NEET 2025 — Chemistry

$\frac{1}{\lambda }\propto \frac{1}{n_1^2}-\frac{1}{n_2^2}$. For $2\to 3$: $\frac{1}{4}-\frac{1}{9}=\frac{5}{36}$. For $4\to 6$: $\frac{1}{16}-\frac{1}{36}=\frac{5}{144}$. So $\frac{{\lambda }_{2\to 3}}{{\lambda }_{4\to 6}}=\frac{5/144}{5/36}=\frac{1}{4}$.

Ar, ${\mathrm{K}}^{+}$, $\mathrm{C}{\mathrm{l}}^{-}$, $\mathrm{C}{\mathrm{a}}^{2+}$ and ${\mathrm{S}}^{2-}$ all have 18 electrons (C true); atomic radius grows down Group 1, so Cs > Rb > Li (E true). Ga in fact has a low melting point (A false); N and Cl share the Pauling value 3.0 (B false); Mg > Al in first ionisation enthalpy (D false). Hence C and E.

$\mathrm{P}{\mathrm{b}}^{2+}$ precipitates as the chloride in Group I; $\mathrm{A}{\mathrm{l}}^{3+}$ as the hydroxide in Group III; $\mathrm{C}{\mathrm{o}}^{2+}$ as the sulphide in Group IV; $\mathrm{M}{\mathrm{g}}^{2+}$ is in Group VI. So A-III, B-IV, C-I, D-II.

HBr with peroxide adds anti-Markovnikov, placing Br on the less substituted ring carbon (C-2) of 1-methylcyclopentene. KCN substitutes Br by –CN, and Na/ethanol reduces the nitrile to $–C{\mathrm{H}}_2\mathrm{N}{\mathrm{H}}_2$. P is 1-methyl-2-(aminomethyl)cyclopentane — structure (1).

$E_n=-R_H\frac{Z^2}{n^2}$ and $r_n=a_0\frac{n^2}{Z}$. $\mathrm{H}{\mathrm{e}}^{+}$ ($Z=2$): $E=-4\times 2.18\times 10^{-18}=-8.72\times 10^{-18}$ J, $r=52.9/2=26.4$ pm. $\mathrm{L}{\mathrm{i}}^{2+}$ ($Z=3$): $E=-9\times 2.18\times 10^{-18}=-19.62\times 10^{-18}$ J, $r=52.9/3=17.6$ pm.

$[\mathrm{NiC}{\mathrm{l}}_4{]}^{2-}$ ($\mathrm{N}{\mathrm{i}}^{2+}$, $d^8$, weak-field tetrahedral) has two unpaired electrons; so does $[\mathrm{Ni}({\mathrm{H}}_2\mathrm{O}{)}_6{]}^{2+}$ (octahedral $d^8$). $\mathrm{Ni}(\mathrm{CO}{)}_4$ and $\mathrm{Ni}(\mathrm{PP}{\mathrm{h}}_3{)}_4$ contain Ni(0), $d^{10}$; $[\mathrm{Ni}(\mathrm{CN}{)}_4{]}^{2-}$ is square planar $d^8$, all paired. Paramagnetic: A and D.

Arsenic forms arsine, $\mathrm{As}{\mathrm{H}}_3$ (I correct). Antimony does form the pentoxide $\mathrm{S}{\mathrm{b}}_2{\mathrm{O}}_5$ (II incorrect).

Main-group elements have their outermost electrons in $s$ or $p$ orbitals: $[\mathrm{Ne}]3{\mathrm{s}}^1$ (Na) and $[\mathrm{Kr}]4{\mathrm{d}}^{10}5{\mathrm{s}}^{2}5{\mathrm{p}}^5$ (I). B and D are $d$-block, E is $f$-block (a $6d$ actinoid configuration). So A and C.

Dalton's theory treated atoms as indivisible and gave no account of molecules, so it could not explain Gay-Lussac's law of gaseous volumes — that needed Avogadro's hypothesis.

In $\mathrm{K}{\mathrm{O}}_2$ the superoxide ion ${\mathrm{O}}_2^{-}$ gives oxygen $-\frac{1}{2}$ each, i.e. K is $+1$ (the element asked); in ${\mathrm{H}}_2{\mathrm{O}}_2$ oxygen is $-1$; in ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$ sulphur is $+6$. Matching the underlined elements gives $+1,-1,+6$.

99.9% completion leaves $\frac{1}{1000}$ of the reactant, i.e. about $2^{10}$-fold reduction — ten half-lives. $t\approx 10\times 1=10$ minutes.

Wavelength absorbed is inversely related to ${\Delta }_0$. $\mathrm{C}{\mathrm{N}}^{-}$ (strongest field) gives the largest splitting, then $\mathrm{N}{\mathrm{H}}_3$; $\mathrm{T}{\mathrm{i}}^{3+}$ with water splits less, and $\mathrm{C}{\mathrm{u}}^{2+}$ with water least (absorbs in the red, ~800 nm). So B < A < D < C.

Cis–trans isomerism needs two different groups on each of two ring carbons (or double-bond carbons). 1,2-Dimethylcyclohexane can place the two methyls on the same or opposite faces of the ring. Pent-1-ene and 2-methylhex-2-ene have identical groups on one double-bond carbon; 1,1-dimethylcyclopropane has both methyls on one carbon.

The overall constant is the product $K=K_{a_1}{K}_{a_2}{K}_{a_3}$, so the logs add (A). Removing successive protons from an increasingly negative ion gets harder, so $K_{a_1}>K_{a_2}>K_{a_3}$ (C) and ${\mathrm{H}}_3\mathrm{P}{\mathrm{O}}_4$ is the strongest acid of the three (B). D is baseless. Hence A, B and C.

Sodium benzoate with soda lime decarboxylates to benzene; n-hexane over $\mathrm{M}{\mathrm{o}}_2{\mathrm{O}}_3$ at 773 K aromatises to benzene; ethyne passed through a red-hot iron tube trimerises to benzene. Benzenediazonium chloride warmed with water gives phenol, not benzene — reaction (4).

${\Lambda }_m^{\circ }=349.6+50.4=400\mathrm{S}\mathrm{c}{\mathrm{m}}^2\mathrm{mo}{\mathrm{l}}^{-1}$. Degree of dissociation $\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^{\circ }}=\frac{90}{400}=0.225$.

Bond order zero means no net bonding — the molecule does not form (I false). Higher bond order means a shorter, stronger bond, so bond length decreases (II false).

$\mathrm{Xe}{\mathrm{O}}_3$: $s{p}^3$ with one lone pair, pyramidal (II). $\mathrm{Xe}{\mathrm{F}}_2$: $s{p}^{3}d$ with three lone pairs, linear (I). $\mathrm{XeO}{\mathrm{F}}_4$: $s{p}^3{d}^2$ with one lone pair, square pyramidal (IV). $\mathrm{Xe}{\mathrm{F}}_6$: $s{p}^3{d}^3$, distorted octahedral (III).

The reaction is exothermic ($\Delta H=-74.8$ kJ/mol), so the products must lie 74.8 kJ below the reactants, and there must still be an activation barrier between them. Only diagram (1) shows a hump followed by products lower than reactants.

Humidity is water vapour (liquid) dispersed in air (gas) — II. Alloys are solid in solid — I. Amalgams are mercury (liquid) in a solid metal — IV. Smoke is solid particles in gas — III. Hence A-II, B-I, C-IV, D-III.

Aliphatic amines are stronger bases than aromatic ones because the aryl lone pair is delocalised into the ring. Among aliphatic amines, the secondary N-ethylethanamine beats primary ethanamine (inductive effect plus solvation balance); among aromatic ones, the N-methyl group makes N-methylaniline slightly stronger than aniline. Order: N-ethylethanamine > ethanamine > N-methylaniline > benzenamine.

Atoms $=$ moles $\times$ atoms per formula unit. A: $2\times 6=12$ mol atoms. B: $4\times 3=12$. C: $6\times 3=18$. D: $6\times 2=12$. E: $5\times 3=15$. A, B and D are equal.

Vitamin $B_{12}$ deficiency causes pernicious anaemia (IV); vitamin D, rickets (III); vitamin $B_2($riboflavin), cheilosis (I); vitamin $B_6($pyridoxine), convulsions (II).

Alkyl groups are electron-releasing and destabilise the carboxylate anion, so acidity falls as alkyl substitution grows: $\mathrm{HCOOH}>\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}>(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{CHCOOH}>(\mathrm{C}{\mathrm{H}}_3{)}_3\mathrm{CCOOH}$.

Ferromagnetism is indeed treated as an extreme form of paramagnetism (I true). $\mathrm{C}{\mathrm{r}}^{2+}$ is $3d^4$ with 4 unpaired electrons; $\mathrm{N}{\mathrm{d}}^{3+}$ is $4f^3$ with 3 — not the same (II false).

Chloroform (b.p. 61 °C) and aniline (184 °C) differ widely — simple distillation (IV). Crude oil fractions — fractional distillation (III). Glycerol decomposes at its boiling point — distillation under reduced pressure (I). Aniline from water — steam distillation (II). Hence A-IV, B-III, C-I, D-II.

$K_c=\frac{k_f}{k_b}=\frac{1}{2500}=4\times 10^{-4}$. $\Delta n_g=2-1=1$, so $K_p=K_c(RT)=4\times 10^{-4}\times 0.0831\times 1000\approx 0.033$.

Diazotisation of aniline with $\mathrm{HN}{\mathrm{O}}_2$ at 273–278 K gives benzenediazonium chloride, which is unstable when dry (I correct). Direct iodination of benzene is reversible and poor, so iodobenzene is made from the diazonium salt and KI (II correct).

Monochlorination of 2-methylbutane at each distinct hydrogen: C-1 (the two equivalent methyls) gives 1-chloro-2-methylbutane, chiral at C-2 (2 enantiomers); C-2 gives 2-chloro-2-methylbutane (1); C-3 gives 2-chloro-3-methylbutane, chiral (2); C-4 gives 1-chloro-3-methylbutane (1). Total $2+1+2+1=6$.

Bond strength rises with the $s$-character of the carbon orbital. The terminal alkyne C–H (II, $sp$) is strongest, then the aromatic C–H (I, $s{p}^2$), then the cyclopropane C–H (III, roughly $s{p}^{2.5}$ but weaker than aryl). II > I > III.

Bromine water is decolourised by addition to C=C (styrene) and by electrophilic bromination of activated rings (phenol, aniline give tribromo products). Cyclohexane (1) is saturated and unactivated, so it does not react.

Excess $\mathrm{C}{\mathrm{H}}_3\mathrm{MgBr}$ attacks both electrophilic sites: the ketone becomes a tertiary alcohol (–C(OH)(CH${}_3)$Ph), and the nitrile adds one equivalent to form an imine salt that hydrolyses to a methyl ketone on work-up. Product (2): the hydroxy-ketone.

Boiling-point elevation depends on the total particle concentration $iC$. Urea: $0.01$. $\mathrm{KN}{\mathrm{O}}_3$: $2\times 0.01=0.02$. $\mathrm{N}{\mathrm{a}}_2\mathrm{S}{\mathrm{O}}_4$: $3\times 0.01=0.03$. Glucose: $0.015$. $\mathrm{N}{\mathrm{a}}_2\mathrm{S}{\mathrm{O}}_4$ boils highest.

The Haber process uses iron (I); Wacker oxidation uses $\mathrm{PdC}{\mathrm{l}}_2$ (II); Wilkinson's catalyst is $[(\mathrm{PP}{\mathrm{h}}_3{)}_3\mathrm{RhCl}]$ (III); Ziegler's catalyst is $\mathrm{TiC}{\mathrm{l}}_4$ with triethyl/trimethylaluminium (IV).

Raoult's law predicts $p=\frac{5}{15}(63)+\frac{10}{15}(78)=21+52=73$ torr. The observed 70 torr is lower, so the solution shows negative deviation (stronger A–B interactions).

Fructose is the keto-hexose found in honey, exists in $\alpha$ and $\beta$ anomeric (furanose) forms, and is laevorotatory — hence 'laevulose'.

Reducing an ester only as far as the aldehyde needs DIBAL-H, $\mathrm{AlH}(\mathrm{iBu}{)}_2$, at low temperature followed by water. $\mathrm{LiAl}{\mathrm{H}}_4$ would go on to benzyl alcohol, $\mathrm{NaB}{\mathrm{H}}_4$ does not reduce esters, and ${\mathrm{H}}_2/Pd-\mathrm{BaS}{\mathrm{O}}_4$ is for acid chlorides (Rosenmund).

Iodide is the best leaving group among the halides — the large, polarisable ion spreads its charge and is a weak base — so 1-iodobutane reacts faster than 1-chlorobutane in ${\mathrm{S}}_{\mathrm{N}}2$. Both statements are true and R explains A.

${\Delta }_f{H}^{\circ }(\mathrm{BaS}{\mathrm{O}}_4)={\Delta }_f{H}^{\circ }(\mathrm{B}{\mathrm{a}}^{2+})+{\Delta }_f{H}^{\circ }(\mathrm{S}{\mathrm{O}}_4^{2-})+\Delta H_{cryst}$: $-349=x-216-4.5\Rightarrow x=-128.5$ kcal/mol.

Six structural isomers: oxolane (THF), 2-methyloxetane, 3-methyloxetane, 2-ethyloxirane, 2,2-dimethyloxirane and 2,3-dimethyloxirane. Adding stereoisomers: 2-methyloxetane and 2-ethyloxirane are each a pair of enantiomers, and 2,3-dimethyloxirane exists as the cis (meso) form plus a trans pair. Total $1+2+1+2+1+3=10$.

Dipole moments: ${\mathrm{H}}_2\mathrm{O}$ (1.85 D) > $\mathrm{N}{\mathrm{H}}_3$ (1.47 D) > $\mathrm{CHC}{\mathrm{l}}_3$ (1.04 D) — A correct. Bond enthalpy: N≡N (945) > O=O (498) > H–H (436 kJ/mol) — D correct. $\mathrm{Xe}{\mathrm{F}}_2$ has three lone pairs on Xe, more than $\mathrm{Xe}{\mathrm{F}}_4$'s two — B wrong. Bond lengths run N–O (~140 pm) > C–H (109 pm) > O–H (96 pm), the reverse of C — C wrong. Hence A and D.

The reaction is endothermic, so higher temperature shifts it forward (A). Adding either reactant also pushes it forward (C, D). Lower temperature would reduce the yield. Hence A, C and D.

A rate constant in ${\mathrm{s}}^{-1}$ means first order. $t=\frac{2.303}{k}\log \frac{[A{]}_0}{[A]}=\frac{2.303}{0.03}\log 8=\frac{2.303\times 3\times 0.301}{0.03}\approx 69.3$ s.

Fusing sodium with the compound gives NaCN, $\mathrm{N}{\mathrm{a}}_2\mathrm{S}$ and NaX from nitrogen, sulphur and halogen respectively — the basis of Lassaigne's test. Oxidising carbon with CuO to $\mathrm{C}{\mathrm{O}}_2$ is the Liebig combustion method for estimating carbon, not Lassaigne's.