CUET 2024 Chemistry Question Paper with Answers & Solutions

Correct answer: B

Camphor (solid) dispersed in nitrogen (gas) is a solid solute in a gaseous solvent, i.e. a solid–gas solution.

Correct answer: A

$\ce{CH3Br}$ (B) reacts with Mg in dry ether to give Grignard reagent $\ce{CH3MgBr}$ (A); hydrolysis with $\ce{H2O}$ gives $\ce{CH4}$ (D); chlorination of $\ce{CH4}$ with $\ce{Cl2}$/heat gives $\ce{CH3Cl}$ (C). Order: B, A, D, C.

Correct answer: C

All four statements are correct per NCERT: osmotic pressure is used for molar mass of macromolecules (A); $\Pi = CRT$, so proportional to molarity (B); reverse osmosis occurs when external pressure exceeds osmotic pressure on the concentrated side (C); edema is water retention in tissues due to osmosis (D).

Correct answer: B

By Raoult's law $P = P^\circ_A x_A + P^\circ_D x_D = 700$. Let $x_D=y$: $500(1-y)+800y=700 \Rightarrow 500+300y=700 \Rightarrow y=0.667$, i.e. 66.67 mole percent.

Correct answer: A

Rate $= k[\ce{X}][\ce{A2}]^2$. Doubling volume halves each concentration. New rate $= k(\tfrac{1}{2}[\ce{X}])(\tfrac{1}{2}[\ce{A2}])^2 = \tfrac{1}{8}$ of original. So rate decreases eight times.

Correct answer: C

$\ce{P4O10}$ has 4 P atoms in a tetrahedron with 6 bridging O (each forming 2 P–O sigma bonds = 12) and 4 terminal P=O (each terminal P=O has 1 sigma bond = 4). Total sigma bonds = 12 + 4 = 16.

Correct answer: A

Cryolite ($\ce{Na3AlF6}$) is added in the Hall–Heroult process mainly to lower the melting point (fusion temperature) of alumina and to increase conductivity; the primary stated purpose in NCERT is to lower the melting point and make the mixture conducting.

Correct answer: B

The unit of rate constant $\text{s}^{-1}$ (i.e. $\text{mol}^0\,\text{L}^0\,\text{s}^{-1}$) corresponds to a first order reaction ($k$ unit = $\text{conc}^{1-n}\text{time}^{-1}$, giving $n=1$).

Correct answer: D

For a complex (multistep) reaction the overall rate is governed by the slowest (rate-determining) step, so the order equals the molecularity of the slowest step.

Correct answer: B

For association $i = 1 - \alpha(1 - \tfrac{1}{n})$. $0.80 = 1 - 0.3(1-\tfrac{1}{n}) \Rightarrow 0.3(1-\tfrac{1}{n}) = 0.20 \Rightarrow 1-\tfrac{1}{n} = 0.667 \Rightarrow \tfrac{1}{n}=0.333 \Rightarrow n=3$.

Correct answer: A

Ethylenediamine (en) is neutral. The complex is $\ce{[Co(en)3]2(SO4)3}$; 3 sulphate ions carry $-6$ charge balanced by 2 Co cations, so each cation is $+3$. Oxidation number of Co = +3.

Correct answer: A

In Gly-Ala, glycine provides the free $-\ce{NH2}$ end and its $-\ce{COOH}$ forms the peptide bond with alanine's $-\ce{NH2}$, leaving alanine's $-\ce{COOH}$ free: $\ce{H2N-CH2-CO-NH-CH(CH3)-COOH}$.

Correct answer: C

$\ce{[Cr(NH3)6]Cl3 -> [Cr(NH3)6]^{3+} + 3Cl^-}$, giving 1 complex cation + 3 chloride ions = 4 ions.

Correct answer: B

Moles = mass/molar mass: $\ce{H2}$ (2) = 8 mol; CO (28) = 0.571; $\ce{O2}$ (32) = 0.5; $\ce{CO2}$ (44) = 0.364. Decreasing molecules: D > C > A > B.

Correct answer: A

For fcc, $r = \dfrac{a}{2\sqrt{2}} = \dfrac{361}{2\times1.414} = \dfrac{361}{2.828} \approx 127\,\text{pm}$.

Correct answer: A

75% completion = two half-lives. So $2 t_{1/2} = 32$ min $\Rightarrow t_{1/2} = 16$ min, which is the time for 50% completion.

Correct answer: C

Diamagnetic species (no unpaired electrons) are repelled. $\ce{K4[Fe(CN)6]}$ has Fe(II) low-spin d6 ($t_{2g}^6$), zero unpaired electrons, diamagnetic. ($\ce{[CdCl4]^{2-}}$ Cd2+ d10 is also diamagnetic, but the intended low-spin cyanide answer is $\ce{K4[Fe(CN)6]}$.)

Correct answer: B

$\ce{[Mn(CN)6]^{4-}}$ has Mn(II) d5; $\ce{CN^-}$ is a strong field ligand giving low spin $t_{2g}^5$ with 1 unpaired electron. $\mu = \sqrt{1(1+2)} = \sqrt{3} = 1.73$ BM.

Correct answer: D

Boiling point order: n-Butane (only weak van der Waals) < Ethoxyethane (ether, weak dipole, no H-bond) < Pentanal (polar carbonyl, stronger dipole) < Pentan-1-ol (intermolecular H-bonding, highest). So D.

Correct answer: D

Phenol + Zn dust → benzene (A). Friedel-Crafts methylation → toluene (B). Oxidation with $\ce{K2Cr2O7/H2SO4}$ → benzoic acid (C). $-\ce{COOH}$ is meta-directing, so nitration gives m-nitrobenzoic acid (D).

Correct answer: A

The smaller the gold number, the greater the protective power. A has the lowest gold number range (0.005–0.01), so it is the best protective colloid.

Correct answer: C

In strongly acidic medium aniline is protonated to anilinium ion ($-\ce{NH3+}$), which is meta-directing, giving a substantial amount (~47%) of m-nitroaniline.

Correct answer: B

Aniline is least basic (resonance delocalises lone pair). Among ethylamines in aqueous medium the order is $2^\circ > 1^\circ > 3^\circ$: $\ce{(C2H5)2NH > C2H5NH2 > (C2H5)3N}$. So increasing: $\ce{C6H5NH2 < C2H5NH2 < (C2H5)3N < (C2H5)2NH}$.

Correct answer: A

HVZ reaction requires a carboxylic acid with an alpha-hydrogen. $\ce{R-CH2-COOH}$ has alpha-H atoms, so it undergoes alpha-halogenation. Formic acid has no alpha-C; aldehyde/ketone are not carboxylic acids.

Correct answer: D

Acidity increases with electron-withdrawing power of the alpha-substituent. $\ce{HCOOH}$ (no EWG) is weakest; then Cl, then F (more electronegative), then $\ce{NO2}$ (strongest $-I$). Order: $\ce{HCOOH < ClCH2COOH < FCH2COOH < NO2CH2COOH}$.

Correct answer: C

Nucleophilic addition reactivity increases with electrophilicity of carbonyl C and decreases with steric/+ effects. Acetophenone (ketone, methyl +I + steric) is least; p-tolualdehyde (electron-donating $-\ce{CH3}$) next; benzaldehyde; p-nitrobenzaldehyde (electron-withdrawing $-\ce{NO2}$) most reactive.

Correct answer: C

In the Gatterman-Koch reaction (benzene + CO + HCl with anhydrous $\ce{AlCl3}$/CuCl), the attacking electrophile is the formyl cation $\ce{HCO+}$, which gives benzaldehyde.

Correct answer: A

Cannizzaro reaction occurs for aldehydes lacking alpha-hydrogen. Formaldehyde has no alpha-H (B) and undergoes self oxidation–reduction (disproportionation) with conc. alkali (D). So (B) and (D).

Correct answer: B

After protonation, C–O cleaves to give the more stable tertiary carbocation $\ce{(CH3)3C+}$ (and $\ce{CH3OH}$), which then combines with $\ce{I^-}$. This is an $\ce{S_N1}$ pathway. So (B) and (C).

Correct answer: C

Aniline's lone pair coordinates to Lewis acid $\ce{AlCl3}$ forming a salt (A); the N acquires a positive charge (C); the resulting $\ce{-N+}$ group is strongly deactivating, so substitution fails (D). So (A), (C) and (D).

Correct answer: A

Chlorine withdraws electrons by $-I$ inductive effect (A), which deactivates and destabilises the intermediate carbocation (B), but it donates electron density by resonance/+M (D) stabilising ortho/para positions, making it o,p-directing. So (A), (B) and (D).

Correct answer: A

Etard reaction oxidises toluene with chromyl chloride ($\ce{CrO2Cl2}$), then hydrolysis of the chromium complex gives benzaldehyde, an aromatic aldehyde.

Correct answer: C

(A) sequence of amino acids = primary (I); (B) regular folding via H-bonding = secondary (II); (C) fibrous proteins relate to tertiary structure shape (IV); (D) arrangement of multiple polypeptide chains = quaternary (III). So A-I, B-II, C-IV, D-III.

Correct answer: A

Tollen's = ammoniacal silver nitrate (III); Jones reagent = chromium trioxide–sulphuric acid (IV); Lucas reagent = conc. HCl + $\ce{ZnCl2}$ (II); Fehling solution contains Rochelle salt (sodium potassium tartrate) (I). So A-III, B-IV, C-II, D-I.

Correct answer: D

Swarts = $\ce{RX + AgF -> R-F}$ (III); Finkelstein = $\ce{RX + NaI -> R-I}$ (IV); Sandmeyer uses diazonium + $\ce{Cu2X2}$ (I); Wurtz = $\ce{2RX + 2Na -> R-R}$ (II). So A-III, B-IV, C-I, D-II.

Correct answer: D

Vitamin A deficiency → xerophthalmia (II); Thiamine (B1) deficiency → beri-beri (III); Glucocorticoids deficiency relates to Addison's disease (IV); Estradiol regulates the menstrual cycle (I). So A-II, B-III, C-IV, D-I.

Correct answer: C

Acid hydration follows Markovnikov, giving the more substituted (more stable carbocation) alcohol. (A) 3-methylpent-2-ene → 3-methylpentan-3-ol tertiary alcohol (II); (B) 1-methylcyclohexene → 1-methylcyclohexanol (I); (C) but-2-ene → butan-2-ol (IV); (D) ethylidenecyclohexane → 2-cyclohexylbutan-2-ol/1-(cyclohexyl) tertiary alcohol (III). So A-II, B-I, C-IV, D-III.

Correct answer: D

Morphine, heroin and codeine are narcotic analgesics. Ranitidine is an antacid ($\ce{H2}$-receptor antagonist), not an analgesic.

Correct answer: A

Increasing acidity = decreasing pKa. Order of $-I$ effect: HCOOH (no halogen, weakest) < $\ce{BrCH2COOH}$ < $\ce{ClCH2COOH}$ < $\ce{FCH2COOH}$ (F most electronegative, strongest acid). So D < A < B < C.

Correct answer: D

$\ce{S_N2}$ reactivity decreases with steric hindrance: $1^\circ$ unhindered > $1^\circ$ hindered (neopentyl-like) > $2^\circ$ > $3^\circ$. (C) tertiary slowest, (D) $\ce{(CH3)2CHCH2Br}$ hindered primary, (B) secondary, (A) straight primary fastest. Increasing order: C < D < B < A.

Correct answer: C

Moles of $\ce{H2} = \dfrac{67.2}{22.4} = 3.0$ moles at STP.

Correct answer: D

Each $\ce{H2}$ gives 2 electrons (anode: $\ce{H2 -> 2H2O + 2e^-}$). 3.0 mol $\ce{H2} \times 2 = 6$ moles of electrons.

Correct answer: B

Charge $Q = n F = 6 \times 96500 = 579000$ C.

Correct answer: B

$\ce{Ag+ + e^- -> Ag}$ needs 1 mol e- per mol Ag. 6 mol electrons deposit 6 mol Ag $= 6 \times 108 = 648$ g.

Correct answer: D

As stated in NCERT, the $\ce{H2}$–$\ce{O2}$ fuel cell was used to provide electrical power on the Apollo space programme (and the water produced was used for drinking).

Correct answer: B

Zr and Hf share almost identical atomic radii and chemical properties due to lanthanoid contraction, not nuclear properties — so statement (B) is incorrect. (A is true: removing 2nd e- from Ag breaks stable $4d^{10}$ so its I.E.2 is high; C true; D true.)

Correct answer: C

Cr+ has a stable $3d^5$ configuration, so removing the 2nd electron is very difficult — I.E.2 of Cr is unusually high (highest among V, Cr, Mn). Mn+ ($3d^54s^1$) loses the 4s electron easily, so its I.E.2 is lower. Thus V < Cr > Mn.

Correct answer: B

Colour depends on number of unpaired d-electrons. $\ce{VO^{2+}}$ (V(IV), $d^1$) and $\ce{Cu^{2+}}$ ($d^9$, equivalent to one hole) both give blue colour in aqueous solution. So the pair $\ce{VOCl2}$ and $\ce{CuCl2}$ have the same (blue) colour.

Correct answer: C

Mn shows the maximum oxidation state of +7 (e.g. in $\ce{KMnO4}$), the highest in the first transition series, since it can use all its $3d^5 4s^2$ electrons.

Correct answer: B

In actinoids the 5f, 6d and 7s energy levels are close (small 5f–6d energy gap), so more electrons can participate in bonding, giving a larger range of oxidation states than lanthanoids (where 4f–5d gap is larger).