CUET 2022 Chemistry Question Paper with Answers & Solutions

Correct answer: A

Ionic solids have ions as constituent particles (A correct). They are hard and brittle, not soft (B wrong). They are electrical insulators in the solid state (C correct) but conduct in the molten/aqueous state because ions become mobile (D correct). They have high melting and boiling points due to strong electrostatic forces (E wrong). So correct: A, C and D.

Correct answer: B

In hcp lattice of B, number of B atoms = N, number of tetrahedral voids = 2N. A occupies 2/3 of tetrahedral voids = (2/3)(2N) = 4N/3. Ratio A:B = 4N/3 : N = 4:3, so formula is $\ce{A4B3}$.

Correct answer: C

Elevation in boiling point is a colligative property proportional to van't Hoff factor i. Glucose i=1; NaCl i=2; MgCl2 i=3; AlCl3 i=4; Al2(SO4)3 i=5. Increasing order: A(1) < B(2) < C(3) < D(4) < E(5).

Correct answer: B

Moles of NaOH = 5/40 = 0.125 mol. Volume = 450 mL = 0.45 L. Molarity = 0.125/0.45 = 0.278 M.

Correct answer: D

Ionic (electrolytic) conductance depends on nature of electrolyte (A), temperature (D) and nature of solvent (E). It is due to movement of ions, not electrons (B wrong), and is called electrolytic, not electronic, conductance (C wrong). Correct: A, D and E.

Correct answer: A

By Kohlrausch's law: $\Lambda^\circ_{HOAc} = \Lambda^\circ_{HCl} + \Lambda^\circ_{NaOAc} - \Lambda^\circ_{NaCl} = 425.9 + 91.0 - 126.4 = 390.5$ S cm$^2$ mol$^{-1}$.

Correct answer: B

Mn in $\ce{MnO4^-}$ is +7, reduced to +2 in $\ce{Mn^2+}$, a gain of 5 electrons. For 1 mol, 5 mol electrons = 5 F.

Correct answer: C

In electrolysis of aqueous NaCl (brine), $\ce{H2}$ is liberated at cathode (water reduced preferentially over Na+) and $\ce{Cl2}$ at anode (due to overvoltage of O2). Products: H2 and Cl2.

Correct answer: C

Aspartame is unstable at cooking/baking temperatures (decomposes on heating), so its use is limited to cold foods and soft drinks.

Correct answer: C

Units of rate constant for order n are mol$^{1-n}$ L$^{n-1}$ s$^{-1}$. Units L mol$^{-1}$ s$^{-1}$ = mol$^{-1}$ L s$^{-1}$ correspond to second order (n=2).

Correct answer: B

$\ce{2NH3 -> N2 + 3H2}$. For zero order, rate = k = $2.5\times10^{-4}$. Rate of decomposition of NH3 = $-\frac{1}{2}\frac{d[NH3]}{dt}$... rate of reaction = k. Rate of production of H2 = 3 × rate = $3 \times 2.5\times10^{-4} = 7.5\times10^{-4}$ mol L$^{-1}$ s$^{-1}$.

Correct answer: B

Molecularity = number of reacting species in an elementary reaction. Only one molecule of $\ce{NH4NO2}$ reacts, so molecularity = one (unimolecular).

Correct answer: C

Physisorption has low enthalpy of adsorption (20-40 kJ/mol) because it involves weak van der Waals forces. High enthalpy of adsorption is a characteristic of chemisorption. So C is NOT a characteristic of physisorption.

Correct answer: B

An emulsion is a colloid of liquid dispersed in liquid. Hair cream (oil-in-water/water-in-oil) is an emulsion. Smoke is solid-in-gas (aerosol), paint is sol (solid-in-liquid), cheese is gel (liquid-in-solid).

Correct answer: B

Caprolactam undergoes ring-opening polymerisation to give Nylon-6.

Correct answer: D

Blood is a positively charged sol (negatively charged sols include starch, gum, gold sol, As2S3). Among the options, blood carries positive charge.

Correct answer: B

Siderite ($\ce{FeCO3}$) = Iron (II); Malachite ($\ce{CuCO3.Cu(OH)2}$) = Copper (III); Calamine ($\ce{ZnCO3}$) = Zinc (IV); Bauxite ($\ce{Al2O3.2H2O}$) = Aluminium (I). So A-II, B-III, C-IV, D-I.

Correct answer: B

The Van Arkel method (vapour phase refining via iodide) is used for refining Zirconium (Zr) and Titanium (Ti).

Correct answer: B

Thermal stability of group 16 hydrides decreases down the group: $\ce{H2O > H2S > H2Se > H2Te > H2Po}$. Increasing order: $\ce{H2Po (C) < H2Te (D) < H2Se (B) < H2S (E) < H2O (A)}$, i.e. C < D < B < E < A.

Correct answer: A

Ammonia = Haber's process (IV); Chlorine = Deacon process (III); Sulphuric acid = Contact process (II); Nitric acid = Ostwald's process (I). So A-IV, B-III, C-II, D-I.

Correct answer: C

$\ce{BrO3^-}$ has 3 bond pairs and 1 lone pair on central atom, giving pyramidal (sp3, trigonal pyramidal) shape. $\ce{XeO3}$ also has 3 bond pairs and 1 lone pair, pyramidal shape. So XeO3 is isostructural with BrO3^-.

Correct answer: A

Max oxidation states: Ti = +4 (II); V = +5 (III); Mn = +7 (I); Cu = +2 (IV). So A-II, B-III, C-I, D-IV.

Correct answer: C

Copper has a positive standard reduction potential $E^\circ_{Cu^{2+}/Cu} = +0.34$ V, which is unique in the first transition series (all others are negative). This is because of its high enthalpy of atomisation and low hydration enthalpy.

Correct answer: D

Atomic number 25 is Mn. $\ce{Mn^2+}$ has configuration $3d^5$, so 5 unpaired electrons. $\mu = \sqrt{n(n+2)} = \sqrt{5(7)} = \sqrt{35} = 5.92$ BM.

Correct answer: A

$\ce{Sc^3+}$ has $3d^0$ configuration (no d electrons), so no d-d transitions are possible and it is colourless. The others have partially filled d orbitals and are coloured.

Correct answer: B

A correct: SN2 proceeds with inversion (Walden inversion). B wrong: racemisation is conversion of enantiomer into racemic mixture (the reverse). C correct: racemic mixture has equal proportions of two enantiomers. D wrong: enantiomers are NON-superimposable mirror images. E correct: chirality is the property of non-superimposable mirror images. Correct: A, C and E.

Correct answer: A

Neopentyl chloride is $\ce{(CH3)3C-CH2Cl}$, i.e. 2,2-dimethyl-1-chloropropane = 1-Chloro-2,2-dimethylpropane.

Correct answer: C

The benzylic -CH2OH (a primary benzylic alcohol) reacts with HCl to substitute -OH with -Cl forming -CH2Cl. The phenolic -OH on the aromatic ring does not react with HCl (aryl C-OH is not replaced). So the product retains the phenolic OH and converts CH2OH to CH2Cl, matching option C (CH2Cl and OH).

Correct answer: A

A correct: BP increases with carbon number. B wrong: BP decreases with branching (less surface area). C wrong: alcohols have HIGHER BP than haloalkanes of comparable mass due to H-bonding. D correct: alcohols have higher BP than hydrocarbons. E: the high BP is due to (inter)molecular hydrogen bonding; the statement says intramolecular which is loose but in CUET context taken as correct. Best answer A, D and E.

Correct answer: D

Acid strength: alcohols are weakest, then phenols with electron-donating groups, then phenol, then phenols with electron-withdrawing (nitro) groups (more nitro = stronger). Order: Propanol(A) < 4-methylphenol(E) < phenol(D) < 3-nitrophenol(B) < 3,5-dinitrophenol(C). So A < E < D < B < C.

Correct answer: B

$\ce{NaBH4}$ selectively reduces aldehydes and ketones but does NOT reduce esters (or carboxylic acids). So the ring ketone (C=O) is reduced to a secondary alcohol (OH), while the ester group -CH2-C(=O)-O-CH3 remains intact. This matches option B.

Correct answer: A

Aryl alkyl ethers cleave with HBr/HI to give a phenol and an alkyl halide. The C(alkyl)-O bond breaks: $\ce{C6H5-O-C2H5 + HBr -> C6H5-OH + C2H5Br}$. Phenol + ethyl bromide form. The aryl-O bond is not cleaved. Matches option A.

Correct answer: C

A zwitterion has both a positive ($\ce{-NH3^+}$) and a negative ($\ce{-COO^-}$) charge, so it can react with both acids and bases, showing amphoteric behaviour.

Correct answer: B

Acetophenone (1) = C6H5COCH3 = structure II; Benzaldehyde (2) = C6H5CHO = structure I; Benzoic acid (3) = C6H5COOH = structure IV; Benzophenone (4) = C6H5COC6H5 = structure III. So 1-II, 2-I, 3-IV, 4-III, matching option B (A-II, B-I, C-IV, D-III).

Correct answer: A

Ethanal ($\ce{CH3CHO}$) has the CH3-CO group and gives a positive iodoform test (yellow precipitate), while propanal ($\ce{CH3CH2CHO}$) does not. Both give positive Tollen's and Fehling's tests, so those cannot distinguish them. Iodoform test distinguishes them.

Correct answer: D

Aldol condensation requires at least one alpha-hydrogen. Methanal (HCHO) has none, benzaldehyde has none, 2,2-dimethylbutanal has no alpha-H (the alpha carbon is quaternary). Phenylacetaldehyde ($\ce{C6H5CH2CHO}$) has alpha-hydrogens on the CH2, so it undergoes aldol condensation.

Correct answer: A

A correct: analgesics relieve pain without impairing consciousness. B wrong: tranquilizers are neurologically active drugs affecting the central nervous system. C wrong: morphine is a narcotic (opioid) analgesic, not non-narcotic. D correct: disinfectants for inanimate objects, antiseptics for living tissue. E correct: e.g. phenol acts as antiseptic at 0.2% and disinfectant at 1%. Correct: A, D and E.

Correct answer: B

Relative sweetness (vs sucrose=1): Saccharin ~550, Aspartame ~100, Sucralose ~600, Alitame ~2000. Alitame is the sweetest, about 2000 times cane sugar.

Correct answer: D

A copolymer is made from two or more different monomers. Glyptal is formed from ethylene glycol and phthalic acid (two monomers), so it is a copolymer. Polypropene, polystyrene and PVC are homopolymers (single monomer).

Correct answer: C

Maltose (glucose + glucose) is a disaccharide. Glucose is a monosaccharide; glycogen and starch are polysaccharides.

Correct answer: A

Ethylamine $\ce{C2H5NH2}$ (a primary amine) accepts one proton from HCl to form the ethylammonium chloride salt $\ce{C2H5-NH3^+ Cl^-}$.

Correct answer: C

In aqueous solution, basicity of ethylamines follows: secondary > primary > tertiary > ammonia (due to combined inductive, steric and solvation effects). So $\ce{(C2H5)2NH}$ (diethylamine, secondary) is the most basic.

Correct answer: D

In the gas phase (no solvation effect), only the inductive effect operates, so basicity increases with number of alkyl groups: tertiary > secondary > primary. Increasing order: 1° < 2° < 3°.

Correct answer: D

Highest pKb = weakest base. Aniline ($\ce{C6H5NH2}$) is the weakest base because the lone pair on nitrogen is delocalised into the aromatic ring. So aniline has the highest pKb.

Correct answer: C

Highest Kb = strongest base. Diethylamine $\ce{(C2H5)2NH}$ (secondary aliphatic amine) is the strongest base among the options; aromatic amines (B) are weak. So $\ce{(C2H5)2NH}$ has the highest Kb.

Correct answer: C

In $\ce{[Co(NH3)5Cl]Cl2}$, ligands named alphabetically: ammine (5) before chlorido = pentaamminechlorido. Co oxidation state: complex ion charge is +2 (two Cl- counter ions), 5 NH3 neutral + 1 Cl(-1), so Co = +3. The complex cation is named with metal name + (III). Counter ion = chloride. So Pentaamminechloridocobalt(III) chloride. Note: the suffix should be 'cobalt' (cation) not 'cobaltate', but among given options C is the closest/intended answer (correct oxidation state III, correct ligand order, single chloride). Selecting C.

Correct answer: B

Co is +3, $3d^6$. With NH3 (strong field) ligands, the complex is low spin: all 6 d electrons pair up in t2g (d2sp3), giving 0 unpaired electrons. $\mu = \sqrt{0(0+2)} = 0.0$ BM (diamagnetic).

Correct answer: B

Co(III), $3d^6$ low spin with strong field NH3 ligands, uses inner d orbitals: $d^2sp^3$ hybridisation, giving octahedral geometry.

Correct answer: D

In $\ce{[Co(NH3)5Cl]^2+}$, Co is bonded to 5 NH3 ligands and 1 Cl ligand = 6 coordinating atoms. Coordination number = 6.

Correct answer: C

Primary valence corresponds to the oxidation state of the metal. In $\ce{[Co(NH3)5Cl]Cl2}$, Co is in +3 oxidation state, so primary valence = 3.