NEET 2025 Chemistry Questions with Solutions

Explanation

Using the Rydberg formula: 1/λ = R[(1/n1²) − (1/n2²)] For n = 2 → 3: 1/λ1 = R[(1/2²) − (1/3²)] = 5R/36 For n = 4 → 6: 1/λ2 = R[(1/4²) − (1/6²)] = R/18 Therefore, λ1/λ2 = (R/18)/(5R/36) = 2/5 Comparing the given options and the official key, the correct option is (4).

Explanation

Statement C is correct because $\mathrm{Ar}$, $\mathrm{K^+}$, $\mathrm{Cl^-}$, $\mathrm{Ca^{2+}}$, and $\mathrm{S^{2-}}$ each contain 18 electrons and are therefore isoelectronic species. Statement E is also correct because atomic radius increases down Group 1, hence: $$\mathrm{Cs > Rb > Li}$$ Statement A is incorrect because gallium does not have a very high melting point. Statement B is incorrect because on the Pauling scale, nitrogen and chlorine have nearly equal electronegativity values. Statement D is incorrect because the correct order of first ionization enthalpy is: $$\mathrm{Si > Mg > Al > Na}$$ Therefore, the correct answer is: $$\mathrm{C\ and\ E\ only}$$

Explanation

In qualitative inorganic analysis: • Co2+ belongs to Group IV • Mg2+ belongs to Group VI • Pb2+ belongs to Group I • Al3+ belongs to Group III Thus: A → III B → IV C → I D → II Hence the correct option is (2).

Explanation

In the presence of peroxide, HBr adds to the alkene via anti-Markovnikov addition. Bromine attaches to the less substituted carbon atom. Treatment with KCN substitutes Br by CN through nucleophilic substitution, forming a nitrile. Reduction with Na(Hg)/C₂H₅OH converts the nitrile group $(-CN)$ into a primary amine $(-CH₂NH₂)$. Therefore, the final major product corresponds to option (A).

Explanation

[NiCl₄]²⁻ is tetrahedral with two unpaired electrons and is paramagnetic. [Ni(H₂O)₆]²⁺ is octahedral high-spin d⁸ and paramagnetic. Ni(CO)₄, [Ni(CN)₄]²⁻ and Ni(PPh₃)₄ are diamagnetic.

Explanation

Arsenic forms arsine (AsH₃), analogous to ammonia. Antimony can form antimony pentoxide (Sb₂O₅), so Statement II is incorrect.

Explanation

Main group elements belong to s-block and p-block. [Ne]3s¹ corresponds to sodium (s-block) and [Kr]4d¹⁰5s²5p⁵ corresponds to iodine (p-block). The remaining configurations belong to transition or inner transition elements.

Explanation

Dalton’s atomic theory successfully explained the laws of conservation of mass, constant proportions, and multiple proportions. However, it failed to explain Gay-Lussac’s law of gaseous volumes.

Explanation

In KO₂, potassium has oxidation state +1. In H₂O₂, oxygen has oxidation state −1 due to peroxide linkage. In H₂SO₄, sulfur has oxidation state +6.

Explanation

For a first order reaction: k = 0.693 / t₁/₂ = 0.693 min⁻¹ For 99.9% completion, 0.1% reactant remains: 0.001 = e^(−kt) t = 6.9 / 0.693 ≈ 10 minutes. Hence, the correct answer is 10 minutes.

Explanation

The wavelength of absorbed light is inversely proportional to the crystal field splitting energy $\Delta_o$: $$\lambda \propto \frac{1}{\Delta_o}$$ Using the spectrochemical series: $$\mathrm{CN^- > NH_3 > H_2O}$$ Hence, the crystal field splitting follows: $$[\mathrm{Co(CN)_6}]^{3-} > [\mathrm{Co(NH_3)_6}]^{3+} > [\mathrm{Ti(H_2O)_6}]^{3+} > [\mathrm{Cu(H_2O)_4}]^{2+}$$ Therefore, the wavelength absorbed follows the reverse order: $$[\mathrm{Co(CN)_6}]^{3-} < [\mathrm{Co(NH_3)_6}]^{3+} < [\mathrm{Ti(H_2O)_6}]^{3+} < [\mathrm{Cu(H_2O)_4}]^{2+}$$ Thus, the correct option is: $$\text{B < A < D < C}$$

Explanation

Cis-trans isomerism is possible when restricted rotation exists and substituents are arranged differently in space. 1,2-Dimethylcyclohexane can exist in cis and trans forms due to the cyclic structure.

Explanation

For stepwise ionization of phosphoric acid: $$K = K_{a1}K_{a2}K_{a3}$$ Taking logarithm: $$\log K = \log K_{a1} + \log K_{a2} + \log K_{a3}$$ Hence statement A is true. $H_3PO_4$is stronger than its conjugate acids$H_2PO_4^{-}$and$HPO_4^{2-}$, so statement B is true. For polyprotic acids:$$K_{a1} > K_{a2} > K_{a3}$$ Hence statement C is true. Statement D is incorrect because there is no such relation between the dissociation constants. Therefore, the correct option is (4): A, B and C only.

Explanation

Sodium benzoate with soda lime undergoes decarboxylation to form benzene. n-Hexane undergoes aromatization in presence of catalyst to form benzene. Acetylene polymerizes in a red hot iron tube to produce benzene. However, benzene diazonium chloride with warm water forms phenol, not benzene. Therefore option D is correct.

Explanation

For a weak electrolyte, $$\alpha = \frac{\Lambda_m}{\Lambda_m^{\circ}}$$ The limiting molar conductivity is: $$\Lambda_m^{\circ} = \Lambda_+^{\circ} + \Lambda_-^{\circ}$$ $$\Lambda_m^{\circ} = 349.6 + 50.4 = 400\ \mathrm{S\ cm^2\ mol^{-1}}$$ Therefore, $$\alpha = \frac{90}{400} = 0.225$$ Hence, the degree of dissociation is $0.225$.

Explanation

A molecule having bond order zero is unstable and generally does not exist independently, so Statement I is false. Bond length decreases as bond order increases because stronger bonding pulls atoms closer together. Hence Statement II is also false. Therefore, both statements are false.

Explanation

Conductance in solution depends on the number of ions produced. $[\mathrm{Co}(\mathrm{NH}_3)_3\mathrm{Cl}_3]$ is a neutral complex and does not ionise in solution, hence it shows minimum conductance. $[\mathrm{Co}(\mathrm{NH}_3)_4\mathrm{Cl}_2]$ gives ions in solution to some extent, while: $$[\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{Cl}]\mathrm{Cl} \rightarrow [\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{Cl}]^+ + \mathrm{Cl}^-$$ and $$[\mathrm{Co}(\mathrm{NH}_3)_6]\mathrm{Cl}_3 \rightarrow [\mathrm{Co}(\mathrm{NH}_3)_6]^{3+} + 3\mathrm{Cl}^-$$ produce more ions. Therefore, the minimum conductance is shown by option (A).

Explanation

XeO3 has sp3 hybridization with pyramidal shape. XeF2 has sp3d hybridization with linear geometry. XeOF4 has sp3d2 hybridization with square pyramidal shape. XeF6 has sp3d3 hybridization with distorted octahedral geometry. Therefore, the correct matching is A-II, B-I, C-IV, D-III, which corresponds to option (A).

Explanation

The reaction has ΔH = -74.8 kJ mol⁻¹, which means it is exothermic. Therefore, the products must lie at a lower energy level than the reactants by 74.8 kJ mol⁻¹. Option (A) correctly represents this energy profile.

Explanation

Humidity is an example of liquid in gas solution. Alloys are solid in solid solutions. Amalgams are liquid in solid solutions because mercury is dissolved in a metal. Smoke is an example of solid in gas colloidal system. Hence, the correct matching is A-II, B-I, C-IV, D-III, which corresponds to option (B).

Explanation

Aliphatic amines are more basic than aromatic amines because the lone pair on nitrogen in aromatic amines is delocalized into the benzene ring. Among aliphatic amines, secondary amines are generally more basic than primary amines due to the +I effect of alkyl groups. Therefore, the decreasing order is: N-ethylethanamine > ethanamine > N-methylaniline > benzenamine.

Explanation

Calculate the number of moles and total atoms: A: 212 g Na₂CO₃ = 2 mol, each molecule has 6 atoms → 12 mol atoms B: 248 g Na₂O = 4 mol, each molecule has 3 atoms → 12 mol atoms C: 240 g NaOH = 6 mol, each molecule has 3 atoms → 18 mol atoms D: 12 g H₂ = 6 mol, each molecule has 2 atoms → 12 mol atoms E: 220 g CO₂ = 5 mol, each molecule has 3 atoms → 15 mol atoms Hence A, B, and D have equal number of atoms.

Explanation

Electron-donating alkyl groups decrease the acidity of carboxylic acids due to the +I effect, which destabilizes the carboxylate ion. Formic acid has no alkyl group and is therefore the strongest acid among these. Increasing alkyl substitution decreases acidity. Hence: HCOOH > CH₃COOH > (CH₃)₂CHCOOH > (CH₃)₃CCOOH.

Explanation

Ferromagnetism is an extreme form of paramagnetism, so Statement I is true. Cr²⁺ has electronic configuration [Ar] 3d⁴, containing 4 unpaired electrons. Nd³⁺ has configuration [Xe] 4f³, containing 3 unpaired electrons. Therefore, the number of unpaired electrons is not the same, so Statement II is false.

Explanation

CHCl3 and aniline can be separated by simple distillation due to sufficient difference in boiling points. Crude oil is separated into fractions using fractional distillation. Glycerol from spent-lye is purified by distillation under reduced pressure because glycerol decomposes at high temperature. Aniline and water are separated using steam distillation since aniline is steam volatile. Hence the correct matching is A-IV, B-III, C-I, D-II.

Explanation

For a reversible reaction: Kc = kf / kb Given that the backward rate constant is 2500 times the forward rate constant: Kc = 1 / 2500 = 0.0004 For the reaction: A(g) ⇌ 2B(g) Δn = 2 − 1 = 1 Using: Kp = Kc(RT)^Δn Kp = 0.0004 × (0.0831 × 1000) Kp = 0.0004 × 83.1 = 0.03324 ≈ 0.033 Hence, option (C) is correct.

Explanation

Benzenediazonium salts are prepared by diazotisation of aniline using nitrous acid at low temperature (273–278 K). These salts are unstable and decompose readily in the dry state. Direct iodination of benzene is difficult. Therefore, iodobenzene is commonly prepared by treating benzenediazonium salt with KI. Hence, both Statement I and Statement II are correct, so option (A) is correct.

Explanation

The compound is 2-methylbutane. Monochlorination can occur at four different types of hydrogen positions. Considering stereoisomerism, chlorination at the CH₂ carbon produces two enantiomers. Thus total products obtained are 6.

Explanation

Bond dissociation energy increases with increasing s-character of the hybrid orbital. In compound II, the marked C–H bond is on an sp carbon, so it has highest bond dissociation energy. In compound I, the carbon is sp² hybridised, while in compound III the bond is on an sp³ carbon. Hence the order is II > I > III.

Explanation

Cyclohexane is a saturated hydrocarbon and does not react with bromine water under normal conditions. Phenol, styrene and aniline decolourize bromine water due to electrophilic substitution or addition reactions.

Explanation

Excess CH₃MgBr reacts first with the ketone group and also with the nitrile group. After acidic hydrolysis, the nitrile converts into a ketone while the original ketone becomes a tertiary alcohol. Therefore, the major product formed is the hydroxy ketone shown in option (B).

Explanation

Elevation in boiling point depends on i × C, where i is van’t Hoff factor and C is concentration. For urea: i × C = 1 × 0.01 = 0.01 For KNO₃: i × C = 2 × 0.01 = 0.02 For Na₂SO₄: i × C = 3 × 0.01 = 0.03 For glucose: i × C = 1 × 0.015 = 0.015 Na₂SO₄ gives the highest effective particle concentration, hence highest boiling point.

Explanation

Haber process uses Fe catalyst. Wacker oxidation uses PdCl₂ catalyst. Wilkinson catalyst is [(PPh₃)₃RhCl]. Ziegler catalyst consists of TiCl₄ and trialkyl aluminium compounds such as Al(CH₃)₃. Hence the correct matching is A-I, B-II, C-III, D-IV.

Explanation

According to Raoult's law, the ideal vapour pressure is: Pideal = (5/15 × 63) + (10/15 × 78) = 21 + 52 = 73 torr Observed vapour pressure is 70 torr, which is less than the ideal value. Hence the solution shows negative deviation from Raoult's law.

Explanation

D-Fructose is present in honey, is a ketohexose sugar, exists in α and β anomeric forms, and is laevorotatory. Therefore, the correct answer is D-Fructose.

Explanation

The reaction converts an ester into an aldehyde. Diisobutylaluminium hydride (DIBAL-H), represented as AlH(iBu)₂, selectively reduces esters to aldehydes under controlled conditions followed by hydrolysis. Hence option B is correct.

Explanation

In SN2 reactions, the rate depends on the leaving group ability. Iodide ion is a better leaving group than chloride ion because of its larger size and weaker C-I bond. Therefore, iodoalkanes undergo SN2 reactions faster than chloroalkanes.

Explanation

Using Hess’s law: For crystallisation: $$\mathrm{Ba^{2+}(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s)}$$ $$\Delta H = -4.5\ \text{kcal/mol}$$ Applying enthalpy relation: $$\Delta H = \Delta H_f^\circ[\mathrm{BaSO_4(s)}] - \left(\Delta H_f^\circ[\mathrm{Ba^{2+}}] + \Delta H_f^\circ[\mathrm{SO_4^{2-}}]\right)$$ Substituting values: $$-4.5 = -349 - \left(x + (-216)\right)$$ $$-4.5 = -349 - x + 216$$ $$-4.5 = -133 - x$$ $$x = -128.5\ \text{kcal/mol}$$ Hence, the correct answer is option (A).

Explanation

Considering all possible cyclic ether structures for molecular formula C4H8O along with their stereoisomers, the total number of isomers obtained is 10.

Explanation

A is correct because dipole moment order is H2O > NH3 > CHCl3. D is also correct because bond enthalpy order is N2 > O2 > H2. Statement B is incorrect since XeF2 has more lone pairs than XeO3. Statement C is incorrect because actual bond length order differs from the given sequence.

Explanation

The forward reaction is endothermic because ΔH is positive. According to Le Chatelier’s principle, increasing temperature favors the forward reaction and increases NO yield. Increasing the concentration of reactants $$N_2$$ or $$O_2$$ also shifts equilibrium towards product formation. Hence statements A, C and D are correct.

Explanation

For a first order reaction: $$t = \frac{2.303}{k} \log \frac{[R]_0}{[R]}$$ Substituting values: $$t = \frac{2.303}{0.03} \log \frac{7.2}{0.9}$$ $$= \frac{2.303}{0.03} \log 8$$ Since $$\log 8 = 3 \log 2 = 3 \times 0.301 = 0.903$$ $$t = \frac{2.303 \times 0.903}{0.03} \approx 69.3\ s$$ Hence the correct answer is 69.3 s.

Explanation

In Lassaigne’s test, sodium fusion converts covalently bonded elements like nitrogen, sulphur, and halogens into ionic compounds such as NaCN, Na2S, and NaX respectively. The reaction: $$2CuO + C \rightarrow 2Cu + CO_2$$ is related to oxidation/reduction involving copper oxide and carbon, and does not belong to Lassaigne’s test. Hence, option (D) is correct.