JEE Chemistry The Solid State MCQs Set 19

Answer: (b) Electrons trapped in anionic vacancies

Explanation:
1. Heating an alkali halide in the vapour of the alkali metal gives a metal-excess crystal.
2. The extra metal atoms ionise and the electrons diffuse into anion vacancies; these trapped electrons absorb visible light and give colour.

Teacher's Note:
1. KCl turns violet and LiCl pink on this treatment.
2. F comes from the German word Farbe (colour).

Answer: (a) Neutral

Explanation:
1. A p-type semiconductor is silicon doped with a group 13 element, which leaves electron holes.
2. The holes are caused by missing electrons but the crystal as a whole contains equal numbers of protons and electrons, so it is electrically neutral.

Teacher's Note:
1. The same applies to n-type semiconductors.
2. Holes behave as positive carriers, but the crystal is not charged.

Answer: (d) 5.2 g/cm\(^3\)

Explanation:
1. bcc: Z = 2; \( a=4\times10^{-8} \) cm.
2. \( \rho=\dfrac{ZM}{N_Aa^3}=\dfrac{2\times100}{6.022\times10^{23}\times6.4\times10^{-23}}=5.19\approx5.2 \) g/cm\(^3\).

Teacher's Note:
1. Convert pm to cm: 1 pm = \( 10^{-10} \) cm.
2. Density of an fcc element of the same M and a would be twice this value.

Answer: (c) \( \sqrt{\dfrac{2}{3}}\,a \)

Explanation:
1. Spacing between adjacent close-packed layers \( h=\sqrt{\dfrac{8}{3}}\,r \).
2. With \( r=a/2 \): \( h=\sqrt{\dfrac{8}{3}}\cdot\dfrac a2=\sqrt{\dfrac{2}{3}}\,a \).

Teacher's Note:
1. Check: \( \sqrt{2/3}\times2r=1.633r \), the same as \( \sqrt{8/3}\,r \).
2. This is the perpendicular layer spacing, not the nearest-neighbour distance between layers (which is 2r).

Answer: (a) 2.108 Å

Explanation:
1. Anions touch along the edge, so \( 2r=a=2.88 \) Å.
2. Body diagonal = \( \sqrt{3}a=4.988 \) Å, so \( d_{void}=\sqrt{3}a-a=0.732a=2.108 \) Å.

Teacher's Note:
1. Cubical void radius ratio is 0.732.
2. The void is the largest sphere that fits at the cell centre.

Answer: (d) There are twice as many tetrahedral voids as close-packed atoms

Explanation:
1. N close-packed spheres give N octahedral voids and 2N tetrahedral voids.

Teacher's Note:
1. In fcc: 4 atoms, 4 octahedral and 8 tetrahedral voids per cell.
2. In hcp: 6 atoms, 6 octahedral and 12 tetrahedral voids per cell.

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

Explanation:
1. Per 4 O ions: 8 tetrahedral voids, 4 octahedral voids.
2. A = 8 x 1/6 = 4/3; B = 4 x 1/3 = 4/3; O = 4.
3. A : B : O = 4/3 : 4/3 : 4 = 1 : 1 : 3, so ABO3.

Teacher's Note:
1. Always compute per cell with the unit cell counts and then reduce.
2. The same compound with the same fractions for A2B2O6 would reduce to ABO3.

Answer: (c) 21.45 g/cm\(^3\), 1.387 Å

Explanation:
1. fcc: Z = 4. \( \rho=\dfrac{4\times195}{6.022\times10^{23}\times(3.9231\times10^{-8})^3}=21.45 \) g/cm\(^3\).
2. \( r=\dfrac{a}{2\sqrt{2}}=\dfrac{3.9231}{2.828}=1.387 \) Å.

Teacher's Note:
1. 1 Å = \( 10^{-8} \) cm.
2. For fcc, \( a=2\sqrt{2}\,r \).

Answer: (b) 2.878 Å

Explanation:
1. In fcc the closest atoms lie along a face diagonal; distance \( =\dfrac{a}{\sqrt{2}} \).
2. \( =\dfrac{4.070}{1.414}=2.878 \) Å.

Teacher's Note:
1. This equals 2r.
2. a/2 = 2.035 Å is the distance between a corner and the next lattice point along the edge, not the closest approach.

Answer: (a) The coordination number of each cation and anion is 6

Explanation:
1. Each \( \ce{Na+} \) is surrounded by 6 \( \ce{Cl-} \) and each \( \ce{Cl-} \) by 6 \( \ce{Na+} \) (6:6 coordination).

Teacher's Note:
1. Formula units per cell: 4.
2. Radius ratio of NaCl is 0.52.

Answer: (d) CsCl structure

Explanation:
1. 8:8 coordination is the caesium chloride structure (simple cubic arrangement of one ion with the other at the body centre).
2. NaCl is 6:6 and ZnS is 4:4.

Teacher's Note:
1. Radius ratio 0.732-1.0 gives CN 8.
2. CsCl has one formula unit per unit cell.

Answer: (c) 8 : 8

Explanation:
1. Pressure favours a more compact structure with higher coordination number.
2. The 6:6 rock-salt structure converts to the 8:8 CsCl structure.

Teacher's Note:
1. Heating (instead of pressure) tends to decrease the coordination number.
2. This is the same change as NaCl to CsCl type under pressure.

Answer: (a) Schottky defect

Explanation:
1. Equal numbers of cations and anions missing from lattice points is a Schottky defect.
2. It lowers the density but keeps the crystal electrically neutral.

Teacher's Note:
1. Schottky defects are common in highly ionic compounds with similar ion sizes (NaCl, KCl, CsCl).
2. They are stoichiometric defects.

Answer: (d) Motion of ions and not the motion of electrons

Explanation:
1. The defects create vacant sites (or interstitial positions) into which ions can hop.
2. Conduction is therefore ionic, not electronic.

Teacher's Note:
1. Electronic conduction occurs in metals and metal-excess crystals (F-centres).
2. The conductivity is far less than that of a metal.

Answer: (b) \( \dfrac{1}{2},\ \dfrac{1}{8} \)

Explanation:
1. This is the spinel \( \ce{MgAl2O4} \). For 4 O atoms there are 4 octahedral and 8 tetrahedral holes.
2. Al = 2 of 4 octahedral holes, m = 1/2.
3. Mg = 1 of 8 tetrahedral holes, n = 1/8.

Teacher's Note:
1. Spinel formula: \( \ce{AB2O4} \), ratio 1 : 2 : 4.
2. In normal spinels the divalent ion is in tetrahedral holes and the trivalent in octahedral holes.

Answer: (c) 2, 3 and 4 only

Explanation:
1. An atom in the topmost layer touches 6 in its own layer and 3 below, so 9 neighbours (not 12): statement 1 is false.
2. Packing efficiency is 74%, octahedral : tetrahedral voids are 1 : 2 per atom, and \( a=2\sqrt{2}\,r \): statements 2, 3, 4 are true.

Teacher's Note:
1. 12 is the coordination number only for an atom with layers on both sides.
2. The surface atom of a close-packed lattice has fewer neighbours than a bulk atom.

Answer: (b) 2

Explanation:
1. Volume of one cell = \( (4\times10^{-8})^3=6.4\times10^{-23} \) cm\(^3\). Mass of a cell = 8 x 6.4 x 10^-23 = 5.12 x 10^-22 g.
2. Cells in 256 g = \( 256/5.12\times10^{-22}=5\times10^{23} \).
3. Atoms = 4 per cell x 5 x 10^23 = 2 x 10^24, so N = 2.

Teacher's Note:
1. Each fcc cell has 4 atoms.
2. The molar mass is 8 x 6.022 x 10^23 x 6.4 x 10^-23 / 4 = 77.0 g/mol, a consistent value.

Answer: (a) 3

Explanation:
1. NaCl type MX: cations at corners and face centres, anions at edge centres and the body centre.
2. After (i) one X remains (centre). (ii) 6 face cations become X: 6 x 1/2 = 3 anions. (iii) corner cations removed. (iv) the central X becomes M.
3. Final: 3 anions (faces) and 1 cation (centre). Ratio = 3.

Teacher's Note:
1. Keep a running count for each step instead of the final picture.
2. The resulting formula is MX3.

Answer: (d) \( \ce{A2B5} \)

Explanation:
1. A = 8 x 1/8 = 1. B = 5 x 1/2 = 5/2.
2. A : B = 1 : 5/2 = 2 : 5, so A2B5.

Teacher's Note:
1. A perfect fcc AB3 cell loses 1/2 B for each missing face.
2. Multiply both numbers by 2 to remove fractions.

Answer: (c) F-centres

Explanation:
1. Alkali halide heated in alkali metal vapour gets anion vacancies filled by electrons (F-centres).
2. The trapped electrons absorb visible light and impart colour.

Teacher's Note:
1. Schottky and Frenkel defects do not give colour.
2. The colour depends on the halide.