SOLUTIONS
2025 BOARD EXAM
CBSE CLASS 12-PCM CHEMISTRY Board Paper 2025 — Set 2
2025
CHEMISTRY
CLASS 12-PCM
CBSE EXAMINATION PAPER-2025
CHEMISTRY
(Solved)
General Instructions :
Read the following instructions carefully and follow them :
- This question paper contains 30 questions. All questions are compulsory.
- This question paper is divided into 5 sections.
- Section A – questions number 1 to 2 are case based questions
- Section B – questions number 3 to 17 are multiple choice questions
- Section C – questions number 18 to 22 are very short answer
- Section D – questions number 23 to 27 are short answer
- Section E – questions number 28 to 30 are long 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 spontaneous flow of the solvent through a semipermeable membrane from a pure solvent to a solution or from a dilute solution to a concentrated solution is called osmosis. The phenomenon of osmosis can be demonstrated by taking two eggs of the same size. In an egg, the membrane below the shell and around the egg material is semipermeable. The outer hard shell can be removed by putting the egg in dilute hydrochloric acid. After removing the hard shell, one egg is placed in distilled water and the other in a saturated salt solution. After some time, the egg placed in distilled water swells-up while the egg placed in salt solution shrinks. The external pressure applied to stop the osmosis is termed as osmotic pressure (a colligative property). Reverse osmosis takes place when the applied external pressure becomes larger than the osmotic pressure.
(1) Define reverse osmosis. Name one SPM which can be used in the process of reverse osmosis .
(2) Which one of the following will have higher osmotic pressure in1 M KCI or 1 M urea solution. Justify your answer.
(3) What do you expect to happen when red blood corpuscles (RBC's) are placed in 0.5% NaCl solution ?
(4) Why osmotic pressure is a colligative property ?
This means osmotic pressure increases with solute concentration, as explained in the case of eggs swelling or shrinking due to osmosis.
It is one of the properties like lowering of vapour pressure, elevation of boiling point, and depression of freezing point that depend only on solute quantity, not type.
Section B
The charge required for the reduction of 1 mol of Mn0⁻₄ to Mn0₂ is
The correct answer is 3F. The reduction of MnO₄⁻ to MnO₂ involves a change in the oxidation state of manganese from +7 in MnO₄⁻ to +4 in MnO₂, requiring the transfer of 3 electrons per Mn atom. This means that for one mole of MnO₄⁻, 3 moles of electrons are needed. Since 1 mole of electrons corresponds to one Faraday (F), the total charge required for the reduction of 1 mole of MnO₄⁻ to MnO₂ is 3F.
Which among the following is false statement?
The false statement is 'Molecularity of a reaction may be zero.' This is incorrect because molecularity refers to the number of reactant molecules involved in an elementary reaction, and it is always a positive integer (1 for unimolecular, 2 for bimolecular, etc.). It can never be zero, as that would imply that the reaction occurs without any reactant molecules, which is not possible in chemical reactions.
The number of molecules that react with each other in an elementary reaction is a measure of the:
The correct answer is 'molecularity of the reaction' because molecularity refers to the number of reacting species (atoms, ions, or molecules) that must collide simultaneously to initiate a chemical reaction, which is specifically defined for elementary reactions.
The element having [Ar]3d¹⁰4s¹ electronic configuration is:
The complex ions [Co(NH₃)₅(NO₂)]²⁺ and [Co(NH₃)₅(ONO)]²⁺ are called:
The correct option is 'Linkage isomers' because the two complex ions differ in how the nitrite ligand (NO₂) coordinates to the cobalt ion; it can bind either through the nitrogen atom or the oxygen atom, resulting in different linkage in the coordination complex.
Which is the correct IUPAC name for
The correct IUPAC name is 1-Chloro-4-Methylbenzene. This name indicates that there is a chlorine atom (−Cl) attached to the benzene ring at position 1 and a methyl group (−CH₃) attached at position 4. This naming aligns with IUPAC rules, where substituents are numbered to give the lowest locants, and alphabetical order is used to determine the priority when both substituents are at equal positions. Hence, "chloro" comes before "methyl" in the name.
What will be formed after oxidation reaction of secondary alcohol with chromic anhydride (CrO₃)?
The correct answer is 'Ketone' because secondary alcohols are oxidized to ketones when treated with chromic anhydride (CrO₃), as indicated in the provided context.
The conversion of phenol to salicylic acid can be accomplished by:
The correct answer is Kolbe's reaction.In this reaction, phenol is first treated with sodium hydroxide (NaOH) to form sodium phenoxide, which is then heated with carbon dioxide (CO₂) under pressure (around 125°C and 6–7 atm). This introduces a –COOH group at the ortho position of the aromatic ring, forming sodium salicylate, which upon acidification gives salicylic acid.
Which of the following is/are examples of denaturation of protein?
The correct options are Curdling of milk and Coagulation of egg white, which are examples of denaturation of proteins as explained in the context. Denaturation occurs when proteins lose their secondary and tertiary structures due to changes in temperature or pH, while the primary structure remains intact. The context specifically mentions curdling of milk due to lactic acid from bacteria and coagulation of egg white upon boiling as common examples.
Nucleotides are joined together by:
The correct answer is 'Phosphodiester linkage' because nucleotides are linked together via phosphodiester bonds that form between the 5' and 3' carbon atoms of the pentose sugar in the nucleotide structure. This linkage is essential in forming the backbone of nucleic acids such as DNA and RNA.
Scurvy is caused due to deficiency of:
Scurvy is caused by a deficiency of Vitamin C, also known as Ascorbic acid, which is highlighted in the context as leading to bleeding gums. The other options listed do not cause Scurvy.
Assertion (A) : In a first order reaction, if the concentration of the reactant is doubled, its half-life is also doubled.
Reason (R) : The half-life of a reaction does not depend upon the initial concentration of the reactant in a first order reaction.
Assertion (A) is false, but Reason (R) is true. The half-life of a first order reaction is independent of the initial concentration of the reactant, meaning that doubling the concentration does not affect the half-life.
Assertion (A) : In a first order reaction, if the concentration of the reactant is doubled, its half-life is also doubled
Reason (R) : The half-life of a reaction does not depend upon the initial concentration of the reactant in a first order reaction.
Assertion (A) is false, because in a first order reaction, the half-life is constant and does not change with the concentration of the reactant; it is independent of initial concentration. Reason (R) is true, as it accurately describes that the half-life for first order reactions remains the same regardless of changes in reactant concentration.
Assertion (A) : Aromatic primary amines cannot be prepared by Gabriel Phthalimide synthesis.
Reason (R) : Aryl halides do not undergo nucleophilic substitution reaction with the anion formed by phthalimide.
Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A). Aromatic primary amines cannot be synthesized through the Gabriel Phthalimide method because aryl halides do not react with the nucleophilic anion derived from phthalimide, thereby confirming Assertion (A) with Reason (R).
Assertion (A) : Vitamin D cannot be stored in our body.
Reason (R) : Vitamin D is fat soluble vitamin and is not excreted from the body in urine.
Assertion (A) is false, but Reason (R) is true. Vitamin D is a fat-soluble vitamin, meaning it can actually be stored in the body, primarily in the liver and adipose tissues, which contradicts Assertion (A). Reason (R) is true as fat-soluble vitamins are not readily excreted in urine, but it does not support the incorrect assertion.
Section C
The rate constant for a zero order reaction A→ P is 0.0030 mol L⁻¹s⁻¹. How long will it take for the initial concentration of A to fall from 0.10 M to 0.075 M ?
The integrated rate law for zero order reaction is: [A] = [A]0 - kt.
Given:
Initial concentration, [A]0 = 0.10 M
Final concentration, [A] = 0.075 M
Rate constant, k = 0.0030 mol L⁻¹ s⁻¹
Substitute values in the equation: 0.075 = 0.10 - 0.0030 * t
0.10 - 0.075 = 0.0030 * t
0.025 = 0.0030 * t
t = 0.025 / 0.0030 = 8.33 seconds.
Therefore, it will take 8.33 seconds for the concentration of A to fall from 0.10 M to 0.075 M.
The decomposition of NH₃ on platinum surface is zero order reaction. what are the rates of production of N₂ and H₂ if k = 2.5 x 10⁻⁴ mol L⁻¹ s⁻¹ ?
Step 1: Write the balanced reaction: 2 NH₃ → N₂ + 3 H₂.
Step 2: Rate of reaction (r) = k since it is zero order.
Step 3: The rate of disappearance of NH₃ = k = 2.5 x 10⁻⁴ mol L⁻¹ s⁻¹.
Step 4: Using stoichiometry, rate of production of N₂ = (1/2) * rate of NH₃ = (1/2)* 2.5 x 10⁻⁴ = 1.25 x 10⁻⁴ mol L⁻¹ s⁻¹.
Step 5: Rate of production of H₂ = (3/2) * rate of NH₃ = (3/2)* 2.5 x 10⁻⁴ = 3.75 x 10⁻⁴ mol L⁻¹ s⁻¹.
Answer: Rate of formation of N₂ is 1.25 x 10⁻⁴ mol L⁻¹ s⁻¹ and rate of formation of H₂ is 3.75 x 10⁻⁴ mol L⁻¹ s⁻¹.
Define the following terms :
(a) Pseudo first order reaction
(b) Half-life period of reaction(t½)
(b) Half-life period of reaction (t½): It is the time required for the concentration of a reactant to decrease to half of its initial value during the reaction.
Examine the following observations :
(a) Transition elements generally form coloured compounds.
(b) Zinc is not regarded as a transition element.
(b) Zinc is not a transition element because its 3d orbitals are fully filled (3d10) in both ground and common oxidation states. Since it does not have partially filled d orbitals, it does not exhibit typical properties of transition elements such as variable oxidation states and coloured compounds.
Name the following coordination compounds according to IUPAC norms :
(a) [Co(NH₃)₄(H₂O)Cl]Cl₂
(b) [CrC1₂(en)₂ ] Cl
(b) The complex ion is [CrCl₂(en)₂]+ with one chloride ion as counter ion. The ligands are ethane-1,2-diamine abbreviated as en and chloro. Alphabetically, chloro comes before ethane-1,2-diamine. The oxidation state of chromium is +3. So, the name is dichloridobis(ethane-1,2-diamine)chromium(III) chloride.
Section D
At 25 ⁰C the saturated vapour pressure of water is 24 mm Hg. Find the saturated vapour pressure of a 5% aqueous solution of urea at the same temperature. (Molar mass of urea = 60 g mol⁻¹)
Step 1: Calculate moles of urea = 5 / 60 = 0.0833 mol
Step 2: Calculate moles of water = 95 / 18 = 5.28 mol
Step 3: Calculate mole fraction of water, Xwater = moles of water / (moles of water + moles of urea) = 5.28 / (5.28 + 0.0833) = 0.9845
Step 4: Use Raoult's law: vapour pressure of solution = Xwater * vapour pressure of pure water = 0.9845 * 24 = 23.63 mm Hg
Answer: The saturated vapour pressure of 5% aqueous urea solution at 25 deg C is approximately 23.63 mm Hg.
The electrical resistance of a column of 0.05 M NaOH solution of area 0.8 cm² and length 40 cm is 5 x 10³ ohm. Calculate its resistivity, conductivity and molar conductivity.
(a) Resistivity (\rho):
Resistivity \rho = R * A / l = (5 x 103) * (0.8 * 10-4) / 0.4 = 1 ohm m.
(b) Conductivity (\kappa):
Conductivity \kappa = 1 / \rho = 1 / 1 = 1 S m-1.
(c) Molar conductivity (\Lambdam):
First, convert concentration to mol/m3: c = 0.05 mol L-1 = 0.05 * 1000 = 50 mol m-3.
Molar conductivity \Lambdam = \kappa / c = 1 / 50 = 0.02 S m2 mol-1 = 20 S cm2 mol-1 (since 1 m2 = 104 cm2).
Summary:
Resistivity = 1 ohm m, Conductivity = 1 S m-1, Molar conductivity = 20 S cm2 mol-1.
Using valence bond theory, explain the hybridization and magnetic character of the following :
(a) [Co(NH₃)₆]³⁺
(b) [Ni(CO)₄]
[At. no. : Co = 27, Ni = 28]
(b) In [Ni(CO)₄], Ni is in 0 oxidation state with 3d₈ 4s₂ electronic configuration. CO is a strong field ligand leading to pairing of electrons. Hybridization occurs involving one 3d, one 4s and two 4p orbitals resulting in dsp² hybridization giving a square planar geometry. It is diamagnetic because all electrons are paired.
(a) Define the following :
(i) Enantiomers
(ii) Racemic mixture
(b) Why is chlorobenzene resistant to nucleophilic substitution reaction ?
(a)(ii) A racemic mixture is an equimolar mixture of two enantiomers of a chiral compound which shows no optical activity because the rotations caused by each enantiomer cancel each other.
(b) Chlorobenzene is resistant to nucleophilic substitution because the lone pair of electrons on chlorine delocalises into the benzene ring through resonance, making the C-Cl bond stronger and less reactive. Also, the aromatic ring is electron-rich, repelling nucleophiles and stabilising the molecule, thus hindering the attack of nucleophiles and making nucleophilic substitution difficult.
Define the following terms :
(a) Glycosidic linkage
(b) Invert sugar
(c) Oligosaccharides
(b) Invert sugar is a mixture of glucose and fructose obtained by the hydrolysis of sucrose. When sucrose is treated with acid or enzyme invertase, it breaks down into equal parts of glucose and fructose. This sugar mixture is called invert sugar because it inverts the direction of rotation of plane-polarized light from dextrorotatory (sucrose) to levorotatory (invert sugar).
(c) Oligosaccharides are carbohydrates composed of 2 to 10 monosaccharide units linked together by glycosidic linkages. They are intermediate in size between monosaccharides and polysaccharides and are commonly found on the surface of cells, playing important roles in cell recognition and signaling.
Section E
(a) Give the IUPAC name of CH₃−CH=CH−CHO.
(b) Give a simple chemical test to distinguish between propanal and propanone.
(c) How will you convert the following :
(i) Toluene to benzoic acid
(ii) Ethanol to propan—2—ol
(iii) Propanal to 2—hydroxy propanoic acid
(b) To distinguish between propanal and propanone, perform the Tollen's test. Aldehydes like propanal react with Tollen's reagent to give a silver mirror on the inner surface of the test tube, while ketones like propanone do not give this test.
(c) (i) Toluene to benzoic acid: Oxidize toluene using KMnO4 in alkaline medium with heat. The methyl group is oxidized to the carboxylic acid group, forming benzoic acid.
(ii) Ethanol to propan-2-ol: First convert ethanol to bromoethane using PBr3, then carry out a Grignard reaction with formaldehyde followed by acidic hydrolysis to get propan-2-ol.
(iii) Propanal to 2-hydroxy propanoic acid: React propanal with HCN to form cyanohydrin, then hydrolyze the nitrile group (-CN) using dilute acid to get 2-hydroxy propanoic acid (lactic acid).
An organic compound 'A', molecular formula C₂H₆0 oxidises with Cr0₃ to form a compound 'B'. Compound 'B' on warming with iodine and aqueous solution of NaOH gives a yellow precipitate of compound 'C'. When compound 'A' is heated with conc. H₂S0₄ at 413 K gives a compound 'D', which on reaction with excess HI gives compound 'E'. Identify compounds 'A', 'B', 'C', 'D' and 'E' and write chemical equations involved.
(a) Compound A with molecular formula C2H6O is ethanol (CH3CH2OH).
(b) On oxidation with CrO3, ethanol (A) forms compound B which is ethanal (acetaldehyde, CH3CHO).
Equation: CH3CH2OH + [O] --CrO3--> CH3CHO + H2O
(c) Compound B (ethanal) on warming with iodine and NaOH gives a yellow precipitate (compound C) of iodoform (CHI3) indicating a methyl ketone or aldehyde with methyl group.
Equation: CH3CHO + 3I2 + 4NaOH --> CHI3 (yellow ppt) + HCOONa + 3NaI + 3H2O
(d) When compound A (ethanol) is heated with concentrated H2SO4 at 413 K, dehydration occurs producing compound D which is ethene (CH2=CH2).
Equation: CH3CH2OH --conc H2SO4, 413 K--> CH2=CH2 + H2O
(e) Compound D (ethene) on reaction with excess HI gives compound E which is ethyl iodide (C2H5I).
Equation: CH2=CH2 + HI --> CH3CH2I
Summary:
A - Ethanol (CH3CH2OH)
B - Ethanal (CH3CHO)
C - Iodoform (CHI3)
D - Ethene (CH2=CH2)
E - Ethyl iodide (CH3CH2I)
These reactions show typical oxidation, haloform test, dehydration and addition reaction of ethanol and its derivatives.
(a) Write chemical equations of the following reactions :
(i) Phenol is treated with conc. HNO3
(ii) Propene is treated with B H followed by oxidation by H202/OH-.
(iii) Sodium t-butoxide is treated with CH3 Cl.
(b) Give a simple chemical test to distinguish between butan—1—ol and butan—2—ol.
(c) Arrange the following in increasing order of acid strength : phenol, ethanol, water.
(i) When phenol is treated with concentrated nitric acid, nitration occurs mainly at the ortho and para positions forming a mixture of 2-nitrophenol and 4-nitrophenol. The reaction can be written as:
C6H5OH + HNO3 (conc.) -> o-nitrophenol + p-nitrophenol + H2O
(ii) Propene reacts with borane (B H3) to give an organoborane intermediate which on oxidation with hydrogen peroxide (H2O2) in alkaline medium (OH-) gives propanol. The hydroboration-oxidation reaction gives anti-Markovnikov addition of water.
CH3-CH=CH2 + B H3 -> organoborane intermediate
Organoborane + H2O2/OH- -> CH3-CH2-CH2OH
(iii) Sodium t-butoxide reacts with methyl chloride (CH3 Cl) via nucleophilic substitution to give t-butyl methyl ether.
((CH3)3CO-)Na+ + CH3Cl -> (CH3)3COCH3 + NaCl
(b)
To distinguish between butan-1-ol and butan-2-ol, use Lucas test. When treated with Lucas reagent (ZnCl2 in concentrated HCl), butan-2-ol reacts faster forming turbidity due to formation of alkyl chloride, while butan-1-ol reacts slowly. Thus, appearance of turbidity immediately indicates butan-2-ol, while delayed turbidity or no turbidity indicates butan-1-ol.
(c)
The increasing order of acid strength among phenol, ethanol and water is:
ethanol < water < phenol
This is because phenol has resonance stabilization of its phenoxide ion, making it more acidic than water and ethanol. Water is more acidic than ethanol due to better stabilization of its conjugate base.
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