Solid-State and Surface Chemistry previous year questions with solutions

1.0Introduction to Solids

Solids: Substances with definite shape and volume.

Properties:
Strong intermolecular forces.
Incompressible and rigid.
Particles are closely packed.

2.0Classification of Solids

3.0Amorphous vs Crystalline Solids

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)

Number of Atoms in Unit Cells

Formula Used:

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

Packing Efficiency

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

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

$\text{Density (d)}=\frac{Z\times M}{a^3\times N_A}$

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:

Electrical Properties

Doping increases conductivity:

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

Magnetic Properties

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

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:

$xm=kP1/n,logxm=logk+1nlogP\frac{x}{m}=k{P}^{1/n},\log \frac{x}{m}=\log k+\frac{1}{n}\log Pmx=kP1/n,logmx=logk+n1logP$

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   Molecular weight0.79 \frac {Molecular weight}{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 = 3-20093 (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⁺ =0.790.93 \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–  

(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 

(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:

$$\begin{gathered} \left|\frac{x}{m}\right|=K_p^{1/n} \\ \log \frac{x}{m}=\log k+\frac{1}{n}\log P \\ Y=3x+2.505,\frac{1}{n}=3, \\ \log K=2.505 \end{gathered}$$

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

Choose the correct answer from the options given below:

(1) B, C, D only         

(2) A, B, D only

(3) A, B only             

(4) A, C, D only

Ans. (3)

Solution 

$$\begin{gathered} \frac{x}{m}=k_p^{1/n} \\ \text{and}\log \frac{x}{m}=\log k+\frac{1}{n}\log p \end{gathered}$$

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)

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

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 .  

$$\begin{gathered} \text{Mass per unit cell}=\frac{Z\times M.M}{N_A}\text{ g} \\ =\frac{4\times 58.5}{N_A}\text{ g} \\ {d}_{\text{unit cell}}=\frac{m}{V}=\frac{m}{a^3} \\ \Rightarrow \frac{4\times 58.5}{N_A\cdot a^3}=43.1 \\ \Rightarrow a^3=9.02\times 10^{-24}{\text{ cm}}^3 \\ \Rightarrow a=2.08\times 10^{-8}\text{ cm} \\ \Rightarrow a=2.08\times 10^{-10}\text{ m} \\ a=2\left(r_{{\text{Na}}^{+}}+r_{{\text{Cl}}^{-}}\right) \\ \Rightarrow r_{{\text{Na}}^{+}}+r_{{\text{Cl}}^{-}}=1.04\times 10^{-10}\text{ m} \end{gathered}$$

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