NEET 2024 — Chemistry

Glucose has an aldehyde group, so it reduces Tollens' reagent and adds HCN and NH₂OH. It does not give the Schiff's test and does not form a bisulphite adduct with NaHSO₃ — anomalies explained by its cyclic hemiacetal structure. So B and E.

$E_n = -\frac{Z^2}{n^2} \times 13.6$ eV. For $\mathrm{He^+}$ ($Z=2$, $n=1$): factor $4$. For $\mathrm{Be^{3+}}$ ($Z=4$, $n=2$): factor $\frac{16}{4} = 4$. Same factor, so the energy is $-x$ J.

$\mathrm{BaCl_2 + Na_2SO_4 \rightarrow BaSO_4 + 2NaCl}$ is a double displacement (precipitation) with no change in any oxidation number. The other three involve electron transfer.

Isothermal: constant temperature (II). Isochoric: constant volume (III). Isobaric: constant pressure (IV). Adiabatic: no heat exchange (I).

$Q_c = \frac{[\mathrm{B}][\mathrm{C}]}{[\mathrm{A}]^2} = \frac{(2\times10^{-3})^2}{(2\times10^{-3})^2} = 1$, which is greater than $K_c = 4\times10^{-3}$. The reaction proceeds backward to reduce $Q_c$.

$\mathrm{NO_2^-}$ can bind through N or O — linkage isomerism (A-II). Exchanging $\mathrm{SO_4^{2-}}$ and $\mathrm{Br^-}$ between coordination sphere and counter-ion — ionisation isomerism (B-III). Two complex ions can swap ligands — coordination isomerism (C-IV). Water inside versus outside the sphere — solvate/hydrate isomerism (D-I).

Entropy rises when disorder rises: liquid → vapour (A), a solid giving two gases (C), and one gas molecule becoming two (D). Cooling a crystal towards 0 K lowers entropy (B). So A, C and D.

Converting a terminal alkene to the aldehyde with one more carbon at the chain end needs anti-Markovnikov hydration: $\mathrm{BH_3}$ then $\mathrm{H_2O_2/OH^-}$ gives the primary alcohol, and PCC oxidises it gently to the aldehyde without going on to the acid — option (1).

A: cleaving a C=C to two ketones is ozonolysis, (i) $\mathrm{O_3}$ (ii) Zn–$\mathrm{H_2O}$ (IV). B: benzophenone from benzene is Friedel–Crafts acylation with benzoyl chloride/$\mathrm{AlCl_3}$ (I). C: cyclohexanol to cyclohexanone uses $\mathrm{CrO_3}$ (II). D: ethylbenzene to benzoate needs $\mathrm{KMnO_4/KOH}$, $\Delta ($III). A-IV, B-I, C-II, D-III.

$K_p = K_c (RT)^{\Delta n_g}$; they are equal only when $\Delta n_g = 0$. For $\mathrm{PCl_5 \rightleftharpoons PCl_3 + Cl_2}$, $\Delta n_g = 2 - 1 = 1$, so $K_p \neq K_c$. The other three have equal moles of gas on both sides.

The Lucas test turns turbid instantly only with tertiary alcohols, which form a stable tertiary carbocation. 2-Methylpropan-2-ol, $\mathrm{(CH_3)_3C{-}OH}$ (structure 3), is tertiary; (1) is secondary and (2), (4) are primary.

Branching makes a molecule more compact and spherical, reducing surface contact and hence the van der Waals forces, so boiling point falls from n-pentane to isopentane to neopentane. Both statements are correct and II explains I.

Aniline forms a complex with the Lewis-acid catalyst $\mathrm{AlCl_3}$ through its nitrogen lone pair, so Friedel–Crafts fails. Gabriel synthesis needs an alkyl halide to undergo $S_N2$ with phthalimide; aryl halides do not, so aniline cannot be made this way. Both true.

$\mathrm{Mn^{3+}}$ is $d^4$; gaining an electron gives the stable half-filled $d^5$ configuration of $\mathrm{Mn^{2+}}$. This extra stability makes the reduction unusually favourable, i.e. $E^\circ$ unusually positive.

A solid that passes directly to vapour on heating (e.g. camphor, naphthalene, ammonium chloride) is purified by sublimation: the vapour is condensed back to pure solid, leaving non-volatile impurities behind.

Fehling's solution A is aqueous copper(II) sulphate; Fehling's B is alkaline sodium potassium tartrate. They are mixed just before use.

Ethane: one C–$C \sigma$ bond (III). Ethene: one $\sigma$ and one $\pi ($IV). Ethyne: one $\sigma$ and two $\pi (I)$. By molecular orbital theory the $\mathrm{C_2}$ molecule has a double bond made of two $\pi$ bonds only (II). Hence A-III, B-IV, C-II, D-I.

Only in o-nitrophenol are the –OH and –NO₂ groups close enough to form a hydrogen bond within the same molecule (a six-membered chelate ring). The m- and p-isomers and HF hydrogen-bond only between molecules.

One mole contains $6.022\times10^{23}$ atoms, so 4 mol has the most. 4 g of He is 1 mol; 22.71 L at STP is 1 mol, so 2.271 L is 0.1 mol; 4 u is a single atom.

Faradays needed equal electrons transferred. $\mathrm{H_2O \to \tfrac12 O_2}$: 2 e⁻ per mole (II). $\mathrm{MnO_4^- \to Mn^{2+}}$: 5 e⁻ (IV). $\mathrm{Ca^{2+} \to Ca}$: 2 e⁻ per mole, so 1.5 mol needs 3 F (I). $\mathrm{FeO \to Fe_2O_3}$: Fe goes +2 → +3, 1 e⁻ per Fe, 1 F (III). Hence A-II, B-IV, C-I, D-III.

Polonium, the heaviest and most metallic member, does not show the −2 state; O, S, Se and Te all form −2 ions or hydrides.

Spin-only moment depends on unpaired electrons: $\mathrm{Ti^{3+}}$ ($d^1$) 1, $\mathrm{Cr^{2+}}$ ($d^4$) 4, $\mathrm{Mn^{2+}}$ ($d^5$) 5, $\mathrm{Fe^{2+}}$ ($d^6$) 4, $\mathrm{Sc^{3+}}$ ($d^0$) 0. B and D both have 4 unpaired electrons, $\mu = \sqrt{24} \approx 4.9$ BM.

A tertiary carbon is bonded to three other carbons. 2,3-Dimethylbutane, $\mathrm{(CH_3)_2CH{-}CH(CH_3)_2}$, has two such carbons (C2 and C3). The others have at most one, or none.

Henry's law $p = K_H x$: a larger $K_H$ means a lower mole fraction dissolved at a given pressure. Solubility therefore increases as $K_H$ falls: B ($2\times10^{-5}$) > C (35) > A (145).

$\mathrm{S_N1}$ rate depends on carbocation stability. 1-Bromo-1-phenylethane (3) ionises to a secondary benzylic cation stabilised by resonance with the ring — far more stable than the secondary cyclohexyl (1) or the primary (4) cations; bromobenzene (2) does not undergo $\mathrm{S_N1}$ at all.

The 1-methylcyclohexyl cation (3) is tertiary — three alkyl groups stabilise it by hyperconjugation and +I effect. (1) and (4) are secondary and (2) is primary.

Both are octahedral $\mathrm{Co^{3+}}$ ($d^6$) complexes. $\mathrm{NH_3}$ is a strong-field ligand, giving low-spin $t_{2g}^6$ — diamagnetic. $\mathrm{F^-}$ is weak-field, giving high-spin with four unpaired electrons — paramagnetic. Both statements are true.

1 g NaOH $= \frac{1}{40} = 0.025$ mol. HCl supplied $= 0.025 \times 0.75 = 0.01875$ mol. NaOH left $= 0.025 - 0.01875 = 0.00625$ mol $= 0.25$ g $= 250$ mg.

Boiling points rise with molar mass down the group ($\mathrm{H_2S < H_2Se < H_2Te}$), but water sits above all of them because of extensive hydrogen bonding. Both statements are true.

Ionisation enthalpy generally rises across the period, but Be ($2s^2$, filled subshell) is higher than B (whose $2p$ electron is easier to remove). Order: Li < B < Be < C < N.

The Arrhenius equation in two-temperature form, $\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$, gives $E_a$ from rate constants measured at two temperatures.

Electronegativity rises across a period and falls down a group: Si < C < N < O < F.

$n$ fixes orbital size (II), $l$ its shape (I), $m_l$ its orientation (III), and $m_s$ the spin orientation of the electron (IV). Hence A-III, B-IV, C-I, D-II.

$\mathrm{NH_3}$: $sp^3$ with one lone pair — trigonal pyramidal (I). $\mathrm{BrF_5}$: $sp^3d^2$ with one lone pair — square pyramidal (IV). $\mathrm{XeF_4}$: $sp^3d^2$ with two lone pairs — square planar (II). $\mathrm{SF_6}$: octahedral (III).

Arrhenius: $\ln k = \ln A - \dfrac{E_a}{R}\cdot\dfrac1T$. A plot of $\ln k$ against $1/T$ is a straight line with negative slope $-E_a/R$ and positive intercept $\ln A$ — graph (3).

$w = -2.303\,nRT\log\frac{p_1}{p_2} = -2.303 \times 1 \times 2 \times 298 \times \log 2 = -413.14$ cal. Work done by the gas on expansion is negative in the sign convention used.

$\mathrm{CO_3^{2-}}$ has three equivalent resonance structures with the double bond on each oxygen in turn. $\mathrm{BF_3}$ is trigonal planar with zero dipole; $\mathrm{NH_3}$ has a larger dipole than $\mathrm{NF_3}$; ozone has two resonance structures.

$\mathrm{PBr_3}$ replaces –OH by –Br without rearrangement, giving 1-bromo-2-methylcyclohexane (A). Alcoholic KOH then eliminates HBr by the Saytzeff rule to the more substituted alkene, 1-methylcyclohexene (B) — option (4).

Diamagnetism needs no unpaired $f$ electrons. $\mathrm{Ce^{4+}}$ is $4f^0$ and $\mathrm{Yb^{2+}}$ is $4f^{14}$ — both diamagnetic. $\mathrm{Ce^{3+}}$ ($f^1$), $\mathrm{Eu^{2+}}$/$\mathrm{Gd^{3+}}$ ($f^7$), $\mathrm{Eu^{3+}}$ ($f^6$), $\mathrm{Pm^{3+}}$ ($f^4$), $\mathrm{Sm^{3+}}$ ($f^5$) are all paramagnetic.

Moles per 100 g: A $\frac{32}{64} = 0.5$, B $\frac{20}{40} = 0.5$, C $\frac{48}{32} = 1.5$. Ratio $1 : 1 : 3$, so $\mathrm{ABC_3}$.

Qualitative-analysis groups: $\mathrm{Cu^{2+}}$ Group II, $\mathrm{Al^{3+}}$ Group III, $\mathrm{Co^{2+}}$ Group IV, $\mathrm{Ba^{2+}}$ Group V, $\mathrm{Mg^{2+}}$ Group VI. Increasing group order: B, A, D, C, E.

$K_c = \frac{[\mathrm{N_2}][\mathrm{O_2}]}{[\mathrm{NO}]^2} = \frac{3.0\times10^{-3} \times 4.2\times10^{-3}}{(2.8\times10^{-3})^2} \approx 1.607$. Starting from $0.1$ M NO: $K_c = \frac{(0.05\alpha)^2}{(0.1(1-\alpha))^2} = \frac{\alpha^2}{4(1-\alpha)^2}$, so $\frac{\alpha}{2(1-\alpha)} = 1.268 \Rightarrow \alpha \approx 0.717$.

$\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$ with $k_2/k_1 = 4$, $T_1 = 300$ K, $T_2 = 330$ K: $0.6021 = \frac{E_a}{2.303 \times 8.314} \times \frac{30}{300 \times 330} \Rightarrow E_a \approx 38.04$ kJ/mol.

Dilute sulphuric acid suppresses hydrolysis of $\mathrm{Fe^{2+}}$ (which would otherwise give a turbid basic salt) without oxidising it; nitric acid would oxidise $\mathrm{Fe^{2+}}$ to $\mathrm{Fe^{3+}}$, and HCl introduces chloride.

$\mathrm{CH_3CH_2CH_2I}$ with NaCN gives butanenitrile (A). Partial hydrolysis converts the nitrile to the amide, butanamide (B). Hofmann bromamide degradation with $\mathrm{Br_2/NaOH}$ removes one carbon, giving propylamine (C).

Charge $= It = 9.6487 \times 100 = 964.87$ C $= 0.01$ F. Copper needs 2 F per mole, so $0.005$ mol is deposited: $0.005 \times 63 = 0.315$ g.

Hot acidic $\mathrm{KMnO_4}$ cleaves the C=C of stilbene completely and oxidises each fragment fully: both ends carry one H, so each becomes benzoic acid — product (1).

A homoleptic complex has only one kind of ligand ($\mathrm{[Co(NH_3)_6]^{3+}}$); a heteroleptic one has more than one kind ($\mathrm{[Co(NH_3)_4Cl_2]^+}$). Both statements are true and II is the definition behind I.

$\Pi = CRT$, so the slope of $\Pi$ against $C$ is $RT$: $T = \frac{25.73}{0.083} = 310$ K $= 37^\circ$C.

$\mathrm{PCl_3}$ with an alcohol gives the alkyl chloride and phosphorous acid $\mathrm{H_3PO_3}$; $\mathrm{PCl_5}$ gives the alkyl chloride, HCl and phosphoryl chloride $\mathrm{POCl_3}$.