CBSE Chemistry Subjective Questions Answer 2025
SECTION: B
17. (A) The rate constant of 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 0.10M to 0.075M?2
For zero order reaction,
k = 1/t [ [A0] − [A] ]
[A]0 = Initial concentration = 0.10M
[A] = concentration after time t = 0.075M
k = 0.0030 mol L−1s−1
Putting these values in the above equation, we get-
0.0030 = 1/t [0.10 − 0.075]
or, t = 0.025/0.0030 = 8.33 seconds.
17. (B) The decomposition of NH3 on platinum surface is a zero order reaction. What would be the rate of production of N2 and H2 if k = 2.5 × 10−4mol L−1s−1?2
2NH3 → N2 + 3H2
− d[NH3]/2dt = d[N2]/dt = d[H2]/3dt
d[NH3]/dt = rate = k × [NH3]0 = 2.5 × 10−4 mol L−1s−1
d[N2]/dt = − d[NH3]/2dt
= 1/2 × 2.5 × 10−4 mol L−1s−1
d[H2] = − 3d[NH3]/2dt
= 3/2 × 2.5 × 10−4
Rate of production of H2 = 7.5 × 10−4 mol L−1 s−1
= 3.75 × 10−4 mol L−1s−1
Rate= −d[NH3]/dt = k × [NH3]0
= 2.5 × 10−4 mol L−1s−1
Rate of production of N2 = 2.5 × 10−4 mol L−1s−1
18. Define the following terms
a. Pseudo-First Order Reaction
b. Half Life Period of a Reaction(t1/2)2 x 1
Pseudo-First Order Reaction: A reaction that appears to be first order, but is actually second order or higher is a pseudo-first order reaction. This happens when one reactant is in excess and its concentration remains relatively constant during the reaction. Example:
CH3COOC2H5 + H2O → CH3COOH + C2H5OH
Half Life Period of a Reaction(t1/2): The time during which the initial concentration of reactant reduce to half is called half life period of the reaction.
For zero order reaction, the half life perriod is
t1/2 = [R]/2K and
For 1st order reaction, the half life perriod is
t1/2 = 0.693/k
19. Explain the following observation
a. Transition elements generally forms colored compounds
b. Zn is not regarded as transition element2 x 1
a. Transition elements generally forms colored compounds because they have partially filled d-orbitals, which allows for electronic transitions between these orbitals when exposed to visible light, absorbing specific wavelengths and reflecting the complementary color.
b. Zinc is not considered a transition element because it has completely filled d-orbital (3d10) in its ground state and oxidation states.
20. Name the following coordination compounds according to IUPAC norms
a. [Co(NH3)4(H2O)Cl]Cl2
b. [CrCl2(en)2]Cl2 x 1
a. Tetraammineaquachlorocobalt(III) chloride
b. Dichlorobis(ethylenediamine)chromium(III) chloride
21. (a) In the following pair of halogen compounds, which compound undergoes SN1 reaction faster and why ? 1
2-methyl-2-chloropropane undergoes an SN1 reaction faster than 3-chloropentane because it forms a more stable tertiary carbocation compared to the secondary carbocation formed by 3-chloropentane as we know that a more stable carbocation leads to a faster SN1 reaction rate.
21. (b) Arrangen the following compounds in increasing order of their reactivity towards SN2 displacement.
2-bromo 2-methyl butane, 1-bromopentane, 2-bromopentane 1
2-bromo 2-methyl butane < 2-bromopentane < 1-bromopentane
SECTION: C
22.
23.
24. Complete and balance the following chemical equations3 x 1
25. Using the valence bond theory, explain the hybridization and magnetic character of the following:2 x 1.5
(a) Co[(NH3)6]3+
(b) Ni[(CO)4]
[At. No. Co = 27, Ni = 28]
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26. (a) Define the following:2 + 1
(i) Enantiomers
(ii) Racemic Mixture
Enantiomers: Enantiomers are a pair of molecules that exist in two forms that are mirror images of one another but cannot be superimposed one upon the other. Enantiomers are chemically identical.
Racemic Mixture: Racemic mixture is a mixture of equal quantities (50:50) of two enantiomers, or substances that have dissymmetric molecular structures that are mirror images of one another.
26. (b) Why is chlorobenzene resist to nucleophilic substitution reaction ?
Chlorobenzene is resistant to nucleophilic substitution reactions because of resonance and the electronegativity of the carbon atom.
27. (A) Explain the following reactions and write chemical equation involved3 x 1
(a) Wolf-Kishner Reduction
(b) Etard Reaction
(c) Cannizzaro Reaction
Wolff-Kishner Reduction: Wolff-Kishner reduction is used to convert carbonyl group into methylene groups by reacting with hydrazine followed by heating with sodium or potassium hydroxide in high boiling solvent such as ethylene glycol.
Etard Reaction: Toluene oxidizes into benzaldehyde by chromyl chloride followed by hydrolysis. This reaction is called Etard reaction.
Cannizzaro Reaction: Aldehydes which do not have an α-hydrogen atom, undergo self oxidation and reduction (disproportionation) reaction on heating with concentrated alkali. In this reaction, one molecule of the aldehyde is reduced to alcohol while another is oxidised to carboxylic acid salt.
27. (B) Write the structure of A, B and C in the following sequence of reactions:2 x 1.5
28. Define the following terms:3 x 1
(a) Glycosidic linkage
(b) Invert Sugar
(c) Oligosaccharides
Glycosidic linkage: Glycosidic linkage is a type of covalent bond that connects a carbohydrate molecule to another carbohydrate molecule.
Invert Sugar: An equimolar mixture of glucose and fructose are obtained from sucrose by hydrolysis in the presence of an acid such as dil. HCI or the enzyme invertase or sucrase, is called invert sugar.
Oligosaccharides: Oligosaccharides are carbohydrates composed of 2 to 10 monosaccharide units linked together by glycosidic bonds.
SECTION: D
29. (b) (i) What do you expect to happen when red blood corpuscles (RBS's) are placed in 0.5% NaCl solution ? 1
When red blood corpuscles (RBC's) are placed in a 0.5% NaCl solution, they will swell and burst due to osmosis, as the solution is considered hypotonic to the cell.
29. (b) (ii) Which one of the following will have higher osmotic pressure in 1M KCl or 1M urea solution. Justify your answer. 1
1M KCl solution will have a higher osmotic pressure than a 1M urea solution because KCl dissociates into two ions (K+ and Cl-) when dissolved in water, while urea does not dissociate and only contributes one particle per molecule. We know that osmotic pressure depends on the number of solute particles, so the KCl solution will exert a higher osmotic pressure.
29. (c) Why osmotic pressure is colligative property ? 1
Osmotic pressure is a colligative property because it depends on the number of solute particles in a solution, rather than the identity of the solute.
30. (b) Why pKb of aniline is more than that of methylamine ? 1
The pKb of aniline is higher than that of methylamine because aniline is a weaker base than methylamine due to the delocalization of electrons in aniline, which reduces the availability of electrons to donate.
30. (c) (i) Arrange the following in the increasing order of their basic character in an aquesous solution
(CH3)3N, (CH3)2NH, NH3, CH3NH2 1
In aqueous solution, the order of basicity is (CH3)2NH > CH3NH2 > (CH3)3N > NH3.
30. (c) (ii) Why ammonolysis of alkyl halides is not a good method to prepare to pure amines ? 1
Ammonolysis of alkyl halides is not a good method to prepare pure amines because it produces a mixture of primary, secondary, and tertiary amines, along with quaternary ammonium salts, making it difficult to isolate a single pure amine product.
SECTION: E
31. (A) (a) Give IUPAC name of CH3–CH=CH–CHO 1
(b) Give a simple chemical test to distinguish between propanal and propanone. 1
(c) How will you convert the following 3 x 1
(i) Toluene to benzoic acid
(ii) Ethanol to propan-2-ol
(iii) Propanal to 2-hydroxypropanoic acid
(a) IUPAC name of CH3–CH=CH–CHO is but-2-enal.
(b) Tollen's test is a simple chemical test that can distinguish between propanal and propanone.
(c) (i) Toluene is oxidize to benzoic acid by a strong oxidizing agent like potassium permanganate (KMnO4)
(ii) Ethanol is oxidize to ethanal, then react with a methyl magnesium bromide followed by hydrolysis gives propan-2-ol.
(iii) Propanal reacts with hydrogen cyanide followed by hydrolysis yields 2-hydroxypropanoic acid.
31. (B) Complete each synthesis by giving missing starting material, reagent or products: 5 x 1
32.
33. (A) An organic compound 'A', molecular formula C2H6O oxidises with CrO3 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. H2SO4 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. 5
Compound A (C2H6O) oxidizes with CrO3 to form compound B.
C2H6O may be either ethanol (CH3CH2OH) or methoxymethane (CH3OCH3).
Since 'B' gives a positive iodoform test, 'A' must be a secondary alcohol or a primary alcohol with a CH3CH(OH)- or CH3C(O)- group.
Therefore, 'A' must be ethanol (CH3CH2OH).
Ethanol oxidizes with CrO3 to form ethanal (CH3CHO). So, 'B' is ethanal.
Compound B (CH3CHO) on warming with iodine and aqueous NaOH gives a yellow precipitate of compound C.
This is iodoform test, which is positive for methyl ketones and secondary alcohols or primary alcohols containing CH3CH(OH)- group.
The yellow precipitate is iodoform (CHI3). So, 'C' is iodoform.
Compound A (CH3CH2OH) heated with conc. H2SO4 at 413 K gives compound D.
Ethanol undergoes dehydration in the presence of conc. H2SO4 at 413 K to form an ether.
2CH3CH2OH → CH3CH2OCH2CH3 + H2O
So, 'D' is diethyl ether (CH3CH2OCH2CH3).
Compound D (CH3CH2OCH2CH3) on reaction with excess HI gives compound E.
Ethers react with excess HI to form alkyl halides.
CH3CH2OCH2CH3 + 2HI → 2CH3CH2I + H2O
So, 'E' is ethyl iodide (CH3CH2I).
33. (B) (a) Write chemical equations of the following reactions: 3 + 1 + 1
(i) Phenol is treated with conc. HNO3.
(ii) Propene is treated with B2H6 followed by oxidation by H2O2/OH−.
(iii) Sodium t-butoxide is treated with CH3Cl.
(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
(B) (a) (i) Phenol is treated with conc. HNO3 forms 2,4,6-trinitrophenol called picric acid.
(ii) When propene is treated with B2H6 (diborane) followed by oxidation with H2O2/OH−, propan-1-ol (n-propanol) obtained.
(iii) Sodium t-butoxide is treated with CH3Cl to form methyltert butylether.
(b) Chemical test to distinguish between butan-1-ol and butan-2-ol
Butan-1-ol and butan-2-ol can be distinguish by iodoform test. Butan-2-ol gives positive iodoform test while butan-1-ol does not. Lucas test can also be use to distinguish.
(c) The increasing order of acid strength is Ethanol < Water < Phenol
