NEET 2026 re-exam Chemistry answer key with solutions: questions 46–90

Friedel-Crafts acetylation gives acetophenone, then haloform reaction (NaOCl) cleaves the methyl ketone to sodium benzoate (P) + CHCl3 (Q). Acidifying P gives benzoic acid; Q (chloroform) is slowly air-oxidised in light to phosgene, a poisonous gas.

• (1) P is an aromatic carboxylate salt, not giving a primary alcohol on any step shown
• (2) P is sodium benzoate, an aromatic carboxylate salt, but Q is chloroform (CHCl3), a non-aromatic haloalkane byproduct of the haloform cleavage — not both aromatic.
• (4) P is a carboxylate salt and Q is a haloalkane, neither is a carbonyl compound

[Fe(ox)3]3- is an octahedral tris-chelate; its two mirror images are non-superimposable, so it is chiral (optically active). trans-[Cr(H2O)2(ox)2]- has a plane of symmetry (the two H2O and the plane containing them), so it is achiral.

• (1) trans-[Cr(H2O)2(ox)2]- has a plane of symmetry, so it is NOT chiral
• (2) This calls both statements incorrect, but [Fe(ox)3]3- is a genuinely chiral octahedral tris-chelate (no mirror plane) — Statement-I is true, so 'both incorrect' already fails there.
• (4) This flips which statement is right: Statement-I ([Fe(ox)3]3- is chiral) is actually correct, while Statement-II is the false one — the trans isomer has a mirror plane through the two water ligands, so it is achiral.

The empty p orbital on the benzylic carbon is stabilised by resonance with the filled pi system of the ring (filled pi orbital) and by hyperconjugation with the filled C-H sigma bonds of the adjacent CH3 group (filled sigma orbitals). Both are filled-orbital donations into the empty p orbital.

• (2) The donating orbitals must be filled (occupied), not empty, to stabilise a cation
• (3) Sigma* is an empty antibonding orbital and cannot donate electron density
• (4) Neither donor orbital can be empty; stabilisation requires filled donor orbitals

Potash alum is K2SO4·Al2(SO4)3·24H2O. One formula unit contains 2 K+ ions and 4 SO4²⁻ ions (1 from K2SO4 plus 3 from Al2(SO4)3). So the ratio of K+ to SO4²⁻ is 2:4, which simplifies to 1:2.

• (2) This inverts the true 2:4 (=1:2) ratio, swapping the counts so SO4²⁻ outnumbers K+ by 2:1 instead of the reverse.
• (3) This comes from counting only the 3 SO4²⁻ from Al2(SO4)3 and forgetting the 1 SO4²⁻ contributed by K2SO4, giving 2:3 instead of the correct 2:4 (1:2).
• (4) This is the reverse of the 2:3 miscount (itself from omitting K2SO4's SO4²⁻) — flipping the numbers around does not fix the missing ion, so it is still wrong.

In the beta-pleated sheet, adjacent extended polypeptide chains are held side-by-side by intermolecular H-bonds between C=O and N-H of different chains. Quaternary structure (subunit association) is not the same as tertiary structure; many single-chain proteins are biologically active; the alpha-helix is right-handed, stabilised by intramolecular H-bonds.

• (1) Two or more subunits define quaternary structure, not tertiary structure
• (2) Many proteins without quaternary structure are biologically active
• (4) This confuses the alpha-helix with a left-handed screw; its intramolecular H-bonds (C=O to N-H, 4 residues apart) actually form a right-handed coil, not a left-handed one.

Rounding-off rule: if the digit to be dropped is exactly 5 (with nothing beyond, or effectively 5), the preceding digit is left unchanged if even, and increased by 1 if odd. 17.0145 -> the digit after the 3rd decimal (4, even) stays: 17.014. 21.0235 -> the digit after the 3rd decimal (3, odd) rounds up: 21.024.

• (1) For 21.0235, the digit before the dropped 5 is 3 (odd), which must round up to 4 giving 21.024; keeping it at 21.023 misapplies the round-half-to-even rule to an odd digit.
• (2) For 17.0145, the digit before the dropped 5 is 4 (even) and stays unchanged at 17.014; rounding it up to 17.015 wrongly treats an even digit as odd, and also keeps 21.023's odd-digit error.
• (4) 17.015 incorrectly rounds the even preceding digit 4 up to 5; the round-half-to-even rule keeps an even digit unchanged, so 17.0145 should give 17.014, not 17.015.

For AB type AgBr: S=sqrt(Ksp)=7.07e-7 M. For AB2 type Zn(OH)2: Ksp=4S^3 -> S=6.3e-6 M. For A2B2 type Hg2Cl2: Ksp=4S^3 -> S=6.9e-7 M. Comparing molar solubilities: Zn(OH)2 (6.3e-6) > AgBr (7.07e-7) > Hg2Cl2 (6.9e-7).

• (1) Places Hg2Cl2 highest, but its computed solubility is the lowest
• (2) Places AgBr highest, but Zn(OH)2 has the largest computed solubility
• (3) Ranks Hg2Cl2 above AgBr, but the computed molar solubilities are AgBr ≈ 7.1×10⁻⁷ M and Hg2Cl2 ≈ 6.9×10⁻⁷ M — AgBr is slightly larger, and Zn(OH)2 (≈6.3×10⁻⁶ M) is largest overall, not smallest.

Electron gain enthalpy (magnitude of negative value) is highest for Cl, not F, because the small size of F causes strong inter-electronic repulsion in its compact 2p subshell when an extra electron is added, lowering the effective energy release. So the order by magnitude is Cl > F > Br > I.

• (1) Places F highest, but F's small size makes its electron gain enthalpy less negative than Cl's
• (2) Puts Br above Cl and F, contrary to the actual anomaly (only F is anomalously low, not Br)
• (4) This fully reverses the true Cl > F > Br > I order, putting I (least exothermic electron gain) first and Cl (most exothermic) last — the opposite of the actual group trend.

A is the standard defining condition of an ideal solution (mix-H=0, mix-V=0) and is correct. R is false: interactions DO occur between P and Q in an ideal solution; the condition is that A-B interactions are of the same strength as A-A and B-B interactions, not that no interaction occurs.

• (1) Calls R correct, but R claims zero interaction between P and Q; an ideal solution needs A-B interactions equal in strength to A-A and B-B, not their absence — so R is false and cannot explain A.
• (2) Treats R as merely a non-explanation, but R is not just unrelated — it is factually false: ideal solutions still have intermolecular interactions, just equal-strength ones, not zero.
• (4) Calls A false, but ΔmixH=0 and ΔmixV=0 IS the standard defining condition of an ideal solution, so A is correct; it is R (claiming zero interactions) that is wrong.

The test is the Lassaigne (sodium fusion) test. Fusing with sodium turns covalently bound nitrogen into NaCN and sulfur into Na₂S; when a compound contains both, the extract holds sodium thiocyanate (NaSCN), and that is what gives the blood-red colour (ferric thiocyanate). Sulfur alone gives a violet colour with sodium nitroprusside. Every amino acid contains nitrogen, so the deciding atom is sulfur: of the four, only methionine (side chain –CH₂–CH₂–S–CH₃) has it. The question's wording pairs nitroprusside with the red colour, but either way methionine is the only option with the element being tested for.

• (1) Leucine's side chain is an isobutyl group: C and H only, no sulfur, so its extract has no sulfide or thiocyanate to give a colour.
• (2) Threonine has an –OH in its side chain, which is easily mistaken for a 'reactive' group, but it has no sulfur.
• (4) Serine (–CH₂OH side chain) is another oxygen-containing amino acid with no sulfur, so the sulfur tests stay negative.

Use Gibbs energy additivity: -3F(-0.04) = -1F.E(Fe3+/Fe2+) + -2F(-0.44). So 0.12F = -E.F + 0.88F, giving E = 0.88-0.12 = 0.76 V.

• (1) Arithmetic slip in combining the two half-cell free energies
• (3) -0.48 V simply adds the two given potentials (-0.04 + -0.44) without weighting by electron number; combining the correct ΔG values (nFE) instead gives +0.76 V, not their raw sum.
• (4) +0.92 V comes from adding 2×0.44 V to the raw -0.04 V instead of 3×0.04 V (0.88+0.04=0.92); the Fe3+/Fe couple must be weighted by its 3 electrons first, giving +0.76 V.

Balanced redox equation: 2KMnO4 + 5H2C2O4 + 3H2SO4 -> ... giving mole ratio KMnO4:oxalic acid = 2:5. Moles oxalic acid = 0.01L x 0.25M = 2.5e-3 mol. Moles KMnO4 = (2/5)x2.5e-3 = 1.0e-3 mol in 10 mL, so molarity = 1.0e-3/0.01 = 0.10 M.

• (2) 0.20 M follows from using a 4:5 KMnO4-to-oxalic-acid mole ratio (double the correct value); the balanced equation (2MnO4⁻ + 5C2O4²⁻ + 16H+ → 2Mn2+ + 10CO2 + 8H2O) fixes that ratio at 2:5, giving 0.10 M.
• (3) Assumes a 1:1 mole ratio, ignoring the balanced stoichiometry
• (4) 0.15 M follows from a 3:5 KMnO4-to-oxalic-acid mole ratio; the correct balanced ratio is 2:5 (5 mol oxalic acid, each losing 2e-, balances 2 mol MnO4-, each gaining 5e-), giving 0.10 M, not 0.15 M.

From the spectrochemical series, field strength increases: weak field (halides) < H2O < NH3 < strong field pi-acceptors like CO. So decreasing order is CO > NH3 > H2O > Cl-.

• (2) Places H2O above NH3, reversing their actual relative field strength
• (3) Cl- > H2O > NH3 > CO reverses the entire spectrochemical series; the weak-field halide Cl- actually sits at the bottom, and the strong pi-acceptor CO at the top, not the other way round.
• (4) Cl- > NH3 > H2O > CO both reverses the series and swaps NH3 and H2O's positions; the true order (weakest to strongest field) is Cl- < H2O < NH3 < CO.

Free expansion is against zero external pressure (w=0) and isothermal so dU=0, hence q=0. Since q=0 for the surroundings, S(surroundings)=0. S(system)=nR ln(V2/V1)=2x2.303xRxlog(100/10)=4.606R (positive, since the process is spontaneous and irreversible).

• (1) Sets both to zero, but free expansion from 10 L to 100 L is an irreversible volume increase with ΔS(system)=nR ln(V2/V1)=2×2.303×Rlog10=4.606R, not zero — only ΔS(surroundings) is truly zero, since q=0.
• (2) Surroundings exchange no heat (q=0) in free expansion, so their entropy change is zero, not -4.606R
• (3) System entropy is not zero; it is the surroundings' entropy change that is zero

Rate of consumption of A: -(1/2)d[A]/dt = k, so [A]0-[A]t = 2kt. 75% of A is consumed means [A]t = 0.5 M. So 2-0.5 = 2(1.0)t, giving t = 1.5/2 = 0.75 min.

• (1) Omits the factor of 2 from the stoichiometric coefficient of A
• (3) 1.0 min corresponds to only 50% of A being consumed ([A] falling from 2 M to 1 M), not the 75% asked for; at 75% consumption [A]=0.5 M, and 2kt=[A]0-[A]t needs t=0.75 min.
• (4) Overestimates the time by not applying the stoichiometric factor correctly

3d transition metals (except end members) do have high melting points, and this is directly explained by the participation of (n-1)d electrons together with ns electrons in strong metallic/covalent-type interatomic bonding, so R correctly explains A.

• (2) R is in fact the standard textbook explanation for A, not an unrelated true statement
• (3) This claims R is false, but R states the standard fact that (n-1)d electrons join the ns electrons in interatomic metallic bonding — a genuine, textbook-correct reason for the strong bonding.
• (4) This claims A is false, but 3d transition metals do characteristically have high melting points because their (n-1)d electrons join the ns electrons in strong metallic bonding, so A is correct.

Using the Kohlrausch/Debye-Huckel-Onsager form, lambda(m) = lambda0(m) - A.sqrt(c). 145 = lambda0(m) - 90x0.1, so lambda0(m) = 154 S cm2 mol^-1. By Kohlrausch's law, lambda0(m) = lambda0(X+) + lambda0(Y-) = 74 + lambda0(Y-), giving lambda0(Y-) = 80 S cm2 mol^-1.

• (2) 100.0 results from misreading the concentration as 0.1 M instead of 0.01 M — using √0.1≈0.316 in the correction gives λ0m≈173, and 173-74≈100, instead of the correct λ0m=154 from √0.01=0.1.
• (3) Confuses the slope value with the ion's limiting conductivity
• (4) 76.0 comes from an addition slip while extrapolating λ0m: 145+90×0.1 correctly gives 154, not 150; totalling it as 150 by mistake and then subtracting the cation's 74 lands at 76 instead of 80.

C4H10 + 13/2 O2 -> 4CO2 + 5H2O. Molar mass of C4H10 = 58 g/mol, so 116 g = 2 mol, giving 2x4 = 8 mol CO2 = 8x44 = 352 g.

• (2) 322 g does not follow from any consistent stoichiometric path: 116 g of butane (MW 58) is exactly 2 mol, and the balanced equation fixes 4 mol CO2 per mol butane (8 mol → 352 g), not a ratio of 3.66.
• (3) Corresponds to only 1 mole of butane's CO2, ignoring that 116 g is 2 moles
• (4) 362 g is similarly inconsistent: 2 mol of butane can only give a clean multiple of 44 g CO2 under the 4:1 CO2-to-butane ratio (8 mol → 352 g); 362 g implies about 8.23 mol CO2, which the equation does not give.

ln k = ln A - Ea/(RT). Reading the plot's intercept, ln A = 6 (from the y-intercept). Setting ln(e^2)=2: 2 = 6 - 6640/(8.3T), giving 6640/(8.3T)=4, so T = 6640/(8.3x4) = 200 K.

• (1) 125 K doesn't match a simple misreading of the plot: with intercept ln A=6, Ea=6.64 kJ/mol, and ln(e²)=2, solving 2=6-6640/(8.3T) gives exactly T=200 K; 125 K fits no natural single substitution error here.
• (2) 150 K follows from misreading 'k=e²' as literally k=2, using ln k=ln2≈0.693 instead of 2; solving 0.693=6-6640/(8.3T) gives T≈151 K, close to this option, instead of the correct T=200 K.
• (4) 250 K follows if the plot's intercept is misread as ln A≈5.2 instead of 6; solving 2=5.2-6640/(8.3T) gives T=250 K, but the plot's actual y-intercept is 6, giving the correct T=200 K.

In both reactions MnO2 (Mn +4) oxidises the halide ion to the halogen while Mn is itself reduced (+4 to +2): 2NaX + MnO2 + 2H2SO4 -> Na2SO4 + MnSO4 + X2 + 2H2O. So Statement-I (claiming oxidation of Mn) is false, and Statement-II (reduction of Mn) is true.

• (1) Statement-I incorrectly claims oxidation of Mn, when Mn is reduced in both cases
• (2) Statement-II is actually correct (Mn is reduced), so both cannot be incorrect
• (3) This calls Statement-I correct and Statement-II incorrect — backwards: MnO2 always acts as the oxidant, so Mn is reduced (+4→+2) in both reactions, making Statement-I false and Statement-II true.

Ti3+ (d1, 1 unpaired e-), Mn3+ (d4, strong-field CN- gives low spin t2g^4 -> 2 unpaired e-), Fe3+ (d5, strong-field CN- gives low spin t2g^5 -> 1 unpaired e-), Co3+ (d6, strong-field NH3 gives low spin t2g^6 -> 0 unpaired e-). The maximum unpaired electrons (2) and hence the highest spin-only magnetic moment is for [Mn(CN)6]3-.

• (2) [Fe(CN)6]3- is indeed low spin with 1 unpaired electron, but that gives a spin-only moment of √3≈1.73 BM — far below [Mn(CN)6]3-'s 2 unpaired electrons (√8≈2.83 BM).
• (3) [Co(NH3)6]3+ is low spin d6 with 0 unpaired electrons (diamagnetic)
• (4) [Ti(H2O)6]3+ (d1) does have 1 unpaired electron, giving the smallest moment (√3≈1.73 BM) of the four species, not the largest — [Mn(CN)6]3- with 2 unpaired electrons is highest.

Ce3+ is 4f1 (1 unpaired), Nd3+ is 4f3 (3 unpaired), Tb3+ is 4f8 (6 unpaired), Ho3+ is 4f10 (4 unpaired, by Hund's rule filling 7 orbitals: 3 doubly + 4 singly occupied gives 4 unpaired). So Ho3+ has exactly 4 unpaired electrons.

• (1) Nd3+ is 4f3 (atomic number 60, minus 3 electrons); by Hund's rule this singly fills 3 of the 7 f-orbitals, giving only 3 unpaired electrons, short of the 4 needed.
• (2) Ce3+ is 4f1, a single electron in one of the seven 4f orbitals — just 1 unpaired electron, far fewer than the 4 found in Ho3+ (4f10).
• (3) Tb3+ is 4f8: the first 7 electrons singly occupy all seven orbitals (Hund's rule) and the 8th pairs up with one of them, leaving 6 unpaired electrons — more than the 4 in Ho3+.

The name lists four ammine, one aqua, one chlorido ligand in the coordination sphere of Co(III): [Co(NH3)4(H2O)Cl]. The complex ion charge = +3 (Co) + 0(NH3x4) + 0(H2O) + (-1)(Cl) = +2, so it needs 2 Cl- counter-ions outside the bracket to balance charge: [Co(NH3)4(H2O)Cl]Cl2.

• (1) Places both Cl and H2O incorrectly (one Cl inside bracket as ligand, one as water of crystallisation) and drops the ionisable chloride
• (2) Omits the coordinated aqua and chlorido ligands from the bracket entirely
• (3) Uses only one counter-ion Cl-, which does not balance the +2 charge on the complex ion

For isothermal reversible processes, w = -nRT ln(Vf/Vi). So w1 = -RT1 ln(V2/V1) and w3 = -RT2 ln(V4/V3). Adding: w1+w3 = -RT1 ln(V2/V1) - RT2 ln(V4/V3) = -2T1 ln(V2/V1) - 2T2 ln(V4/V3) using R=2.

• (2) For adiabatic processes q=0, so w2 = -DeltaU2 and w4 = -DeltaU4, giving w2+w4 = -(DeltaU2+DeltaU4), not DeltaU2-DeltaU4
• (3) Mixes an isothermal work term with an adiabatic one incorrectly, and drops T2 entirely
• (4) The net cyclic work is generally non-zero unless net q for the full cycle is zero, which is not established

Nitrogen (2p3, half-filled, extra stable) has an anomalously high first ionisation enthalpy compared to O; O's first IE (1314) is lower than N's (1402) and F's (1681). Removing an electron from O (2p4) gives O+ (2p3), a stable half-filled configuration, which lowers the energy needed to ionise O, correctly explaining A.

• (2) R is precisely the textbook explanation for A, so it IS the correct explanation
• (3) This calls R false, but removing an electron from O's 2p4 configuration does give a stable, half-filled 2p3 arrangement (like N's own ground state) — a real energetic gain that correctly explains A.
• (4) This calls A false, but O's first ionisation enthalpy (1314 kJ/mol) genuinely is lower than both N's (1402 kJ/mol) and F's (1681 kJ/mol) — a well-documented anomaly, so A is correct.

A is the definition of an ideal solution (true). Chloroform-acetone forms stronger unlike-molecule H-bonds than like-like interactions, so it shows negative deviation (B true). Aniline-phenol actually shows NEGATIVE deviation (stronger hetero H-bonding between phenolic O-H and aniline N lone pair), not positive, so C is false.

• (2) C is false: aniline-phenol shows negative, not positive, deviation
• (3) This selects only A, but B (chloroform-acetone shows negative deviation from Raoult's law, from strong hetero H-bonding) is also a true statement that should be included.
• (4) This pairs A with C, but aniline and phenol actually show negative deviation — stronger phenol O-H···N-aniline hydrogen bonding lowers the vapour pressure below the ideal value, not raises it.

Grignard reagents are strong bases; the acidic N-H proton of the primary amine is removed by the carbanion of cyclohexylmagnesium bromide, converting the Grignard's alkyl group directly into the corresponding hydrocarbon (cyclohexane) while forming a magnesium amide salt as the other product.

• (1) No C-N bond-forming substitution occurs here; the Grignard only removes the acidic proton
• (2) There is no C-C coupling between the two rings in this acid-base reaction
• (3) Bicyclohexyl would require a C-C coupling reaction, not an acid-base proton transfer

Boron (2s,2p valence shell, 4 orbitals total, only 3 valence electrons) can accept one more electron pair (e.g. in BH4-, BF4-) giving a maximum covalency of 4, since only 4 orbitals (one 2s + three 2p) are available. Be, Mg, Al are s- and p-block elements whose valence-orbital counts and covalencies differ from these claims (Be has 4 valence orbitals; Mg, having accessible 3d orbitals, can exceed covalency 4; Al has 4, not 5, valence orbitals in the n=3 shell relevant here).

• (2) Beryllium (2s, 2p) has four valence orbitals available, not three
• (3) Magnesium can expand its covalency beyond four using 3d orbitals
• (4) Aluminium's valence shell (3s, 3p) offers four orbitals, not five, at the level tested

Equal concentrations of N and D at equilibrium means K=1, so Delta G deg = -RT ln(1) = 0, i.e. Delta H = T.Delta S at this temperature. T = 333 K (60+273). Delta S = Delta H / T = 666/333 = 2.0 kJ K^-1 mol^-1.

• (2) 2000.0 comes from computing ΔH/T in J K⁻¹ mol⁻¹ (666,000 J ÷ 333 K = 2000) and then mislabelling that number as kJ K⁻¹ mol⁻¹ — the correct value in kJ units is 666÷333=2.0.
• (3) Mistakenly reports the temperature value itself instead of the computed ratio
• (4) Uses an incorrect temperature (60, uncorrected to Kelvin) in the division

PCl5: sp3d, trigonal bipyramidal (III). BrF5: sp3d2 with one lone pair, square pyramidal (IV). BF4-: sp3, tetrahedral (I). [Ni(CN)4]2-: dsp2 (strong-field CN-, low spin d8 Ni2+), square planar (II). So A-III, B-IV, C-I, D-II.

• (1) This swaps PCl5 and BrF5's geometries: PCl5 (sp3d, no lone pairs) is trigonal bipyramidal (III), while BrF5 (sp3d2, one lone pair) is square pyramidal (IV), not the reverse.
• (3) This keeps PCl5 correct (III) but assigns BrF5 tetrahedral (I, actually BF4-'s shape), BF4- square planar (II, actually [Ni(CN)4]2-'s shape), and [Ni(CN)4]2- square pyramidal (IV, actually BrF5's).
• (4) Assigns square planar to BrF5 and square pyramidal to [Ni(CN)4]2-, which is backwards

Work out what each reaction really gives, then check whether the drawing is that compound. Br₂ adds anti (through a bromonium ion); anti addition to trans-but-2-ene gives meso-2,3-dibromobutane. In the meso compound, putting Br anti to Br forces Me anti to Me and H anti to H (the molecule has a centre of symmetry in that conformation). The drawing for Statement I has Br anti but the methyls gauche, which is one enantiomer of the (±) pair, so Statement I is incorrect. Cold alkaline KMnO₄ adds two OH groups syn; syn addition to cis-but-2-ene gives meso-butane-2,3-diol. Rotating the back carbon of Statement II's drawing by 60° eclipses OH with OH, Me with Me and H with H, which shows a mirror plane: the drawing is meso, so Statement II is correct.

• (1) Accepts Statement I because trans-but-2-ene + Br₂ does give a meso product, without checking that the drawn projection (methyls gauche while the bromines are anti) is actually the chiral isomer.
• (2) Rejects Statement II by assuming KMnO₄ adds anti like Br₂; cold alkaline KMnO₄ adds both OH groups syn, which turns cis-but-2-ene into the meso diol.
• (3) Swaps the two stereochemical outcomes: the drawing in Statement I is not meso, while the drawing in Statement II is.

Lindlar's catalyst stops at the alkene and adds H₂ syn, so K is (Z)-1-phenylprop-1-ene; sodium in liquid ammonia adds the two hydrogens anti, so L is (E)-1-phenylprop-1-ene. K and L differ only in the arrangement about the C=C, so they are geometrical isomers. HBr alone adds through the more stable benzylic carbocation, giving M = 1-bromo-1-phenylpropane. HBr with a peroxide adds through the more stable benzylic radical, so Br ends up on the other carbon: N = 2-bromo-1-phenylpropane. M and N have Br on different carbons, so they are position isomers, not stereoisomers.

• (2) Enantiomers must be non-superimposable mirror images with a chiral centre; K and L are achiral alkenes that differ only in cis/trans geometry.
• (3) M and N are saturated bromides with no C=C, so they cannot be geometrical isomers; they differ in which carbon carries Br.
• (4) Stereoisomers have the same connectivity; the peroxide effect moves Br to a different carbon, so M and N are structural (position) isomers.

Facial (fac) and meridional (mer) isomerism arises specifically in octahedral complexes of the type [Ma3b3] (three identical pairs of two different monodentate ligands). [Cr(py)3Cl3] fits this Ma3b3 pattern, so it shows fac/mer isomers. The other complexes are Ma6, Ma4b2 or with a bidentate ligand, which do not show fac/mer isomerism.

• (2) [Cr(H2O)6]3+ is Ma6 type, with only one possible arrangement
• (3) [Co(NH3)4(H2O)2]3+ is Ma4b2 type, which shows cis/trans, not fac/mer, isomerism
• (4) [Ni(en)2(H2O)2]2+ contains a bidentate ligand (en), not the Ma3b3 monodentate pattern needed for fac/mer isomerism

B: Hofmann bromamide degradation converts PhCONH2 (benzamide) directly to aniline (PhNH2) with loss of one carbon, via KOH/Br2. D: Acid hydrolysis of acetanilide (PhNHCOCH3) with HCl/H2O on heating cleaves the amide to give aniline + acetic acid. A gives benzylamine (PhCH2NH2), not aniline, since LiAlH4 reduces the nitrile carbon into the amine's CH2 group. C: NaBH4 is a mild reducing agent and does NOT reduce the nitro group, so PhNO2 remains largely unreacted.

• (1) This includes A, but LiAlH4 reduces the nitrile carbon of PhCN into a CH2 group, giving benzylamine (PhCH2NH2), not aniline — only B (Hofmann degradation) gives aniline here.
• (3) Neither A nor C give aniline; C fails because NaBH4 cannot reduce -NO2
• (4) This includes C, but NaBH4 is too mild to reduce an aromatic nitro group — PhNO2 is largely recovered unchanged; only D (acetanilide hydrolysis) reliably gives aniline, alongside B.

Vitamin A (carotene precursor) is sourced from carrots (IV). Vitamin B12 (cobalamin) is found in meat/animal products (I), since it is not synthesised by plants. Vitamin E (tocopherol, an antioxidant fat) is rich in vegetable/sunflower oil (II). Vitamin K is abundant in green leafy vegetables (III). So A-IV, B-I, C-II, D-III.

• (1) This assigns vitamin A to sunflower oil (II) and vitamin E to carrots (IV) — swapped; vitamin A (carotene) actually comes from carrots, and vitamin E (tocopherol) from oils like sunflower oil.
• (3) This keeps vitamin A correct (carrots) but swaps B and C: vitamin B12 comes from meat (I), not sunflower oil (II), and vitamin E comes from sunflower oil (II), not meat (I).
• (4) Assigns green leafy vegetables to A instead of D, and meat to B correctly but mismatches C and D

Assign oxidation states by standard rules (O=-2 except peroxide, H=+1 except metal hydride): N2O5 -> N=+5; Al2O3 -> Al=+3; H2S -> S=-2. So decreasing order N(+5) > Al(+3) > S(-2), matching option 2.

• (1) P in P4O10=+5, S in SO3=+6, O in H2O=-2, giving order S>P>O, not P>S>O as claimed
• (3) Pb in PbO2=+4, N in N2O3=+3, S in SO3=+6, giving actual order S>Pb>N, not Pb>N>S
• (4) P in P4O6=+3, Cl in Cl2O7=+7, Al in AlH3=-3 (hydride), giving actual order Cl>P>Al, not P>Cl>Al

The ketone has α-hydrogens only on C-5 (C-2 carries the two methyls). Its enolate can attack benzaldehyde (cross-aldol) or a second molecule of the ketone (self-aldol). On heating with base the aldol dehydrates by removing the acidic α-hydrogen, so the new C=C always forms conjugated with the C=O. Cross-aldol therefore gives 5-benzylidene-2,2-dimethylcyclopentan-1-one as its E and Z forms (options 2 and 4), and self-aldol gives the cyclopentylidene ketone of option 3. Option 1 puts the double bond inside the second ring, away from the carbonyl, which this dehydration does not produce.

• (2) This is the normal cross-aldol product (benzylidene conjugated with C=O), so it can be obtained.
• (3) Self-condensation followed by dehydration toward the carbonyl gives exactly this conjugated alkylidene ketone.
• (4) The benzylidene C=C can form with either geometry; the Z isomer is also a product (usually the minor one).

Conjugated double bonds are separated by exactly one single bond (alternating double-single-double). In hepta-1,3-diene (CH2=CH-CH=CH-CH2-CH2-CH3), the C1=C2 and C3=C4 double bonds are separated by only the C2-C3 single bond, making them conjugated. The 1,4-, 1,5- and 1,6-dienes have two or more intervening single bonds (isolated dienes).

• (2) Hepta-1,4-diene has two single bonds (C2-C3, C3-C4... ) between the double bonds, making it an isolated (non-conjugated) diene
• (3) Hepta-1,5-diene has three intervening single bonds, clearly isolated
• (4) Hepta-1,6-diene has the double bonds at opposite ends of the chain, fully isolated

Oxidising the methyl group of p-nitrotoluene with acidic KMnO4 gives p-nitrobenzoic acid; the electron-withdrawing -NO2 group (-I, -M) stabilises the conjugate base, making it a stronger acid than benzoic acid (Statement I true). Reducing the -NO2 to -NH2 with Sn/HCl (then neutralising) gives p-toluidine (p-methylaniline); the electron-donating methyl group (+I, hyperconjugation) increases electron density on N, making it more basic than aniline (Statement II true).

• (2) This calls both statements false, but both are correct: the -NO2 group makes p-nitrobenzoic acid a stronger acid than benzoic acid, and reducing -NO2 to -NH2 gives p-toluidine, more basic than aniline.
• (3) This calls Statement-II false, but p-toluidine (from reducing p-nitrotoluene) has an electron-donating methyl group that raises electron density on nitrogen, making it more basic than aniline.
• (4) This calls Statement-I false, but the electron-withdrawing -NO2 group in p-nitrobenzoic acid stabilises the conjugate base, genuinely making it a stronger acid than benzoic acid — Statement-I is true.

On strong heating (513 K), 2KMnO4 -> K2MnO4 + MnO2 + O2. The dark green potassium manganate (K2MnO4, Mn in +6 state, d1) formed is paramagnetic due to its one unpaired electron.

• (2) Mn3O4 is not the species formed by this specific thermal decomposition and is not the green product
• (3) MnO (manganese(II) oxide) is not a thermal-decomposition product of KMnO4 at 513 K and is not green; the actual product is dark-green K2MnO4 (Mn6+, d1), formed alongside MnO2 and O2.
• (4) KO2 (potassium superoxide) is unrelated to KMnO4's thermal decomposition

The Etard reaction (CrO2Cl2/CS2 then hydrolysis) converts the methyl group of toluene directly to a -CHO group, giving P = benzaldehyde (PhCHO). Benzaldehyde is also independently obtained by the Rosenmund reduction: hydrogenation of benzoyl chloride over Pd on BaSO4 (poisoned catalyst prevents over-reduction). Statement 4 is a correct alternative synthesis of the same product P.

• (1) Benzaldehyde has no -COOH group, so it does not effervesce with NaHCO3
• (2) Benzene + CH3COCl/AlCl3 (Friedel-Crafts acylation) gives acetophenone (PhCOCH3), not benzaldehyde
• (3) Aldehydes do not typically give a white precipitate with bromine water the way phenols do; this is not benzaldehyde's characteristic test

A 1:3 electrolyte dissociates into 1 complex cation and 3 simple anions. [Co(NH3)6]Cl3 dissociates into [Co(NH3)6]3+ and 3Cl-, giving a 1:3 electrolyte. [CoCl2(NH3)4]Cl is 1:1, [CoCl(NH3)5]Cl2 is 1:2, and [Co(NH3)3(NO2)3] is neutral (non-electrolyte, all ligands are inside the coordination sphere).

• (1) [CoCl2(NH3)4]Cl gives only 1 Cl- ion outside the bracket, a 1:1 electrolyte
• (2) [CoCl(NH3)5]Cl2 gives 2 Cl- ions outside the bracket, a 1:2 electrolyte
• (4) [Co(NH3)3(NO2)3] is a neutral complex with no ionisable counter-ions at all

The number of radial nodes for an orbital equals n-l-1. Each crossing of psi(r) through zero (excluding the origin and infinity) is one radial node. Curve A (1s-like) shows no crossing = 0 nodes; curve B shows one crossing = 1 node (like 2s); curve C shows two crossings = 2 nodes (like 3s); curve D also crosses but with a different pattern corresponding to 1 node (like 3p). So the curve with exactly two radial nodes is C.

• (1) Curve A shows a smooth monotonic decay with zero crossings, i.e. 0 radial nodes
• (2) Curve B crosses the ψ=0 axis only once (like a 2s orbital), giving 1 radial node, not the 2 radial nodes the question asks for.
• (4) Curve D's pattern corresponds to just 1 radial node (analogous to a 3p orbital, n-l-1=3-1-1=1), not the 2 radial nodes shown by curve C (like a 3s orbital, n-l-1=3-0-1=2).

Ethers have the weakest polarity (no H-bond donor, only a C-O-C dipole). Ketones (C=O) are more polar than ethers due to the strong carbonyl dipole (no resonance donation weakening it, unlike esters), but ketones cannot H-bond as donors. Alcohols can both donate and accept H-bonds, making them more polar than ketones. Carboxylic acids, with an even more polarised O-H and strong dimerising H-bonds, are the most polar. So A < C < B < D.

• (1) Places C (ketone) above B (alcohol), reversing their actual polarity order
• (2) Places D (acid) below B (alcohol), which is incorrect since acids are the most polar
• (3) Places B (alcohol) above C incorrectly relative position but keeps ether lowest -- still wrongly orders C below A

Ne2 has 20 electrons total, filling the MO sequence up through sigma1s, sigma*1s, sigma2s, sigma*2s, sigma2pz, (pi2px=pi2py), (pi*2px=pi*2py), sigma*2pz. All bonding and antibonding orbitals up to and including sigma*(2pz) are completely filled, so the HOMO is sigma*(2p) (also giving bond order 0, consistent with Ne2 not existing as a stable molecule).

• (1) pi(2p) is filled but is not the highest energy occupied orbital; sigma*(2p) lies above it and is also filled
• (2) sigma(2p) is filled but several higher orbitals (pi, pi*, sigma*) are also filled above it
• (3) pi*(2p) is filled but sigma*(2p), which lies higher in energy, is also completely filled