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Chapter Tests for Class 12 Chemistry
One test per Chemistry chapter, made for Class 12 and matched to the CBSE 2026-27 marking scheme. Click any chapter below to begin. No login needed, and you can retake tests freely.
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Quick Practice - Class 12 Chemistry (NCERT Core) Select any chapter below to test your understanding with 5 high-yield multiple-choice questions, instant scoring, and step-by-step verified explanations. Q1.Which of the following concentration units is temperature-independent? Answer: (c) Molality (m) and Mole Fraction. Molality and mole fraction depend purely on mass, whereas molarity and normality depend on liquid volume, which expands or contracts with temperature.
Q2.What is the value of the van 't Hoff factor (i) for a completely dissociated aqueous solution of potassium sulphate [K2SO4]? Answer: (b) 3. K2SO4 dissociates into 2 K+ ions and 1 SO42- ion, making total particles n = 3. For 100% dissociation, i = 3.
Q3.According to Henry's Law, when the temperature of a liquid solvent increases, the solubility of gases in it: Answer: (a) Decreases, because dissolution of gas is an exothermic process. Gas dissolution releases heat. Following Le Chatelier's principle, higher temperature drives gas out of solution, which is why aquatic species prefer colder water.
Q4.An azeotropic mixture of two liquids boils at a lower temperature than either of its components when the solution shows: Answer: (b) Large positive deviation from Raoult's Law (Minimum boiling azeotrope). Weaker intermolecular forces between unlike molecules increase vapor pressure, causing the mixture to boil at a lower temperature (e.g., 95% ethanol + 5% water).
Q5.What is the osmotic pressure (Π) of a 0.1 M glucose solution at 300 K? (R = 0.0821 L atm K-1 mol-1) Answer: (c) 2.463 atm. Π = CRT = 0.1 × 0.0821 × 300 = 2.463 atm. (Glucose is non-electrolyte, so i = 1).
Q1.What is the limiting molar conductivity of an electrolyte according to Kohlrausch's Law? Answer: (b) Sum of individual limiting ionic conductivities of its constituent cations and anions. At infinite dilution, each ion migrates independently and makes a definite contribution toward the total molar conductivity.
Q2.How much charge in Faradays is required for the complete reduction of 1 mole of Cr2O72- to Cr3+ in acidic medium? Answer: (d) 6 F. Reaction: Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O. Since 6 moles of electrons are transferred per mole of dichromate, charge required = 6 Faradays.
Q3.What is the unit of the cell constant (G*) of a conductivity cell? Answer: (a) cm-1 or m-1. Cell constant is the ratio of the distance between electrodes (l) to their cross-sectional area (A): G* = l/A = cm / cm2 = cm-1.
Q4.In a secondary Lead-Acid storage battery, what chemical transformation occurs at both electrodes during discharging? Answer: (c) Both anode (Pb) and cathode (PbO2) are converted into PbSO4. Discharging consumes H2SO4 and deposits lead sulfate on both plates, which is reversed during recharging.
Q5.The standard reduction potential for zinc is E° = -0.76 V and copper is E° = +0.34 V. The standard EMF of the Daniel Cell is: Answer: (b) 1.10 V. E°cell = E°cathode - E°anode = +0.34 V - (-0.76 V) = +1.10 V.
Q1.What are the units of the rate constant (k) for a first-order chemical reaction? Answer: (d) s-1 (time-1). General formula: (mol L-1)1-n s-1. For first order (n = 1), (mol L-1)0 s-1 = s-1.
Q2.If the initial concentration of a reactant in a first-order reaction is doubled, its half-life (t1/2) will: Answer: (b) Remain constant and unchanged. For a first-order reaction, t1/2 = 0.693 / k, which is independent of initial concentration.
Q3.A catalyst increases the speed of a chemical reaction primarily by: Answer: (a) Providing an alternative pathway with lower Activation Energy (Ea). A catalyst lowers the energy barrier without altering ΔH, ΔG, or the chemical equilibrium state.
Q4.The rate of a reaction is given by Rate = k [A]1/2 [B]2. What is the overall order of the reaction? Answer: (c) 2.5 (or 5/2). Overall order is the sum of the powers of reactant concentrations in the rate law: 1/2 + 2 = 5/2 = 2.5.
Q5.In the Arrhenius plot of ln k versus 1/T, what does the slope of the straight line equal? Answer: (a) -Ea / R. From Arrhenius equation ln k = ln A - (Ea / RT), plotting y = ln k against x = 1/T gives slope = -Ea/R and intercept = ln A.
Q1.Why are transition metal ions frequently colored in aqueous solutions? Answer: (b) Due to d-d electronic transitions in the presence of ligands. Degenerate d-orbitals split in energy. Unpaired electrons absorb visible light to transition between split d-orbitals, giving the ion its complementary color.
Q2.What causes the steady, gradual decrease in atomic and ionic radii across the Lanthanoid series (Lanthanoid Contraction)? Answer: (c) Poor shielding effect of 4f electrons combined with increasing nuclear charge. Diffuse 4f orbitals shield nuclear charge poorly, pulling outer electrons closer as atomic number increases.
Q3.Which 3d transition series element exhibits the highest possible oxidation state of +7? Answer: (d) Manganese (Mn). Manganese has electronic configuration [Ar] 3d5 4s2. In KMnO4, it uses all 7 valence electrons to exhibit a +7 oxidation state.
Q4.Why is Sc3+ (Z = 21) diamagnetic and colorless, while Ti3+ (Z = 22) is paramagnetic and purple? Answer: (a) Sc3+ has a 3d0 empty configuration, whereas Ti3+ has one unpaired electron (3d1). With no d-electrons (3d0), Sc3+ cannot undergo d-d transitions and has no unpaired magnetic spins.
Q5.What is the calculated 'spin-only' magnetic moment (μ) of a Fe2+ ion (Z = 26)? Answer: (b) 4.90 BM. Fe2+ has configuration 3d6, which contains n = 4 unpaired electrons. μ = √[n(n+2)] = √[4(6)] = √24 ≈ 4.90 Bohr Magnetons (BM).
Q1.What is the IUPAC name of the complex [Co(NH3)5(CO3)]Cl? Answer: (c) Pentaamminecarbonatocobalt(III) chloride. Ligands are listed alphabetically: 'ammine' before 'carbonato'. Cobalt is in the +3 oxidation state [x + 0 + (-2) - 1 = 0 &implies; x = +3].
Q2.Ligands that can coordinate to a central metal atom through two different donor atoms (e.g., NO2- / ONO- and SCN- / NCS-) give rise to: Answer: (a) Linkage Isomerism. Ambidentate ligands possess two distinct donor atoms, producing linkage isomers depending on which atom bonds to the metal.
Q3.What is the hybridization and magnetic geometry of the inner orbital complex [Fe(CN)6]3-? Answer: (b) d2sp3, Paramagnetic (1 unpaired electron). CN- is a strong field ligand that forces pairing in Fe3+ (3d5), leaving 1 unpaired electron and two empty 3d orbitals for d2sp3 inner orbital hybridization.
Q4.Which synthetic chelating ligand is widely used to treat acute Lead poisoning in humans? Answer: (d) EDTA (Ethylenediaminetetraacetic acid). Calcium-disodium EDTA binds heavy metal lead ions (Pb2+) into a stable, water-soluble chelate that is excreted safely in urine.
Q5.How many moles of AgCl precipitate are formed when 1 mole of [Co(NH3)5Cl]Cl2 reacts with excess silver nitrate (AgNO3)? Answer: (c) 2 moles. Only the two ionizable chloride ions outside the coordination sphere react with Ag+ to form 2 moles of AgCl precipitate.
Q1.What is the stereochemical outcome of an SN2 nucleophilic substitution reaction at a chiral carbon centre? Answer: (a) Complete Inversion of Configuration (Walden Inversion). SN2 occurs via a single concerted step with backside nucleophilic attack, inverting the spatial orientation around the carbon.
Q2.Why are haloarenes (e.g., chlorobenzene) much less reactive toward nucleophilic substitution compared to haloalkanes? Answer: (c) Resonance gives partial double bond character to the C-Cl bond, strengthening it. Delocalization of chlorine's lone pair into the benzene ring shortens and strengthens the C-Cl bond, making it difficult to cleave.
Q3.According to Saytzeff's Rule, the major product in the dehydrohalogenation of 2-bromobutane is: Answer: (b) But-2-ene (more substituted alkene). Elimination favors hydrogen removal from the β-carbon holding fewer hydrogens, producing the more stable, highly alkylated alkene.
Q4.When chlorobenzene is heated with sodium metal in dry ether, the coupling reaction is known as: Answer: (a) Fittig Reaction (forming Diphenyl). Coupling two aryl halides using sodium metal in dry ether is the Fittig Reaction. Coupling an alkyl with an aryl halide is the Wurtz-Fittig Reaction.
Q5.Why is chloroform stored in dark-colored, tightly sealed bottles filled to the brim? Answer: (d) To prevent oxidation by sunlight into poisonous Phosgene gas (COCl2). In the presence of light and oxygen, chloroform slowly oxidizes to carbonyl chloride (phosgene): 2CHCl3 + O2 → 2COCl2 + 2HCl.
Q1.Lucas reagent (anhydrous ZnCl2 + conc. HCl) produces immediate cloudiness at room temperature with: Answer: (c) Tertiary (3°) alcohols (gives immediate turbidity). Tertiary carbocations form rapidly, producing insoluble alkyl chloride cloudiness immediately at room temperature.
Q2.Why is phenol significantly more acidic than aliphatic alcohols like ethanol? Answer: (b) The phenoxide ion is stabilized by resonance delocalization of negative charge over the benzene ring. Losing H+ produces a resonance-stabilized phenoxide ion, whereas the ethoxide ion experiences destabilizing +I inductive effects without resonance.
Q3.In the Kolbe's Reaction, treating sodium phenoxide with carbon dioxide (CO2) followed by acidification yields: Answer: (a) Salicylic acid (2-hydroxybenzoic acid). Electrophilic substitution of phenoxide by weak electrophile CO2 followed by acidification forms ortho-hydroxybenzoic acid (salicylic acid).
Q4.In Williamson Ether Synthesis, preparing tert-butyl methyl ether requires reacting: Answer: (d) Sodium tert-butoxide with primary methyl halide (CH3Br). Primary alkyl halides undergo SN2 substitution smoothly. Using a tertiary halide causes elimination into isobutylene instead of substitution.
Q5.Treating anisole (methoxybenzene) with concentrated HI yields: Answer: (b) Phenol and Methyl iodide (CH3I). The aryl-oxygen bond has partial double bond character and resists cleavage, so I- attacks the smaller methyl carbon to form CH3I and phenol.
Q1.Which of the following organic carbonyl compounds will NOT give a positive Cannizzaro reaction? Answer: (c) Acetaldehyde (CH3CHO). Cannizzaro's reaction occurs only in aldehydes lacking α-hydrogens. Acetaldehyde contains three α-hydrogens, so it undergoes Aldol condensation instead.
Q2.What characteristic yellow crystalline precipitate is formed in the Iodoform test for compounds containing the CH3C=O group? Answer: (b) Triiodomethane / Iodoform (CHI3). Methyl ketones and ethanol react with I2/NaOH to precipitate yellow CHI3 crystals with an antiseptic odor.
Q3.The Rosenmund Reduction converts an acyl chloride (R-COCl) to an aldehyde using: Answer: (a) H2 in the presence of Pd on BaSO4 (poisoned with sulfur or quinoline). BaSO4 acts as a catalyst poison, preventing further reduction of the aldehyde into an alcohol.
Q4.Which reagent quantitatively reduces both aldehydes and ketones into corresponding hydrocarbons (Clemmensen Reduction)? Answer: (d) Zinc amalgam and concentrated hydrochloric acid (Zn-Hg / conc. HCl). Clemmensen reduction deoxygenates carbonyl groups into methylene (-CH2-) units using acidic Zn-Hg.
Q5.What is the correct decreasing order of acidic strength among substituted carboxylic acids? Answer: (a) CF3COOH > CCl3COOH > CH2ClCOOH > CH3COOH. Stronger electron-withdrawing groups (-I effect) stabilize the carboxylate anion, increasing acid strength.
Q1.The Carbylamine Test (Isocyanide test) used to distinguish primary amines produces a foul smell of isocyanide when heated with: Answer: (b) Chloroform (CHCl3) and alcoholic Potassium Hydroxide (KOH). Only primary (1°) aliphatic and aromatic amines form foul-smelling isocyanides (R-NC). Secondary and tertiary amines do not react.
Q2.What is the correct order of basic strength of methyl-substituted amines in aqueous solution? Answer: (c) (CH3)2NH > CH3NH2 > (CH3)3N > NH3 (2° > 1° > 3° > NH3). Basic strength in aqueous medium is governed by the combined effects of inductive (+I), hydration energy, and steric hindrance. For methyl groups, 2° > 1° > 3°. For ethyl groups, 2° > 3° > 1°.
Q3.Gabriel Phthalimide Synthesis is strictly used for the preparation of: Answer: (a) Pure primary (1°) aliphatic amines. Aryl halides cannot undergo nucleophilic substitution with the phthalimide anion, so aniline cannot be prepared by this method.
Q4.Hoffmann Bromamide Degradation reaction of an amide with Br2 and NaOH produces an amine with: Answer: (d) One less carbon atom than the starting parent amide. R-CONH2 + Br2 + 4NaOH → R-NH2 + Na2CO3 + 2NaBr + 2H2O. The carbonyl carbon is eliminated as carbonate.
Q5.What is the chemical identity of Hinsberg's reagent used to differentiate 1°, 2°, and 3° amines? Answer: (b) Benzenesulphonyl chloride (C6H5SO2Cl). Primary amines form sulfonamides soluble in alkali; secondary amines form sulfonamides insoluble in alkali; tertiary amines do not react.
Q1.Which of the following nitrogenous bases is present in RNA but absent in DNA? Answer: (c) Uracil. DNA contains Adenine, Guanine, Cytosine, and Thymine. In RNA, Thymine is replaced by Uracil.
Q2.During protein denaturation (e.g., boiling an egg or curdling milk), which structural levels of the protein are destroyed? Answer: (b) Secondary and tertiary structures, leaving the primary peptide chain intact. Heat disrupts hydrogen and ionic bonds, causing globules to unfold while covalent peptide bonds (primary structure) remain intact.
Q3.Sucrose is a non-reducing sugar because its glycosidic linkage involves: Answer: (a) C1 of α-D-glucose and C2 of β-D-fructose, locking both reducing functional groups. Because the reducing hemiacetal and hemiketal groups are tied up in the glycosidic bond, sucrose cannot reduce Fehling's or Tollens' reagent.
Q4.Deficiency of which water-soluble vitamin causes the disease Beriberi? Answer: (b) Vitamin B1 (Thiamine). Vitamin B1 deficiency causes Beriberi (loss of appetite and nerve disorders). Vitamin C deficiency causes Scurvy, and Vitamin A causes night blindness.
Q5.Amino acids exist in aqueous solution predominantly as dipolar internal salts known as: Answer: (d) Zwitterions (+H3N-CHR-COO-). The carboxylic -COOH group transfers a proton to the amino -NH2 group, forming an internal dipolar zwitterion with amphoteric properties.
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