JEE Chemistry The Solid State MCQs Set 22

Answer: (b) Only tetrahedral voids are found on the line

Explanation:
1. A tetrahedral void lies directly above (or below) each sphere of the adjacent layer, with that sphere at its apex.
2. Between layers I and II there is a tetrahedral void just above the layer I atom; between layers II and III another just below the layer III atom.
3. Octahedral voids lie where a hollow of one layer is exactly over a hollow of the other, which does not occur on this vertical line.

Teacher's Note:
1. The line passes through the empty hollow of layer II.
2. So two tetrahedral voids lie on it and no octahedral void.

Answer: (b) Ag in 3/8 of the tetrahedral voids; Au in 1/4 of the octahedral voids

Explanation:
1. ccp Cu4 has 4 octahedral and 8 tetrahedral voids per cell.
2. Ag = 3 needs 3 of 8 tetrahedral voids = 3/8.
3. Au = 1 of 4 octahedral voids = 1/4.

Teacher's Note:
1. Check the others: the second option gives Ag 2, Au 4; the third Ag 4, Au 8.
2. Only the first option reproduces 4 : 3 : 1.

Answer: (a) 3.439 Å

Explanation:
1. Use additivity of ionic radii: d(RbBr) = d(RbCl) + d(KBr) - d(KCl).
2. = 3.285 + 3.293 - 3.139 = 3.439 Å.

Teacher's Note:
1. Check with sodium: d(NaBr) - d(NaCl) would equal d(KBr) - d(KCl) = 0.154 Å.
2. The data for NaBr are not needed.

Answer: (d) Zinc blende has a ccp (fcc) structure and wurtzite an hcp structure

Explanation:
1. In zinc blende S\(^{2-}\) ions form a ccp (ABCABC) arrangement with Zn\(^{2+}\) in half the tetrahedral voids.
2. In wurtzite the S\(^{2-}\) ions are in hcp (ABAB) stacking with Zn\(^{2+}\) in half of the tetrahedral voids.

Teacher's Note:
1. Both are examples of 4 : 4 coordination.
2. The difference is only in the stacking sequence.

Answer: (c) 62%

Explanation:
1. Atoms per mL = \( \dfrac{1.65}{40}\times6.022\times10^{23}=2.48\times10^{22} \).
2. Volume of one atom = \( \dfrac43\pi(1.54\times10^{-8})^3=1.53\times10^{-23} \) cm\(^3\).
3. Occupied volume = 0.38 mL per mL; empty = 62%.

Teacher's Note:
1. The packing efficiency of 38% is lower than for close-packed solids because the data are only approximate.
2. Always subtract the occupied fraction from 100.

Answer: (a) 12

Explanation:
1. An hcp unit cell has 6 atoms effectively, so 12 tetrahedral voids in total (and 6 octahedral).
2. All 12 tetrahedral voids lie inside the hexagonal prism.

Teacher's Note:
1. Count 12 = 2 x 6.
2. The 12 tetrahedral voids are each wholly inside the cell, as are the 6 octahedral ones.

Answer: (d) 4

Explanation:
1. A forms ccp and B fills all tetrahedral voids (fluorite type): each A has 8 B neighbours (x = 8).
2. Each B has 4 A neighbours (y = 4).
3. x - y = 4.

Teacher's Note:
1. This is the CaF2 structure, where cation CN is 8 and anion CN is 4.
2. The ratio of coordination numbers is inverse to the ratio of ions in the formula.

Answer: (c) 4

Explanation:
1. In square close packing each sphere touches its four neighbours (north, south, east, west).
2. The hexagonal close packing in 2-D gives 6.

Teacher's Note:
1. Coordination number in 3-D square-layer stacking is 6 (simple cubic).
2. Packing efficiency of a square layer is 78.5%.

Answer: (b) \( \ce{X7Y24} \)

Explanation:
1. X = 7 x 1/8 = 7/8; Y = 6 x 1/2 = 3.
2. X : Y = 7/8 : 3 = 7 : 24.

Teacher's Note:
1. Perfect cell: XY3.
2. Multiply by 8 to remove fractions.

Answer: (a) \( \ce{X15Y16Z} \)

Explanation:
1. Perfect cell: X4Y4.
2. Two corners replaced: Z = 2 x 1/8 = 1/4; X = 4 - 1/4 = 15/4; Y = 4.
3. X : Y : Z = 15/4 : 4 : 1/4 = 15 : 16 : 1.

Teacher's Note:
1. Multiply by 4 to get whole numbers.
2. Total of X and Z is unchanged at 4.

Answer: (d) +8 electronic charge

Explanation:
1. ccp cell: 4 O\(^{2-}\), 8 tetrahedral and 4 octahedral voids.
2. Mg\(^{2+}\) = 8 x 1/8 = 1; Al\(^{3+}\) = 4 x 1/2 = 2.
3. Positive charge = 1 x 2 + 2 x 3 = +8 (balancing 4 x -2 = -8).

Teacher's Note:
1. The cell has the formula MgAl2O4 once.
2. Always check that charges balance.

Answer: (c) The unit cell volume is \( 1.214\times10^{-22} \) cm\(^3\) and the atomic radius is about 175 pm

Explanation:
1. Mass of a cell = \( \dfrac{4\times207.2}{6.022\times10^{23}}=1.376\times10^{-21} \) g.
2. V = 1.376e-21 / 11.34 = \( 1.214\times10^{-22} \) cm\(^3\), so a = 4.95 Å.
3. \( r=\dfrac{a}{2\sqrt2}=1.75 \) Å = 175 pm.

Teacher's Note:
1. A volume of 10^-19 cm3 would be an edge of about 460 nm, far too large.
2. fcc: r = a/(2 root 2).

Answer: (a) \( r_{Na^+}/r_{F^-}\approx0.7 \), coordination numbers 6 : 6 and rock-salt structure

Explanation:
1. r(Na+) = 2.31 - 1.36 = 0.95 Å; ratio = 0.95/1.36 = 0.70.
2. 0.70 lies between 0.414 and 0.732, so octahedral voids: 6 : 6 coordination and rock-salt structure.

Teacher's Note:
1. The limit for 8 : 8 is 0.732 and above.
2. The effective nuclear charges of the two ions are not equal.

Answer: (b) (i), (ii) and (iii)

Explanation:
1. Each Zn\(^{2+}\) is in a tetrahedral void surrounded by 4 S\(^{2-}\) and each S\(^{2-}\) by 4 Zn\(^{2+}\).
2. The bond angle of a regular tetrahedron is 109 degrees 28'.
3. For cation-anion contact with no anion-anion contact in a tetrahedral hole, r+/r- = 0.225.

Teacher's Note:
1. The ratio 0.225 is the lower limit for CN 4.
2. ZnS has r+/r- = 0.40, above the limit.

Answer: (c) 56

Explanation:
1. Fe : O = 72.36/56 : 27.64/16 = 1.292 : 1.7275 = 3 : 4, so the oxide is Fe3O4 (232 g/mol).
2. Mass of the cell = 5.2 x (8.39 x 10^-8)^3 = 3.07 x 10^-21 g; molar mass per cell = 1849.
3. Formula units per cell = 1849 / 232 = 8, so atoms = 8 x 7 = 56.

Teacher's Note:
1. Fe3O4 (magnetite) has the inverse spinel structure with 8 formula units per cell.
2. Count atoms (ions) not formula units.

Answer: (b) 30%

Explanation:
1. With cation-anion contact \( a=2(r_++r_-)=3r_- \).
2. Atoms per cell: 4 cations + 4 anions. Volume = \( \tfrac43\pi r_-^3\,(4+4\times\tfrac18)=6\pi r_-^3 \).
3. Fraction filled = \( 6\times3.15/27=0.70 \), so void = 30%.

Teacher's Note:
1. The calculation needs a^3 = 27 r_-^3.
2. The result is lower than for fcc because the cations are small.

Answer: (a) 4.0 Å

Explanation:
1. Ratio = 1.6/1.864 = 0.858, which lies in 0.732 - 1.0, so the structure is CsCl type (CN 8).
2. Cation-anion contact along the body diagonal: \( \sqrt3a=2(r_++r_-)=6.928 \).
3. a = 6.928 / 1.732 = 4.0 Å.

Teacher's Note:
1. For rock salt the edge would be 2(r+ + r-) = 6.93 Å, but the ratio is too high for that structure.
2. CsCl ratio is 0.93.

Answer: (d) 7

Explanation:
1. For 4 O\(^{2-}\): Zn = 8 x 1/8 = 1; Al = 4 x 1/2 = 2.
2. Formula ZnAl2O4, so x + y + z = 1 + 2 + 4 = 7.

Teacher's Note:
1. Charge check: +2 + 6 = +8 = 4 x 2.
2. MgAl2O4 is the mineral spinel.

Answer: (c) Square close packing

Explanation:
1. In type (I) the spheres of successive rows lie directly above each other horizontally and vertically.
2. Each sphere touches four neighbours, which is square close packing.

Teacher's Note:
1. Type (II) gives hexagonal close packing in the plane.
2. The 3-D analogue of type (I) is the simple cubic lattice.

Answer: (a) Is better in type II

Explanation:
1. Type (I): square arrangement, 2-D efficiency 78.5%.
2. Type (II): hexagonal arrangement, 2-D efficiency 90.7%.
3. Type II is better.

Teacher's Note:
1. Hexagonal layers are the densest 2-D packing.
2. 3-D: simple cubic 52.4% versus ccp/hcp 74%.

Answer: (d) Square, triangular

Explanation:
1. Four touching spheres of square packing enclose a square void.
2. Three touching spheres of the staggered arrangement enclose a triangular void.

Teacher's Note:
1. Tetrahedral and octahedral voids arise only in 3-D packing.
2. Each sphere in the hexagonal layer has 6 triangular voids around it.

Answer: (b) P-3, Q-4, R-1, S-2

Explanation:
1. Limiting radius ratios: triangular 0.155, tetrahedral 0.225, octahedral / square planar 0.414, cubic 0.732.
2. So P-3, S-2, Q-4, R-1.

Teacher's Note:
1. Memorise the sequence 0.155, 0.225, 0.414, 0.732.
2. CN 3, 4, 6, 8 respectively.