NEET 2024 — Chemistry

Glucose has an aldehyde group, so it reduces Tollens' reagent and adds HCN and NH${}_2$OH. It does not give the Schiff's test and does not form a bisulphite adduct with NaHSO${}_3$ — anomalies explained by its cyclic hemiacetal structure. So B and E.

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

$\mathrm{BaC}{\mathrm{l}}_2+\mathrm{N}{\mathrm{a}}_2\mathrm{S}{\mathrm{O}}_4\to \mathrm{BaS}{\mathrm{O}}_4+2NaCl$ 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).

$Q_c=\frac{[\mathrm{B}][\mathrm{C}]}{[\mathrm{A}{]}^2}=\frac{(2\times 10^{-3}{)}^2}{(2\times 10^{-3}{)}^2}=1$, which is greater than $K_c=4\times 10^{-3}$. The reaction proceeds backward to reduce $Q_c$.

$\mathrm{N}{\mathrm{O}}_2^{-}$ can bind through N or O — linkage isomerism (A-II). Exchanging $\mathrm{S}{\mathrm{O}}_4^{2-}$ and $\mathrm{B}{\mathrm{r}}^{-}$ 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: $\mathrm{B}{\mathrm{H}}_3$ then ${\mathrm{H}}_2{\mathrm{O}}_2/O{\mathrm{H}}^{-}$ 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–${\mathrm{H}}_2\mathrm{O}$ (IV). B: benzophenone from benzene is Friedel–Crafts acylation with benzoyl chloride/$\mathrm{AlC}{\mathrm{l}}_3$ (I). C: cyclohexanol to cyclohexanone uses $\mathrm{Cr}{\mathrm{O}}_3$ (II). D: ethylbenzene to benzoate needs $\mathrm{KMn}{\mathrm{O}}_4/KOH$, $\Delta ($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 $\mathrm{PC}{\mathrm{l}}_5\rightleftharpoons \mathrm{PC}{\mathrm{l}}_3+\mathrm{C}{\mathrm{l}}_2$, $\Delta n_g=2-1=1$, so $K_p\ne K_c$. 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, $(\mathrm{C}{\mathrm{H}}_3{)}_3\mathrm{C}-\mathrm{OH}$ (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 $\mathrm{AlC}{\mathrm{l}}_3$ through its nitrogen lone pair, so Friedel–Crafts fails. Gabriel synthesis needs an alkyl halide to undergo $S_{N}2$ with phthalimide; aryl halides do not, so aniline cannot be made this way. Both true.

$\mathrm{M}{\mathrm{n}}^{3+}$ is $d^4$; gaining an electron gives the stable half-filled $d^5$ configuration of $\mathrm{M}{\mathrm{n}}^{2+}$. 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 –NO${}_2$ 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 $6.022\times 10^{23}$ 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. ${\mathrm{H}}_2\mathrm{O}\to \frac{1}{2}{\mathrm{O}}_2$: 2 e⁻ per mole (II). $\mathrm{Mn}{\mathrm{O}}_4^{-}\to \mathrm{M}{\mathrm{n}}^{2+}$: 5 e⁻ (IV). $\mathrm{C}{\mathrm{a}}^{2+}\to \mathrm{Ca}$: 2 e⁻ per mole, so 1.5 mol needs 3 F (I). $\mathrm{FeO}\to \mathrm{F}{\mathrm{e}}_2{\mathrm{O}}_3$: 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: $\mathrm{T}{\mathrm{i}}^{3+}$ ($d^1$) 1, $\mathrm{C}{\mathrm{r}}^{2+}$ ($d^4$) 4, $\mathrm{M}{\mathrm{n}}^{2+}$ ($d^5$) 5, $\mathrm{F}{\mathrm{e}}^{2+}$ ($d^6$) 4, $\mathrm{S}{\mathrm{c}}^{3+}$ ($d^0$) 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, $(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{CH}-\mathrm{CH}(\mathrm{C}{\mathrm{H}}_3{)}_2$, 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 ($2\times 10^{-5}$) > C (35) > A (145).

${\mathrm{S}}_{\mathrm{N}}1$ 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 ${\mathrm{S}}_{\mathrm{N}}1$ 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 $\mathrm{C}{\mathrm{o}}^{3+}$ ($d^6$) complexes. $\mathrm{N}{\mathrm{H}}_3$ 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 (${\mathrm{H}}_2\mathrm{S}<{\mathrm{H}}_2\mathrm{Se}<{\mathrm{H}}_2\mathrm{Te}$), 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.

$\mathrm{N}{\mathrm{H}}_3$: $s{p}^3$ with one lone pair — trigonal pyramidal (I). $\mathrm{Br}{\mathrm{F}}_5$: $s{p}^3{d}^2$ with one lone pair — square pyramidal (IV). $\mathrm{Xe}{\mathrm{F}}_4$: $s{p}^3{d}^2$ with two lone pairs — square planar (II). $\mathrm{S}{\mathrm{F}}_6$: octahedral (III).

Arrhenius: $\ln k=\ln A-\frac{E_a}{R}\cdot \frac{1}{T}$. 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).

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

$\mathrm{C}{\mathrm{O}}_3^{2-}$ has three equivalent resonance structures with the double bond on each oxygen in turn. $\mathrm{B}{\mathrm{F}}_3$ is trigonal planar with zero dipole; $\mathrm{N}{\mathrm{H}}_3$ has a larger dipole than $\mathrm{N}{\mathrm{F}}_3$; ozone has two resonance structures.

$\mathrm{PB}{\mathrm{r}}_3$ 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. $\mathrm{C}{\mathrm{e}}^{4+}$ is $4f^0$ and $\mathrm{Y}{\mathrm{b}}^{2+}$ is $4f^{14}$ — both diamagnetic. $\mathrm{C}{\mathrm{e}}^{3+}$ ($f^1$), $\mathrm{E}{\mathrm{u}}^{2+}$/$\mathrm{G}{\mathrm{d}}^{3+}$ ($f^7$), $\mathrm{E}{\mathrm{u}}^{3+}$ ($f^6$), $\mathrm{P}{\mathrm{m}}^{3+}$ ($f^4$), $\mathrm{S}{\mathrm{m}}^{3+}$ ($f^5$) 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 $\mathrm{AB}{\mathrm{C}}_3$.

Qualitative-analysis groups: $\mathrm{C}{\mathrm{u}}^{2+}$ Group II, $\mathrm{A}{\mathrm{l}}^{3+}$ Group III, $\mathrm{C}{\mathrm{o}}^{2+}$ Group IV, $\mathrm{B}{\mathrm{a}}^{2+}$ Group V, $\mathrm{M}{\mathrm{g}}^{2+}$ Group VI. Increasing group order: B, A, D, C, E.

$K_c=\frac{[{\mathrm{N}}_2][{\mathrm{O}}_2]}{[\mathrm{NO}{]}^2}=\frac{3.0\times 10^{-3}\times 4.2\times 10^{-3}}{(2.8\times 10^{-3}{)}^2}\approx 1.607$. 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 $\mathrm{F}{\mathrm{e}}^{2+}$ (which would otherwise give a turbid basic salt) without oxidising it; nitric acid would oxidise $\mathrm{F}{\mathrm{e}}^{2+}$ to $\mathrm{F}{\mathrm{e}}^{3+}$, and HCl introduces chloride.

$\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{I}$ with NaCN gives butanenitrile (A). Partial hydrolysis converts the nitrile to the amide, butanamide (B). Hofmann bromamide degradation with $\mathrm{B}{\mathrm{r}}_2/NaOH$ 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 $\mathrm{KMn}{\mathrm{O}}_4$ 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 ($[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_6{]}^{3+}$); a heteroleptic one has more than one kind ($[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_4\mathrm{C}{\mathrm{l}}_2{]}^{+}$). 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.

$\mathrm{PC}{\mathrm{l}}_3$ with an alcohol gives the alkyl chloride and phosphorous acid ${\mathrm{H}}_3\mathrm{P}{\mathrm{O}}_3$; $\mathrm{PC}{\mathrm{l}}_5$ gives the alkyl chloride, HCl and phosphoryl chloride $\mathrm{POC}{\mathrm{l}}_3$.