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NEET (UG) · revision cheat sheet · neetlogic.in

Chemistry

22 chapters · 63 topics · ~180 marks · 44 solved previous-year questions

Class 11

1. Some Basic Concepts of Chemistry

Mole concept and molar mass: NCERT's current STP is 1 bar, giving an ideal gas a molar volume of 22.7 L — not the older 22.4 L (at 1 atm) many default to by habit. NTP (25°C, 1 atm) gives yet a third value, 24.45 L, so match the number to whichever condition the question actually states.
Stoichiometry and limiting reagent: The limiting reagent is found by dividing each reactant's available moles by its own stoichiometric coefficient and comparing — not by comparing raw moles or masses directly, which is why the reagent taken in larger quantity can still be the one that runs out first.
Concentration terms: Molarity is defined per unit volume, so it changes with temperature as the solution expands or contracts; molality, mole fraction and mass percentage are defined per unit mass and don't — a distinction NEET tests by asking which concentration term stays constant on heating.

Dalton's atomic theory could not explain which of the following?

Ans: Law of gaseous volume·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.

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Among the following, choose the ones with an equal number of atoms. A. 212 g of

N a_{2} C O_{3} (s)

[molar mass = 106 g] B. 248 g of

N a_{2} O (s)

[molar mass = 62 g] C. 240 g of NaOH(s) [molar mass = 40 g] D. 12 g of

H_{2} (g)

[molar mass = 2 g] E. 220 g of

C O_{2} (g)

[molar mass = 44 g] Choose the correct answer from the options given below:

Ans: A, B and D only·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.

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2. Structure of Atom

Bohr model and hydrogen spectrum: Bohr's energy and radius formulas are exact only for one-electron species (H, He⁺, Li²⁺) — applying them to a multi-electron atom without adjustment is the usual overreach. Only the Balmer series (transitions down to n=2) falls in the visible range; Lyman is UV and the rest are infrared.
Quantum numbers, orbitals and their shapes: Given a quantum number set, check l ≤ n−1 and |m| ≤ l first — that's where most "is this set possible" options actually fail. Total nodes equal n−1, split into l angular nodes and n−l−1 radial nodes, a formula NEET applies directly rather than just defining.
Electronic configuration, Aufbau, Pauli, Hund: Cr ([Ar]3d⁵4s¹) and Cu ([Ar]3d¹⁰4s¹) break the Aufbau filling order for the extra stability of a half-filled or filled d-subshell — NEET's standard exceptions. And when a transition metal ion forms, the 4s electrons are removed first, not 3d, even though 4s filled first — write Fe²⁺ with a 3d⁴4s² tail instead of the correct 3d⁶ and the trap has worked.

The energy and radius of the first Bohr orbit of

H e^{+}

and

L i^{2 +}

are (given

R_{H} = 2.18 \times 1 0^{- 18}

a_{0} = 52.9

pm):

Ans:

E (L i^{2 +}) = - 19.62 \times 1 0^{- 18}

r (L i^{2 +}) = 17.6

pm;

E (H e^{+}) = - 8.72 \times 1 0^{- 18}

r (H e^{+}) = 26.4

pm·

E_{n} = - R_{H} \frac{Z ^{2}}{n ^{2}}

and

r_{n} = a_{0} \frac{n ^{2}}{Z}

H e^{+}

(

Z = 2

E = - 4 \times 2.18 \times 1 0^{- 18} = - 8.72 \times 1 0^{- 18}

r = 52.9/2 = 26.4

pm.

L i^{2 +}

(

Z = 3

E = - 9 \times 2.18 \times 1 0^{- 18} = - 19.62 \times 1 0^{- 18}

r = 52.9/3 = 17.6

pm.

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The ratio of the wavelengths of the light absorbed by a hydrogen atom when it undergoes the

n = 2 \to n = 3

and

n = 4 \to n = 6

transitions, respectively, is:

Ans:

\frac{1}{4}

\frac{1}{λ} \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{λ _{2 \to 3}}{λ _{4 \to 6}} = \frac{5/144}{5/36} = \frac{1}{4}

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3. Classification of Elements and Periodicity in Properties

Modern periodic table and electronic basis: The modern periodic law orders elements by atomic number, not atomic mass — the switch from Mendeleev's basis that fixed anomalies like the Co/Ni and Ar/K ordering. Hydrogen's dual placement (loses one electron like an alkali metal, needs one like a halogen) is a recurring "which group" trap.
Periodic trends: radius, ionisation enthalpy, electron gain enthalpy, electronegativity: Chlorine, not fluorine, has the most negative electron gain enthalpy in its group — fluorine's small size packs the incoming electron into an already-crowded 2p subshell, offsetting its higher electronegativity. Ionisation enthalpy also dips at boron (below Be) and at oxygen (below N), because removing an electron from a filled 2s² or half-filled 2p³ costs extra — the "always rises across a period" rule breaks at exactly these two points.

Which among the following electronic configurations belong to main-group elements? A.

[Ne] 3 s^{1}

· B.

[Ar] 3 d^{3} 4 s^{2}

· C.

[Kr] 4 d^{10} 5 s^{2} 5 p^{5}

· D.

[Ar] 3 d^{10} 4 s^{1}

· E.

[Rn] 5 f^{0} 6 d^{2} 7 s^{2}

Choose the correct answer from the options given below:

Ans: A and C only·Main-group elements have their outermost electrons in

s

p

orbitals:

[Ne] 3 s^{1}

(Na) and

[Kr] 4 d^{10} 5 s^{2} 5 p^{5}

(I). B and D are

d

-block, E is

f

-block (a

6 d

actinoid configuration). So A and C.

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Which of the following statements are true? A. Unlike Ga, which has a very high melting point, Cs has a very low melting point. B. On the Pauling scale, the electronegativity values of N and Cl are not the same. C. Ar,

K^{+}

C l^{-}

C a^{2 +}

and

S^{2 -}

are all isoelectronic species. D. The correct order of the first ionisation enthalpies of Na, Mg, Al and Si is Si > Al > Mg > Na. E. The atomic radius of Cs is greater than that of Li and Rb. Choose the correct answer from the options given below:

Ans: C and E only·Ar,

K^{+}

C l^{-}

C a^{2 +}

and

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.

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4. Chemical Bonding and Molecular Structure

Ionic and covalent bonds, Lewis structures: Fajans' rules explain why some "ionic" compounds behave covalently: a small, highly charged cation next to a large, polarisable anion pulls electron density back toward itself, which is why AlCl₃ and BeCl₂ don't act like typical ionic solids. Octet exceptions — BF₃ incomplete, PCl₅/SF₆ expanded, NO odd-electron — are tested directly, not as a footnote.
VSEPR and molecular shapes: Shape isn't the same as electron-pair geometry: NH₃ and H₂O both have a tetrahedral arrangement of electron pairs, but lone pairs are dropped when naming the shape (pyramidal, bent). The repulsion order lone pair–lone pair > lone pair–bond pair > bond pair–bond pair is why NH₃'s bond angle (107°, one lone pair) is squeezed less than H₂O's (104.5°, two lone pairs).
Valence bond theory and hybridisation: Hybridisation is decided by the steric number — sigma bonds plus lone pairs on the central atom — not by the number of bonds alone; forgetting a lone pair is the most common way to misassign sp³ to something that's actually sp³d or sp³d² (XeF₄ is sp³d², with two lone pairs on Xe alongside its four bonds).
Molecular orbital theory: MOT's headline NEET question is O₂'s paramagnetism — two unpaired electrons in its degenerate π*2p orbitals, which the paired-electron double-bond picture from Lewis structures or VBT can't show at all. Watch the filling order too: from Li₂ to N₂, σ2p fills after π2p (s–p mixing pushes it up); from O₂ onward σ2p drops back below π2p.
Hydrogen bonding and dipole moment: Net dipole moment is a vector sum of bond dipoles, not a scalar one — CO₂'s two C=O dipoles cancel exactly in its linear geometry (μ=0), while bent H₂O's don't. The classic trap: o-nitrophenol's intramolecular hydrogen bond makes it more volatile than p-nitrophenol, whose hydrogen bonding is intermolecular — having H-bonds isn't enough, the question is whether they're within or between molecules.

Given below are two statements: Statement I: A hypothetical diatomic molecule with bond order zero is quite stable. Statement II: As bond order increases, the bond length increases. In the light of the above statements, choose the most appropriate answer from the options given below:

Ans: Both Statement I and Statement II are false·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).

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Identify the correct orders against the property mentioned: A.

H_{2} O > N H_{3} > CHC l_{3}

— dipole moment B.

Xe F_{4} > Xe O_{3} > Xe F_{2}

— number of lone pairs on the central atom C. O–H > C–H > N–O — bond length D.

N_{2} > O_{2} > H_{2}

— bond enthalpy Choose the correct answer from the options given below:

Ans: A, D only·Dipole moments:

H_{2} O

(1.85 D) >

N H_{3}

(1.47 D) >

CHC l_{3}

(1.04 D) — A correct. Bond enthalpy: N≡N (945) > O=O (498) > H–H (436 kJ/mol) — D correct.

Xe F_{2}

has three lone pairs on Xe, more than

Xe 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.

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5. Thermodynamics

First law, enthalpy and Hess's law: NCERT's sign convention is ΔU = q + w with w the work done ON the system — mix that up with the older "w = work done BY the system" convention and every work term flips sign. For gas-phase reactions, ΔH and ΔU differ by Δn(gas)×RT, a correction it's easy to forget even when Δn(gas) isn't zero.
Entropy and Gibbs energy, spontaneity: Spontaneity is decided by ΔG = ΔH − TΔS, never by ΔH or ΔS alone — an endothermic process (ΔH positive) can still be spontaneous if ΔS is large enough and T is high, which is exactly the case NEET uses to catch anyone pattern-matching "exothermic means spontaneous."
Bond enthalpy and thermochemistry: Bond enthalpies used in ΔH = Σ(bonds broken) − Σ(bonds formed) are average values for a polyatomic molecule — each successive C–H bond broken in methane actually takes a different amount of energy, and the tabulated figure is a mean. That's why a bond-enthalpy calculation only ever gives an approximate ΔH, not the exact value Hess's law gives from standard enthalpies of formation.

The standard heat of formation, in kcal/mol, of

B a^{2 +}

is: [Given: standard heat of formation of the

S O_{4}^{2 -}

ion (aq) = –216 kcal/mol, standard heat of crystallisation of

BaS O_{4} (s)

= –4.5 kcal/mol, standard heat of formation of

BaS O_{4} (s)

= –349 kcal/mol]

Ans: –128.5·

Δ_{f} H^{\circ} (BaS O_{4}) = Δ_{f} H^{\circ} (B a^{2 +}) + Δ_{f} H^{\circ} (S O_{4}^{2 -}) + Δ H_{cr y s t}

- 349 = x - 216 - 4.5 \Rightarrow x = - 128.5

kcal/mol.

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The work done during the reversible isothermal expansion of one mole of hydrogen gas at

2 5^{\circ}

C from a pressure of 20 atmosphere to 10 atmosphere is (given

R = 2.0 cal K^{- 1} mo l^{- 1}

Ans: –413.14 calories·

w = - 2.303 n R T lo g \frac{p _{1}}{p _{2}} = - 2.303 \times 1 \times 2 \times 298 \times lo g 2 = - 413.14

cal. Work done by the gas on expansion is negative in the sign convention used.

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6. Equilibrium

Law of mass action, Kc and Kp, Le Chatelier: Kp and Kc are related by a factor of (RT) raised to the power Δn — moles of gaseous product minus reactant — and getting that exponent's sign backwards inverts the whole conversion. Separately, adding an inert gas at constant volume leaves equilibrium untouched, but adding it at constant pressure shifts it toward the side with more gas moles, because the container must expand to keep pressure fixed.
Ionic equilibrium: acids, bases, pH, buffers: For a strong acid diluted to around 10⁻⁷–10⁻⁸ M, pH = −log(C) alone gives a value above 7 — impossible for an acid — because it ignores water's own contribution of H⁺ at that concentration. This "pH of an extremely dilute strong acid" question is a NEET standard precisely because the naive formula gives a physically impossible answer.
Solubility product and common-ion effect: The Ksp expression's exponents come straight from dissolution stoichiometry — Ag₂CrO₄ gives [Ag⁺]²[CrO₄²⁻], and dropping that square is the most common setup error. A common ion lowers molar solubility but never changes Ksp itself, which stays constant at a given temperature — a distinction NEET tests by asking what actually changes when a common ion is added.

For the reaction

A (g) ⇌ 2B (g)

, the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500 at 1000 K. [Given

R = 0.0831 L atm mo l^{- 1} K^{- 1}

K_{p}

for the reaction at 1000 K is:

Ans: 0.033·

K_{c} = \frac{k _{f}}{k _{b}} = \frac{1}{2500} = 4 \times 1 0^{- 4}

Δ n_{g} = 2 - 1 = 1

, so

K_{p} = K_{c} (R T) = 4 \times 1 0^{- 4} \times 0.0831 \times 1000 \approx 0.033

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A higher yield of NO in

N_{2} (g) + O_{2} (g) ⇌ 2NO (g)

can be obtained at [

Δ H

of the reaction

= + 180.7 kJ mo l^{- 1}

]: A. Higher temperature · B. Lower temperature · C. Higher concentration of

N_{2}

· D. Higher concentration of

O_{2}

Choose the correct answer from the options given below:

Ans: A, C, D only·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.

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7. Redox Reactions

Oxidation number and balancing redox equations: Oxygen isn't always −2: it's −1 in peroxides (H₂O₂), −1/2 in superoxides (KO₂), and +2 in OF₂, where fluorine is the more electronegative atom — NEET tests exactly these exceptions, not the default. Also watch for disproportionation, where the same element is simultaneously oxidised and reduced in one reaction, such as Cl₂ with cold dilute NaOH.
Types of redox reactions: Whether one species can oxidise another is decided by E°cell = E°(reduction of the oxidiser) − E°(reduction of the reducing agent) coming out positive — that's how "will X displace Y" questions are actually meant to be solved, not from a memorised reactivity list. Disproportionation (one element both oxidised and reduced) and its reverse, comproportionation, are the classification names NEET most often asks you to apply correctly.

Consider the following compounds:

K O_{2}

H_{2} O_{2}

and

H_{2} S O_{4}

. The oxidation states of the underlined elements (O in

K O_{2}

, O in

H_{2} O_{2}

, S in

H_{2} S O_{4}

) in them are, respectively:

Ans: +1, –1 and +6·In

K O_{2}

the superoxide ion

O_{2}^{-}

gives oxygen

- \frac{1}{2}

each, i.e. K is

+ 1

(the element asked); in

H_{2} O_{2}

oxygen is

- 1

; in

H_{2} S O_{4}

sulphur is

+ 6

. Matching the underlined elements gives

+ 1, - 1, + 6

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Which reaction is not a redox reaction?

Ans:

BaC l_{2} + N a_{2} S O_{4} \to BaS O_{4} + 2NaCl

BaC l_{2} + N a_{2} S O_{4} \to BaS O_{4} + 2NaCl

is a double displacement (precipitation) with no change in any oxidation number. The other three involve electron transfer.

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8. Organic Chemistry: Some Basic Principles and Techniques

IUPAC nomenclature: When two numbering directions both give substituents low locants, the tie breaks at the first point of difference, not the lowest sum — a frequent numbering slip. The functional-group seniority order (roughly: acid > ester > amide > nitrile > aldehyde > ketone > alcohol > amine) decides which group becomes the suffix and which get demoted to prefixes like oxo- or hydroxy-.
Isomerism: structural and stereo: Cis/trans and E/Z don't always agree — E/Z is assigned by CIP priority, cis/trans by which groups simply look alike, so a compound can be "cis" by eye but "E" by priority. Meso compounds are the other classic trap: chiral centres present, but an internal plane of symmetry cancels the optical rotation, so 2ⁿ overcounts the stereoisomers whenever a meso form exists — tartaric acid is the standard example.
Electronic effects: inductive, resonance, hyperconjugation: Hyperconjugation needs a C–H (or C–C) sigma bond directly attached to an sp² carbon — more such adjacent H's means more stability, which is why carbocation and alkene stability both rise with the number of alkyl-substituted carbons. Inductive effect fades with distance through sigma bonds; resonance doesn't fade the same way, but does need the participating orbitals to be coplanar — twist them out of plane and resonance stops working even though the atoms are still technically conjugated.
Reaction intermediates and types of reactions: Carbocation stability (3° > 2° > 1° > methyl, from hyperconjugation and +I) is exactly reversed for carbanions (methyl > 1° > 2° > 3°) — alkyl groups that stabilise a positive centre by donating electron density destabilise a centre that's already electron-rich. Mixing up which intermediate benefits from alkyl substitution is the trap.

Which one of the following compounds can exist as cis–trans isomers?

Ans: 1,2-Dimethylcyclohexane·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.

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The total number of possible isomers (both structural as well as stereoisomers) of cyclic ethers of molecular formula

C_{4} H_{8} O

is:

Ans: 10·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

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9. Hydrocarbons

Alkanes: conformations and reactions: Ethane's anti-staggered conformation is the most stable (zero torsional strain) and the fully eclipsed one the least — tested via potential-energy-vs-dihedral-angle graphs, not just the word "staggered." In free-radical halogenation, bromination is far more selective for 3° C–H than chlorination — "both give the same product ratio" is the trap.
Alkenes and alkynes: addition reactions, Markovnikov: The peroxide (anti-Markovnikov) effect works only for HBr — not HCl or HI — because only the bromine free-radical chain steps are both energetically favourable; with HCl and HI, normal Markovnikov addition holds regardless of peroxide. That HBr-only exclusivity, not just "peroxide flips the rule," is what NEET actually tests.
Aromatic hydrocarbons: electrophilic substitution, directing effects: Halogens are the one substituent that's ortho/para-directing yet still deactivating — their lone pair donates into the ring by resonance (directing) while their electronegativity withdraws inductively (deactivating), a combination no other common group shows. Friedel–Crafts reactions also simply fail on an already strongly deactivated ring, such as nitrobenzene, no matter how much catalyst is used.

Given below are two statements: Statement I: The boiling points of the three isomeric pentanes follow the order n-pentane > isopentane > neopentane. Statement II: When branching increases, the molecule attains the shape of a sphere. This results in a smaller surface area for contact, due to which the intermolecular forces between the spherical molecules are weak, thereby lowering the boiling point. In the light of the above statements, choose the most appropriate answer from the options given below:

Ans: Both Statement I and Statement II are correct·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.

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Which one of the following compounds (shown) does not decolourise bromine water?

Ans: Structure (1)·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.

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10. Purification and Characterisation of Organic Compounds

Purification methods: crystallisation, distillation, chromatography: Steam distillation only works for a substance that's virtually insoluble in water and has some vapour pressure at 373 K — it lets something like aniline distil below its own, much higher, boiling point instead of decomposing there. In chromatography, Rf is the distance moved by the compound divided by the distance moved by the solvent front, always less than 1 — a value NEET expects you to compute, not just define.
Qualitative and quantitative analysis: Kjeldahl's method for %N fails for nitro, azo and diazo compounds, and for nitrogen locked in a ring like pyridine — that nitrogen doesn't convert to ammonium sulphate the way amine nitrogen does, so the method silently underestimates. In Lassaigne's test, if both N and S are present but sodium is insufficient, NaSCN can form instead of separate NaCN and Na₂S — giving a false-negative Prussian-blue test for nitrogen.

Match List I with List II.

List I (Mixture)	List II (Method of separation)
A. $CHC l_{3} + C_{6} H_{5} N H_{2}$	I. Distillation under reduced pressure
B. Crude oil in the petroleum industry	II. Steam distillation
C. Glycerol from spent-lye	III. Fractional distillation
D. Aniline–water	IV. Simple distillation

Choose the correct answer from the options given below:

Ans: A-IV, B-III, C-I, D-II·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.

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Which one of the following reactions does not belong to Lassaigne's test?

Ans:

2CuO + C Δ 2Cu + C O_{2}

·Fusing sodium with the compound gives NaCN,

N a_{2} S

and NaX from nitrogen, sulphur and halogen respectively — the basis of Lassaigne's test. Oxidising carbon with CuO to

C O_{2}

is the Liebig combustion method for estimating carbon, not Lassaigne's.

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Class 12

11. Solutions

Concentration terms, Henry's and Raoult's laws: A higher Henry's law constant means lower gas solubility, not higher — an inverse relationship (p = KH·x) that's easy to flip under pressure. For non-ideal solutions, positive deviation from Raoult's law (weaker A–B interactions, e.g. ethanol–acetone) gives a minimum-boiling azeotrope, while negative deviation (stronger A–B interactions from H-bonding, e.g. chloroform–acetone) gives a maximum-boiling one — a pairing students often reverse.
Colligative properties: All four colligative properties scale with the number of solute particles, not the solute's identity, which is why every formula carries a van't Hoff factor i. A dissociating electrolyte gives i > 1 and inflates the effect; a solute that associates, like benzoic acid dimerising in benzene, gives i < 1 and shrinks it — the direction of that correction is the usual trap.
Abnormal molar mass and van 't Hoff factor: i > 1 means the molar mass calculated from the observed colligative property comes out lower than the true value — dissociation, more particles than expected. i < 1 means it comes out higher — association, like acetic acid dimerising in benzene, fewer particles than expected — and getting which direction the "abnormal" molar mass shifts is exactly what these questions test.

Which of the following aqueous solutions will exhibit the highest boiling point?

Ans: 0.01 M

N a_{2} S O_{4}

·Boiling-point elevation depends on the total particle concentration

i C

. Urea:

0.01

KN O_{3}

2 \times 0.01 = 0.02

N a_{2} S O_{4}

3 \times 0.01 = 0.03

. Glucose:

0.015

N a_{2} S O_{4}

boils highest.

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5 moles of liquid X and 10 moles of liquid Y make a solution having a vapour pressure of 70 torr. The vapour pressures of pure X and Y are 63 torr and 78 torr respectively. Which of the following is true regarding the described solution?

Ans: The solution shows negative deviation.·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).

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12. Electrochemistry

Electrochemical cells and Nernst equation: At equilibrium, E(cell) = 0, not E°(cell) = 0 — that's the condition linking E°(cell) to K through 0 = E° − (0.0591/n)log K. In the standard cell notation, oxidation (anode) is written on the left and reduction (cathode) on the right — reverse that and every EMF you calculate flips sign.
Conductance and Kohlrausch's law: A weak electrolyte's molar conductivity shoots up sharply on extreme dilution and its plot doesn't extrapolate to a finite value the way a strong electrolyte's does — which is exactly why the weak electrolyte's limiting molar conductivity is obtained indirectly, via Kohlrausch's law of independent migration of ions, never by extrapolating its own graph.
Electrolysis and Faraday's laws; batteries and corrosion: In aqueous NaCl electrolysis, Cl⁻ is oxidised at the anode instead of water, even though water's thermodynamic oxidation potential is lower — a real overpotential effect examiners like precisely because it contradicts the "always predict from E° alone" shortcut. Faraday's second law compares deposits by equivalent mass, not molar mass, so equal charge deposits different moles of ions of different charge.

If the molar conductivity (

Λ_{m}

) of a

0.050 mol L^{- 1}

solution of a monobasic weak acid is

90 S c m^{2} mo l^{- 1}

, its extent (degree) of dissociation will be [assume

λ_{+}^{\circ} = 349.6 S c m^{2} mo l^{- 1}

and

λ_{-}^{\circ} = 50.4 S c m^{2} mo l^{- 1}

Ans: 0.225·

Λ_{m}^{\circ} = 349.6 + 50.4 = 400 S c m^{2} mo l^{- 1}

. Degree of dissociation

α = \frac{Λ _{m}}{Λ _{m}^{\circ}} = \frac{90}{400} = 0.225

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The mass in grams of copper deposited by passing a current of 9.6487 A through a voltameter containing copper sulphate solution for 100 seconds is (given molar mass of Cu =

63 g mo l^{- 1}

, 1 F = 96487 C):

Ans: 0.315 g·Charge

= I t = 9.6487 \times 100 = 964.87

= 0.01

F. Copper needs 2 F per mole, so

0.005

mol is deposited:

0.005 \times 63 = 0.315

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13. Chemical Kinetics

Rate law, order and molecularity: Order is an experimentally determined exponent in the rate law — it can be zero, fractional, even negative — while molecularity is a theoretical count of colliding species in one elementary step and must be a positive integer; the two coincide only for a genuine single-step reaction. You cannot read a complex, multi-step reaction's rate law off its balanced equation the way you can for an elementary step.
Integrated rate equations and half-life: First-order half-life (t½ = 0.693/k) doesn't depend on starting concentration — the signature fact used to identify first order from a data table. Zero-order half-life is the opposite: directly proportional to initial concentration (t½ = [A]₀/2k), and mixing up which order does which is the standard trap.
Arrhenius equation and activation energy: A catalyst speeds up a reaction by lowering the activation energy through an alternate pathway — it doesn't change ΔH, and because it accelerates the forward and reverse reactions equally, it never shifts the equilibrium position or the value of K. That "catalysts don't touch equilibrium, only kinetics" point is a recurring assertion-reason question.

If the half-life (

t_{1/2}

) for a first-order reaction is 1 minute, then the time required for 99.9% completion of the reaction is closest to:

Ans: 10 minutes·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.

NEET 2025 · open →

If the rate constant of a reaction is

0.03 s^{- 1}

, how much time does it take for a

7.2 mol L^{- 1}

concentration of the reactant to get reduced to

0.9 mol L^{- 1}

? (Given

lo g 2 = 0.301

)

Ans: 69.3 s·A rate constant in

s^{- 1}

means first order.

t = \frac{2.303}{k} lo g \frac{[ A ] _{0}}{[ A ]} = \frac{2.303}{0.03} lo g 8 = \frac{2.303 \times 3 \times 0.301}{0.03} \approx 69.3

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14. The p-Block Elements

Group 13 and 14 elements: The inert pair effect makes the lower oxidation state more stable going down groups 13 and 14 — Tl⁺ is more stable than Tl³⁺, Pb²⁺ more stable than Pb⁴⁺ — the reverse of what "higher group, higher favoured state" intuition suggests. Diborane is the other signature trap: B₂H₆ has too few electrons for conventional 2-electron bonds at every position, forcing two 3-centre-2-electron bridge bonds at the bridging hydrogens.
Group 15 and 16: nitrogen, oxygen families: N₂'s triple bond makes it exceptionally unreactive, while P4 in the same group is so reactive it's stored underwater — a contrast NEET uses to test whether "same group, similar reactivity" has been over-applied. Catenation is actually stronger for sulfur than oxygen despite O sitting above S, because lone-pair repulsion between the small oxygen atoms weakens the O–O single bond.
Group 17 and 18: halogens, noble gases: F–F bond dissociation enthalpy is anomalously low, even lower than Cl–Cl, because the small fluorine atoms pack their lone pairs close enough to repel each other strongly — bond strength doesn't track atomic size monotonically down this group. Among noble gases, only xenon forms an established range of compounds; helium, neon and argon's very high ionisation enthalpy keeps them essentially compound-free, a fact often tested as assertion-reason.

Match List I with List II.

List I	List II
A. $Xe O_{3}$	I. $s p^{3} d$ ; linear
B. $Xe F_{2}$	II. $s p^{3}$ ; pyramidal
C. $XeO F_{4}$	III. $s p^{3} d^{3}$ ; distorted octahedral
D. $Xe F_{6}$	IV. $s p^{3} d^{2}$ ; square pyramidal

Choose the correct answer from the options given below:

Ans: A-II, B-I, C-IV, D-III·

Xe O_{3}

s p^{3}

with one lone pair, pyramidal (II).

Xe F_{2}

s p^{3} d

with three lone pairs, linear (I).

XeO F_{4}

s p^{3} d^{2}

with one lone pair, square pyramidal (IV).

Xe F_{6}

s p^{3} d^{3}

, distorted octahedral (III).

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Given below are two statements: Statement I: Like nitrogen, which can form ammonia, arsenic can form arsine. Statement II: Antimony cannot form antimony pentoxide. In the light of the above statements, choose the most appropriate answer from the options given below:

Ans: Statement I is correct but Statement II is incorrect·Arsenic forms arsine,

As H_{3}

(I correct). Antimony does form the pentoxide

S b_{2} O_{5}

(II incorrect).

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15. The d- and f-Block Elements

Transition elements: trends and properties: Zn, Cd and Hg are d-block elements but not true transition elements — their d-subshell is completely filled (d¹⁰) in both the metal and its common ions, so they miss the defining traits: variable oxidation states, coloured ions, catalytic activity. Also, second- and third-row transition metals in the same group (Zr and Hf, for instance) have almost identical atomic radii, because lanthanoid contraction cancels out the increase you'd otherwise expect going down a group.
Potassium dichromate and permanganate: MnO₄⁻'s reduction product depends entirely on the medium: Mn²⁺ (colourless, 5 electrons gained) in acid, MnO₂ (brown precipitate, 3 electrons) in neutral or faintly alkaline, MnO₄²⁻ (green, 1 electron) in strongly alkaline — equivalent-weight calculations hinge on picking the right one, and acidic is the default unless stated otherwise. Orange Cr₂O₇²⁻ and yellow CrO₄²⁻ interconvert with pH, not by any redox change.
Lanthanoids and actinoids: Lanthanoid contraction — steady radius shrinkage across the series from poor 4f shielding — is why second- and third-row transition metals end up almost the same size element for element (Zr≈Hf, Nb≈Ta), which is exactly why those pairs are so hard to separate chemically. +3 is the dominant lanthanoid oxidation state, but watch for Ce⁴⁺ and Eu²⁺, both reaching the extra stability of an empty or half-filled f-subshell — the same logic as Cr and Cu in the d-block.

Match List I with List II.

List I	List II
A. Haber process	I. Fe catalyst
B. Wacker oxidation	II. $PdC l_{2}$
C. Wilkinson catalyst	III. $[(PP h_{3})_{3} RhCl]$
D. Ziegler catalyst	IV. $TiC l_{4}$ with $Al (C H_{3})_{3}$

Choose the correct answer from the options given below:

Ans: A-I, B-II, C-III, D-IV·The Haber process uses iron (I); Wacker oxidation uses

PdC l_{2}

(II); Wilkinson's catalyst is

[(PP h_{3})_{3} RhCl]

(III); Ziegler's catalyst is

TiC l_{4}

with triethyl/trimethylaluminium (IV).

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Given below are two statements: Statement I: Ferromagnetism is considered an extreme form of paramagnetism. Statement II: The number of unpaired electrons in a

C r^{2 +}

ion (

Z = 24

) is the same as that of an

N d^{3 +}

ion (

Z = 60

). In the light of the above statements, choose the correct answer from the options given below:

Ans: Statement I is true but Statement II is false·Ferromagnetism is indeed treated as an extreme form of paramagnetism (I true).

C r^{2 +}

3 d^{4}

with 4 unpaired electrons;

N d^{3 +}

4 f^{3}

with 3 — not the same (II false).

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16. Coordination Compounds

Nomenclature and Werner's theory: Werner's key distinction is primary valence (ionisable, the oxidation state) versus secondary valence (non-ionisable, the fixed coordination number and geometry) — a complex's conductivity in solution tests exactly this, since only ions outside the coordination sphere carry current. In naming, ligands are listed alphabetically by name regardless of charge or multiplying prefix (bis, tris), and an anionic complex ion always ends in "-ate."
Isomerism in coordination compounds: Ambidentate ligands like NO₂⁻/ONO⁻ (nitro/nitrito) and SCN⁻/NCS⁻ give linkage isomers — same formula, different donor atom — a distinct category from ionisation isomerism, where a different ion sits inside versus outside the coordination sphere. For an MA₃B₃ octahedral complex, the geometrical isomers are called fac and mer, not cis/trans, which is reserved for the MA₄B₂ and MA₂B₂ patterns.
Valence bond and crystal field theory: Tetrahedral crystal-field splitting is always smaller than octahedral for the same metal and ligands (roughly 4/9 the size) — small enough that tetrahedral complexes are almost always high-spin regardless of the ligand, unlike octahedral complexes where a strong-field ligand like CN⁻ or CO forces low-spin pairing. VBT explains a complex's geometry and magnetic moment through hybridisation, but not why it's coloured or why one ligand counts as "strong field" — that gap is exactly what crystal field theory fills.

The correct order of the wavelength of light absorbed by the following complexes is: A.

[Co (N H_{3})_{6}]^{3 +}

· B.

[Co (CN)_{6}]^{3 -}

· C.

[Cu (H_{2} O)_{4}]^{2 +}

· D.

[Ti (H_{2} O)_{6}]^{3 +}

Choose the correct answer from the options given below:

Ans: B < A < D < C·Wavelength absorbed is inversely related to

Δ_{0}

C N^{-}

(strongest field) gives the largest splitting, then

N H_{3}

;

T i^{3 +}

with water splits less, and

C u^{2 +}

with water least (absorbs in the red, ~800 nm). So B < A < D < C.

NEET 2025 · open →

Which of the following are paramagnetic? A.

[NiC l_{4}]^{2 -}

· B.

Ni (CO)_{4}

· C.

[Ni (CN)_{4}]^{2 -}

· D.

[Ni (H_{2} O)_{6}]^{2 +}

· E.

Ni (PP h_{3})_{4}

Choose the correct answer from the options given below:

Ans: A and D only·

[NiC l_{4}]^{2 -}

(

N i^{2 +}

d^{8}

, weak-field tetrahedral) has two unpaired electrons; so does

[Ni (H_{2} O)_{6}]^{2 +}

(octahedral

d^{8}

Ni (CO)_{4}

and

Ni (PP h_{3})_{4}

contain Ni(0),

d^{10}

;

[Ni (CN)_{4}]^{2 -}

is square planar

d^{8}

, all paired. Paramagnetic: A and D.

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17. Haloalkanes and Haloarenes

Nucleophilic substitution: SN1 and SN2: SN2 is a single backside attack that gives complete inversion of configuration (Walden inversion) at the stereocentre; SN1 goes through a planar carbocation attacked from either face, giving near-racemisation instead — "inversion versus racemisation" is the fastest way to tell the two mechanisms apart in a question stem. SN1 favours 3° substrates and polar protic solvents that stabilise the carbocation; SN2 favours 1° substrates and is blocked by bulky groups at the reacting carbon.
Haloarenes: reactions and directing effects: Aryl C–X bonds are shorter and stronger than alkyl C–X bonds, because the halogen's lone pair conjugates into the ring, which is exactly why chlorobenzene resists the SN1/SN2 chemistry that works readily on an alkyl halide. Nucleophilic substitution on a haloarene needs either forcing conditions (high temperature and pressure with NaOH) or a strong electron-withdrawing group like NO₂ at the ortho/para position to make the ring susceptible.

The major product of the reaction shown in the figure is:

Ans: Structure (2)·Excess

C H_{3} 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.

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Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): The first compound shown undergoes the

S_{N} 2

reaction faster than the second compound shown. Reason (R): Iodine is a better leaving group because of its large size. In the light of the above statements, choose the correct answer from the options given below:

Ans: Both A and R are true and R is the correct explanation of A·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

S_{N} 2

. Both statements are true and R explains A.

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18. Alcohols, Phenols and Ethers

Preparation and reactions of alcohols: The Lucas test tells 1°, 2° and 3° alcohols apart by how fast turbidity appears with ZnCl₂/HCl — instantly for 3° (via a stable carbocation), within minutes for 2°, not at all at room temperature for 1°. Acidity among simple alcohols runs 1° > 2° > 3°, the opposite of carbocation stability, because the same electron-donating alkyl groups that stabilise a cation destabilise the alkoxide anion instead.
Acidity and reactions of phenols: Phenol is more acidic than an alcohol because the phenoxide ion delocalises its negative charge into the ring by resonance, but it's still far less acidic than a carboxylic acid, whose carboxylate spreads over two equivalent oxygens instead. An electron-withdrawing group at the ortho or para position (like NO₂) raises phenol's acidity further; an electron-donating one (CH₃, OCH₃) lowers it — ranking substituted phenols by acidity is a recurring question type.
Ethers: preparation and cleavage: Williamson synthesis is an SN2 reaction, so the alkyl halide half must be primary or unhindered — pair a bulky 3° alkyl halide with an alkoxide instead and you get an eliminated alkene, not the ether, because the strongly basic alkoxide just deprotonates it. In cleavage of an alkyl aryl ether by HI, the aryl–oxygen bond never breaks — the products are always phenol plus an alkyl iodide, never a haloarene.

The products A and B obtained in the following reactions, respectively, are:

3ROH + PC l_{3} \to 3RCl + A

ROH + PC l_{5} \to RCl + HCl + B

Ans:

H_{3} P O_{3}

and

POC l_{3}

PC l_{3}

with an alcohol gives the alkyl chloride and phosphorous acid

H_{3} P O_{3}

;

PC l_{5}

gives the alkyl chloride, HCl and phosphoryl chloride

POC l_{3}

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Which one of the following alcohols (three drawn as structures; the fourth is

C H_{3} C H_{2} C H_{2} C H_{2} OH

) reacts instantaneously with Lucas reagent?

Ans: Structure (3)·The Lucas test turns turbid instantly only with tertiary alcohols, which form a stable tertiary carbocation. 2-Methylpropan-2-ol,

(C H_{3})_{3} C - OH

(structure 3), is tertiary; (1) is secondary and (2), (4) are primary.

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19. Aldehydes, Ketones and Carboxylic Acids

Nucleophilic addition reactions: Reactivity toward nucleophilic addition runs aldehydes above ketones, and an aryl group directly on the carbonyl carbon lowers it further still, since the ring donates electron density into the C=O by resonance and makes that carbon less electrophilic. That steric-plus-electronic combination is what "arrange by reactivity toward HCN/NaHSO₃" questions are actually testing, not a memorised list.
Named reactions: aldol, Cannizzaro, Clemmensen, Wolff–Kishner: Cannizzaro reaction only happens on an aldehyde with no alpha-hydrogen — with an alpha-H present, concentrated base drives aldol condensation instead, so checking for alpha-H is the first move. Clemmensen reduction (Zn-Hg/HCl) and Wolff–Kishner reduction (NH₂NH₂/KOH) both take C=O to CH₂, but the choice between them depends on the rest of the molecule — Clemmensen's acid would attack acid-sensitive groups, Wolff–Kishner's base would attack base-sensitive ones.
Carboxylic acids: acidity and reactions: A carboxylic acid liberates CO₂ gas from NaHCO₃; phenol, despite also being acidic, doesn't, because phenol (pKa around 10) is weaker than carbonic acid while carboxylic acids are stronger — this is the standard test for telling the two functional groups apart experimentally. Electron-withdrawing substituents raise acidity most from the alpha position, fading fast with distance: chloroacetic acid is markedly more acidic than acetic acid, but the boost shrinks quickly as the Cl moves further away.

The correct order of decreasing acidity of the following aliphatic acids is:

Ans:

HCOOH > C H_{3} COOH > (C H_{3})_{2} CHCOOH > (C H_{3})_{3} CCOOH

·Alkyl groups are electron-releasing and destabilise the carboxylate anion, so acidity falls as alkyl substitution grows:

HCOOH > C H_{3} COOH > (C H_{3})_{2} CHCOOH > (C H_{3})_{3} CCOOH

NEET 2025 · open →

Fehling's solution 'A' is:

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

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20. Amines

Basicity of amines: In water, aliphatic amine basicity doesn't rise monotonically with substitution — the usual order is 2° > 1° > 3° > NH₃, because a 3° amine's extra alkyl groups add electron density but also add steric hindrance that weakens solvation of the resulting ammonium ion, and that loss outweighs the electronic gain. Aniline is far less basic than ammonia for a different reason: its nitrogen lone pair delocalises into the ring by resonance, and a ring nitro group weakens it further still.
Preparation and reactions; diazonium salts: Diazonium salts are stable in solution only at 0–5°C — warm it and it hydrolyses to phenol with loss of N₂ gas, which is why every diazonium reaction is carried out cold and used immediately rather than stored. Azo-coupling with phenol needs mildly alkaline conditions and with aniline needs mildly acidic conditions, and it always occurs para to the activating group.

Given below are two statements: Statement I: Benzenediazonium salt is prepared by the reaction of aniline with nitrous acid at 273–278 K. It decomposes easily in the dry state. Statement II: Insertion of iodine into the benzene ring is difficult and hence iodobenzene is prepared through the reaction of benzenediazonium salt with KI. In the light of the above statements, choose the most appropriate answer from the options given below:

Ans: Both Statement I and Statement II are correct·Diazotisation of aniline with

HN 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).

NEET 2025 · open →

The correct order of decreasing basic strength of the given amines is:

Ans: N-ethylethanamine > ethanamine > N-methylaniline > benzenamine·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.

NEET 2025 · open →

21. Biomolecules

Carbohydrates: Sucrose is the classic non-reducing sugar because its glycosidic bond ties up both anomeric carbons — C1 of glucose and C2 of fructose — leaving no free anomeric OH to open into a reactive aldehyde or ketone, unlike maltose and lactose, which keep one free and do reduce Fehling's or Tollens'. This is also why sucrose shows no mutarotation, while glucose's optical rotation visibly drifts as its α and β anomers interconvert through the open-chain form.
Proteins, enzymes and vitamins: Denaturation destroys a protein's secondary and tertiary structure, and its biological activity, but leaves the primary structure — the peptide-bonded amino acid sequence — untouched. Fat-soluble vitamins (A, D, E, K) get stored in body fat and can reach toxic levels on overdose; water-soluble ones (B-complex, C) aren't stored and must come from the diet regularly, which is why their deficiencies (scurvy for C, beriberi for B₁) show up faster.
Nucleic acids: Chargaff's rule lets you find all four base percentages in double-stranded DNA from just one — %A = %T and %G = %C, so purines always equal pyrimidines overall, a calculation NEET poses directly. Structurally: DNA's sugar is deoxyribose (no 2'-OH) against RNA's ribose, DNA pairs A with T using 2 hydrogen bonds where RNA substitutes uracil for thymine, and G pairs with C using 3 — so a strand richer in G-C is more thermally stable.

Match List I with List II.

List I (Name of vitamin)	List II (Deficiency disease)
A. Vitamin $B_{12}$	I. Cheilosis
B. Vitamin D	II. Convulsions
C. Vitamin $B_{2}$	III. Rickets
D. Vitamin $B_{6}$	IV. Pernicious anaemia

Choose the correct answer from the options given below:

Ans: A-IV, B-III, C-I, D-II·Vitamin

B_{12}

deficiency causes pernicious anaemia (IV); vitamin D, rickets (III); vitamin

B_{2} (

riboflavin), cheilosis (I); vitamin

B_{6} (

pyridoxine), convulsions (II).

NEET 2025 · open →

Sugar 'X': A. is found in honey B. is a keto sugar C. exists in

α

and

β

anomeric forms D. is laevorotatory 'X' is:

Ans: D-Fructose·Fructose is the keto-hexose found in honey, exists in

α

and

β

anomeric (furanose) forms, and is laevorotatory — hence 'laevulose'.

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22. Principles Related to Practical Chemistry

Detection of functional groups and elements: The iodoform test isn't exclusive to methyl ketones — any compound with a CH₃–CH(OH)– group, such as ethanol or isopropanol, also gives it, because it oxidises to the methyl ketone or aldehyde in situ under the test conditions. A positive iodoform test on an unknown alcohol therefore doesn't by itself prove a ketone is present — that's the standard trap.
Titrimetric analysis and salt analysis: Indicator choice depends on where the equivalence point's pH actually falls: phenolphthalein suits weak acid–strong base titrations (equivalence point is basic), methyl orange suits strong acid–weak base (equivalence point is acidic), and either works for strong acid–strong base. Reaching for phenolphthalein out of habit on a strong-acid–weak-base titration — where the colour change happens well past the actual equivalence point — is exactly the error this tests.

Match List I with List II.

List I (Ion)	List II (Group number in cation analysis)
A. $C o^{2 +}$	I. Group I
B. $M g^{2 +}$	II. Group III
C. $P b^{2 +}$	III. Group IV
D. $A l^{3 +}$	IV. Group VI

Choose the correct answer from the options given below:

Ans: A-III, B-IV, C-I, D-II·

P b^{2 +}

precipitates as the chloride in Group I;

A l^{3 +}

as the hydroxide in Group III;

C o^{2 +}

as the sulphide in Group IV;

M g^{2 +}

is in Group VI. So A-III, B-IV, C-I, D-II.

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Given below are certain cations. Using inorganic qualitative analysis, arrange them in increasing group number from 0 to VI. A.

A l^{3 +}

· B.

C u^{2 +}

· C.

B a^{2 +}

· D.

C o^{2 +}

· E.

M g^{2 +}

Choose the correct answer from the options given below:

Ans: B, A, D, C, E·Qualitative-analysis groups:

C u^{2 +}

Group II,

A l^{3 +}

Group III,

C o^{2 +}

Group IV,

B a^{2 +}

Group V,

M g^{2 +}

Group VI. Increasing group order: B, A, D, C, E.

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