SOLUTIONS
2025 BOARD EXAM
CBSE CLASS 12-PCB CHEMISTRY Board Paper 2025 — Set 3
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 26 questions. All questions are compulsory.
- This question paper is divided into 4 sections.
- Section A – questions number 1 to 3 are case based questions
- Section B – questions number 4 to 17 are multiple choice questions
- Section C – questions number 18 to 21 are very short answer
- Section D – questions number 22 to 26 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
According to the generally accepted definition of the ideal solution there are equal interaction forces acting between molecules belonging to the same or different species. (This is equivalent to the statement that the activity of the components equals the concentration.) Strictly speaking, this condition is fulfilled only in exceptional cases for mixtures (optical isomers, isotopic mixtures of an element, hydrocarbon mixtures). It is still usual to talk about ideal solutions as limiting cases in reality since very dilute solutions behave ideally with respect to the solvent. This view is further supported by the fact that Raoult’s law empirically found for describing the behaviour of the solvent in dilute solutions can be deduced thermodynamically via the assumption of ideal behaviour of the solvent.
Answer the following questions :
(1) Give one example of miscible liquid pair which shows negative deviation from Raoult’s law. What is the reason for such deviation ?
(2) Raoult’s law is a special case of Henry’s law. Comment.
(3) State Raoult’s law for a solution containing volatile components.
(4) Write two characteristics of an ideal solution.
Ribose and 2-deoxyribose have an important role in biology. Among the most important derivatives are those with phosphate groups attached at the 5 position. Mono-, di- and tri-phosphate forms are important, as well as 3-5 cyclic monophosphates. Purines and pyrimidines form an important class of compounds with ribose and deoxyribose. When these purine and pyrimidine derivatives are coupled to a ribose sugar, they are called nucleosides.
Answer the following questions :
(1) Differentiate between nucleotide and nucleoside.
(2) Mention two important functions of nucleic acid.
(3) What products would be formed when DNA is hydrolysed ? How is DNA different from RNA with reference to a structure ?
(4) Name the linkage which joins two nucleotides. Name the base that is found in nucleotide of RNA but not in DNA.
Section B
In an electrochemical cell, the following reaction takes place :
2Cu⁺(aq) + Zn(s) → 2Cu(s) + Zn²⁺(aq)
E°cell = 1·28 V
As the reaction progresses, what will happen to the overall voltage of the cell?
Out of Fe³⁺, Sc³⁺, Cr³⁺ and Co³⁺ ions, the one which is colourless in aqueous solution is:
[Atomic number : Fe = 26, Sc = 21, Cr = 24, Co = 27]
Hoffmann Bromamide degradation reaction is given by:
The value of Henry’s constant Kₕ is:
Out of the following statements, the incorrect statement is:
he incorrect statement is 'Lanthanoids are radioactive in nature.'
This is incorrect because most lanthanoids are not radioactive. Only a few elements like Promethium (Pm) exhibit radioactivity. The majority of lanthanoids are stable and non-radioactive. The context emphasizes the chemical and physical properties of lanthanoids, such as the lanthanoid contraction, which affects atomic and ionic radii—not radioactivity, which is a key feature of actinoids, not lanthanoids.
Out of 2-Bromobutane, 1-Bromobutane, 2-Bromopropane and 1-Bromopropane, the molecule which is chiral in nature is:
The product of the oxidation of I⁻ with MnO₄⁻ in alkaline medium is:
Polyhalogen compounds have wide application in industries and agriculture. DDT is also a very important polyhalogen compound. It is a:
What amount of electric charge is required for the reduction of 1 mole of MnO₄⁻ into Mn²⁺?
Alkenes are formed by heating alcohols with conc. H₂SO₄. The first step in the reaction is:
Assertion (A) : Cuprous salts are diamagnetic.
Reason (R) : Cuprous ion has completely filled 3d-orbitals.
Assertion (A) : n-Butyl chloride has higher boiling point than n-Butyl bromide.
Reason (R) : C – Cl bond is more polar than C – Br bond.
Assertion (A) : Acetanilide is less basic than aniline.
Reason (R) : Acetylation of aniline results in decrease of electron density on nitrogen.
Assertion (A) : Electrolysis of aqueous NaCl gives H₂ at cathode and Cl₂ at anode.
Reason (R) : Chlorine has higher oxidation potential than H₂O.
Section C
The reaction between H₂ (g) and I₂ (g) was carried out in a sealed isothermal container. The rate law for the reaction was found to be :
Rate =Rate = k[H₂][I₂]
If 1 mole of H₂ (g) was added to the reaction chamber and the temperature was kept constant, then predict the change in rate of the reaction and the rate constant.
The rate law is Rate = k[H₂][I₂]. If 1 mole of H₂ is added, the concentration of H₂ increases, so the rate of reaction will increase proportionally because rate depends directly on [H₂].
Since the temperature is constant in the isothermal container, the rate constant k does not change.
Therefore,
- Rate of reaction increases
- Rate constant k remains unchanged.
PtCl₄ . 2KCl doesn’t give precipitate of AgCl with AgNO₃ solution. Write the structural formula and IUPAC name of the complex.
IUPAC name: Potassium hexachloroplatinate(IV)
Explanation: PtCl4 . 2KCl is actually potassium hexachloroplatinate(IV) containing the complex ion [PtCl6]2-. Here, all chloride ions are coordinated to Pt(IV) as ligands and no free chloride ions exist. Therefore, no AgCl precipitate forms with AgNO3.
Define fuel cell. Give two advantages of fuel cell over ordinary cell.
Advantages of Fuel Cell over Ordinary Cell:
(1) Fuel cells can produce electricity continuously as long as fuel is supplied, whereas ordinary cells have limited stored energy.
(2) Fuel cells are more efficient and pollution-free compared to ordinary cells which may involve harmful chemicals and produce waste.
What is meant by essential amino acids ? Why are amino acid amphoteric in nature ?
(b) Amino acids are amphoteric in nature because they contain both acidic carboxyl group (-COOH) and basic amino group (-NH2) in the same molecule. Thus, they can react with both acids and bases, behaving as either acid or base.
Section D
Calculate the cell voltage of the voltaic cell which is set up by joining following half-cells at 25⁰C: Al/Al³⁺ (0·001 M) and Ni/Ni²⁺ (0·1 M)
Given : EᵒNi²⁺/ Ni = – 0·25 V, EᵒAl³⁺/Al = – 1·66 V
Oxidation (Anode): Al → Al3+ + 3e-, Eᵒ = –1.66 V
Reduction (Cathode): Ni2+ + 2e- → Ni, Eᵒ = –0.25 V
Step 2: Identify the anode and cathode. Since Al has more negative Eᵒ, it will be oxidized and Ni will be reduced.
Step 3: Calculate the standard cell potential, Eᵒcell = Eᵒcathode – Eᵒanode = (–0.25) – (–1.66) = 1.41 V
Step 4: Use the Nernst equation to calculate the cell voltage at given concentrations:
Ecell = Eᵒcell – (0.0591 / n) * log(Q), where n = number of electrons transferred, Q = reaction quotient.
Balanced equation: 2Al + 3Ni2+ → 2Al3+ + 3Ni, n = 6 electrons.
Q = [Al3+]2 / [Ni2+]3 = (0.001)2 / (0.1)3 = 1e–6 / 1e–3 = 0.001
Step 5: Calculate Ecell:
Ecell = 1.41 – (0.0591 / 6) * log(0.001) = 1.41 – 0.00985 * (–3) = 1.41 + 0.02955 = 1.4395 V
Final Answer: The cell voltage of the voltaic cell is approximately 1.44 V at 25 deg C.
Give explanation for each of the following observations :
(a) With the same d-orbital configuration (d4), Mn3+ ion is an oxidising agent whereas Cr2+ ion is a reducing agent.
(b) Actinoid contraction is greater from element to element than that among lanthanoids.
(c) Transition metals form large number of interstitial compounds with H, B, C and N.
(b) Actinoid contraction is greater than lanthanoid contraction because actinoids have poor shielding effect due to the involvement of 5f orbitals which penetrate closer to the nucleus. This results in a greater decrease in atomic and ionic sizes across the series compared to lanthanoids where 4f orbitals shield electrons more effectively.
(c) Transition metals have small atomic sizes and large surface areas with vacant d-orbitals. They can accommodate smaller atoms like H, B, C, and N in their interstitial spaces forming interstitial compounds. These compounds exhibit high hardness and stability due to strong metal-nonmetal bonds.
An aqueous solution of NaOH was made and its molar mass from the measurement of osmotic pressure at 27⁰C was found to be 25 g mol⁻¹. Calculate the percentage dissociation of NaOH in this solution.
[Atomic mass : Na = 23 u, O = 16 u, H = 1 u]
Arrange the following compounds as asked : 3
(a) in decreasing order of pKb values C₂H₅NH₂, (C₂H₅)₂NH, C₆H₅NHCH₃, C₆H₅NH₂
(b) increasing order of boiling point C₂H₅OH, C₂H₅NH₂, (CH₃)₂NH
(c) increasing order of solubility in water C₆H₅NH₂, (C₂H₅)₂NH, C₂H₅NH₂
Decreasing order of pKb (increasing basic strength) is: C₆H₅NH₂ > C₆H₅NHCH₃ > C₂H₅NH₂ > (C₂H₅)₂NH
(b) Boiling point depends on intermolecular forces like hydrogen bonding and molecular weight.
C₂H₅OH has strongest hydrogen bonding due to O-H group, so highest boiling point.
C₂H₅NH₂ and (CH₃)₂NH have N-H bonds and can hydrogen bond, but C₂H₅NH₂ has stronger hydrogen bonding due to two N-H bonds compared to one in (CH₃)₂NH.
Increasing order of boiling point is: (CH₃)₂NH < C₂H₅NH₂ < C₂H₅OH
(c) Solubility in water depends on the ability to form hydrogen bonds and polarity.
C₂H₅NH₂ is most soluble due to strong hydrogen bonding and smaller size.
(C₂H₅)₂NH is less soluble due to bulkier alkyl groups reducing hydrogen bonding.
C₆H₅NH₂ is least soluble due to hydrophobic phenyl ring.
Increasing order of solubility is: C₆H₅NH₂ < (C₂H₅)₂NH < C₂H₅NH₂
An aromatic compound ‘A’ with molecular formula C₈H₈O gives positive 2,4-DNP test. It gives yellow precipitate. of compound ‘B’ on treatment with sodium hypoiodite. Compound ‘A’ does not react with Tollen’s or Fehling’s reagent; on drastic oxidation with KMnO₄ it forms a carboxylic acid ‘C’. Elucidate the structures of A, B and C. Also give their IUPAC names.
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