acetylation of glucose with acetic anhydride, which produces glucose pentaacetate, showing five hydroxyl groups were present to react.
Section D
Give reasons for the following :
(a) The pH of aqueous NaCl increases when it is electrolysed.
(b) Unlike dry cell, mercury cell has a constant cell potential through its lifetime.
(c) Conductivity of solution decreases with dilution.
(b) Mercury cell has a constant cell potential because it uses liquid mercury as one electrode which maintains constant concentration of Hg2+ ions, resulting in stable emf throughout its life.
(c) The conductivity decreases on dilution because the number of ions per unit volume decreases, so fewer charge carriers are present, reducing the ability of the solution to conduct electricity.
Answer the following about the complexes [FeF₆]³⁻ and [Fe(CN)₆]⁴⁻ :
(i) Write the hybridization involved in each case.
(ii) Which of them is the outer orbital complex and which one is the inner orbital complex ?
(iii) Compare their magnetic behaviour.
[Atomic number : Fe = 26]
(ii) [FeF6]3- is an outer orbital complex because it involves sp3d2 hybridization using 4s and 4p orbitals and 4d orbital. [Fe(CN)6]4- is an inner orbital complex as it involves d2sp3 hybridization using 3d orbitals of Fe.
(iii) Magnetic behaviour: [FeF6]3- has five unpaired electrons (high spin), so it is paramagnetic with high magnetic moment. [Fe(CN)6]4- has all electrons paired (low spin), so it is diamagnetic with no magnetic moment. Thus, [FeF6]3- is paramagnetic and [Fe(CN)6]4- is diamagnetic.
(i) What happens to the colour of complex [Ti(H₂O)₆]³⁺ when heated gradually ?
(ii) Write the electronic configuration for d⁵ ion if Δₒ < P.
(iii) Write the hybridization and magnetic behaviour of the complex [Ni(CO)₄].
[Atomic number : Ni = 28]
(b) For a d5 ion with crystal field splitting energy (Δo) less than pairing energy (P), the configuration will be high spin. Therefore, the electronic configuration is t2g3 eg2 with five unpaired electrons.
(c) In the complex [Ni(CO)4], nickel has oxidation state zero and electronic configuration of Ni is 3d8 4s2. The hybridization is dsp2 resulting in a tetrahedral geometry. Since CO is a strong field ligand, it causes pairing of electrons, making the complex diamagnetic (all electrons are paired).
A compound (A) with molecular formula C₄H₅N on reduction with DIBAL-H followed by hydrolysis, gives a compound (B). Compound (B) gives positive Tollens’ test but does not give iodoform test. Compound (B) can also be obtained when ethanal is treated with dilute NaOH followed by heating. Identify (A) and (B). Write the reactions of (A) with DIBAL-H followed by hydrolysis.
CH3CN + DIBAL-H (at low temperature) → intermediate imine aluminum complex
Intermediate + H2O (hydrolysis) → CH3CHO (acetaldehyde)
Thus, reduction of nitrile (A) with DIBAL-H followed by hydrolysis produces the aldehyde (B).
How will you obtain the following from aniline ? Give chemical equations only.
(a) Sulphanilic acid
(b) Phenylisocyanide
(c) Acetanilide
C6H5NH2 + H2SO4 (conc.) \u2192 C6H5NH3+ HSO4- (anilinium hydrogensulphate)
C6H5NH3+ HSO4- \u2192 C6H4SO3HNH2 (Sulphanilic acid) (heating at 453-473 K)
(b) Phenylisocyanide:
C6H5NH2 + CHCl3 + KOH \u2192 C6H5NC + KCl + 2H2O
(c) Acetanilide:
C6H5NH2 + (CH3CO)2O \u2192 C6H5NHCOCH3 + CH3COOH
Vapour pressure of pure water at 298 K is 24·8 mm Hg. Calculate the lowering in vapour pressure of an aqueous solution which freezes at– 0·3°C. (Kf of water = 1·86 K kg mol⁻¹)
Freezing point depression, ΔTf = 0.3 deg C = 0.3 K
Freezing point depression constant, Kf = 1.86 K kg mol-1
Step 1: Calculate molality (m) of the solution
The formula for freezing point depression is ΔTf = Kf * m
So, m = ΔTf / Kf = 0.3 / 1.86 = 0.1613 mol kg-1
Step 2: Calculate mole fraction of solute (X2)
Let the solvent be water with molality m means m moles of solute per 1 kg of solvent.
Number of moles of solvent (water) in 1 kg = 1000 g / 18 g mol-1 = 55.56 mol
Mole fraction of solute,
X2 = m / (m + moles of solvent) = 0.1613 / (0.1613 + 55.56) = 0.0029
Step 3: Calculate lowering in vapour pressure (ΔP)
Lowering in vapour pressure is given by Raoult's law,
ΔP = P0 * X2 = 24.8 * 0.0029 = 0.072 mm Hg
Answer: The lowering in vapour pressure of the solution is 0.072 mm Hg.
Other Years — CHEMISTRY