Solid State — JEE Main Questions

1. NaCl is doped with 10⁻³ mol% SrCl₂ and also has 0.01% Schottky defects. What is the total number of cation vacancies per mole of NaCl? (N_A = 6.022 × 10²³)

1. 6.022 × 10¹⁹
2. 6.022 × 10²⁰
3. 6.624 × 10¹⁹
4. 6.624 × 10²⁰

Answer: 6.624 × 10¹⁹

Doping contribution: 10⁻³ mol% means a mole fraction of 10⁻³/100 = 10⁻⁵ Sr²⁺ ions. Each Sr²⁺ replaces one Na⁺ and creates one cation vacancy (to preserve charge), so vacancies from doping = 10⁻⁵ × N_A = 10⁻⁵ × 6.022×10²³ = 6.022×10¹⁸. Schottky contribution: 0.01% = 10⁻⁴ of cation sites are vacant, so = 10⁻⁴ × N_A = 10⁻⁴ × 6.022×10²³ = 6.022×10¹⁹. Total cation vacancies per mole = 6.022×10¹⁸ + 6.022×10¹⁹ = 0.6022×10¹⁹ + 6.022×10¹⁹ = 6.624×10¹⁹.

2. The absorption spectra of F-centres in three alkali halides show peaks at 470 nm (NaCl), 560 nm (KCl), and 600 nm (LiCl). Which statement correctly explains the trend?

1. Larger anion vacancy size leads to higher energy absorption, shifting colour towards blue.
2. The absorption energy is independent of the host lattice and depends only on the trapped electron.
3. Larger anion vacancy size leads to lower energy absorption, shifting colour towards red.
4. The colour is due to d-d transitions of the metal ion, and the trend follows the crystal field splitting.

Answer: Larger anion vacancy size leads to lower energy absorption, shifting colour towards red.

F-centre is an electron trapped in an anion vacancy. The energy levels of the electron depend on the size of the vacancy: a larger vacancy (like in NaCl with Cl⁻) gives a smaller energy gap, so absorption at longer wavelength (lower energy). NaCl absorbs at 470 nm (blue) but appears yellow; KCl absorbs at 560 nm (green) appears lilac; LiCl absorbs at 600 nm (orange) appears pink. Larger vacancy → lower energy → red shift.

3. A sample of NaCl is heated in sodium vapour and turns yellow. The density decreases by 0.1% due to formation of F-centres. Original density is 2.165 g/cm³. What is the concentration of F-centres per cm³? (Na=23, Cl=35.5, N_A=6.022×10²³)

1. 2.2×10¹⁹
2. 4.4×10¹⁹
3. 3.3×10¹⁹
4. 1.1×10¹⁹

Answer: 2.2×10¹⁹

Each F-centre corresponds to one missing Cl⁻ ion (occupied by a trapped electron), so the fractional density decrease equals the fraction of NaCl formula units that are missing a Cl⁻. Number of NaCl formula units per cm³ = (ρ × N_A)/M = (2.165 × 6.022×10²³)/58.5 ≈ 2.23×10²². Density decrease = 0.1% = 10⁻³ of formula units have an F-centre. Number of F-centres per cm³ = 10⁻³ × 2.23×10²² ≈ 2.2×10¹⁹.

4. In a cubic close-packed (ccp) arrangement of N spheres, how many tetrahedral voids are present, and what is the limiting radius ratio r_void / r_sphere for a tetrahedral void (perfect fit, just touching)?

1. N tetrahedral voids; r_void/r_sphere = 0.414
2. 2N tetrahedral voids; r_void/r_sphere = 0.414
3. 2N tetrahedral voids; r_void/r_sphere = 0.225
4. N/2 tetrahedral voids; r_void/r_sphere = 0.155

Answer: 2N tetrahedral voids; r_void/r_sphere = 0.225

In any close-packed structure (ccp or hcp) the number of tetrahedral voids is exactly twice the number of close-packed spheres, so for N spheres there are 2N tetrahedral voids. A tetrahedral void is the gap enclosed by 4 spheres in tetrahedral geometry; the limiting (just-touching) radius ratio is r_void/r_sphere = 0.225.

5. When NaCl is heated in sodium vapour, it turns yellow. What type of defect forms and what is the resulting conductivity?

1. Schottky defect due to missing ion pairs; insulator
2. Metal deficiency defect due to cation vacancies; p-type semiconductor
3. Frenkel defect due to cation displacement; ionic conductor
4. Metal excess defect due to F-centres; n-type semiconductor

Answer: Metal excess defect due to F-centres; n-type semiconductor

Heating NaCl in Na vapour causes Na atoms to deposit, lose electrons, and form Na⁺ ions. The electrons are trapped in anion vacancies, creating F-centres (metal excess defect). These trapped electrons can be excited to the conduction band, making the crystal an n-type semiconductor.

6. Which type of defect involves a missing atom from a lattice site, creating a vacancy?

1. Surface defect
2. Line defect
3. Volume defect
4. Point defect

Answer: Point defect

A vacancy is a point defect (0D) because it involves a single lattice point. Point defects include vacancies, interstitials, and impurity atoms. Line defects (dislocations) are 1D, surface defects (grain boundaries) are 2D, and volume defects (pores) are 3D.

7. Given ionic radii of Na⁺ = 95 pm and Cl⁻ = 181 pm, what is the density of NaCl? (M = 58.5 g/mol, N_A = 6.022 × 10²³)

1. 2.17 × 10³ kg/m³
2. 2.17 g/cm³
3. 2.17 × 10⁻³ g/cm³
4. 2.17 × 10⁻⁶ g/cm³

Answer: 2.17 g/cm³

Radius ratio = 95/181 ≈ 0.52, between 0.414 and 0.732, so coordination number 6 (octahedral) → NaCl-type fcc with Z = 4. Edge length a = 2(r⁺ + r⁻) = 552 pm = 5.52 × 10⁻⁸ cm. Density ρ = ZM/(a³N_A) = (4 × 58.5) / ((5.52 × 10⁻⁸)³ × 6.022 × 10²³) ≈ 2.17 g/cm³.

8. A metal crystallises in a cubic lattice with density 19.3 g cm⁻³ and edge length 408 pm. Its atomic mass (g mol⁻¹) is closest to:

1. 98.5
2. 197
3. 295.5
4. 49.25

Answer: 197

For fcc, Z=4. Using ρ = ZM/(a³N_A), M = ρ a³ N_A / Z. a = 408 pm = 4.08×10⁻⁸ cm, a³ = 6.79×10⁻²³ cm³. ρ = 19.3 g cm⁻³, N_A = 6.022×10²³. M = (19.3 × 6.79×10⁻²³ × 6.022×10²³)/4 = (19.3 × 40.88)/4 = 789/4 = 197.25 ≈ 197 g mol⁻¹, identifying gold.

9. A sample of FeO has the formula Fe_{0.93}O. What percentage of iron ions are in the +3 oxidation state?

1. 7%
2. 14%
3. 21%
4. 28%

Answer: 14%

Let x = fraction of Fe²⁺ and y = fraction of Fe³⁺. Total Fe = 0.93, so x + y = 0.93. Charge balance: 2x + 3y = 2 (since O²⁻). Substitute x = 0.93 - y: 2(0.93 - y) + 3y = 2 → 1.86 - 2y + 3y = 2 → y = 0.14. So 14% of Fe ions are Fe³⁺.

10. NaCl has a theoretical density of 2.165 g/cm³. If 2% Schottky defects are present, what is the new density?

1. 2.165 g/cm³
2. 2.122 g/cm³
3. 2.143 g/cm³
4. 2.208 g/cm³

Answer: 2.122 g/cm³

Schottky defects remove equal numbers of Na⁺ and Cl⁻ ions, reducing the number of formula units per unit cell. The effective Z becomes Z × (1 - 0.02). Density is proportional to Z, so new density = 2.165 × 0.98 = 2.122 g/cm³.

11. MgFe₂O₄ is an inverse spinel ferrite. Its magnetic behaviour and net moment are best described as:

1. Ferrimagnetic with small net moment due to unequal opposing Fe³⁺ spins
2. Paramagnetic with no net moment due to unpaired electrons in Fe³⁺
3. Antiferromagnetic with zero net moment due to equal opposing Fe³⁺ spins
4. Ferromagnetic with large net moment due to parallel Fe³⁺ spins

Answer: Ferrimagnetic with small net moment due to unequal opposing Fe³⁺ spins

In MgFe₂O₄ inverse spinel, tetrahedral sites have Fe³⁺ and octahedral sites have Mg²⁺ and Fe³⁺. The two Fe³⁺ ions have opposite spins but unequal crystal fields, giving a small net moment — characteristic of ferrimagnetism.

12. How many Bravais lattices are possible in the cubic crystal system?

Answer: 3.0

The cubic system allows primitive (P), body-centred (I), and face-centred (F) centring, giving 3 Bravais lattices. End-centred (C) is not possible in cubic because it would be equivalent to a primitive tetragonal lattice.