JEE Chemistry The Solid State MCQs Set 25

Answer: (c) \( 0.025\times N_A \)

Explanation:
1. Moles in 8 g = \( \frac{8}{80}=0.1 \), so atoms = \( 0.1N_A \).
2. An fcc unit cell contains 4 atoms, so the number of unit cells = \( \frac{0.1N_A}{4}=0.025N_A \).

Teacher's Note:
1. Divide the number of atoms by the atoms per unit cell (4 for fcc).

Answer: (a) \( \ce{AB2O4} \)

Explanation:
1. For N oxide ions there are 2N tetrahedral and N octahedral voids.
2. A = \( \frac18\times2N=\frac N4 \); B = \( \frac12\times N=\frac N2 \); O = N.
3. Ratio A : B : O = 1 : 2 : 4, so the formula is \( \ce{AB2O4} \).

Teacher's Note:
1. Count voids per ion: 2 tetrahedral and 1 octahedral.

Answer: (d) 12 \( \ce{Na+} \) ions

Explanation:
1. Each Na+ has 6 Cl- nearest neighbours (octahedral).
2. The next nearest neighbours are the Na+ ions at face-diagonal distance, and there are 12 of them.

Teacher's Note:
1. Nearest = 6 Cl-; next nearest = 12 Na+.

Answer: (b) 11.11%

Explanation:
1. Let x be Cu+ and y be Cu2+ per S: \( x+y=1.8 \) and, for neutrality, \( x+2y=2 \).
2. Hence \( y=0.2 \) and \( x=1.6 \).
3. Mole percent of Cu2+ = \( \frac{0.2}{1.8}\times100=11.11\% \).

Teacher's Note:
1. Use both the atom count and charge balance.

Answer: (b) 2A, C, 2D

Explanation:
1. The body diagonal passes through two corners (A), two tetrahedral voids (D) lying at one quarter and three quarters of the diagonal, and the body-centre octahedral void (C).
2. So it contains 2A, 1C and 2D; no face-centre atoms lie on it.

Teacher's Note:
1. Tetrahedral voids sit on the body diagonal at 1/4 and 3/4 positions.

Answer: (d) Substitutional cation vacancy

Explanation:
1. Each \( \ce{Cd^{2+}} \) replaces two \( \ce{K+} \) ions to maintain electrical neutrality.
2. One site is occupied by Cd2+ and the other becomes a cation vacancy, an impurity defect.

Teacher's Note:
1. Aliovalent impurity gives vacancies.

Answer: (a) 45 pm

Explanation:
1. For a tetrahedral void, \( r^+/r^-=0.225 \).
2. \( r^+=0.225\times200=45 \) pm.

Teacher's Note:
1. Memorise limiting ratios: 0.155 triangular, 0.225 tetrahedral, 0.414 octahedral.

Answer: (c) 9

Explanation:
1. Corners: \( 8\times\frac18=1 \).
2. There are 4 body diagonals with 2 atoms each (inside the cell): 8 atoms.
3. Total = 1 + 8 = 9.

Teacher's Note:
1. Corner atoms count one-eighth each; internal atoms count fully.

Answer: (c) \( \ce{A2BC2} \)

Explanation:
1. Per unit cell: A = 4, octahedral voids = 4, tetrahedral voids = 8.
2. B = 50% of 4 = 2 and C = 50% of 8 = 4.
3. A : B : C = 4 : 2 : 4 = 2 : 1 : 2.

Teacher's Note:
1. ccp lattice with N atoms has N octahedral and 2N tetrahedral voids.

Answer: (a) \( \ce{Al2O3} \)

Explanation:
1. For 6 oxide ions there are 6 octahedral voids, of which two thirds are 4 filled by Al3+.
2. Al : O = 4 : 6 = 2 : 3, giving \( \ce{Al2O3} \).

Teacher's Note:
1. Reduce the ratio to its simplest form for the empirical formula.

Answer: (d) ABC

Explanation:
1. A at the body centre: 1.
2. B at corners: \( 8\times\frac18=1 \).
3. C at two face centres: \( 2\times\frac12=1 \), so the formula is ABC.

Teacher's Note:
1. Face-centre atoms are shared by 2 cells, corners by 8.

Answer: (b) \( \ce{A7B24} \)

Explanation:
1. A = \( 7\times\frac18=\frac78 \); B = \( 6\times\frac12=3 \).
2. Ratio \( \frac78:3=7:24 \), so the formula is \( \ce{A7B24} \).

Teacher's Note:
1. Convert fractional numbers to whole numbers.

Answer: (b) \( \ce{ABO3} \)

Explanation:
1. Per 4 oxide ions: tetrahedral voids = 8, octahedral = 4.
2. A = \( \frac16\times8=\frac43 \) and B = \( \frac13\times4=\frac43 \).
3. A : B : O = 4/3 : 4/3 : 4 = 1 : 1 : 3.

Teacher's Note:
1. Fix 4 anions and count the voids.

Answer: (d) 8, 4

Explanation:
1. A occupies the ccp positions (4 per cell) and B fills all 8 tetrahedral voids, giving \( \ce{AB2} \) (fluorite type).
2. Each B has 4 A neighbours, and each A has 8 B neighbours.

Teacher's Note:
1. In fluorite, cation CN = 8 and anion CN = 4.

Answer: (a) 4, 4

Explanation:
1. Rock salt: 4 Mg2+ and 4 O2-, so 4 formula units.
2. Zinc blende: S2- at the fcc positions (4) and Zn2+ in 4 of the 8 tetrahedral voids, so 4 formula units.

Teacher's Note:
1. Both fcc-based structures have 4 formula units.

Answer: (c) 8, 6

Explanation:
1. In bcc, each atom touches the 8 atoms at the corners (or centre) of the surrounding cube: 8 nearest neighbours.
2. The next nearest are the 6 atoms in the adjacent cells along the cube axes, at distance a.

Teacher's Note:
1. Coordination number of bcc is 8.

Answer: (c) 0.73

Explanation:
1. \( a^3=\frac{168.36}{3.97\times6.023\times10^{23}} \), so \( a=4.13 \) Å.
2. Cs+ touches Cl- along the body diagonal: \( r^++r^-=\frac{\sqrt3a}{2}=3.58 \) Å.
3. Cl- ions touch along the edge: \( r^-=\frac a2=2.06 \) Å, so \( r^+=1.51 \) Å and \( r^+/r^-=0.73 \).

Teacher's Note:
1. Body diagonal gives r+ + r-; the cell edge gives 2r-.

Answer: (a) AB

Explanation:
1. The plane contains 4 corner B (\( 4\times\frac18 \)), 2 face-centre B (\( 2\times\frac12 \)), the body-centre A (1) and 2 edge-centre A (\( 2\times\frac14 \)).
2. Removed: B = 1.5, A = 1.5. Remaining: B = 4 - 1.5 = 2.5 and A = 4 - 1.5 = 2.5.
3. Ratio 1 : 1, so the formula is AB.

Teacher's Note:
1. Edge-centre octahedral voids are shared by 4 cells.