JEE Chemistry The Solid State MCQs Set 12

Answer: (a) 200 pm

Explanation:
1. In bcc, the closest atoms touch along the body diagonal, and distance of closest approach = \( \frac{\sqrt3}{2}a \).
2. \( a=\frac{2\times1.73}{1.732}=2.00\ \text{Å}=200\ \text{pm} \).

Teacher's Note:
1. Distance of closest approach equals 2r.
2. bcc: 2r = (√3/2) a.

Answer: (b) 2 unit cells

Explanation:
1. A face of a cube is common to two adjacent cubes.
2. So a face-centre atom belongs equally to 2 unit cells and contributes 1/2 to each.

Teacher's Note:
1. Corner atoms are shared by 8 cells, edge atoms by 4, face atoms by 2.
2. Body-centre atoms belong wholly to one cell.

Answer: (b) 3.62 Å

Explanation:
1. \( a=2\sqrt2\,r=2\times1.414\times1.28=3.62\ \text{Å} \).

Teacher's Note:
1. fcc: 4r = √2 a, so a = 2√2 r.
2. Check: 2.828 × 1.28 = 3.62.

Answer: (a) 6

Explanation:
1. In bcc each atom has 8 nearest neighbours (the body-diagonal contacts).
2. The next-nearest neighbours lie a distance a away along the cell edges: 6 of them.

Teacher's Note:
1. Nearest neighbours: bcc 8, fcc 12, simple cubic 6.
2. Next-nearest in bcc: 6 at distance a.

Answer: (b) NaWO₃

Explanation:
1. W: 8 corners × 1/8 = 1.
2. O: 12 edge centres × 1/4 = 3.
3. Na: 1 at the body centre = 1.
4. Formula: NaWO₃.

Teacher's Note:
1. Use the contribution rule: corner 1/8, edge 1/4, face 1/2, body 1.
2. Write the ratio of atoms in simplest whole numbers.

Answer: (b) 190.68 pm

Explanation:
1. In a bcc-type (CsCl) cell the ions touch along the body diagonal: \( \sqrt3\,a=2(r^++r^-) \).
2. \( r^++r^-=\frac{1.732\times480}{2}=415.7\ \text{pm} \).
3. \( r^+=415.7-225=190.7\ \text{pm} \).

Teacher's Note:
1. The body diagonal of the cube passes through the corner anion, the central cation and the opposite corner anion.
2. Do not use the edge length alone.

Answer: (c) Cu₄Ag₃Au

Explanation:
1. Cu in ccp: 4 atoms per cell.
2. Ag at 12 edge centres: 12 × 1/4 = 3.
3. Au at body centre: 1.
4. Formula Cu₄Ag₃Au.

Teacher's Note:
1. ccp is the same as fcc, with 4 atoms per cell.
2. Edge-centre atoms contribute 1/4 each.

Answer: (a) X₇Y₂₄

Explanation:
1. Normal cell: X = 8 × 1/8 = 1, Y = 6 × 1/2 = 3.
2. With one corner missing: X = 7 × 1/8 = 7/8, Y = 3.
3. Ratio X : Y = 7/8 : 3 = 7 : 24, so X₇Y₂₄.

Teacher's Note:
1. Multiply both numbers by the lowest common multiple to get whole numbers.
2. A missing corner removes 1/8 of an atom.

Answer: (b) X₇Y₂₄Z

Explanation:
1. X: 7 × 1/8 = 7/8; Z: 1 × 1/8 = 1/8; Y: 6 × 1/2 = 3.
2. Multiply by 8: X₇ Z₁ Y₂₄.
3. Formula X₇Y₂₄Z.

Teacher's Note:
1. Treat Z at one corner as contributing 1/8 of an atom.
2. Multiply by 8 to remove the fractions.

Answer: (a) 1.002 g cm⁻³

Explanation:
1. \( d=\frac{ZM}{N_Aa^3}=\frac{2\times23}{6.022\times10^{23}\times(4.24\times10^{-8})^3} \).
2. Denominator = \( 6.022\times10^{23}\times7.62\times10^{-23}=45.9 \).
3. d = 46/45.9 = 1.002 g cm⁻³.

Teacher's Note:
1. bcc has Z = 2.
2. Cube the edge after converting it to cm.

Answer: (b) 6

Explanation:
1. Corner atoms: 12 × 1/6 = 2.
2. End-face centre atoms: 2 × 1/2 = 1.
3. Enclosed atoms: 3.
4. Total = 2 + 1 + 3 = 6.

Teacher's Note:
1. In a hexagonal prism each corner atom is shared by 6 cells.
2. The hcp unit cell therefore has 6 atoms.

Answer: (c) \( 1.80\times10^{-22} \)

Explanation:
1. \( a=2\sqrt2\,r=2\times1.414\times2.00=5.657\ \text{Å} \).
2. \( V=a^3=(5.657\times10^{-8})^3=1.81\times10^{-22}\ \text{cm}^3 \).

Teacher's Note:
1. First find a from r, then cube it.
2. Convert Å to cm only once, before cubing.

Answer: (c) X₂Z

Explanation:
1. If Z atoms number N, the tetrahedral sites number 2N.
2. All are occupied by X, so X : Z = 2N : N = 2 : 1.
3. Formula X₂Z.

Teacher's Note:
1. Tetrahedral voids = 2 × atoms, octahedral voids = 1 × atoms.
2. Antifluorite type compounds follow this pattern.

Answer: (a) An fcc arrangement along with 50% occupancy of tetrahedral holes

Explanation:
1. Carbon atoms occupy fcc positions (4 per cell) and alternate tetrahedral voids (4 of the 8) are also occupied by carbon.
2. So 4 of 8 tetrahedral voids, that is 50%, are occupied. Total 8 C atoms per unit cell.

Teacher's Note:
1. Zinc blende has the same arrangement with Zn and S in place of the two sets of carbon atoms.
2. Each carbon is tetrahedrally bonded to four others.

Answer: (c) C₂A₃

Explanation:
1. If there are N anions there are N octahedral voids.
2. Cations = 2/3 N, so C : A = 2/3 : 1 = 2 : 3.
3. Formula C₂A₃.

Teacher's Note:
1. Octahedral voids equal the number of close-packed atoms.
2. Corundum Al₂O₃ is a real example (Al³⁺ in 2/3 of octahedral holes).

Answer: (d) N

Explanation:
1. Each sphere in a close packed layer arrangement is associated with one octahedral void.
2. N spheres give N octahedral and 2N tetrahedral voids.

Teacher's Note:
1. Check with fcc: 4 atoms, 4 octahedral voids (12 edge centres × 1/4 + 1 body centre).
2. Tetrahedral voids are twice as many.

Answer: (b) 100 pm

Explanation:
1. ZnS-type structure has coordination number 4 (tetrahedral), for which the limiting radius ratio is 0.225.
2. \( r^-=\frac{22.5}{0.225}=100\ \text{pm} \).

Teacher's Note:
1. Tetrahedral voids: r⁺/r⁻ from 0.225 to 0.414; the 'ideal' value is the lower limit 0.225.
2. Octahedral limit is 0.414 and cubic limit is 0.732.

Answer: (b) 136.6 to 241.6 pm

Explanation:
1. For the NaCl structure (CN 6) the radius ratio lies between 0.414 and 0.732.
2. \( r^-\ \text{max}=\frac{100}{0.414}=241.5\ \text{pm},\quad r^-\ \text{min}=\frac{100}{0.732}=136.6\ \text{pm} \).
3. So r⁻ = 136.6 to 241.6 pm.

Teacher's Note:
1. A larger r⁺/r⁻ means a smaller r⁻, so the upper ratio limit gives the lower r⁻.
2. Be careful to divide the cation radius by the ratio.

Answer: (a) 1.815 Å

Explanation:
1. When anions touch, they do so along the face diagonal: \( \sqrt2\,a=4r^- \).
2. \( r^-=\frac{1.414\times5.14}{4}=1.817\ \text{Å}\approx1.815\ \text{Å} \).

Teacher's Note:
1. In the NaCl structure Cl⁻ forms an fcc lattice, so Cl⁻ ions touch along the face diagonal in a very small cation case.
2. The face diagonal is √2 a and carries 4 radii.

Answer: (c) 8:4 and 8:8

Explanation:
1. In CaF₂ each Ca²⁺ is surrounded by 8 F⁻ and each F⁻ by 4 Ca²⁺, so 8:4.
2. In CsCl each ion has 8 oppositely charged neighbours, so 8:8.

Teacher's Note:
1. The coordination number ratio matches the formula ratio: CaF₂ is 1 : 2, so CN ratio is 8 : 4.
2. NaCl is 6 : 6, ZnS is 4 : 4.