Class 12 Chemistry Important Questions with Solutions

Class 12 Chemistry Important Questions provide chapter-wise practice across Physical, Inorganic and Organic Chemistry. The questions cover calculations, chemical equations, conversions, nomenclature, reasoning and concept-based answers from the current NCERT course.

Class 12 Chemistry includes numerical concepts such as colligative properties, electrochemical cells and reaction rates. It also covers transition elements, coordination compounds, organic reactions and biomolecules.

The Class 12 Chemistry Important Questions given below help students revise the complete course through direct questions, numericals, conversions and reasoning-based problems. Students should first understand the governing law or reaction and then practise writing complete, balanced answers.

Key Takeaways

• The current Class 12 Chemistry course contains 10 units.
• Physical Chemistry includes Solutions, Electrochemistry and Chemical Kinetics.
• Inorganic Chemistry covers d- and f-Block Elements and Coordination Compounds.
• Organic Chemistry begins with Haloalkanes and Haloarenes and ends with Biomolecules.
• Numerical answers should include the formula, substitution and correct unit.
• Organic reactions should mention the reagent, condition and major product.
• Inorganic reasoning answers should connect electronic configuration with the observed property.

Important Class 12 Chemistry Formulas

Solutions

Molarity:

M = Number of moles of solute/Volume of solution in litres

Molality:

m = Number of moles of solute/Mass of solvent in kg

Mole fraction:

χᵢ = Number of moles of component i/Total number of moles

Raoult’s law:

p₁ = x₁p₁°

Relative lowering of vapour pressure:

(p₁° − p₁)/p₁° = x₂

Elevation in boiling point:

ΔTᵦ = Kᵦm

Depression in freezing point:

ΔT𝒇 = K𝒇m

Osmotic pressure:

π = CRT

Electrochemistry

Cell potential:

Ecell = Ecathode − Eanode

Standard Gibbs energy:

ΔG° = −nFE°cell

Nernst equation at 298 K:

Ecell = E°cell − (0.0591/n)log Q

Conductance:

G = 1/R

Conductivity:

κ = G(l/A)

Molar conductivity:

Λₘ = κ × 1000/C

Chemical Kinetics

Rate of reaction:

Rate = −Δ[R]/Δt

First-order rate equation:

k = (2.303/t)log([R]₀/[R])

Half-life of a first-order reaction:

t₁/₂ = 0.693/k

Arrhenius equation:

k = Ae^(−Eₐ/RT)

Coordination Compounds

Coordination number is the number of donor atoms directly bonded to the central metal atom or ion.

Oxidation number is calculated by balancing the total charge of the complex.

Access Class 12 Chemistry Important Questions in 30 Minutes

• First 10 minutes: Revise formulas from Solutions, Electrochemistry and Chemical Kinetics.
• Next 10 minutes: Review oxidation states, colours, nomenclature and isomerism from Inorganic Chemistry.
• Final 10 minutes: Practise named reactions, conversions, chemical tests and Biomolecules definitions.

Need help with Chemistry numericals, reactions and conversions?

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Very Short Answer Questions – 1 Mark

Q1. What is molality?

Answer: Molality is the number of moles of solute present in one kilogram of solvent.

m = Moles of solute/Mass of solvent in kg

Q2. What is an ideal solution?

Answer: An ideal solution obeys Raoult’s law over the complete range of concentration and has:

ΔHmix = 0

ΔVmix = 0

Q3. What are colligative properties?

Answer: Colligative properties depend on the number of solute particles present and not on their chemical identity.

Q4. Define osmotic pressure.

Answer: Osmotic pressure is the minimum pressure that must be applied to a solution to prevent osmosis.

Q5. What is a galvanic cell?

Answer: A galvanic cell converts the chemical energy of a spontaneous redox reaction into electrical energy.

Q6. What is the function of a salt bridge?

Answer: A salt bridge completes the electrical circuit and maintains electrical neutrality in the two half-cells.

Q7. What is corrosion?

Answer: Corrosion is the gradual deterioration of a metal due to its chemical or electrochemical reaction with the environment.

Q8. Define the order of a reaction.

Answer: The order of a reaction is the sum of the powers of reactant concentrations in its experimentally determined rate law.

Q9. What is molecularity?

Answer: Molecularity is the number of reacting species taking part in a single elementary step.

Q10. What is activation energy?

Answer: Activation energy is the minimum energy that reacting molecules must possess for an effective collision.

Q11. Why do transition elements show variable oxidation states?

Answer: The energies of the ns and (n − 1)d orbitals are comparable, so electrons from both can participate in bonding.

Q12. Why are many transition-metal compounds coloured?

Answer: They are coloured because electrons absorb visible light and undergo transitions between split d-orbitals.

Q13. What is lanthanoid contraction?

Answer: Lanthanoid contraction is the gradual decrease in the atomic and ionic radii of lanthanoids with increasing atomic number.

Q14. Define a coordination entity.

Answer: A coordination entity consists of a central metal atom or ion bonded to a fixed number of ligands.

Q15. What is a ligand?

Answer: A ligand is an ion or molecule that donates one or more electron pairs to a central metal atom or ion.

Q16. What is coordination number?

Answer: Coordination number is the number of donor atoms directly bonded to the central metal atom or ion.

Q17. What is an ambidentate ligand?

Answer: An ambidentate ligand can coordinate through either of two different donor atoms. NO₂⁻ and SCN⁻ are examples.

Q18. What is an SN1 reaction?

Answer: An SN1 reaction is a unimolecular nucleophilic substitution reaction that proceeds through a carbocation intermediate.

Q19. What is denatured alcohol?

Answer: Denatured alcohol is ethanol made unfit for drinking by adding substances such as methanol, pyridine or dyes.

Q20. What is a peptide bond?

Answer: A peptide bond is the –CO–NH– linkage formed between the carboxyl group of one amino acid and the amino group of another.

Short Answer Questions – 2 or 3 Marks

Q21. Distinguish between molarity and molality.

Answer:

Q22. Why does the boiling point of a solution increase when a non-volatile solute is added?

Answer: A non-volatile solute lowers the vapour pressure of the solvent. The solution must therefore be heated to a higher temperature before its vapour pressure becomes equal to atmospheric pressure.

Hence, its boiling point increases.

Q23. Calculate the molarity of a solution containing 5.85 g of NaCl in 500 mL of solution.

Solution:

Molar mass of NaCl:

= 23 + 35.5

= 58.5 g mol⁻¹

Number of moles:

= 5.85/58.5

= 0.1 mol

Volume:

= 500 mL

= 0.5 L

Molarity:

M = 0.1/0.5

M = 0.2 mol L⁻¹

Q24. Calculate the mole fraction of ethanol in a solution containing 46 g ethanol and 54 g water.

Solution:

Molar mass of ethanol = 46 g mol⁻¹

Moles of ethanol:

= 46/46

= 1 mol

Molar mass of water = 18 g mol⁻¹

Moles of water:

= 54/18

= 3 mol

Total moles:

= 1 + 3

= 4 mol

Mole fraction of ethanol:

χethanol = 1/4

χethanol = 0.25

Q25. Write the cell notation for a Daniell cell and identify its anode and cathode.

Answer:

Cell notation:

Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)

At the anode:

Zn → Zn²⁺ + 2e⁻

At the cathode:

Cu²⁺ + 2e⁻ → Cu

Zinc is the anode and copper is the cathode.

Q26. Calculate E°cell for the cell:

Zn | Zn²⁺ || Cu²⁺ | Cu

Given:

E°(Cu²⁺/Cu) = +0.34 V

E°(Zn²⁺/Zn) = −0.76 V

Solution:

E°cell = E°cathode − E°anode

E°cell = 0.34 − (−0.76)

E°cell = 1.10 V

Q27. Distinguish between conductivity and molar conductivity.

Answer:

Q28. Why does molar conductivity increase on dilution?

Answer: On dilution, the ions move more freely because interionic attraction decreases. Weak electrolytes also ionise to a greater extent.

Therefore, molar conductivity increases.

Q29. Distinguish between order and molecularity.

Answer:

Q30. The half-life of a first-order reaction is 20 minutes. Calculate its rate constant.

Solution:

For a first-order reaction:

t₁/₂ = 0.693/k

Therefore:

k = 0.693/t₁/₂

k = 0.693/20

k = 0.03465 min⁻¹

Q31. Explain why zinc is not regarded as a transition element.

Answer: Zinc and Zn²⁺ have completely filled d-orbitals.

Zn: 3d¹⁰4s²

Zn²⁺: 3d¹⁰

A transition element must have an incomplete d-subshell in its atom or common oxidation state. Therefore, zinc is not regarded as a transition element.

Q32. Why do transition metals form complex compounds?

Answer: Transition-metal ions are small, have high charge density and contain vacant orbitals that can accept electron pairs from ligands.

Therefore, they readily form coordination compounds.

Q33. Explain why Cu⁺ is unstable in aqueous solution.

Answer: Cu⁺ undergoes disproportionation in aqueous solution:

2Cu⁺ → Cu²⁺ + Cu

Cu²⁺ is more strongly hydrated than Cu⁺, which makes the disproportionation favourable.

Q34. Find the oxidation state and coordination number of cobalt in [Co(NH₃)₅Cl]Cl₂.

Solution:

Let the oxidation state of cobalt be x.

NH₃ is neutral and coordinated Cl has a charge of −1.

The charge on the complex ion is +2.

Therefore:

x − 1 = +2

x = +3

There are five NH₃ ligands and one Cl ligand.

Coordination number = 6

Q35. Write the IUPAC name of K₄[Fe(CN)₆].

Answer: Potassium hexacyanidoferrate(II).

Q36. Distinguish between double salts and coordination compounds.

Answer:

Double salts dissociate completely into simple ions in water and lose their identity.

Coordination compounds retain the complex ion in solution and do not dissociate completely into all constituent ions.

Q37. Why are aryl halides less reactive towards nucleophilic substitution than alkyl halides?

Answer: In aryl halides, resonance gives the carbon-halogen bond partial double-bond character. The carbon atom is also sp² hybridised, making the bond shorter and stronger.

Therefore, aryl halides are less reactive.

Q38. What happens when bromoethane is heated with aqueous KOH?

Answer: Bromoethane undergoes nucleophilic substitution to form ethanol.

C₂H₅Br + KOH(aq) → C₂H₅OH + KBr

Q39. How can ethanol be converted into ethene?

Answer: Ethanol is heated with concentrated H₂SO₄ at 443 K.

CH₃CH₂OH → CH₂=CH₂ + H₂O

Q40. Why is phenol more acidic than ethanol?

Answer: The phenoxide ion is stabilised by resonance, whereas the ethoxide ion has no such resonance stabilisation.

Therefore, phenol loses H⁺ more readily than ethanol.

Q41. How will you distinguish between ethanol and phenol?

Answer: Add neutral ferric chloride solution.

Phenol gives a violet colour, while ethanol does not produce this colour.

Q42. How will you distinguish between an aldehyde and a ketone?

Answer: Use Tollens’ reagent.

An aldehyde gives a silver mirror, while a ketone generally does not react.

Q43. Why are carboxylic acids more acidic than alcohols?

Answer: The carboxylate ion formed after loss of H⁺ is stabilised by resonance over two oxygen atoms. The alkoxide ion formed from an alcohol does not have similar resonance stabilisation.

Q44. Why is aniline less basic than ammonia?

Answer: In aniline, the nitrogen lone pair is delocalised into the benzene ring through resonance. It is therefore less available for protonation.

Hence, aniline is less basic than ammonia.

Q45. What is the carbylamine reaction?

Answer: Primary amines react with chloroform and alcoholic KOH to form foul-smelling isocyanides.

RNH₂ + CHCl₃ + 3KOH → RNC + 3KCl + 3H₂O

Q46. Distinguish between reducing and non-reducing sugars.

Answer: Reducing sugars contain a free aldehydic or ketonic group and reduce Tollens’ or Fehling’s reagent.

Non-reducing sugars do not contain a free reducing group. Sucrose is a non-reducing sugar.

Long Answer and Numerical Questions – 4 or 5 Marks

Q47. Calculate the depression in freezing point when 9.0 g of glucose is dissolved in 100 g of water. K𝒇 for water is 1.86 K kg mol⁻¹.

Solution:

Molar mass of glucose = 180 g mol⁻¹

Moles of glucose:

= 9/180

= 0.05 mol

Mass of water:

= 100 g

= 0.1 kg

Molality:

m = 0.05/0.1

m = 0.5 mol kg⁻¹

Using:

ΔT𝒇 = K𝒇m

ΔT𝒇 = 1.86 × 0.5

ΔT𝒇 = 0.93 K

Q48. Calculate the osmotic pressure of a solution containing 0.1 mol of a solute in 2 L of solution at 300 K. Take R = 0.0821 L atm K⁻¹ mol⁻¹.

Solution:

Concentration:

C = 0.1/2

C = 0.05 mol L⁻¹

Using:

π = CRT

π = 0.05 × 0.0821 × 300

π = 1.2315 atm

Therefore, the osmotic pressure is approximately 1.23 atm.

Q49. Calculate the cell potential at 298 K for:

Zn | Zn²⁺(0.1 M) || Cu²⁺(1.0 M) | Cu

Given E°cell = 1.10 V.

Solution:

Cell reaction:

Zn + Cu²⁺ → Zn²⁺ + Cu

Number of electrons:

n = 2

Reaction quotient:

Q = [Zn²⁺]/[Cu²⁺]

Q = 0.1/1.0

Q = 0.1

Using the Nernst equation:

Ecell = E°cell − (0.0591/2)log 0.1

Since:

log 0.1 = −1

Ecell = 1.10 + 0.02955

Ecell = 1.12955 V

Therefore:

Ecell ≈ 1.13 V

Q50. A first-order reaction has a rate constant of 6.93 × 10⁻³ min⁻¹. Calculate its half-life.

Solution:

t₁/₂ = 0.693/k

t₁/₂ = 0.693/(6.93 × 10⁻³)

t₁/₂ = 100 minutes

Q51. Explain the variation in oxidation states of transition elements.

Answer: Transition elements show variable oxidation states because the energies of their ns and (n − 1)d orbitals are similar.

Both s and d electrons can participate in bond formation. Early transition elements show higher oxidation states because they have more unpaired electrons available.

The stability of particular oxidation states also depends on:

• Electronic configuration
• Lattice enthalpy
• Hydration enthalpy
• Nature of ligands
• Reaction conditions

Q52. Explain crystal field splitting in an octahedral complex.

Answer: In an octahedral complex, six ligands approach the central metal ion along the coordinate axes.

The d-orbitals do not experience equal repulsion. The dₓ²₋ᵧ² and d𝓏² orbitals point towards the ligands and experience greater repulsion. They form the higher-energy e𝓰 set.

The dₓᵧ, dᵧ𝓏 and d𝓏ₓ orbitals lie between the axes and experience less repulsion. They form the lower-energy t₂𝓰 set.

The energy difference between the two sets is called octahedral crystal field splitting, represented by Δ₀.

Q53. Explain the SN1 mechanism of hydrolysis of tert-butyl bromide.

Answer:

The reaction takes place in two main steps.

Step 1: Formation of carbocation

This is the slow, rate-determining step.

(CH₃)₃CBr → (CH₃)₃C⁺ + Br⁻

Step 2: Attack by water

(CH₃)₃C⁺ + H₂O → (CH₃)₃COH₂⁺

The protonated alcohol then loses H⁺:

(CH₃)₃COH₂⁺ → (CH₃)₃COH + H⁺

The rate depends only on the concentration of tert-butyl bromide.

Q54. Explain the acidic nature of phenol.

Answer: Phenol ionises to produce a phenoxide ion:

C₆H₅OH ⇌ C₆H₅O⁻ + H⁺

The phenoxide ion is stabilised because its negative charge is delocalised over the oxygen atom and the benzene ring.

The phenol molecule has less resonance stabilisation than the phenoxide ion. This greater stability of the conjugate base makes phenol acidic.

However, phenol is weaker than carboxylic acids because the negative charge in a carboxylate ion is shared equally between two electronegative oxygen atoms.

Q55. Write the reactions involved in the aldol condensation of ethanal.

Answer:

Ethanal containing an alpha-hydrogen reacts in the presence of dilute NaOH.

Two molecules of ethanal combine:

2CH₃CHO → CH₃CH(OH)CH₂CHO

The product is 3-hydroxybutanal, also called aldol.

On heating, it loses water:

CH₃CH(OH)CH₂CHO → CH₃CH=CHCHO + H₂O

The final product is but-2-enal.

Q56. Explain the basic strength of amines in gaseous and aqueous states.

Answer: Alkyl groups increase electron density on nitrogen through the +I effect. This makes the lone pair more available for protonation.

In the gaseous state, basic strength generally increases with the number of alkyl groups:

3° amine > 2° amine > 1° amine > NH₃

In aqueous solution, hydration of the protonated amine also affects basic strength. Steric hindrance can reduce hydration of tertiary ammonium ions.

Therefore, the order in aqueous solution may differ and commonly depends on the alkyl group and solvent conditions.

Q57. Explain the primary, secondary and tertiary structures of proteins.

Answer:

Primary structure: The specific sequence of amino acids in a polypeptide chain.

Secondary structure: Regular folding of the polypeptide chain into structures such as an alpha helix or beta-pleated sheet, mainly due to hydrogen bonding.

Tertiary structure: Further folding of the polypeptide chain into a three-dimensional shape due to interactions between side chains.

The biological activity of a protein depends on its correct three-dimensional structure.

Organic Chemistry Conversions

Q58. Convert chlorobenzene into phenol.

Solution:

Heat chlorobenzene with aqueous NaOH at high temperature and pressure:

C₆H₅Cl + NaOH → C₆H₅ONa + HCl

Acidification gives phenol:

C₆H₅ONa + HCl → C₆H₅OH + NaCl

Q59. Convert propene into propan-2-ol.

Solution:

Add water to propene in the presence of an acid catalyst:

CH₃CH=CH₂ + H₂O → CH₃CH(OH)CH₃

The reaction follows Markovnikov addition.

Q60. Convert ethanol into ethanoic acid.

Solution:

Oxidise ethanol using acidified potassium dichromate or potassium permanganate:

CH₃CH₂OH + 2[O] → CH₃COOH + H₂O

Q61. Convert ethanal into ethanol.

Solution:

Reduce ethanal using H₂/Ni or NaBH₄:

CH₃CHO + H₂ → CH₃CH₂OH

Q62. Convert nitrobenzene into aniline.

Solution:

Reduce nitrobenzene using Sn/HCl or Fe/HCl:

C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O

Q63. Convert aniline into benzenediazonium chloride.

Solution:

Treat aniline with sodium nitrite and hydrochloric acid at 273–278 K:

C₆H₅NH₂ + NaNO₂ + 2HCl

→ C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O

Case-Based Questions

Q64. A student prepares two solutions by dissolving equal masses of glucose and NaCl separately in equal masses of water. Which solution will show a greater depression in freezing point?

Answer: The NaCl solution will generally show a greater depression in freezing point.

NaCl dissociates into Na⁺ and Cl⁻ ions and produces more solute particles than glucose, which does not dissociate.

Since depression in freezing point depends on the number of particles, the NaCl solution shows a larger effect.

Q65. A galvanic cell stops working after some time because charge builds up in the half-cells. Which component is missing and what is its role?

Answer: The missing component is the salt bridge.

It allows the movement of ions, maintains electrical neutrality and completes the circuit without directly mixing the two solutions.

Q66. A reaction becomes faster when the temperature is increased, although the concentrations remain unchanged. Explain.

Answer: At a higher temperature, a greater fraction of molecules possesses energy equal to or greater than the activation energy.

The number of effective collisions therefore increases, raising the reaction rate.

Q67. A coordination compound has the formula [Pt(NH₃)₂Cl₂] and exists in two forms. Identify the type of isomerism.

Answer: It shows geometrical isomerism.

The two forms are cis and trans isomers, depending on whether similar ligands are adjacent or opposite.

Q68. An organic compound gives a silver mirror with Tollens’ reagent but does not react with sodium hydrogen carbonate. Identify the likely functional group.

Answer: The compound likely contains an aldehyde group.

Aldehydes reduce Tollens’ reagent, while the absence of reaction with sodium hydrogen carbonate indicates that it is not a carboxylic acid.

Practice Questions for Class 12 Chemistry

1. Calculate the molality of a solution containing 18 g of glucose in 500 g of water.
2. Explain positive and negative deviations from Raoult’s law.
3. Calculate ΔG° for a cell with E°cell = 1.10 V and n = 2.
4. State Kohlrausch’s law and give one application.
5. Derive the integrated rate equation for a first-order reaction.
6. Explain the effect of temperature on the rate constant using the Arrhenius equation.
7. Explain why transition metals act as catalysts.
8. Compare lanthanoids and actinoids.
9. Write the IUPAC names of [Co(NH₃)₆]Cl₃ and K₂[PtCl₆].
10. Explain ionisation and linkage isomerism with examples.
11. Compare the SN1 and SN2 mechanisms.
12. Explain the preparation of haloarenes from diazonium salts.
13. Write the mechanism of acid-catalysed dehydration of ethanol.
14. Explain the Reimer–Tiemann reaction.
15. Write the mechanism of nucleophilic addition to a carbonyl group.
16. Distinguish between methanal and propanone using chemical tests.
17. Explain the preparation of amines by Gabriel phthalimide synthesis.
18. Write the reactions of benzenediazonium chloride with water and phenol.
19. Distinguish between glucose, fructose and sucrose.
20. Explain the denaturation of proteins.

How to Prepare Class 12 Chemistry Important Questions

Begin with the NCERT definitions, examples and exercise questions. Divide the syllabus into Physical, Inorganic and Organic Chemistry instead of revising all chapters in the same way.

For Physical Chemistry, maintain a formula sheet and solve questions with complete units. For Inorganic Chemistry, connect each property with electronic configuration or structure.

For Organic Chemistry, prepare reaction maps showing the starting compound, reagent, condition and product. Practise conversions without checking the answer until the complete sequence is written.

Class 12 Chemistry Exam Preparation Tips

• Write balanced chemical equations.
• Mention reagents and reaction conditions.
• Use IUPAC names correctly.
• Show every step in numerical problems.
• Include units in the final answer.
• Revise exceptions separately.
• Practise structures of coordination compounds and organic products.
• Use tables for differences.
• Learn the observation and inference for chemical tests.
• Recheck charges and oxidation states before completing an answer.

Class 12 Chemistry Important Links