Chemistry — NEET Complete Yearwise Papers MCQs with Solutions
Free NEET Complete Yearwise Papers Chemistry MCQs with step-by-step solutions (50 questions). Part of NEET 2024 Paper. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Chemistry · medium · numerical
Match List-I with List-II.
List-I (Conversion)
(A) 1 mol of H₂O to O₂
(B) 1 mol of MnO₄⁻ to Mn²⁺
(C) 1.5 mol of Ca from molten CaCl₂
(D) 1 mol of FeO to Fe₂O₃
List-II (Number of Faraday required)
(I) 3 F
(II) 2 F
(III) 1 F
(IV) 5 F
Choose the correct answer from the options given below:
A. A-II, B-III, C-I, D-IV
B. A-III, B-IV, C-II, D-I
C. A-II, B-IV, C-I, D-III ✓ Correct
D. A-III, B-IV, C-I, D-II
Solution: (A) H₂O → ½O₂ + 2e⁻ + 2H⁺ — 2 electrons, so 2 F (II).
(B) MnO₄⁻ + 5e⁻ + 8H⁺ → Mn²⁺ + 4H₂O — 5 electrons, so 5 F (IV).
(C) Ca²⁺ + 2e⁻ → Ca, so 1.5 mol of Ca needs 3 F (I).
(D) Fe²⁺ → Fe³⁺ + e⁻ — 1 electron, so 1 F (III).
Q2 — Chemistry · medium · theory
Which reaction is NOT a redox reaction?
A. H₂ + Cl₂ → 2HCl
B. BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl ✓ Correct
C. Zn + CuSO₄ → ZnSO₄ + Cu
D. 2KClO₃ + I₂ → 2KIO₃ + Cl₂
Solution: A redox reaction needs a change in oxidation state.
In BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl every element keeps the same oxidation state (Ba +2, Cl −1, Na +1, S +6, O −2) — it is a simple double displacement (precipitation) reaction, not a redox one.
The other three all involve oxidation-state changes.
Q3 — Chemistry · medium · theory
Intramolecular hydrogen bonding is present in
A.
B. HF
C. ✓ Correct
D.
Solution: Intramolecular hydrogen bonding needs the donor and acceptor groups close enough to meet within the same molecule — that is, in the ortho position.
Only ortho-nitrophenol has the −OH and −NO₂ groups adjacent, so its hydrogen bond forms inside the molecule (forming a six-membered chelate ring). In the meta and para isomers the groups are too far apart, so their hydrogen bonding is intermolecular. HF also hydrogen bonds intermolecularly.
Q4 — Chemistry · medium · theory
Fehling's solution 'A' is
A. alkaline solution of sodium potassium tartrate (Rochelle's salt)
B. aqueous sodium citrate
C. aqueous copper sulphate ✓ Correct
D. alkaline copper sulphate
Solution: Fehling's reagent is made up fresh from two separately stored solutions: Fehling's A is aqueous copper sulphate, and Fehling's B is the alkaline solution of sodium potassium tartrate (Rochelle's salt).
Q5 — Chemistry · medium · numerical
1 gram of sodium hydroxide was treated with 25 mL of 0.75 M HCl solution. The mass of sodium hydroxide left unreacted is equal to
A. Zero mg
B. 200 mg
C. 750 mg
D. 250 mg ✓ Correct
Solution: NaOH + HCl → NaCl + H₂O (1 : 1)
Moles of NaOH $= \dfrac{1}{40} = 0.025\ \text{mol} = 25$ millimoles.
Millimoles of HCl $= M \times V = 0.75 \times 25 = 18.75$.
NaOH left $= 25 - 18.75 = 6.25$ millimoles $= 6.25 \times 40 = 250\ \text{mg}$.
Q6 — Chemistry · medium · theory
Match List-I with List-II.
List-I (Compound)
(A) NH₃
(B) BrF₅
(C) XeF₄
(D) SF₆
List-II (Shape / geometry)
(I) Trigonal Pyramidal
(II) Square Planar
(III) Octahedral
(IV) Square Pyramidal
Choose the correct answer from the options given below:
A. A-III, B-IV, C-I, D-II
B. A-II, B-III, C-IV, D-I
C. A-I, B-IV, C-II, D-III ✓ Correct
D. A-II, B-IV, C-III, D-I
Solution: NH₃ — three bond pairs and one lone pair on N, so trigonal pyramidal (I).
BrF₅ — five bond pairs and one lone pair, so square pyramidal (IV).
XeF₄ — four bond pairs and two lone pairs, so square planar (II).
SF₆ — six bond pairs and no lone pair, so octahedral (III).
Q7 — Chemistry · medium · theory
The E° value for the Mn³⁺/Mn²⁺ couple is more positive than that of Cr³⁺/Cr²⁺ or Fe³⁺/Fe²⁺ due to change of
A. d⁴ to d⁵ configuration
B. d³ to d⁵ configuration
C. d⁵ to d⁴ configuration ✓ Correct
D. d⁵ to d² configuration
Solution: Going the other way, Mn²⁺ → Mn³⁺ means changing the half-filled, extra-stable d⁵ configuration into d⁴. That costs a lot of energy — Mn has a much larger third ionisation enthalpy — so the reduction of Mn³⁺ back to Mn²⁺ is strongly favoured and E° is very positive.
It is also why the +3 state of manganese is of little importance.
Q8 — Chemistry · medium · theory
Match List-I with List-II.
List-I (Process)
(A) Isothermal process
(B) Isochoric process
(C) Isobaric process
(D) Adiabatic process
List-II (Conditions)
(I) No exchange of heat takes place
(II) Carried out at constant temperature
(III) Carried out at constant volume
(IV) Carried out at constant pressure
Choose the correct answer from the options given below:
A. A-I, B-II, C-III, D-IV
B. A-II, B-III, C-IV, D-I ✓ Correct
C. A-IV, B-III, C-II, D-I
D. A-IV, B-II, C-III, D-I
Solution: Isothermal — constant temperature (II).
Isochoric — constant volume (III).
Isobaric — constant pressure (IV).
Adiabatic — no heat is exchanged with the surroundings (I).
Q9 — Chemistry · medium · theory
Activation energy of any chemical reaction can be calculated if one knows the value of
A. orientation of reactant molecules during collision.
B. rate constant at two different temperatures. ✓ Correct
C. rate constant at standard temperature.
D. probability of collision.
Solution: From the Arrhenius equation in its two-temperature form,
$\log\dfrac{k_1}{k_2} = \dfrac{E_a}{2.303R}\left[\dfrac{1}{T_2} - \dfrac{1}{T_1}\right]$
knowing the rate constant at two different temperatures is enough to work out $E_a$.
Q10 — Chemistry · medium · theory
A compound with a molecular formula of C₆H₁₄ has two tertiary carbons. Its IUPAC name is:
A. 2,3-dimethylbutane ✓ Correct
B. 2,2-dimethylbutane
C. n-hexane
D. 2-methylpentane
Solution: In 2,3-dimethylbutane, CH₃−CH(CH₃)−CH(CH₃)−CH₃, both C-2 and C-3 are attached to three other carbon atoms, so there are exactly two tertiary carbons.
The other C₆H₁₄ isomers have at most one.
Q11 — Chemistry · medium · numerical
'Spin only' magnetic moment is same for which of the following ions?
A. Ti³⁺
B. Cr²⁺
C. Mn²⁺
D. Fe²⁺
E. Sc³⁺
Choose the most appropriate answer from the options given below:
A. B and C only
B. A and D only
C. B and D only ✓ Correct
D. A and E only
Solution: Spin-only moment $\mu = \sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons.
Ti³⁺ (d¹): $n = 1$, $\mu = 1.73$
Cr²⁺ (d⁴): $n = 4$, $\mu = 4.90$
Mn²⁺ (d⁵): $n = 5$, $\mu = 5.92$
Fe²⁺ (d⁶): $n = 4$, $\mu = 4.90$
Sc³⁺ (d⁰): $n = 0$, $\mu = 0$
So Cr²⁺ and Fe²⁺ — B and D — share the same value.
Q12 — Chemistry · medium · theory
Arrange the following elements in increasing order of electronegativity:
N, O, F, C, Si
Choose the correct answer from the options given below:
A. O < F < N < C < Si
B. F < O < N < C < Si
C. Si < C < N < O < F ✓ Correct
D. Si < C < O < N < F
Solution: On the Pauling scale: Si = 1.8, C = 2.5, N = 3.0, O = 3.5, F = 4.0.
So the increasing order is Si < C < N < O < F.
Q13 — Chemistry · medium · theory
Which one of the following alcohols reacts instantaneously with Lucas reagent?
A.
B. ✓ Correct
C. CH₃−CH₂−CH₂−CH₂OH
D.
Solution: With Lucas reagent (conc. HCl + anhydrous ZnCl₂):
3° alcohols react instantaneously (turbidity appears at once).
2° alcohols react in about 5 minutes.
1° alcohols do not react at ordinary temperature.
Option (b) is the tertiary alcohol, so it is the one that reacts instantaneously.
Q14 — Chemistry · medium · theory
Given below are two statements:
Statement I: Both [Co(NH₃)₆]³⁺ and [CoF₆]³⁻ complexes are octahedral but differ in their magnetic behaviour.
Statement II: [Co(NH₃)₆]³⁺ is diamagnetic whereas [CoF₆]³⁻ is paramagnetic.
In the light of the above statements, choose the correct answer from the options given below:
A. Statement I is true but Statement II is false.
B. Statement I is false but Statement II is true.
C. Both Statement I and Statement II are true. ✓ Correct
D. Both Statement I and Statement II are false.
Solution: Both complexes have coordination number six, so both are octahedral.
Co³⁺ is [Ar]3d⁶. NH₃ is a strong field ligand, so pairing occurs, giving d²sp³ hybridisation with no unpaired electron — [Co(NH₃)₆]³⁺ is diamagnetic.
F⁻ is a weak field ligand, so no pairing occurs, giving sp³d² hybridisation with four unpaired electrons — [CoF₆]³⁻ is paramagnetic.
So both statements are true.
Q15 — Chemistry · medium · theory
Given below are two statements:
Statement I: The boiling point of hydrides of Group 16 elements follow the order H₂O > H₂Te > H₂Se > H₂S.
Statement II: On the basis of molecular mass, H₂O is expected to have lower boiling point than the other members of the group but due to the presence of extensive H-bonding in H₂O, it has higher boiling point.
In the light of the above statements, choose the correct answer from the options given below:
A. Statement I is true but Statement II is false.
B. Statement I is false but Statement II is true.
C. Both Statement I and Statement II are true. ✓ Correct
D. Both Statement I and Statement II are false.
Solution: The order H₂O > H₂Te > H₂Se > H₂S is correct: from H₂S onwards the boiling point rises with molecular mass, but water sits right at the top.
That anomaly is exactly what Statement II explains — extensive intermolecular hydrogen bonding in water. So both statements are true and II explains I.
Q16 — Chemistry · medium · theory
Match List I with List II.
List I (Quantum Number)
(A) mₗ
(B) mₛ
(C) l
(D) n
List II (Information provided)
(I) shape of orbital
(II) size of orbital
(III) orientation of orbital
(IV) orientation of spin of electron
Choose the correct answer from the options given below:
A. A-III, B-IV, C-II, D-I
B. A-II, B-I, C-IV, D-III
C. A-I, B-III, C-II, D-IV
D. A-III, B-IV, C-I, D-II ✓ Correct
Solution: mₗ, the magnetic quantum number, gives the orientation of the orbital (III).
mₛ, the spin quantum number, gives the orientation of the electron spin (IV).
l, the azimuthal quantum number, gives the shape of the orbital (I).
n, the principal quantum number, gives the size of the orbital (II).
Q17 — Chemistry · medium · theory
Match List I (Reaction) with List II (Reagents / Condition). Choose the correct answer from the options given below:
A. A-IV, B-I, C-II, D-III ✓ Correct
B. A-I, B-IV, C-II, D-III
C. A-IV, B-I, C-III, D-II
D. A-III, B-I, C-II, D-IV
Solution: (A) The alkene is cleaved into two ketone molecules — that is reductive ozonolysis, (i) O₃ (ii) Zn–H₂O (IV).
(B) Benzene → benzophenone is a Friedel-Crafts acylation with benzoyl chloride and anhydrous AlCl₃ (I).
(C) Cyclohexanol → cyclohexanone is an oxidation by CrO₃ (II).
(D) Ethylbenzene → potassium benzoate needs the side-chain oxidation KMnO₄/KOH, Δ (III).
Q18 — Chemistry · medium · theory
Identify the correct reagents that would bring about the following transformation.
A. (i) BH₃ (ii) H₂O₂/OH⁻ (iii) alk. KMnO₄ (iv) H₃O⁺
B. (i) H₂O/H⁺ (ii) PCC
C. (i) H₂O/H⁺ (ii) CrO₃
D. (i) BH₃ (ii) H₂O₂/OH⁻ (iii) PCC ✓ Correct
Solution: The double bond has to become an aldehyde at the TERMINAL carbon, so the addition must be anti-Markovnikov: hydroboration-oxidation, (i) BH₃ then (ii) H₂O₂/OH⁻, gives the primary alcohol.
The primary alcohol is then oxidised only as far as the aldehyde, which needs a mild oxidising agent — (iii) PCC. A stronger oxidant such as KMnO₄ or CrO₃ would take it on to the carboxylic acid.
Q19 — Chemistry · medium · theory
The reagents with which glucose does not react to give the corresponding tests/products are
A. Tollen's reagent
B. Schiff's reagent
C. HCN
D. NH₂OH
E. NaHSO₃
Choose the correct options from the given below:
A. B and E ✓ Correct
B. E and D
C. B and C
D. A and D
Solution: Glucose exists mostly in the cyclic hemiacetal form, so its free −CHO group is not available for the reactions that need a genuine free aldehyde: it does NOT give the Schiff's test and does NOT form the bisulphite addition product with NaHSO₃.
It does still reduce Tollen's reagent, and it does react with HCN and NH₂OH.
Q20 — Chemistry · medium · theory
Match List I with List II.
List-I (Molecule)
(A) ethane
(B) ethene
(C) carbon molecule, C₂
(D) ethyne
List-II (Number and types of bond/s between two carbon atoms)
(I) one σ-bond and two π-bonds
(II) two π-bonds
(III) one σ-bond
(IV) one σ-bond and one π-bond
Choose the correct answer from the options given below:
A. A-III, B-IV, C-II, D-I ✓ Correct
B. A-III, B-IV, C-I, D-II
C. A-I, B-IV, C-II, D-III
D. A-IV, B-III, C-II, D-I
Solution: Ethane, H₃C−CH₃ — a single bond, so one σ-bond (III).
Ethene, H₂C=CH₂ — one σ-bond and one π-bond (IV).
The C₂ molecule is unusual: molecular orbital theory gives it two π-bonds and no σ-bond (II).
Ethyne, HC≡CH — one σ-bond and two π-bonds (I).
Q21 — Chemistry · medium · theory
Among Group 16 elements, which one does NOT show −2 oxidation state?
A. Te
B. Po ✓ Correct
C. O
D. Se
Solution: Going down group 16 the elements become steadily more metallic (more electropositive), so the tendency to pick up two electrons falls away.
Polonium, the heaviest and most metallic member, does not show the −2 state.
Q22 — Chemistry · medium · numerical
For the reaction $2A \rightleftharpoons B + C$, $K_c = 4 \times 10^{-3}$. At a given time, the composition of the reaction mixture is $[A] = [B] = [C] = 2 \times 10^{-3}\ M$. Then, which of the following is correct?
A. Reaction has a tendency to go in backward direction. ✓ Correct
B. Reaction has gone to completion in forward direction.
C. Reaction is at equilibrium.
D. Reaction has a tendency to go in forward direction.
Solution: The reaction quotient at that moment is
$Q_c = \dfrac{[B][C]}{[A]^2} = \dfrac{(2 \times 10^{-3})(2 \times 10^{-3})}{(2 \times 10^{-3})^2} = 1$
Since $Q_c = 1$ is much larger than $K_c = 4 \times 10^{-3}$, there is too much product for equilibrium. The system relieves that by consuming products, i.e. the reaction has a tendency to go in the BACKWARD direction.
Q23 — Chemistry · medium · theory
Which plot of $\ln k$ vs $\dfrac{1}{T}$ is consistent with the Arrhenius equation?
A.
B. ✓ Correct
C.
D.
Solution: The Arrhenius equation in logarithmic form is
$\ln k = -\dfrac{E_a}{R}\cdot\dfrac{1}{T} + \ln A$
That is a straight line in $\ln k$ against $\dfrac{1}{T}$ with intercept $\ln A$ and slope $-\dfrac{E_a}{R}$ — a NEGATIVE slope, since $E_a$ is positive. Only graph (b) has it.
Q24 — Chemistry · medium · numerical
In which of the following equilibria are $K_p$ and $K_c$ NOT equal?
A. CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g)
B. 2BrCl(g) ⇌ Br₂(g) + Cl₂(g)
C. PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) ✓ Correct
D. H₂(g) + I₂(g) ⇌ 2HI(g)
Solution: $K_p = K_c (RT)^{\Delta n_g}$, so the two are equal only when $\Delta n_g = 0$.
(a) $\Delta n_g = 2 - 2 = 0$
(b) $\Delta n_g = 2 - 2 = 0$
(c) $\Delta n_g = 2 - 1 = 1$ — here $K_p = K_c(RT) \neq K_c$
(d) $\Delta n_g = 2 - 2 = 0$
Q25 — Chemistry · medium · theory
Given below are two statements:
Statement I: The boiling point of three isomeric pentanes follows the order n-pentane > isopentane > neopentane.
Statement II: When branching increases, the molecule attains a shape of sphere. This results in 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:
A. Statement I is correct but Statement II is incorrect.
B. Statement I is incorrect but Statement II is correct.
C. Both Statement I and Statement II are correct. ✓ Correct
D. Both Statement I and Statement II are incorrect.
Solution: Both are correct, and the second is the reason for the first. Branching makes a molecule more compact and ball-like, cutting the surface area available for van der Waals contact, so the intermolecular forces weaken and the boiling point falls.
Hence n-pentane (unbranched) > isopentane (one branch) > neopentane (most branched).
Q26 — Chemistry · medium · theory
The compound that will undergo $S_N1$ reaction with the fastest rate is
A.
B. ✓ Correct
C.
D.
Solution: An $S_N1$ reaction goes through a carbocation, so it is fastest where that carbocation is most stable.
Option (b) ionises to a benzylic (secondary) carbocation, which is stabilised by resonance with the ring. None of the others can do that — the aryl and vinyl-type halides in (a) and (d) barely ionise at all, and (c) gives only a primary carbocation.
Q27 — Chemistry · medium · numerical
The energy of an electron in the ground state ($n = 1$) for He⁺ ion is $-x\ \text{J}$. Then that for an electron in the $n = 2$ state for Be³⁺ ion, in J, is:
A. $-4x$
B. $-\dfrac{4}{9}x$
C. $-x$ ✓ Correct
D. $-\dfrac{x}{9}$
Solution: For a hydrogen-like ion, $E = \dfrac{-2.18 \times 10^{-18}\,Z^2}{n^2}\ \text{J}$.
For He⁺ ($Z = 2$, $n = 1$): $-x = -2.18 \times 10^{-18} \times 4$, so $x = 2.18 \times 10^{-18} \times 4$.
For Be³⁺ ($Z = 4$, $n = 2$): $E = \dfrac{-2.18 \times 10^{-18} \times 16}{4} = -2.18 \times 10^{-18} \times 4 = -x\ \text{J}$.
Q28 — Chemistry · medium · theory
In which of the following processes does entropy increase?
A. A liquid evaporates to vapour.
B. Temperature of a crystalline solid lowered from 130 K to 0 K.
C. 2NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g)
D. Cl₂(g) → 2Cl(g)
Choose the correct answer from the options given below:
A. A, C and D ✓ Correct
B. C and D
C. A and C
D. A, B and D
Solution: A — a liquid becoming vapour is a big rise in randomness, so entropy increases.
B — cooling a crystalline solid towards 0 K reduces random motion, so entropy DECREASES.
C — solid gives solid plus two gases, so entropy increases.
D — one gas molecule becomes two gas atoms, so entropy increases.
Hence A, C and D.
Q29 — Chemistry · medium · theory
On heating, some solid substances change from solid to vapour state without passing through liquid state. The technique used for the purification of such solid substances based on the above principle is known as:
A. Distillation
B. Chromatography
C. Crystallization
D. Sublimation ✓ Correct
Solution: A solid that goes straight to vapour on heating sublimes. Sublimation is used to separate a sublimable compound from non-sublimable impurities: the solid is heated, the vapour is collected on a cold surface, and the impurity is left behind.
Q30 — Chemistry · medium · theory
Match List-I with List-II.
List-I (Complex)
(A) [Co(NH₃)₅(NO₂)]Cl₂
(B) [Co(NH₃)₅(SO₄)]Br
(C) [Co(NH₃)₆][Cr(CN)₆]
(D) [Co(H₂O)₆]Cl₃
List-II (Type of isomerism)
(I) Solvate isomerism
(II) Linkage isomerism
(III) Ionization isomerism
(IV) Coordination isomerism
Choose the correct answer from the options given below:
A. A-I, B-IV, C-III, D-II
B. A-II, B-IV, C-III, D-I
C. A-II, B-III, C-IV, D-I ✓ Correct
D. A-I, B-III, C-IV, D-II
Solution: (A) NO₂⁻ can bind through N or O, so [Co(NH₃)₅(NO₂)]Cl₂ shows LINKAGE isomerism — its isomer is [Co(NH₃)₅(ONO)]Cl₂ (II).
(B) [Co(NH₃)₅(SO₄)]Br swaps the ligand and the counter ion with [Co(NH₃)₅(Br)]SO₄ — IONIZATION isomerism (III).
(C) [Co(NH₃)₆][Cr(CN)₆] exchanges ligands between the two metal centres with [Cr(NH₃)₆][Co(CN)₆] — COORDINATION isomerism (IV).
(D) [Co(H₂O)₆]Cl₃ can move water in and out of the coordination sphere, e.g. [Co(H₂O)₅Cl]Cl₂·H₂O — SOLVATE isomerism (I).