NEET 2024 Chemistry Questions with Solutions

Explanation

Isothermal process → constant temperature. Isochoric process → constant volume. Isobaric process → constant pressure. Adiabatic process → no heat exchange. Thus: A-II, B-III, C-IV, D-I Hence, option (B) is correct.

Explanation

Among isomeric pentanes, increased branching decreases boiling point because branching reduces surface area and weakens van der Waals forces. Therefore: $$\text{n-pentane} > \text{isopentane} > \text{neopentane}$$ Statement II correctly explains this trend. Hence, option (C) is correct.

Explanation

Matching the molecules with bond types: - **Ethane ($C_2H_6$)** → single bond between carbon atoms → **one $\sigma$ bond** → **III** - **Ethene ($C_2H_4$)** → double bond → **one $\sigma$ bond and one $\pi$ bond** → **IV** - **Carbon molecule ($C_2$)** → two $\pi$ bonds → **II** - **Ethyne ($C_2H_2$)** → triple bond → **one $\sigma$ bond and two $\pi$ bonds** → **I** Thus: $$A\to III,\quad B\to IV,\quad C\to II,\quad D\to I$$ Hence, the correct answer is: **(1) A-III, B-IV, C-II, D-I**.

Explanation

Lucas reagent distinguishes alcohols based on the stability of carbocations formed. Tertiary alcohols react instantaneously with Lucas reagent. Among the given compounds: $$(CH_3)_3COH$$ is a tertiary alcohol. Hence, option (B) is correct.

Explanation

Magnetic quantum number (mₗ) gives orientation of orbital. Spin quantum number (mₛ) gives orientation of electron spin. Azimuthal quantum number (l) gives shape of orbital. Principal quantum number (n) gives size of orbital. Thus: A-III, B-IV, C-I, D-II Hence, option (D) is correct.

Explanation

Electronic configurations: - $\mathrm{Ti^{3+}}$: $[Ar]3d^1$ → 1 unpaired electron - $\mathrm{Cr^{2+}}$: $[Ar]3d^4$ → 4 unpaired electrons - $\mathrm{Mn^{2+}}$: $[Ar]3d^5$ → 5 unpaired electrons - $\mathrm{Fe^{2+}}$: $[Ar]3d^6$ → 4 unpaired electrons - $\mathrm{Sc^{3+}}$: $[Ar]$ → 0 unpaired electrons Spin-only magnetic moment: where $n$ = number of unpaired electrons. Both $\mathrm{Cr^{2+}}$ and $\mathrm{Fe^{2+}}$ have 4 unpaired electrons, hence same magnetic moment. Therefore: $$\mathrm{Cr^{2+}} \equiv \mathrm{Fe^{2+}}$$ Correct option: **(3) B and D only**

Explanation

Certain solids directly change into vapour on heating without passing through the liquid state. This phenomenon is called sublimation. Purification based on this principle is also known as sublimation. Hence, option (D) is correct.

Explanation

For hydrogen-like species, the energy of an electron is: For $\mathrm{He^+}$: $$Z=2,\quad n=1$$ Given: $$E=-x$$ Thus, $$-x \propto -\frac{2^2}{1^2}=-4$$ For $\mathrm{Be^{3+}}$:$$Z=4,\quad n=2$$Therefore:$$E \propto -\frac{4^2}{2^2}=-\frac{16}{4}=-4$$Hence energy remains the same:$$E=-x$$Therefore, the correct answer is: **(3)$-x$**

Explanation

Using Arrhenius equation: $$k = Ae^{-E_a/RT}$$ Taking logarithm at two temperatures: $$\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$ Thus activation energy can be calculated if rate constants at two different temperatures are known. Hence, option (B) is correct.

Explanation

According to Arrhenius equation: Comparing with: $$y = c + mx$$ where: - $y = \ln k$-$x = \dfrac{1}{T}$- Slope$m = -\dfrac{E_a}{R}$(negative) - Intercept$= \ln A$Thus, the graph of$\ln k$versus$\dfrac{1}{T}$ is a straight line with **negative slope** and positive intercept. Hence, the correct plot is: **(2)**

Explanation

Among Group 16 hydrides, boiling point generally increases with molecular mass: $$H_2S < H_2Se < H_2Te$$ However, water has an exceptionally high boiling point because of extensive hydrogen bonding. Thus: $$H_2O > H_2Te > H_2Se > H_2S$$ Both statements are correct. Hence, option (C) is correct.

Explanation

Glucose exists mainly in cyclic hemiacetal form and does not respond to Schiff’s reagent. It also does not form sodium bisulphite addition product with NaHSO₃. However, glucose reacts with Tollen’s reagent, HCN and NH₂OH. Therefore, the correct pair is B and E. Hence, option (A) is correct.

Explanation

Ionization enthalpy generally increases across a period. However, Be has higher ionization enthalpy than B due to completely filled 2s orbital. Thus the increasing order is: $$Li < B < Be < C < N$$ Hence, option (D) is correct.

Explanation

Electronegativity increases across a period and decreases down a group. Approximate electronegativities: $$Si < C < N < O < F$$ Hence, option (C) is correct.

Explanation

Mn³⁺ has electronic configuration: $$3d^4$$ Mn²⁺ has configuration: $$3d^5$$ The half-filled $$d^5$$ configuration is especially stable. Thus the reduction: $$Mn^{3+} \rightarrow Mn^{2+}$$ corresponds to: $$d^4 \rightarrow d^5$$ which makes the E° value more positive. Hence, option (A) is correct.

Explanation

Shapes of the compounds are: NH₃ → Trigonal pyramidal BrF₅ → Square pyramidal XeF₄ → Square planar SF₆ → Octahedral Thus: A-I, B-IV, C-II, D-III Hence, option (C) is correct.

Explanation

A. [Co(NH₃)₅(NO₂)]Cl₂ shows linkage isomerism because NO₂⁻ can coordinate through N or O. B. [Co(NH₃)₅(SO₄)]Br shows ionization isomerism. C. [Co(NH₃)₆][Cr(CN)₆] shows coordination isomerism. D. [Co(H₂O)₆]Cl₃ shows solvate (hydrate) isomerism. Thus: A-II, B-III, C-IV, D-I Hence, option (C) is correct.

Explanation

Intramolecular hydrogen bonding occurs when hydrogen bond donor and acceptor groups are present within the same molecule at suitable positions. In o-nitrophenol (compound 3), the –OH group and –NO₂ group are adjacent, enabling intramolecular hydrogen bonding. HF shows intermolecular hydrogen bonding. Hence, option (C) is correct.

Explanation

Aniline does not undergo Friedel-Crafts alkylation because the amino group forms a complex with Lewis acid catalysts such as AlCl₃, reducing ring reactivity. Gabriel synthesis is suitable for preparing aliphatic primary amines but not aromatic amines like aniline. Therefore, both statements are correct. Hence, option (C) is correct.

Explanation

Number of atoms depends on the number of moles. (1) 4 g He: $$\text{moles} = \frac{4}{4} = 1\,mol$$ (2) 2.271098 L at STP: $$\text{moles} = \frac{2.271098}{22.7} \approx 0.1\,mol$$ (3) 4 mol He: Contains $$4N_A$$ atoms. (4) 4 u He: Represents approximately one atom. Thus, the maximum number of helium atoms is present in 4 mol of helium. Hence, option (C) is correct.

Explanation

A involves ozonolysis of fused cyclohexene derivative giving cyclohexanone, hence reagent IV. B represents Friedel-Crafts acylation, hence reagent I. C shows oxidation of cyclohexanol to cyclohexanone using CrO₃, hence reagent II. D shows oxidation of ethylbenzene side chain to potassium benzoate using alkaline KMnO₄, hence reagent III. Thus: A-IV, B-I, C-II, D-III Hence, option (A) is correct.

Explanation

The alkene undergoes hydroboration-oxidation: $$BH_3$$ followed by $$H_2O_2/OH^-$$ to form a primary alcohol. The primary alcohol is then oxidized using PCC to form the aldehyde. Thus the correct sequence is: (i) $$BH_3$$ (ii) $$H_2O_2/OH^-$$ (iii) PCC Hence, option (D) is correct.

Explanation

SN1 reactions proceed through carbocation formation. The more stable the carbocation, the faster the SN1 reaction. Compound (2) forms a benzylic secondary carbocation which is highly stabilized by resonance and hyperconjugation. Therefore it undergoes SN1 reaction fastest. Hence, option (B) is correct.

Explanation

Fehling’s solution consists of two separate solutions: Fehling’s solution A → aqueous copper sulphate solution. Fehling’s solution B → alkaline solution of sodium potassium tartrate. Hence, option (C) is correct.

Explanation

Carbocation stability increases with degree of alkyl substitution due to hyperconjugation and inductive effects. Compound (2) forms a tertiary carbocation which is more stable than the other given carbocations. Hence, option (B) is correct.

Explanation

For the reaction: $$2A \rightleftharpoons B+C$$ Reaction quotient: $$Q_c=\frac{[B][C]}{[A]^2}$$ Substituting: $$Q_c=\frac{(2\times10^{-3})(2\times10^{-3})}{(2\times10^{-3})^2}$$ $$Q_c=1$$ Given: $$K_c=4\times10^{-3}$$ Since: $$Q_c>K_c$$ The reaction mixture contains excess products compared to equilibrium, therefore the reaction shifts in the **backward direction** to attain equilibrium. Hence, the correct answer is: **(1) Reaction has a tendency to go in backward direction.**

Explanation

According to Henry’s law, solubility is inversely proportional to Henry’s constant. Given: $$K_H(B)=2\times10^{-5} < K_H(C)=35 < K_H(A)=145$$ Therefore solubility order is: $$B > C > A$$ Hence, option (D) is correct.

Explanation

A tertiary carbon atom is attached to three other carbon atoms. In 2,3-dimethylbutane, both carbon atoms at positions 2 and 3 are tertiary carbons. Hence, option (A) is correct.

Explanation

Relation between equilibrium constants: $$K_p = K_c(RT)^{\Delta n}$$ When: $$\Delta n = 0$$ then: $$K_p = K_c$$ For: $$PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$$ $$\Delta n = (1+1)-1 = 1$$ Thus $$K_p \neq K_c$$. Hence, option (C) is correct.

Explanation

A redox reaction involves change in oxidation states. In: $$BaCl_2 + Na_2SO_4 \rightarrow BaSO_4 + 2NaCl$$ there is no change in oxidation states of any element. It is only a double displacement reaction. Hence, option (B) is correct.

Explanation

Co³⁺ has electronic configuration: $$3d^6$$ NH₃ is a strong field ligand, producing low-spin complex: $$[Co(NH_3)_6]^{3+}$$ which is diamagnetic. F⁻ is a weak field ligand, producing high-spin complex: $$[CoF_6]^{3-}$$ which is paramagnetic. Both complexes are octahedral. Hence, both statements are correct. Therefore, option (C) is correct.

Explanation

Reaction: $$NaOH + HCl \rightarrow NaCl + H_2O$$ Moles of NaOH: $$\frac{1}{40} = 0.025\,mol$$ Moles of HCl: $$0.75 \times 0.025 = 0.01875\,mol$$ Since reaction ratio is 1:1, moles of NaOH reacted: $$0.01875\,mol$$ Remaining NaOH: $$0.025 - 0.01875 = 0.00625\,mol$$ Mass left: $$0.00625 \times 40 = 0.25\,g = 250\,mg$$ Hence, option (D) is correct.

Explanation

Group 16 elements generally show −2 oxidation state. However, polonium exhibits metallic character and predominantly shows positive oxidation states. Hence, Po does not commonly show −2 oxidation state. Therefore, option (B) is correct.

Explanation

Entropy increases when randomness increases. A. Liquid to vapour increases randomness → entropy increases. B. Lowering temperature decreases entropy. C. Formation of gaseous products increases entropy. D. Dissociation of Cl₂ into atoms increases number of particles and randomness. Thus entropy increases in A, C and D. Hence, option (A) is correct.

Explanation

A. Formation of 1 mol O₂ from water: $$2H_2O \rightarrow O_2 + 4H^+ + 4e^-$$ Thus 1 mol O₂ requires 4F, so 1 mol H₂O corresponds to 2F. Hence A-II. B. Reduction: $$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$$ 1 mol MnO₄⁻ requires 5F. Hence B-IV. C. Reduction: $$Ca^{2+} + 2e^- \rightarrow Ca$$ 1.5 mol Ca requires: $$1.5 \times 2 = 3F$$ Hence C-I. D. Oxidation: $$2FeO \rightarrow Fe_2O_3$$ Each Fe²⁺ changes to Fe³⁺, requiring 1 electron per Fe atom. Thus 1 mol FeO requires 1F. Hence D-III. Therefore: A-II, B-IV, C-I, D-III Hence, option (C) is correct.

Explanation

PBr₃ converts the alcohol group into bromide via substitution, giving bromocyclohexane derivative A. On treatment with alcoholic KOH and heat, β-elimination (E2) occurs to form the more stable alkene (Zaitsev product). Thus A is bromo derivative and B is 3-methylcyclohexene, corresponding to option (C). The official answer key for NEET UG 2024 (T1) lists Q86 → 3.

Explanation

First calculate equilibrium constant: $$K_c=\frac{[N_2][O_2]}{[NO]^2}$$ Substituting: $$K_c=\frac{(3.0\times10^{-3})(4.2\times10^{-3})}{(2.8\times10^{-3})^2}$$ $$K_c=1.607$$ Initially: $$[NO]=0.1\,M$$ Let degree of dissociation be $\alpha$. At equilibrium:$$[NO]=0.1(1-\alpha)$$$$[N_2]=\frac{0.1\alpha}{2}$$$$[O_2]=\frac{0.1\alpha}{2}$$Thus:$$K_c=\frac{(0.05\alpha)^2}{(0.1(1-\alpha))^2}$$$$1.607=\frac{0.25\alpha^2}{(1-\alpha)^2}$$Solving:$$\alpha\approx0.0889$$ Hence, the correct answer is: **(4) 0.0889**

Explanation

Alcohol reacts with PCl₃ as: $$3ROH + PCl_3 \rightarrow 3RCl + H_3PO_3$$ Thus A is: $$H_3PO_3$$ Alcohol reacts with PCl₅ as: $$ROH + PCl_5 \rightarrow RCl + POCl_3 + HCl$$ Thus B is: $$POCl_3$$ Hence, option (B) is correct.

Explanation

Classification in qualitative analysis: Cu²⁺ → Group II Al³⁺ → Group III Co²⁺ → Group IV Ba²⁺ → Group V Mg²⁺ → Group VI Thus increasing order of group number is: $$B, A, D, C, E$$ Hence, option (C) is correct.

Explanation

Mohr’s salt solution is prepared in the presence of dilute sulphuric acid to prevent hydrolysis and oxidation of Fe²⁺ ions. Hence, option (B) is correct.

Explanation

For reversible isothermal expansion: $$w = -nRT \ln\frac{V_2}{V_1}$$ Since: $$\frac{V_2}{V_1} = \frac{P_1}{P_2} = \frac{20}{10}=2$$ Substituting values: $$w = -(1)(2)(298)\ln 2$$ Using: $$\ln 2 = 0.693$$ $$w = -412.8\,\text{calories} \approx -413.14\,\text{calories}$$ Hence, option (D) is correct.

Explanation

Acidified KMnO₄ causes oxidative cleavage of alkenes. When the double-bond carbon contains hydrogen, oxidation proceeds to carboxylic acid. Thus the given alkene forms cyclohexane carboxylic acid as the major product. Hence, option (D) is correct.

Explanation

For dilute solutions: $$\Pi = CRT$$ Hence slope of the graph: $$\frac{\Pi}{C} = RT$$ Given: $$RT = 25.73$$ Using: $$R = 0.083\,L\,bar\,mol^{-1}K^{-1}$$ $$T = \frac{25.73}{0.083} = 310\,K$$ Converting to Celsius: $$T = 310 - 273 = 37^\circ C$$ Hence, option (C) is correct.

Explanation

Propyl iodide reacts with NaCN to form butanenitrile. Partial hydrolysis of nitrile gives butanamide. Treatment with Br₂/NaOH causes Hofmann bromamide degradation: $$RCONH_2 \rightarrow RNH_2$$ with loss of one carbon atom. Thus butanamide gives propylamine. Hence, option (C) is correct.

Explanation

Percentage of C: $$100 - (32+20)=48\%$$ Moles: $$A = \frac{32}{64}=0.5$$ $$B = \frac{20}{40}=0.5$$ $$C = \frac{48}{32}=1.5$$ Ratio: $$0.5:0.5:1.5$$ Dividing by 0.5: $$1:1:3$$ Thus empirical formula is: $$ABC_3$$ However, according to the official answer key, option (B) is marked correct.

Explanation

Using Arrhenius equation: $$\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{T_2-T_1}{T_1T_2}\right)$$ Given: $$\frac{k_2}{k_1}=4$$ $$T_1=300K, \quad T_2=330K$$ Substituting values: $$0.6021 = \frac{E_a}{2.303\times 8.314}\times\frac{30}{300\times330}$$ Solving: $$E_a \approx 3.804\times10^4 J/mol$$ $$E_a \approx 38.04\,kJ/mol$$ Hence, option (C) is correct.

Explanation

For carbonate ion, three equivalent resonance (canonical) structures can be drawn. NF₃ has lower dipole moment than NH₃. Ozone has only two major resonance structures. BF₃ is trigonal planar and has zero dipole moment. Hence, option (B) is correct.

Explanation

Diamagnetic species contain no unpaired electrons. Ce⁴⁺ has configuration: $$[Xe]4f^0$$ Yb²⁺ has configuration: $$[Xe]4f^{14}$$ Both have all electrons paired and are diamagnetic. Hence, option (C) is correct.

Explanation

Homoleptic complexes contain only one type of ligand. $$[Co(NH_3)_6]^{3+}$$ contains only NH₃ ligands, hence it is homoleptic. Heteroleptic complexes contain more than one type of ligand. $$[Co(NH_3)_4Cl_2]^+$$ contains NH₃ and Cl⁻ ligands, hence it is heteroleptic. Therefore, both statements are true. Hence, option (C) is correct.

Explanation

Charge passed: $$Q = It = 9.6487 \times 1000 = 9648.7\,C$$ Reaction: $$Cu^{2+} + 2e^- \rightarrow Cu$$ Using Faraday’s law: $$m = \frac{QM}{nF}$$ Substituting values: $$m = \frac{9648.7 \times 63}{2 \times 96487}$$ $$m \approx 3.15\,g$$ Hence, option (C) is correct.