Solid-State and Surface Chemistry Previous Year Questions with Solutions

Join ALLEN!

(Session 2026 - 27)

Name

Mobile Number

Class

Choose class

Goal

Choose your goal

Preferred Programs

Preferred Mode

State

Choose State

I agree to

I authorise ALLEN Career Institute Pvt Ltd to send me regular updates via Phone calls, Whatsapp, SMS, Robocalls (Automated Calls), Emails, or on postal address.

Solid-State and Surface Chemistry Previous Year Questions with Solutions

Solid-State and Surface Chemistry is a crucial branch of chemistry that explores the physical and chemical properties of solids and the interactions occurring at their surfaces. Mastering Solid-State and Surface Chemistry helps in understanding material science, nanotechnology, and catalysis, making it an essential topic for both academic excellence and real-world applications

Practicing past year questions on Solid-State and Surface Chemistry is essential for students preparing for JEE Main and JEE Advanced. This topic consistently appears in the exams, with 1–2 questions typically asked every year in JEE Main and occasional conceptual or application-based questions in JEE Advanced. Regular practice with past questions ensures better conceptual clarity and boosts exam confidence, making it a vital part of your JEE preparation strategy.

JEE Main Previous Year Solved Questions on Solid-State and Surface Chemistry
JEE Adv Previous Year Solved Questions on Solid-State
JEE Adv Previous Year Solved Questions on Surface Chemistry

1.0Introduction to Solids

Solids: Substances with definite shape and volume.
Properties:
a. Strong intermolecular forces.
b. Incompressible and rigid.
c. Particles are closely packed.

2.0Classification of Solids

Type of Solid	Constituent Particles	Bonding Type	Examples
Molecular Solids	Molecules	van der Waals, H-bond	Ice, CO₂, I₂
Ionic Solids	Ions	Electrostatic forces	NaCl, KBr, ZnS
Metallic Solids	Positive ions in sea of electrons	Metallic bonding	Fe, Cu, Ag
Covalent/Network Solids	Atoms	Covalent bonds	Diamond, SiO₂, Graphite

3.0Amorphous vs Crystalline Solids

Property	Crystalline Solid	Amorphous Solid
Shape	Definite geometric	Irregular
Melting Point	Sharp	Gradual
Long-range Order	Present	Absent
Examples	NaCl, Quartz	Glass, Plastic

Crystal Lattices and Unit Cells

Crystal Lattice: 3D arrangement of points showing positions of atoms/ions/molecules.

Unit Cell: Smallest repeating unit of the lattice.

4.0Types of Unit Cells (by Axes and Angles)

Crystal System	Axes	Angles	Example
Cubic	a = b = c	α = β = γ = 90°	NaCl, Cu
Tetragonal	a = b ≠ c	α = β = γ = 90°	White Sn
Orthorhombic	a ≠ b ≠ c	α = β = γ = 90°	Sulphur
Monoclinic	a ≠ b ≠ c	α = γ = 90° ≠ β	Gypsum
Triclinic	a ≠ b ≠ c	α ≠ β ≠ γ ≠ 90°	K₂Cr₂O₇
Hexagonal	a = b ≠ c	α = β = 90°, γ = 120°	Graphite
Rhombohedral	a = b = c	α = β = γ ≠ 90°	Calcite

Number of Atoms in Unit Cells

Type of Unit Cell	Atoms per Unit Cell
Simple Cubic (SC)	1
Body-Centred Cubic (BCC)	2
Face-Centred Cubic (FCC)	4

Formula Used:

Corner atom: $18 \frac{1}{8}$
Face atom: $12 \frac{1}{2}$
Body-centred atom: 1

Packing Efficiency

Packing Efficiency = $\frac{Volume occupied by spheres}{Total volume of unit cell} \times 100$

Type of Packing	Efficiency
SC	52.4%
BCC	68%
FCC/HCP	74% (Maximum)

Packing in Solids

Close Packing in Solids:

2D Packing:
Square (least efficient)
Hexagonal (more efficient)

3D Packing:
Hexagonal Close Packing (hcp) – ABAB… (Mg, Zn)
Cubic Close Packing (ccp) – ABCABC… (Cu, Ag)
Tetrahedral Voids: Surrounded by 4 spheres.
Octahedral Voids: Surrounded by 6 spheres.

Number of voids:

Tetrahedral = 2 × number of atoms
Octahedral = same as number of atoms

Density of a Unit Cell

$Density (d) = \frac{Z \times M}{a ^{3} \times N _{A}}$

Where:

Z = Number of atoms per unit cell
M = Molar mass
a = Edge length in cm
N_A = Avogadro’s number

5.0Imperfections in Solids

Point Defects:

Type	Description
Stoichiometric	No change in ratio or neutrality.
- Vacancy	Missing atoms
- Interstitial	Extra atom in void
Non-Stoichiometric	Change in ratio
- Metal excess	Extra cations/electrons (ZnO turns yellow)
- Metal deficiency	Missing cations (FeO)
Impurity Defect	Different atoms occupy sites

Electrical Properties

Solid Type	Electrical Conductivity	Mechanism
Conductors	High	Free electrons (metals)
Semiconductors	Moderate	Band gap (Si, Ge)
Insulators	Very low	Large band gap

Doping increases conductivity:

n-type: Extra electrons (P in Si)
p-type: Positive holes (B in Si)

Magnetic Properties

Type	Magnetic Behavior	Example
Paramagnetic	Weakly attracted, unpaired electrons	O₂
Diamagnetic	Weakly repelled, no unpaired electrons	NaCl, H₂O
Ferromagnetic	Strongly attracted, permanent magnetism	Fe, Co, Ni
Antiferromagnetic	Opposing spins cancel out	MnO
Ferrimagnetic	Opposing spins don’t fully cancel	Fe₃O₄, MgFe₂O₄

6.0Surface Chemistry

Surface Chemistry deals with chemical processes at interfaces like solid-liquid, solid-gas, and liquid-gas. Since gases are completely miscible, no interface exists between them. For accurate studies, surfaces must be ultra-clean, often requiring a vacuum of 10⁻⁸–10⁻⁹ Pa to prevent contamination from air.

This field is vital for developing new materials, catalysts, corrosion resistance, and improved industrial processes.

Key Concepts

Adsorption

Accumulation of molecules (adsorbate) on a surface (adsorbent).
Surface phenomenon.
Examples: charcoal, silica gel, alumina.

Absorption

Uniform distribution of a substance throughout the bulk.

Sorption

Combined adsorption and absorption (e.g., dye on cotton).

Adsorption vs Absorption

Characteristic	Adsorption	Absorption
Distribution	Only on surface	Uniform throughout
Penetration	No	Yes
Example 1	Ink on chalk surface	Ink solvent in chalk
Example 2	Silica gel & water vapour	CaCl₂ & water vapour
Concentration	High at surface	Uniform throughout
Appearance	White inside (ink on chalk)	Not applicable

Mechanism of Adsorption

Caused by unbalanced forces at the surface.
Exothermic (∆H < 0), entropy decreases (∆S < 0).
Spontaneous when ∆G = ∆H – T∆S < 0.
Higher surface area → greater adsorption.

Types of Adsorption

Physisorption: Weak van der Waals’ forces.
Chemisorption: Strong chemical bonds (covalent/ionic).

Adsorption Isotherms

Show relation between gas pressure and amount adsorbed at constant temperature.

7.0Freundlich Isotherm

Empirical equation:

$x m = k P 1/ n, l o g x m = l o g k + 1 n l o g P \frac{x}{m} = k P^{1/ n}, lo g \frac{x}{m} = lo g k + \frac{1}{n} lo g P m x = k P 1/ n, l o g m x = l o g k + n 1 l o g P$

Where:

x/mx/mx/m: Amount adsorbed per gram of adsorbent
P: Pressure
k: Adsorption capacity
1/n: Adsorption intensity (0 < 1/n < 1)
High k = greater adsorption
1/n<1 = favorable adsorption
Physisorption decreases with rising temperature

8.0Important PYQs on Solid State and Surface Chemistry

Q.1 When Fe_0.93O is heated in the presence of oxygen, it converts to Fe₂O₃. The number of correct statement/s from the following is ________.

A. The equivalent weight of Fe_0.93O is $\frac{M o l ec u l a r w e i g h t}{0.79}$

B. The number of moles of Fe²⁺ and Fe³⁺ in 1 mole of Fe_0.93O is 0.79 and 0.14 respectively.

C. Fe_0.93O is metal deficient with a lattice comprising of cubic closed packed arrangement of O^2– ions.

D. The % composition of Fe²⁺ and Fe³⁺ in Fe_0.93O is 85% and 15% respectively.

Ans. (D)

Solution:

A : Fe_0.93O ® Fe₂O₃

nf = $(3 - \frac{200}{93}) x 0.93$

nf = 0.79

B : 2x + (0.93 – x) × 3 = 2

x = 0.79

Fe²⁺ = 0.79, Fe³⁺ = 0.21

C : Fact

D : %Fe²⁺ = $\frac{0.79}{0.93}$ × 100 = 85%; Fe³⁺ = 15%

Q.2 Which of the following represents the lattice structure of A_0.95O containing A²⁺, A³⁺ and O^2– ions ? A²⁺ A³⁺ O^2–

a.

b.

c.

Sample Question on solid state and surface chemistry

(1) B and C only

(2) B only

(3) A and B only

(4) A only

Ans. (D)

Sol. Applying electrical neutrality principle in metal deficiency defect. 3 A²⁺ are replaced by 2A³⁺, thus one vacant site per pair of A³⁺ is created

Q.3 Adding surfactants in non polar solvent, the micelles structure will look like

A)

B)

C)

Solid-State and Surface Chemistry - Non polar solvent 3

D)

Solid-State and Surface Chemistry: Non polar solvents 4

(1) b

(2) c

(3) a

(4) d

Ans. (3)

Solution: Non-Polar tail towards non-polar solvent

Q.4 In figure, a straight line is given for Freundrich Adsorption (y = 3x + 2.505). The value of1n and log K are respectively.

(1) 0.3 and log 2.505 (2) 0.3 and 0.7033

(3) 3 and 2.505 (4) 3 and 0.7033

Ans. (3)

Solution:

$\frac{x}{m} = K_{p}^{1/ n} lo g \frac{x}{m} = lo g k + \frac{1}{n} lo g P Y = 3 x + 2.505, \frac{1}{n} = 3, lo g K = 2.505$

Q.5 Which of the following represent the Freundlich adsorption isotherms?

Choose the correct answer from the options given below:

A)

Freundlich adsorption isotherms - sample questions

B)

Freundlich adsorption isotherms - problem 2

C)

Freundlich adsorption isotherms - Sample 3

D)

Freundlich adsorption isotherms - Sample 4

(1) B, C, D only

(2) A, B, D only

(3) A, B only

(4) A, C, D only

Ans. (3)

Solution

$\frac{x}{m} = k_{p}^{1/ n} and lo g \frac{x}{m} = lo g k + \frac{1}{n} lo g p$

Q.6 The enthalpy change for the adsorption process and micelle formation respectively are

(1) ΔH_ads< 0 and ΔH_mic＞0

(2) ΔH_ads< 0 and ΔH_mic< 0

(3) ΔH_ads＞0 and ΔH_mic< 0

(4) ΔH_ads＞0 and ΔH_mic＞0

Ans. (1)

Solution:

Adsorption is exothermic process due to decrease in surface energy. Energy is released when molecules adhere to a surface

Micelle formation is an endothermic process. Energy is required to overcome the repulsion between the hydrophobic tails of the surfactant molecules as they aggregate to form micelles.

Q.7 Ionic radii of cation A⁺ and anion B^- are 102 and 181 pm respectively. These ions are allowed to crystallize into an ionic solid. This crystal has cubic close packing for B^‑. A⁺ is present in all octahedral voids. The edge length of the unit cell of the crystal AB is _____ pm. (Nearest Integer)

Ans. (512)

Solution:

a = 2(r₊ + r_–)

a = 2 (102 + 181)

a = 2(283);

a = 566 pm

Q.8 A 42.12% (w/v) solution of NaCl causes precipitation of a certain sol in 10 hours. The coagulating value of NaCl for the sol is

[Given : Molar mass : Na = 23.0 g mol^–1; Cl = 35.5 g mol^–1]

(1) 36 mmol L^–1

(2) 36 mol L^–1

(3) 1440 mol L^–1

(4) 1440 mmol L^–1

Ans. (4)

$coagulation value = \frac{Millimoles of electrolyte}{Volume of solution in L} Molarity of NaCl = \frac{% ( w / v ) \times 10}{M.Wt} Molarity of NaCl = \frac{42.12 \times 10}{58.5} = 7.2 M Millimoles of NaCl electrolyte = 7200 Coagulation value for 10 hours = \frac{Millimoles of electrolyte}{Volume of solution in L} = 7200 For 2 hours coagulation value = \frac{7200 \times 2}{10} = 1440 millimoles/L$

Q.9 The distance between Na⁺ and Cl^– ions in solid NaCl of density 43.1 g cm^–3 is ____ × 10^–10m. (Nearest Integer) (Given : N_A = 6.02 × 10²³ mol^–1)

Ans. (1)

Solution .

$Mass per unit cell = \frac{Z \times M . M}{N _{A}} g = \frac{4 \times 58.5}{N _{A}} g d_{unit cell} = \frac{m}{V} = \frac{m}{a ^{3}} \Rightarrow \frac{4 \times 58.5}{N _{A} \cdot a ^{3}} = 43.1 \Rightarrow a^{3} = 9.02 \times 1 0^{- 24} cm^{3} \Rightarrow a = 2.08 \times 1 0^{- 8} cm \Rightarrow a = 2.08 \times 1 0^{- 10} m a = 2 (r_{Na^{+}} + r_{Cl^{-}}) \Rightarrow r_{Na^{+}} + r_{Cl^{-}} = 1.04 \times 1 0^{- 10} m$

Q.10 A certain element crystallises in a bcc lattice of unit cell edge length 27 Å. If the same element under the same conditions crystallises in the fcc lattice, the edge length of the unit cell in Å will be ________. (Round off to the Nearest Integer).

[Assume each lattice point has a single atom] [Assume3 = 1.73, 2 = 1.41]

Ans.
$[Assume each lattice point has a single atom] [Assume 3 = 1.73, 2 = 1.41] Ans. For BCC 3 = 4 r \Rightarrow r = \frac{3}{4} \times 27 For FCC a = 2 r = 2 \times \frac{3}{4} \times 27 = \frac{3}{4} \times 27 = 33$

Solid-State and Surface Chemistry Previous Year Questions with Solutions

Solid-State and Surface Chemistry is a crucial branch of chemistry that explores the physical and chemical properties of solids and the interactions occurring at their surfaces. Mastering Solid-State and Surface Chemistry helps in understanding material science, nanotechnology, and catalysis, making it an essential topic for both academic excellence and real-world applications

Practicing past year questions on Solid-State and Surface Chemistry is essential for students preparing for JEE Main and JEE Advanced. This topic consistently appears in the exams, with 1–2 questions typically asked every year in JEE Main and occasional conceptual or application-based questions in JEE Advanced. Regular practice with past questions ensures better conceptual clarity and boosts exam confidence, making it a vital part of your JEE preparation strategy.

JEE Main Previous Year Solved Questions on Solid-State and Surface Chemistry
JEE Adv Previous Year Solved Questions on Solid-State
JEE Adv Previous Year Solved Questions on Surface Chemistry

1.0Introduction to Solids

Solids: Substances with definite shape and volume.
Properties:
a. Strong intermolecular forces.
b. Incompressible and rigid.
c. Particles are closely packed.

2.0Classification of Solids

Type of Solid	Constituent Particles	Bonding Type	Examples
Molecular Solids	Molecules	van der Waals, H-bond	Ice, CO₂, I₂
Ionic Solids	Ions	Electrostatic forces	NaCl, KBr, ZnS
Metallic Solids	Positive ions in sea of electrons	Metallic bonding	Fe, Cu, Ag
Covalent/Network Solids	Atoms	Covalent bonds	Diamond, SiO₂, Graphite