Important Questions Class 11 Chemistry

Class 11 Chemistry explains the structure, composition, properties, and reactions of matter. Important questions class 11 chemistry cover mole concept, atomic structure, chemical bonding, thermodynamics, equilibrium, redox reactions, organic chemistry, and hydrocarbons for 2026 exam preparation.

Chemistry becomes easier when students stop treating it as one subject and start seeing its three parts clearly: numerical chemistry, concept chemistry, and reaction chemistry.

Class 11 Chemistry builds the base for Class 12, NEET, JEE, and board-level problem solving. Units like Thermodynamics, Equilibrium, Redox Reactions, Chemical Bonding, and Hydrocarbons need regular practice because they test formulas, reasoning, reaction logic, and applications together.

Key Takeaways

Class 11 Chemistry Units at a Glance

Important Questions Class 11 Chemistry: Unit-Wise Practice

Important questions class 11 chemistry should be revised unit-wise because each unit tests a different skill.

Mole concept needs calculation accuracy. Chemical Bonding needs diagrams and structure logic. Thermodynamics and Equilibrium need formula selection. Organic Chemistry and Hydrocarbons need reaction reasoning.

Chemistry Class 11 Chapter 1 Question Answer: Some Basic Concepts of Chemistry

Some Basic Concepts of Chemistry builds the calculation base for the subject. It includes mole concept, molarity, limiting reagent, stoichiometry, and significant figures.

These chemistry class 11 chapter 1 question answer sets are important because errors in this unit affect later numerical chapters too.

Some Basic Concepts of Chemistry Questions

Q1. Calculate the mass of sodium acetate required to make 500 mL of 0.375 M aqueous solution. Molar mass of sodium acetate is 82.0245 g mol⁻¹.
Ans.

Molarity = 0.375 M

Volume = 500 mL = 0.5 L

Moles required = Molarity × Volume in litres

= 0.375 × 0.5

= 0.1875 mol

Mass = Moles × Molar mass

= 0.1875 × 82.0245

= 15.38 g

So, 15.38 g of sodium acetate is required.

Q2. In the reaction A + B₂ → AB₂, identify the limiting reagent if 2 mol A and 3 mol B₂ are mixed.
Ans.

From the equation:

1 mol A reacts with 1 mol B₂.

For 2 mol A, 2 mol B₂ is required.

But 3 mol B₂ is available.

So, A is completely consumed first.

A is the limiting reagent, and B₂ is in excess.

Q3. How are 0.50 mol Na₂CO₃ and 0.50 M Na₂CO₃ different?
Ans.

0.50 mol Na₂CO₃ means a fixed amount of sodium carbonate.

It refers only to the number of moles.

0.50 M Na₂CO₃ means concentration.

It means 0.50 moles of sodium carbonate are present in 1 litre of solution.

So, 0.50 mol tells amount, while 0.50 M tells concentration.

Structure of Atom Important Questions

Structure of atom important questions test Bohr’s model, quantum numbers, electronic configuration, de Broglie wavelength, and emission spectra.

This unit becomes easier when students connect each formula to the particle or energy level involved.

Structure of Atom Questions

Q1. What is the maximum number of emission lines when an excited electron in a hydrogen atom drops from n = 6 to the ground state?
Ans.

Number of emission lines = n(n - 1)/2

Here, n = 6

Number of lines = 6(6 - 1)/2

= 6 × 5 / 2

= 15

So, 15 emission lines are possible.

Q2. Explain why the following sets of quantum numbers are not possible: (a) n = 0, l = 0, ml = 0, ms = +1/2 (b) n = 1, l = 1, ml = 0, ms = +1/2.
Ans.

(a) This set is not possible because n cannot be 0.

The principal quantum number starts from 1.

(b) This set is not possible because for n = 1, l can only be 0.

Since l = 1 is not allowed for n = 1, the set is invalid.

Q3. Calculate the wavelength of an electron moving with velocity 2.05 × 10⁷ m s⁻¹. Mass of electron = 9.1 × 10⁻³¹ kg, h = 6.626 × 10⁻³⁴ Js.
Ans.

Using de Broglie equation:

λ = h / mv

λ = 6.626 × 10⁻³⁴ / (9.1 × 10⁻³¹ × 2.05 × 10⁷)

λ = 6.626 × 10⁻³⁴ / 1.8655 × 10⁻²³

λ = 3.55 × 10⁻¹¹ m

So, wavelength = 3.55 × 10⁻¹¹ m or 0.0355 nm.

Chemical Bonding Questions: Unit 4 Important Questions

Chemical bonding questions test Lewis structures, VSEPR theory, hybridisation, molecular orbital theory, and hydrogen bonding.

This unit needs diagrams, shapes, bond angles, and reasons. Do not revise it only through definitions.

Chemical Bonding and Molecular Structure Questions

Q1. Although geometries of NH₃ and H₂O are both distorted tetrahedral, the bond angle in water is less than in ammonia. Explain.
Ans.

Both NH₃ and H₂O have tetrahedral electron geometry.

NH₃ has one lone pair and three bond pairs.

H₂O has two lone pairs and two bond pairs.

Lone pair-lone pair repulsion is greater than lone pair-bond pair repulsion.

Since water has two lone pairs, its bond pairs are pushed closer together.

So, the bond angle in H₂O is 104.5°, while the bond angle in NH₃ is about 107°.

Q2. Define hybridisation. Describe the shapes of sp, sp², and sp³ hybrid orbitals.
Ans.

Hybridisation is the mixing of atomic orbitals of similar energy to form new orbitals of equal energy and identical shape.

sp hybridisation:

1. One s and one p orbital mix.
2. Two sp hybrid orbitals form.
3. Shape is linear.
4. Bond angle is 180°.
5. Example: BeCl₂.

sp² hybridisation:

1. One s and two p orbitals mix.
2. Three sp² hybrid orbitals form.
3. Shape is trigonal planar.
4. Bond angle is 120°.
5. Example: BF₃.

sp³ hybridisation:

1. One s and three p orbitals mix.
2. Four sp³ hybrid orbitals form.
3. Shape is tetrahedral.
4. Bond angle is 109.5°.
5. Example: CH₄.

Q3. Use molecular orbital theory to explain why Be₂ molecule does not exist.
Ans.

Electronic configuration of Be is 1s² 2s².

For Be₂, molecular orbital configuration is:

(σ1s)²(σ1s)²(σ2s)²(σ2s)²

Bond order = (Bonding electrons - Antibonding electrons) / 2

= (4 - 4) / 2

= 0

Bond order zero means there is no net bond formation.

Therefore, Be₂ does not exist as a stable molecule.

Thermodynamics Class 11 Important Questions: Unit 5

Thermodynamics class 11 important questions test enthalpy, entropy, Gibbs free energy, spontaneity, Hess’s law, and bond enthalpy.

For thermodynamics, always write the formula first. Then check units before substituting values.

Thermodynamics Class 11 Questions and Answers

Q1. For the reaction 2A + B → C at 298 K, ΔH = 400 kJ mol⁻¹ and ΔS = 0.2 kJ K⁻¹ mol⁻¹. At what temperature will the reaction become spontaneous?
Ans.

For spontaneity:

ΔG < 0

ΔG = ΔH - TΔS

At the threshold of spontaneity:

ΔG = 0

So,

T = ΔH / ΔS

T = 400 / 0.2

T = 2000 K

The reaction becomes spontaneous above 2000 K because ΔH and ΔS are both positive.

Q2. The enthalpy of combustion of methane, graphite, and dihydrogen at 298 K are -890.3, -393.5, and -285.8 kJ mol⁻¹ respectively. Calculate the enthalpy of formation of CH₄(g).
Ans.

Formation reaction:

C(s) + 2H₂(g) → CH₄(g)

Using Hess’s law:

ΔfH(CH₄) = ΔcH(C) + 2ΔcH(H₂) - ΔcH(CH₄)

= -393.5 + 2(-285.8) - (-890.3)

= -393.5 - 571.6 + 890.3

= -74.8 kJ mol⁻¹

So, enthalpy of formation of methane is -74.8 kJ mol⁻¹.

Q3. For the reaction 2Cl(g) → Cl₂(g), what are the signs of ΔH and ΔS?
Ans.

ΔH is negative because bond formation releases energy.

The reaction is exothermic.

ΔS is negative because two gaseous atoms combine to form one gaseous molecule.

Disorder decreases in the system.

Q4. Comment on the thermodynamic stability of NO(g), given: N₂(g) + 1/2O₂(g) → NO(g), ΔrH = 90 kJ mol⁻¹ and NO(g) + 1/2O₂(g) → NO₂(g), ΔrH = -74 kJ mol⁻¹.
Ans.

Formation of NO from nitrogen and oxygen is endothermic.

This means energy is required to form NO.

So, NO is thermodynamically unstable relative to its elements.

NO also reacts with oxygen to form NO₂ in an exothermic reaction.

This shows that NO is reactive and has low thermodynamic stability.

Equilibrium Class 11 Questions and Answers: Unit 6

Equilibrium class 11 questions test Kc, Kp, Le Chatelier’s principle, pH, and ionic equilibrium.

In numerical questions, write the balanced equation and equilibrium expression before substituting values.

Equilibrium Class 11 Important Questions

Q1. At 473 K, Kc for decomposition of PCl₅ is 8.3 × 10⁻³. If [PCl₅] at equilibrium is 0.5 × 10⁻¹ mol L⁻¹, find the concentrations of PCl₃ and Cl₂.
Ans.

Reaction:

PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)

Kc = [PCl₃][Cl₂] / [PCl₅]

Let [PCl₃] = [Cl₂] = x

Given:

Kc = 8.3 × 10⁻³

[PCl₅] = 0.5 × 10⁻¹ = 0.05 mol L⁻¹

So,

8.3 × 10⁻³ = x² / 0.05

x² = 8.3 × 10⁻³ × 0.05

x² = 4.15 × 10⁻⁴

x = 2.04 × 10⁻² mol L⁻¹

Therefore:

[PCl₃] = [Cl₂] = 0.0204 mol L⁻¹

Q2. Why can pure liquids and solids be ignored while writing equilibrium constant expressions?
Ans.

The concentration of a pure solid or liquid remains constant during a reaction.

Their density does not change significantly.

So, including them in the equilibrium expression would only multiply Kc by a constant.

By convention, their activity is taken as 1.

Therefore, pure solids and liquids are not included in Kc expressions.

Q3. A liquid is in equilibrium with its vapour in a sealed container. The volume is suddenly increased. What happens initially and at final equilibrium?
Ans.

Initially, vapour pressure decreases because the same vapour occupies a larger volume.

The rate of evaporation becomes greater than the rate of condensation.

More liquid evaporates until equilibrium is restored.

At final equilibrium, vapour pressure returns to the same value as before, if temperature remains constant.

Vapour pressure depends on temperature, not container volume.

Redox Reaction Class 11 Important Questions: Unit 7

Redox reaction class 11 questions test oxidation number, oxidising agent, reducing agent, ion-electron method, and galvanic cell notation.

Always identify which species loses electrons and which gains electrons.

Redox Reaction Class 11 Questions and Answers

Q1. Assign oxidation numbers to each element in KMnO₄. What is the oxidation state of Mn?
Ans.

In KMnO₄:

K = +1

O = -2

Four oxygen atoms give total charge:

4 × -2 = -8

Let oxidation state of Mn be x.

For a neutral compound:

+1 + x - 8 = 0

x = +7

So, oxidation state of Mn in KMnO₄ is +7.

Q2. Depict the galvanic cell for the reaction Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s). Which electrode is negatively charged?
Ans.

Cell representation:

Zn(s) | Zn²⁺(aq) || Ag⁺(aq) | Ag(s)

Oxidation occurs at zinc electrode:

Zn → Zn²⁺ + 2e⁻

So, zinc is the anode.

In a galvanic cell, the anode is negatively charged.

Electrons flow from zinc to silver through the external circuit.

Q3. Arrange the following metals in increasing order of reducing power: K⁺/K = -2.39 V, Ag⁺/Ag = 0.80 V, Hg²⁺/Hg = 0.79 V, Mg²⁺/Mg = -2.37 V, Cr³⁺/Cr = -0.74 V.
Ans.

Reducing power increases as standard electrode potential becomes more negative.

Given values:

Ag⁺/Ag = +0.80 V
Hg²⁺/Hg = +0.79 V
Cr³⁺/Cr = -0.74 V
Mg²⁺/Mg = -2.37 V
K⁺/K = -2.39 V

Increasing order of reducing power:

Ag < Hg < Cr < Mg < K

Organic Chemistry Class 11 Important Questions: Unit 8

Organic chemistry class 11 tests nomenclature, isomerism, inductive effect, electromeric effect, resonance, electrophiles, nucleophiles, and Lassaigne’s test.

The unit becomes easier when students understand electron movement instead of memorising reactions blindly.

Organic Chemistry Class 11 Important Questions and Answers

Q1. What are electrophiles and nucleophiles? Explain with examples.
Ans.

Electrophiles are electron-deficient species that accept electron pairs.

They are usually positively charged or have partial positive charge.

Examples:

1. H⁺
2. Cl⁺
3. NO₂⁺
4. AlCl₃

Nucleophiles are electron-rich species that donate electron pairs.

They are usually negatively charged or have lone pairs.

Examples:

1. OH⁻
2. CN⁻
3. NH₃
4. H₂O

In organic reactions, nucleophiles often attack electron-deficient carbon atoms.

Q2. Discuss the chemistry of Lassaigne’s test.
Ans.

In Lassaigne’s test, an organic compound is fused with metallic sodium.

This converts covalently bonded elements into ionic compounds.

Nitrogen converts into sodium cyanide.

Sulphur converts into sodium sulphide.

Halogens convert into sodium halides.

The fused mass is dissolved in water to prepare sodium extract.

For nitrogen detection, sodium extract is treated with FeSO₄ and FeCl₃ in acidic medium.

A Prussian blue colour confirms nitrogen.

Q3. Explain inductive effect and electromeric effect. Which effect explains the acidity order Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH?
Ans.

Inductive effect is the permanent displacement of electrons through sigma bonds due to electronegativity difference.

It decreases with distance from the electronegative atom.

Electromeric effect is a temporary electron displacement in a pi bond system during attack by a reagent.

The acidity order is explained by inductive effect.

Chlorine has a strong electron-withdrawing effect.

More chlorine atoms withdraw more electron density from the carboxyl group.

This makes proton release easier.

So, acidity order is:

Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH

Hydrocarbons Class 11 Important Questions: Unit 9

Hydrocarbons class 11 important questions test alkanes, alkenes, alkynes, benzene, Markovnikov’s rule, peroxide effect, ozonolysis, and aromaticity.

Most questions are mechanism-based, not calculation-based.

Hydrocarbons Questions

Q1. Why is benzene extraordinarily stable though it contains three double bonds?
Ans.

Benzene is stable because of resonance and delocalisation of pi electrons.

Its six pi electrons are delocalised over all six carbon atoms.

This forms a uniform electron cloud above and below the ring.

Delocalisation lowers the energy of benzene.

This extra stability is called resonance energy.

Because of this stability, benzene prefers electrophilic substitution over addition reactions.

Addition reactions would destroy aromatic stability.

Q2. Addition of HBr to propene gives 2-bromopropane, while in the presence of benzoyl peroxide it gives 1-bromopropane. Explain.
Ans.

Without benzoyl peroxide, addition follows Markovnikov’s rule.

Hydrogen adds to the carbon already having more hydrogen atoms.

Bromine attaches to the more substituted carbon.

So, propene gives 2-bromopropane.

In the presence of benzoyl peroxide, the reaction follows a free radical mechanism.

This gives anti-Markovnikov addition.

Bromine attaches to the terminal carbon, forming 1-bromopropane.

This is called the peroxide effect or Kharasch effect.

Q3. Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitution with difficulty?
Ans.

Benzene has a delocalised pi electron cloud.

This electron-rich cloud attracts electrophiles.

So, electrophilic substitution occurs easily.

Benzene resists nucleophilic substitution because nucleophiles are electron-rich.

The pi electron cloud repels nucleophiles.

Also, substitution preserves aromaticity, while addition would disturb it.

11th Chemistry Important Questions: Quick Revision List

Use this list before your 2026 exam. These 11th chemistry important questions cover the highest-value concepts from the syllabus.

1. Mole concept and molarity calculations
2. Limiting reagent questions
3. Valid and invalid quantum number sets
4. de Broglie wavelength numerical
5. NH₃ and H₂O bond angle comparison
6. sp, sp², and sp³ hybridisation
7. Be₂ bond order using molecular orbital theory
8. Gibbs free energy and spontaneity
9. Hess’s law calculations
10. Kc numerical for PCl₅ decomposition
11. Le Chatelier’s principle applications
12. Oxidation number calculation
13. Galvanic cell notation
14. Reducing power order from electrode potential
15. Electrophiles and nucleophiles
16. Lassaigne’s test
17. Markovnikov and anti-Markovnikov addition
18. Benzene stability and electrophilic substitution

All Key Formulas for Class 11 Chemistry