NEET Chemistry Mock Test 2026: 45 MCQs with Answers & Solutions

- Dr.Sanjaykumar Pawar

NEET Chemistry Mock Test 2026 with 45 NCERT-based MCQs, answers and detailed solutions

NEET Chemistry Mock Test 2026 – 45 NCERT-based MCQs with Answer Key and Detailed Solutions

NEET-UG Chemistry Mock Test 2026

General Instructions

This paper contains 45 multiple-choice questions.
Each question has four options: A, B, C and D.
Only one option is correct for each question.
Each correct answer carries +4 marks.
Each incorrect answer carries −1 mark.
An unattempted question carries 0 marks.
Questions are designed at NEET-UG level and are based primarily on NCERT concepts.
Do not refer to the answer key while attempting the question paper.

SECTION A — QUESTION PAPER

1. Some Basic Concepts of Chemistry — Easy

A sample contains 4.4 g of CO₂. The number of oxygen atoms present in the sample is:
(Atomic masses: C = 12, O = 16; N_A = 6.022 × 10²³)

A) 6.022 × 10²²

B) 1.2044 × 10²³

C) 1.2044 × 10²⁴

D) 2.4088 × 10²³

2. Atomic Structure — Easy

Which set of quantum numbers is not permissible for an electron?

A) n = 3, l = 2, m_l = −1, m_s = +1/2

B) n = 4, l = 3, m_l = 0, m_s = −1/2

C) n = 2, l = 2, m_l = 0, m_s = +1/2

D) n = 3, l = 1, m_l = +1, m_s = −1/2

3. Classification of Elements and Periodicity — Easy

Which of the following has the largest atomic radius?

A) Na

B) Mg

C) Al

D) Si

4. Chemical Bonding — Moderate

The correct order of bond angle is:

A) NH₃ > H₂O > CH₄

B) CH₄ > NH₃ > H₂O

C) H₂O > NH₃ > CH₄

D) NH₃ > CH₄ > H₂O

5. Chemical Thermodynamics — Moderate

For a reaction at constant temperature and pressure, a negative value of ΔG indicates that the reaction is:

A) Endothermic only

B) Exothermic only

C) Spontaneous in the forward direction

D) At equilibrium

6. Solutions — Easy

The van't Hoff factor for completely dissociated K₂SO₄ in dilute aqueous solution is approximately:

A) 1

B) 2

C) 3

D) 4

7. Equilibrium — Moderate

For the equilibrium:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

an increase in pressure at constant temperature will:

A) Shift equilibrium towards reactants

B) Shift equilibrium towards products

C) Have no effect

D) Stop the reaction

8. Redox Reactions and Electrochemistry — Moderate

In the reaction:

2Fe²⁺ + Cl₂ → 2Fe³⁺ + 2Cl⁻

the reducing agent is:

A) Fe²⁺

B) Fe³⁺

C) Cl₂

D) Cl⁻

9. Chemical Kinetics — Moderate

For a first-order reaction, the half-life is 20 min. The time required for 87.5% completion is:

A) 20 min

B) 40 min

C) 60 min

D) 80 min

10. P-Block Elements — Easy

Which of the following is the strongest reducing halide ion in aqueous solution?

A) F⁻

B) Cl⁻

C) Br⁻

D) I⁻

11. D- and F-Block Elements — Easy

The colour of many transition-metal ions is primarily due to:

A) Nuclear transitions

B) d-d electronic transitions

C) Proton transfer

D) Complete absence of unpaired electrons

12. Coordination Compounds — Moderate

The oxidation state of cobalt and coordination number of cobalt in [Co(NH₃)₅Cl]Cl₂, respectively, are:

A) +2 and 5

B) +3 and 6

C) +3 and 5

D) +2 and 6

13. Purification and Characterisation of Organic Compounds — Easy

The technique most suitable for separating two miscible liquids having sufficiently different boiling points is:

A) Sublimation

B) Distillation

C) Crystallisation

D) Filtration

14. Some Basic Principles of Organic Chemistry — Moderate

The most stable carbocation among the following is:

A) CH₃⁺

B) CH₃CH₂⁺

C) (CH₃)₂CH⁺

D) (CH₃)₃C⁺

15. Hydrocarbons — Moderate

Propene reacts with HBr in the presence of peroxide to predominantly form:

A) 2-bromopropane

B) 1-bromopropane

C) 1,2-dibromopropane

D) Propan-2-ol

16. Organic Compounds Containing Halogens — Easy

Which compound undergoes an S_N1 reaction most readily?

A) Methyl chloride

B) Ethyl chloride

C) Isopropyl chloride

D) tert-Butyl chloride

17. Organic Compounds Containing Oxygen — Moderate

Which compound gives a positive iodoform test?

A) Methanol

B) Ethanol

C) Propan-1-ol

D) Propanoic acid

18. Organic Compounds Containing Nitrogen — Moderate

Aniline is less basic than ethylamine mainly because:

A) Nitrogen in aniline is sp³-hybridised

B) The lone pair on nitrogen in aniline is delocalised into the benzene ring

C) Ethylamine has no lone pair

D) Aniline contains an electron-releasing phenyl group

19. Biomolecules — Easy

The bond linking two monosaccharide units in a disaccharide is called:

A) Peptide bond

B) Glycosidic bond

C) Phosphodiester bond

D) Hydrogen bond

20. Practical Chemistry — Easy

Which gas is evolved when dilute HCl is added to sodium carbonate?

A) H₂

B) O₂

C) CO₂

D) Cl₂

21. Some Basic Concepts of Chemistry — Moderate

A compound contains 40.0% carbon, 6.67% hydrogen and 53.33% oxygen by mass. Its empirical formula is:

A) CHO

B) CH₂O

C) C₂H₄O₂

D) C₂H₂O

22. Atomic Structure — Moderate

An electron and a proton have the same kinetic energy. The ratio of their de Broglie wavelengths, λ_e / λ_p, is approximately:

A) 1

B) √1836

C) 1/√1836

D) 1836

23. Classification of Elements and Periodicity — Moderate

Consider the following statements:

I. Ionisation enthalpy generally increases from left to right across a period.
II. Electron gain enthalpy of chlorine is more negative than that of fluorine.
III. Atomic radius generally decreases across a period.

The correct option is:

A) I only

B) I and II only

C) I and III only

D) I, II and III

24. Chemical Bonding — Difficult

Which of the following species has the highest bond order?

A) O₂

B) O₂⁺

C) O₂⁻

D) O₂²⁻

25. Chemical Thermodynamics — Moderate

Given:

C(graphite) + O₂ → CO₂, ΔH = −393.5 kJ mol⁻¹

CO + 1/2 O₂ → CO₂, ΔH = −283.0 kJ mol⁻¹

The enthalpy change for:

C(graphite) + 1/2 O₂ → CO

is:

A) −676.5 kJ mol⁻¹

B) −110.5 kJ mol⁻¹

C) +110.5 kJ mol⁻¹

D) +676.5 kJ mol⁻¹

26. Solutions — Moderate

A 0.10 molal aqueous solution of a non-electrolyte is prepared. If the boiling-point elevation constant of water is K_b = 0.52 K kg mol⁻¹, the elevation in boiling point is:

A) 0.052 K

B) 0.52 K

C) 5.2 K

D) 52 K

27. Equilibrium — Difficult

For the equilibrium:

A(g) ⇌ 2B(g)

the equilibrium constant in terms of concentration is K_c = 0.25. If initially only A is present at concentration 1 M, the equilibrium concentration of A is closest to:

A) 0.20 M

B) 0.33 M

C) 0.50 M

D) 0.75 M

28. Redox Reactions and Electrochemistry — Moderate

For the galvanic cell:

Zn | Zn²⁺ || Cu²⁺ | Cu

under standard conditions, E^°_Cu²⁺/Cu = +0.34 V and E^°_Zn²⁺/Zn = −0.76 V.

The standard cell potential is:

A) −1.10 V

B) +1.10 V

C) +0.42 V

D) −0.42 V

29. Chemical Kinetics — Difficult

For a reaction, doubling the concentration of A doubles its rate, while doubling the concentration of B increases the rate fourfold. The rate law is:

A) r = k[A][B]

B) r = k[A]²[B]

C) r = k[A][B]²

D) r = k[A]²[B]²

30. P-Block Elements — Moderate

Which statement about NH₃ is correct?

A) It is a planar molecule

B) Nitrogen is sp²-hybridised

C) It has a pyramidal shape with one lone pair on nitrogen

D) Its bond angle is exactly 120°

31. D- and F-Block Elements — Moderate

Which ion is expected to be diamagnetic?

A) Fe²⁺

B) Co²⁺

C) Zn²⁺

D) Ni²⁺

32. Coordination Compounds — Difficult

The number of geometrical isomers possible for [Pt(NH₃)₂Cl₂] is:

A) 1

B) 2

C) 3

D) 4

33. Purification and Characterisation — Moderate

In paper chromatography, the R_f value of a component is defined as:

A) Distance travelled by solvent / distance travelled by solute

B) Distance travelled by solute / distance travelled by solvent front

C) Distance travelled by solute × distance travelled by solvent

D) Distance travelled by solvent − distance travelled by solute

34. Organic Chemistry — Moderate

The correct decreasing order of acidity is:

A) Ethanol > phenol > acetic acid

B) Acetic acid > phenol > ethanol

C) Phenol > acetic acid > ethanol

D) Ethanol > acetic acid > phenol

35. Hydrocarbons — Difficult

Ozonolysis of an alkene gives ethanal and propanone as the only carbonyl products. The alkene is:

A) Pent-1-ene

B) 2-methylbut-2-ene

C) Pent-2-ene

D) 2-methylbut-1-ene

36. Haloalkanes and Haloarenes — Moderate

Which reagent is most appropriate for converting an alkyl chloride into the corresponding alkyl iodide through the Finkelstein reaction?

A) NaI in acetone

B) KI in water

C) I₂/NaOH

D) NaCl in acetone

37. Organic Compounds Containing Oxygen — Difficult

Which reagent combination can distinguish ethanal from propanone?

A) Tollens' reagent

B) NaOH

C) Dilute HCl

D) NaHCO₃

38. Organic Compounds Containing Nitrogen — Moderate

The product formed when aniline is treated with NaNO₂/HCl at 0–5°C is:

A) Nitrobenzene

B) Benzenediazonium chloride

C) Phenol

D) Chlorobenzene

39. Biomolecules — Moderate

Which statement about proteins is correct?

A) Primary structure is determined by hydrogen bonding only

B) Peptide bonds join amino acids in a polypeptide chain

C) All proteins are composed exclusively of β-sheets

D) Denaturation always hydrolyses peptide bonds

40. Practical Chemistry — Moderate

A solution gives a white precipitate with AgNO₃, soluble in dilute ammonia. The anion is most likely:

A) Cl⁻

B) Br⁻

C) I⁻

D) SO₄²⁻

41. Chemical Bonding — Difficult

Which statement is correct regarding BF₃ and NH₃?

A) Both are electron-rich Lewis bases

B) BF₃ is a Lewis acid whereas NH₃ is a Lewis base

C) BF₃ and NH₃ are both Lewis acids

D) Neither can participate in coordinate bonding

42. Equilibrium — Difficult

The pH of a 0.001 M strong monoprotic acid HA is approximately:

A) 1

B) 2

C) 3

D) 4

43. Electrochemistry — Moderate

During electrolysis of molten NaCl, the products at the cathode and anode respectively are:

A) Cl₂ and Na

B) Na and Cl₂

C) H₂ and O₂

D) NaOH and HCl

44. Organic Compounds Containing Oxygen — Difficult

Which compound undergoes the Cannizzaro reaction?

A) Ethanal

B) Propanal

C) Benzaldehyde

D) Acetone

45. Integrated NCERT/NEET Chemistry — Difficult

Consider the following statements:

I. XeF₂ has a linear molecular geometry.
II. NH₃ has a higher boiling point than PH₃ primarily because of hydrogen bonding.
III. BF₃ is electron deficient and acts as a Lewis acid.
IV. CH₄ is polar because C–H bonds are polar.

Which statements are correct?

A) I, II and III only

B) II and IV only

C) I, III and IV only

D) I, II, III and IV

ANSWER KEY

1-B, 2-C, 3-A, 4-B, 5-C, 6-C, 7-B, 8-A, 9-C, 10-D,
11-B, 12-B, 13-B, 14-D, 15-B, 16-D, 17-B, 18-B, 19-B, 20-C,
21-B, 22-B, 23-D, 24-B, 25-B, 26-A, 27-DROPPED, 28-B, 29-C, 30-C,
31-C, 32-B, 33-B, 34-B, 35-B, 36-A, 37-A, 38-B, 39-B, 40-A,
41-B, 42-C, 43-B, 44-C, 45-A

Question 27 Quality-Control Note: The calculation gives an equilibrium concentration of approximately 0.773 M, but the original options do not contain this value. Therefore Q27 should be treated as a dropped question rather than assigning an incorrect option.

DETAILED SOLUTIONS

Question 1 — Correct Answer: B

Chapter: Some Basic Concepts of Chemistry | Difficulty: Easy

Molar mass of CO₂:

M(CO₂) = 12 + 2(16) = 44 g mol⁻¹

Number of moles:

n = 4.4 / 44 = 0.1 mol

Each CO₂ molecule contains two oxygen atoms.

Number of O atoms = 0.1 × 2 × 6.022 × 10²³ = 1.2044 × 10²³

NCERT Concept: Mole concept and Avogadro constant.

Question 2 — Correct Answer: C

Chapter: Structure of Atom | Difficulty: Easy

For a given value of n:

l = 0, 1, 2, ... (n − 1)

For n = 2, l can only be 0 or 1. Therefore l = 2 is not permissible.

Answer: C.

Question 3 — Correct Answer: A

Chapter: Classification of Elements and Periodicity | Difficulty: Easy

Na, Mg, Al and Si belong to the same period. Across a period, effective nuclear charge increases and atomic radius decreases.

Na > Mg > Al > Si

Answer: A.

Question 4 — Correct Answer: B

Chapter: Chemical Bonding | Difficulty: Moderate

Lone pairs exert greater repulsion than bond pairs. Therefore, increasing the number of lone pairs compresses the bond angle.

CH₄ ≈ 109.5° > NH₃ ≈ 107° > H₂O ≈ 104.5°

Answer: B.

Question 5 — Correct Answer: C

Chapter: Chemical Thermodynamics | Difficulty: Moderate

Gibbs free energy is given by:

ΔG = ΔH − TΔS

At constant temperature and pressure, ΔG < 0 indicates a spontaneous process in the forward direction.

Answer: C.

Question 6 — Correct Answer: C

Chapter: Solutions | Difficulty: Easy

Complete dissociation occurs as:

K₂SO₄ → 2K⁺ + SO₄²⁻

One formula unit produces three particles. Hence:

i = 3

Answer: C.

Question 7 — Correct Answer: B

Chapter: Equilibrium | Difficulty: Moderate

Reactant side contains four moles of gaseous species, while the product side contains two.

Increasing pressure shifts equilibrium towards the side having fewer gaseous molecules.

Answer: B.

Question 8 — Correct Answer: A

Chapter: Redox Reactions | Difficulty: Moderate

Fe²⁺ loses an electron:

Fe²⁺ → Fe³⁺ + e⁻

Therefore Fe²⁺ is oxidised and acts as the reducing agent.

Answer: A.

Question 9 — Correct Answer: C

Chapter: Chemical Kinetics | Difficulty: Moderate

87.5% completion means 12.5% remains:

12.5% = 1/8

Three half-lives are required:

100% → 50% → 25% → 12.5%

t = 3 × 20 = 60 min

Answer: C.

Question 10 — Correct Answer: D

Chapter: P-Block Elements | Difficulty: Easy

Reducing character of halide ions increases down the group:

I⁻ > Br⁻ > Cl⁻ > F⁻

I⁻ is therefore the strongest reducing halide ion.

Question 11 — Correct Answer: B

Chapter: D- and F-Block Elements | Difficulty: Easy

Partially filled d-orbitals can undergo electronic transitions between split d-levels on absorption of visible light.

These are called d-d transitions.

Question 12 — Correct Answer: B

Chapter: Coordination Compounds | Difficulty: Moderate

The two chloride ions outside the coordination sphere contribute −2. Therefore the complex ion has charge +2.

x + 5(0) − 1 = +2

x = +3

Coordination number = 5 NH₃ + 1 Cl = 6.

Answer: B.

Question 13 — Correct Answer: B

Chapter: Purification and Characterisation | Difficulty: Easy

Distillation is used for separating miscible liquids based on differences in boiling points.

Answer: B.

Question 14 — Correct Answer: D

Chapter: Some Basic Principles of Organic Chemistry | Difficulty: Moderate

Carbocation stability follows:

3° > 2° > 1° > methyl

The tertiary carbocation is stabilised by the +I effect and hyperconjugation of three alkyl groups.

Answer: D.

Question 15 — Correct Answer: B

Chapter: Hydrocarbons | Difficulty: Moderate

HBr undergoes anti-Markovnikov addition in the presence of peroxide.

CH₃CH=CH₂ + HBr → CH₃CH₂CH₂Br

Product: 1-bromopropane.

Question 16 — Correct Answer: D

Chapter: Haloalkanes and Haloarenes | Difficulty: Easy

S_N1 reactions proceed through carbocation intermediates. Tertiary carbocations are the most stable among simple alkyl carbocations.

Hence tert-butyl chloride reacts most readily by the S_N1 mechanism.

Question 17 — Correct Answer: B

Chapter: Organic Compounds Containing Oxygen | Difficulty: Moderate

Ethanol is oxidised under the conditions of the iodoform test to ethanal, which contains the required methyl carbonyl group.

It consequently gives yellow precipitate of iodoform, CHI₃.

Question 18 — Correct Answer: B

Chapter: Organic Compounds Containing Nitrogen | Difficulty: Moderate

The lone pair on nitrogen in aniline is delocalised into the benzene ring. Consequently, it is less available for accepting a proton.

Ethylamine is more basic because the ethyl group exerts a +I effect.

Question 19 — Correct Answer: B

Chapter: Biomolecules | Difficulty: Easy

Monosaccharide units are joined through glycosidic linkages.

Answer: B.

Question 20 — Correct Answer: C

Chapter: Practical Chemistry | Difficulty: Easy

CO₃²⁻ + 2H⁺ → CO₂ + H₂O

Therefore carbon dioxide gas is evolved.

Question 21 — Correct Answer: B

Chapter: Some Basic Concepts of Chemistry | Difficulty: Moderate

Assume 100 g of compound.

C = 40/12 = 3.33 mol
H = 6.67/1 = 6.67 mol
O = 53.33/16 = 3.33 mol

Dividing by the smallest value gives:

C : H : O = 1 : 2 : 1

Empirical formula = CH₂O.

Question 22 — Correct Answer: B

Chapter: Atomic Structure | Difficulty: Moderate

de Broglie wavelength is:

λ = h / √(2mK)

For equal kinetic energy:

λ ∝ 1/√m

Therefore:

λₑ / λₚ = √(mₚ/mₑ) = √1836

Answer: B.

Question 23 — Correct Answer: D

Chapter: Periodicity | Difficulty: Moderate

All three statements are correct.

Ionisation enthalpy generally increases across a period.
Chlorine has more negative electron gain enthalpy than fluorine.
Atomic radius generally decreases across a period.

Answer: D.

Question 24 — Correct Answer: B

Chapter: Chemical Bonding | Difficulty: Difficult

Removing an electron from the antibonding orbital increases bond order.

BO(O₂) = 2
BO(O₂⁺) = 2.5
BO(O₂⁻) = 1.5
BO(O₂²⁻) = 1

Hence O₂⁺ has the highest bond order.

Question 25 — Correct Answer: B

Chapter: Chemical Thermodynamics | Difficulty: Moderate

Reverse the second equation:

CO₂ → CO + 1/2 O₂
ΔH = +283.0 kJ mol⁻¹

Add it to:

C + O₂ → CO₂
ΔH = −393.5 kJ mol⁻¹

Therefore:

ΔH = −393.5 + 283.0 = −110.5 kJ mol⁻¹

Question 26 — Correct Answer: A

Chapter: Solutions | Difficulty: Moderate

ΔTᵦ = Kᵦm

ΔTᵦ = 0.52 × 0.10 = 0.052 K

Answer: A.

Question 27 — Dropped Question

Chapter: Equilibrium | Difficulty: Difficult

Let x M of A dissociate:

A ⇌ 2B

[A]ₑ = 1 − x
[B]ₑ = 2x

Therefore:

Kc = [B]²/[A]
0.25 = 4x²/(1 − x)

Rearranging:

16x² + x − 1 = 0

The positive solution is approximately:

x ≈ 0.227

Hence:

[A]ₑ = 1 − 0.227 ≈ 0.773 M

Since 0.773 M is absent from the four options, this question should be dropped rather than assigning a false answer.

Question 28 — Correct Answer: B

Chapter: Electrochemistry | Difficulty: Moderate

E°cell = E°cathode − E°anode

E°cell = 0.34 − (−0.76) = 1.10 V

Answer: B.

Question 29 — Correct Answer: C

Chapter: Chemical Kinetics | Difficulty: Difficult

Let:

r = k[A]ᵐ[B]ⁿ

Doubling A doubles the rate:

2ᵐ = 2 → m = 1

Doubling B makes the rate four times:

2ⁿ = 4 → n = 2

Hence:

r = k[A][B]²

Question 30 — Correct Answer: C

Chapter: P-Block Elements / Chemical Bonding | Difficulty: Moderate

NH₃ has three bond pairs and one lone pair on nitrogen. Its molecular geometry is trigonal pyramidal.

The bond angle is approximately 107°.

Question 31 — Correct Answer: C

Chapter: D- and F-Block Elements | Difficulty: Moderate

Zinc has configuration:

Zn = [Ar] 3d¹⁰4s²

Therefore:

Zn²⁺ = [Ar] 3d¹⁰

There are no unpaired electrons, so Zn²⁺ is diamagnetic.

Question 32 — Correct Answer: B

Chapter: Coordination Compounds | Difficulty: Difficult

[Pt(NH₃)₂Cl₂] is square planar and can exist in two geometrical forms:

cis — identical ligands adjacent
trans — identical ligands opposite

Therefore the number of geometrical isomers is 2.

Question 33 — Correct Answer: B

Chapter: Purification and Characterisation | Difficulty: Moderate

Rf = Distance travelled by solute / Distance travelled by solvent front

Answer: B.

Question 34 — Correct Answer: B

Chapter: Organic Compounds Containing Oxygen | Difficulty: Moderate

The conjugate base of a carboxylic acid is strongly resonance stabilised. Phenoxide is also resonance stabilised, while ethoxide is not.

CH₃COOH > C₆H₅OH > C₂H₅OH

Hence the correct decreasing order is acetic acid > phenol > ethanol.

Question 35 — Correct Answer: B

Chapter: Hydrocarbons | Difficulty: Difficult

The required alkene must have one double-bond carbon producing ethanal and the other producing propanone.

CH₃CH=C(CH₃)₂

This compound is 2-methylbut-2-ene.

Question 36 — Correct Answer: A

Chapter: Haloalkanes and Haloarenes | Difficulty: Moderate

The Finkelstein reaction uses sodium iodide in dry acetone:

R–Cl + NaI → R–I + NaCl

Precipitation of NaCl drives the reaction forward.

Question 37 — Correct Answer: A

Chapter: Aldehydes and Ketones | Difficulty: Difficult

Ethanal is an aldehyde and reduces Tollens' reagent to metallic silver. Propanone is a ketone and does not give the test.

Therefore Tollens' reagent distinguishes the two compounds.

Question 38 — Correct Answer: B

Chapter: Organic Compounds Containing Nitrogen | Difficulty: Moderate

Diazotisation of aniline is performed at 0–5°C:

C₆H₅NH₂ →[NaNO₂/HCl, 0–5°C] C₆H₅N₂⁺Cl⁻

Product: benzenediazonium chloride.

Question 39 — Correct Answer: B

Chapter: Biomolecules | Difficulty: Moderate

Amino acids in proteins are joined by peptide bonds:

−CO−NH−

Denaturation generally affects higher-order structures without breaking the primary peptide backbone.

Question 40 — Correct Answer: A

Chapter: Practical Chemistry | Difficulty: Moderate

Chloride ion gives a white precipitate with silver nitrate:

Ag⁺ + Cl⁻ → AgCl↓

AgCl dissolves in dilute ammonia because of formation of the diamminesilver(I) complex:

AgCl + 2NH₃ → [Ag(NH₃)₂]⁺ + Cl⁻

Therefore the anion is chloride.

Question 41 — Correct Answer: B

Chapter: Chemical Bonding | Difficulty: Difficult

Boron in BF₃ is electron deficient and can accept an electron pair. Hence BF₃ is a Lewis acid.

NH₃ has a lone pair and can donate it, making it a Lewis base.

BF₃ + NH₃ → F₃B←NH₃

Question 42 — Correct Answer: C

Chapter: Ionic Equilibrium | Difficulty: Difficult

For a strong monoprotic acid, complete ionisation occurs:

[H⁺] = 0.001 M = 10⁻³ M

pH = −log[H⁺] = −log(10⁻³) = 3

Answer: C.

Question 43 — Correct Answer: B

Chapter: Electrochemistry | Difficulty: Moderate

In molten NaCl, sodium ions are reduced at the cathode:

Na⁺ + e⁻ → Na

Chloride ions are oxidised at the anode:

2Cl⁻ → Cl₂ + 2e⁻

Thus the products are Na and Cl₂.

Question 44 — Correct Answer: C

Chapter: Aldehydes and Ketones | Difficulty: Difficult

Aldehydes without an α-hydrogen undergo Cannizzaro reaction. Benzaldehyde has no α-hydrogen.

2C₆H₅CHO + OH⁻ → C₆H₅CH₂OH + C₆H₅COO⁻

Therefore benzaldehyde undergoes Cannizzaro reaction.

Question 45 — Correct Answer: A

Chapter: Chemical Bonding / P-Block Elements | Difficulty: Difficult

Statement I: Correct. XeF₂ has a linear molecular geometry because the three lone pairs occupy equatorial positions.

Statement II: Correct. NH₃ forms hydrogen bonds, giving it a higher boiling point than PH₃.

Statement III: Correct. BF₃ is electron deficient and accepts an electron pair, so it is a Lewis acid.

Statement IV: Incorrect. Although individual C–H bonds have small bond polarity, the tetrahedral geometry causes their dipoles to cancel. Methane is therefore non-polar.

Hence statements I, II and III are correct.

Answer: A.

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