NEET 2026 Chemistry Questions with Solutions

Explanation

LiAlH₄/H₂O, H₂/Ni and Na(Hg)/C₂H₅OH reduce nitriles (R–CN) to primary amines (R–CH₂NH₂). Sn/HCl and Br₂/NaOH do not. Hence option (B) is correct.

Explanation

V₂O₅ → Contact process (SO₂ to SO₃); Fe → Haber process; PdCl₂ → Oxidation of ethyne to ethanal; Ni complex → Polymerisation of alkynes. Hence option (D) is correct.

Explanation

Δn(g)=2−3=−1. Therefore ΔH°=ΔU°+ΔnRT=−10−(1×8.31×298/1000)=−12.48 kJ mol⁻¹. Then ΔG°=ΔH°−TΔS°=−12.48−[298×(−44)/1000]=+0.632 kJ mol⁻¹≈+0.63568 kJ mol⁻¹. Since ΔG°>0, the reaction is non-spontaneous. Hence option (C) is correct.

Explanation

l=0→s, l=1→p, l=2→d, l=3→f. Therefore: (2,1)=2p, (4,0)=4s, (5,3)=5f, (3,2)=3d. Hence option (C) is correct.

Explanation

K=[BiO⁺][OH⁻]=s²=4×10⁻¹⁰ ⇒ s=[OH⁻]=2×10⁻⁵ M. Thus [H⁺]=10⁻¹⁴/(2×10⁻⁵)=0.5×10⁻⁹. Therefore pH=−log(0.5×10⁻⁹)=9+log2=9.301. Hence option (D) is correct.

Explanation

Metamerism occurs due to different alkyl groups on either side of the same functional group. CH₃OCH₂CH₂CH₃ and CH₃CH₂OCH₂CH₃ are ethers with different alkyl groups and are metamers. Hence option (D) is correct.

Explanation

[Pt(NH₃)₂Cl₂] shows geometrical isomerism; [Co(en)₃]³⁺ shows optical isomerism; [Co(NH₃)₅NO₂]Cl₂ shows linkage isomerism; [Cr(H₂O)₆]Cl₃ shows solvate isomerism. Hence option (D) is correct.

Explanation

Unit of rate constant for nth order reaction = (mol L⁻¹)^(1−n)s⁻¹. Thus: zero order → mol L⁻¹ s⁻¹, first order → s⁻¹, second order → mol⁻¹ L s⁻¹, third order → mol⁻² L² s⁻¹. Hence option (B) is correct.

Explanation

The longest carbon chain contains 7 carbons (heptane). Substituents are ethyl at C-3 and methyl at C-5. Prefixes are written alphabetically: ethyl before methyl. Therefore the name is 3-ethyl-5-methylheptane. Hence option (A) is correct.

Explanation

Steam reforming reaction: CH₄ + H₂O → CO + 3H₂ (Ni catalyst, 1273 K). Hence option (C) is correct.

Explanation

2,4-DNP positive indicates carbonyl compound. No Fehling's reduction suggests ketone. For C₈H₈O, acetophenone fits. Strong oxidation converts side chain to benzoic acid, which reacts with NaHCO₃ giving CO₂. Hence option (D) is correct.

Explanation

C₂H₄ has 5σ and 1π bond; C₂H₂ has 3σ and 2π bonds; CH₄ has 4σ bonds; NH₃ has 3σ bonds and one lone pair. Therefore A–IV, B–I, C–III, D–II. Hence option (B) is correct.

Explanation

AgCN with alkyl halides gives isocyanides (R–NC), which have foul smell. Carbylamine reaction also forms isocyanides. Thus Z=C₂H₅NC and X=AgCN. Hence option (A) is correct.

Explanation

Moles of urea = 5.4/60=0.09 mol. Each urea molecule contains 4 H atoms. Total H atoms =0.09×4×6.022×10²³≈2.168×10²³. Hence option (D) is correct.

Explanation

Nitrogen and oxygen do not have vacant d orbitals, so dπ–pπ bonding is not possible. Nitrogen forms pπ–pπ bonds with itself (N≡N). Therefore statement (4) is incorrect.

Explanation

Ambidentate ligands can coordinate through two different donor atoms. Thiocyanate (SCN⁻) can bind through S or N, making it ambidentate. Hence option (C) is correct.

Explanation

Metallic character decreases across a period and increases down a group. Therefore increasing metallic character: P < Si < Be < Mg < Na. Hence option (A) is correct.

Explanation

Cumene → phenol uses O₂/H₂O; CH₃COOH → CH₃CH₂OH via esterification then hydrogenation; CH₃CH₂CH₂OH → CH₃CH(OH)CH₃ via dehydration then hydration; benzene → phenol through sulfonation, fusion with NaOH and acidification. Hence option (B) is correct.

Explanation

Cerium (Z=58) has configuration [Xe]4f¹5d¹6s². In +4 oxidation state it attains stable empty 4f⁰ configuration, making +4 state possible. Hence option (D) is correct.

Explanation

CH₃CH₂CH₂OH + PCl₅ → CH₃CH₂CH₂Cl + POCl₃ + HCl, so X=POCl₃. Elimination with alcoholic KOH gives propene. Addition of HBr in peroxide follows anti-Markovnikov rule, giving CH₃CH₂CH₂Br. Hence option (B) is correct.

Explanation

[PtCl₂(NH₃)₂] → square planar; [Co(NH₃)₆]³⁺ → octahedral; [NiCl₄]²⁻ → tetrahedral; [Fe(CO)₅] → trigonal bipyramidal. Hence option (B) is correct.

Explanation

Phenols react with phthalic anhydride in presence of concentrated H₂SO₄ to form phenolphthalein, which turns pink in NaOH solution. Hence option (B) is correct.

Explanation

Benzene forms toluene, which on nitration gives ortho- and para-nitrotoluene. These isomers have different boiling points and can be separated by fractional distillation under reduced pressure. Hence option (A) is correct.

Explanation

A: molality=(2.5/60)×1000/75=0.556 m (true). B: molarity=(5/40)/(450/1000)=0.278 M (true). C: gases are more soluble in cold water (true). D is false because gas solubility increases with pressure. E is false since mole fraction of B=xB=nB/(nA+nB). Hence option (A) is correct.

Explanation

In Lassaigne’s test, covalently bonded elements (N, S, halogens) are converted into ionic sodium salts for detection. Hence option (B) is correct.

Explanation

Using Faraday’s law: mass=(MIt)/(nF)=63×1.5×600/(2×96487)=0.2938 g. Hence option (B) is correct.

Explanation

For a zero-order reaction: [R]=[R₀]−kt, which gives a straight line with negative slope (−k) when concentration is plotted against time. Hence option (A) is correct.

Explanation

ClF₃ has five electron pairs around Cl (3 bond pairs + 2 lone pairs). Electron geometry is trigonal bipyramidal and molecular geometry is T-shaped. Hence option (A) is correct.

Explanation

Acetate salts release acetic acid with dilute H₂SO₄, producing vinegar smell and acidic vapours that turn blue litmus red. Hence option (C) is correct.

Explanation

Ka=Kw/Kb=10⁻¹⁴/10⁻¹⁰=10⁻⁴, hence pKa=4. Since [X⁻]=[HX], pH=pKa+log([X⁻]/[HX])=4. Hence option (B) is correct.

Explanation

Using Nernst equation: E=E°−(0.059/2)log([H⁺]²/PH₂). Substituting [H⁺]=0.02 M and PH₂=2 atm gives E≈0.109 V. Hence option (D) is correct.

Explanation

Spin-only magnetic moment: μ=√[n(n+2)] BM. For Ti²⁺ (3d²), n=2 unpaired electrons. Thus μ=√8=2.84 BM. Hence option (C) is correct.

Explanation

Oxygen shows oxidation states of 0 (O₂), −1 (peroxides), −2 (oxides), and positive values in compounds with fluorine. Therefore statement (4) is incorrect.

Explanation

Formal charge = Valence electrons − (Nonbonding electrons + ½×Bonding electrons). O₂ has 0 charge, O₁ has +1 charge and O₃ has −1 charge. Hence the order (2,1,3) is 0, +1, −1. Therefore option (B) is correct.

Explanation

Phenolphthalein is colourless in acidic medium and pink in alkaline medium. Near the alkaline endpoint, the colour changes from colourless to pink. Hence option (D) is correct.

Explanation

Reaction: CO₂ + C → 2CO. Let x dm³ CO₂ react. Final volume=(1−x)+2x=1+x=1.4 ⇒ x=0.4. Therefore remaining CO₂=0.6 dm³ and CO formed=0.8 dm³. Hence option (B) is correct.

Explanation

Ethane gives chloroethane under UV chlorination, which reacts with NH₃ to form ethylamine. Diazotisation followed by hydrolysis converts ethyl diazonium salt to ethanol. Hence Z=C₂H₅OH and option (C) is correct.

Explanation

Arrhenius equation: ln k = ln A − Ea/RT. Comparing, Ea/R=1.25×10⁴. Therefore Ea=(1.25×10⁴)(1.987)=24837 cal mol⁻¹≈24.84 kcal mol⁻¹. Hence option (A) is correct.

Explanation

Kp=Kc(RT)^Δng. Kp≠Kc when Δng≠0. For reaction (2), Δng=2−4=−2, so Kp≠Kc. For the other reactions, Δng=0, therefore Kp=Kc. Hence option (B) is correct.

Explanation

Among Mg, Mg²⁺, Al and Al³⁺, neutral Mg has the largest size while Al³⁺ has the smallest size due to high positive charge causing greater contraction. Therefore statement (1) is incorrect.

Explanation

Acetone and chloroform form intermolecular hydrogen bonding between the oxygen atom of acetone and hydrogen of chloroform. This strengthens attractive forces, lowers vapour pressure and causes negative deviation from Raoult’s law. Hence option (B) is correct.

Explanation

Benzene with excess Cl₂/anhydrous AlCl₃ gives hexachlorobenzene (6 chlorine atoms). Benzene with Cl₂ under UV undergoes addition to form benzene hexachloride (BHC), containing 6 chlorine atoms. Therefore X and Y each contain 6 chlorine atoms. Hence option (C) is correct.