JEE Chemistry The Solid State MCQs Set 11

Answer: (a) Molecular crystal

Explanation:
1. CO₂ molecules in the solid are held together only by weak van der Waals (dispersion) forces.
2. Solids whose units are discrete molecules held by such forces are molecular crystals.

Teacher's Note:
1. Identify the particles at the lattice points: molecules, ions, atoms or a metal-ion lattice with electrons.
2. Molecular solids are soft, have low melting points and do not conduct electricity.

Answer: (d) B and D only

Explanation:
1. Amorphous solids are like supercooled liquids: the arrangement is regular only over a few particle diameters, so they have short range order (B) and no long range order.
2. Because bond strengths differ from place to place, they soften over a range of temperature and have no sharp melting point (D).

Teacher's Note:
1. Crystalline solids have both short and long range order and a sharp melting point.
2. Amorphous solids are isotropic and melt over a range.

Answer: (c) In solid NaCl the ions have no (or very little) freedom to move

Explanation:
1. Solid NaCl is made of Na⁺ and Cl⁻ ions fixed at lattice sites.
2. Electrical conduction needs mobile charge carriers; the ions can only vibrate about their positions, so solid NaCl does not conduct. Molten or aqueous NaCl does conduct.

Teacher's Note:
1. Ionic solids conduct only when the ions become mobile, that is, on melting or dissolving.
2. Do not confuse 'no ions' with 'ions that cannot move'.

Answer: (b) A unit cell with a = b ≠ c, α = β = 90°, γ = 120° is hexagonal

Explanation:
1. (a) AgBr shows both Schottky and Frenkel defects, so it is wrong.
2. (b) The hexagonal system has a = b ≠ c with α = β = 90° and γ = 120°, so it is correct.
3. (c) Frenkel defect occurs when the cation is much smaller than the anion, so r⁺/r⁻ is low.
4. (d) In NaCl each Na⁺ has 6 Cl⁻ neighbours.

Teacher's Note:
1. Memorise the seven crystal systems with their axial lengths and angles.
2. Frenkel defect needs a small cation that can fit into an interstitial site.

Answer: (c) Rhombohedral: α = β = γ ≠ 90°, a = b = c

Explanation:
1. Cubic has a = b = c with all angles 90°, so (a) is wrong.
2. Monoclinic has a ≠ b ≠ c with α = γ = 90° ≠ β, so (b) is wrong.
3. Rhombohedral (trigonal) has a = b = c and α = β = γ ≠ 90°, so (c) is right.
4. Triclinic has a ≠ b ≠ c and α ≠ β ≠ γ ≠ 90°, so (d) is wrong.

Teacher's Note:
1. Rhombohedral is a cube squashed along a body diagonal: all edges equal, all angles equal but not 90°.
2. Triclinic is the least symmetric system: nothing is equal.

Answer: (c) has 4 NaCl units

Explanation:
1. NaCl has a face centred cubic arrangement of Cl⁻ with Na⁺ in all octahedral voids.
2. Cl⁻: 8 × 1/8 + 6 × 1/2 = 4. Na⁺: 12 × 1/4 + 1 = 4.
3. So the cell contains 4 formula units of NaCl and is electrically neutral.

Teacher's Note:
1. Count corner atoms 1/8, face atoms 1/2, edge atoms 1/4 and body atoms 1.
2. A unit cell of an ionic solid is always neutral.

Answer: (b) 3.72 Å

Explanation:
1. In the CsCl-type (bcc-like) cell, Cs⁺ is at the body centre and Br⁻ at the corners, and they touch along the body diagonal.
2. Body diagonal = \( \sqrt3\,a \) = 2(r⁺ + r⁻).
3. \( r^++r^-=\frac{\sqrt3}{2}\times4.3=3.72\ \text{Å} \).

Teacher's Note:
1. In CsCl-type cells the oppositely charged ions touch along the body diagonal, not along the edge.
2. Nearest-neighbour distance = \( \frac{\sqrt3 a}{2} \) for this lattice.

Answer: (c) \( 1.5\times10^{23} \)

Explanation:
1. Molar mass of NaCl = 58.5 g mol⁻¹, so 58.5 g is 1 mol, that is \( 6.022\times10^{23} \) formula units.
2. Each unit cell holds 4 formula units.
3. \( \text{Number of cells}=\frac{6.022\times10^{23}}{4}\approx1.5\times10^{23} \).

Teacher's Note:
1. First convert mass to formula units, then divide by Z.
2. Z = 4 for NaCl (fcc type).

Answer: (d) 911

Explanation:
1. In bcc the nearest neighbour distance is \( \frac{\sqrt3}{2}a=4.52\ \text{Å} \), so \( a=\frac{2\times4.52}{\sqrt3}=5.22\ \text{Å} \).
2. \( d=\frac{Z M}{N_A a^3}=\frac{2\times39}{6.022\times10^{23}\times(5.22\times10^{-8})^3}=0.911\ \text{g cm}^{-3} \).
3. That is 911 kg m⁻³.

Teacher's Note:
1. Always convert Å to cm (1 Å = 10⁻⁸ cm) before using the density formula in g cm⁻³.
2. Multiply g cm⁻³ by 1000 to get kg m⁻³.

Answer: (a) \( 1.857\times10^{-8}\ \text{cm} \)

Explanation:
1. In bcc, atoms touch along the body diagonal: \( \sqrt3\,a=4r \).
2. \( r=\frac{\sqrt3 a}{4}=\frac{1.732\times4.29}{4}=1.857\ \text{Å}=1.857\times10^{-8}\ \text{cm} \).

Teacher's Note:
1. Memorise: simple cubic r = a/2, bcc r = √3a/4, fcc r = a/(2√2).
2. 1 Å = 10⁻⁸ cm.

Answer: (a) \( 4\times10^{25} \)

Explanation:
1. Volume of 100 g = 100/10 = 10 cm³.
2. Edge = 100 pm = \( 10^{-8} \) cm, so volume of one cell = \( 10^{-24}\ \text{cm}^3 \).
3. Number of cells = \( 10/10^{-24}=10^{25} \).
4. fcc has 4 atoms per cell, so atoms = \( 4\times10^{25} \).

Teacher's Note:
1. Total volume divided by volume of one unit cell gives the number of cells.
2. Multiply by Z for the number of atoms.

Answer: (b) 1 : 2

Explanation:
1. For N close-packed atoms there are N octahedral voids and 2N tetrahedral voids.
2. So atoms : tetrahedral holes = N : 2N = 1 : 2.

Teacher's Note:
1. The same ratio holds for hcp.
2. In an fcc cell: 4 atoms, 4 octahedral and 8 tetrahedral voids.

Answer: (a) 32%

Explanation:
1. Packing efficiency of bcc: \( \frac{2\times\frac43\pi r^3}{a^3}\times100 \) with \( a=\frac{4r}{\sqrt3} \).
2. \( =\frac{\sqrt3\pi}{8}\times100=68\% \).
3. Vacant space = 100 − 68 = 32%.

Teacher's Note:
1. Packing efficiencies: simple cubic 52.4%, bcc 68%, fcc/hcp 74%.
2. Vacant space = 100 − packing efficiency.

Answer: (a) 219.25 pm

Explanation:
1. In fcc atoms touch along the face diagonal: \( \sqrt2\,a=4r \).
2. \( r=\frac{a}{2\sqrt2}=\frac{620}{2.828}=219.2\ \text{pm} \).

Teacher's Note:
1. fcc: face diagonal = 4r.
2. Use √2 = 1.414.

Answer: (a) Four Fe²⁺ and four O²⁻

Explanation:
1. Volume of cell = \( (5.0\times10^{-8})^3=1.25\times10^{-22}\ \text{cm}^3 \).
2. Mass of cell = \( 4.0\times1.25\times10^{-22}=5\times10^{-22}\ \text{g} \).
3. \( Z=\frac{d\,a^3N_A}{M}=\frac{5\times10^{-22}\times6.022\times10^{23}}{71.85}\approx4.2\approx4 \).
4. So there are 4 formula units, that is 4 Fe²⁺ and 4 O²⁻ ions.

Teacher's Note:
1. Z must be a whole number; round the calculated value.
2. Use M(FeO) = 56 + 16 = 72 g mol⁻¹.

Answer: (b) CsCl type

Explanation:
1. Radius ratio = \( \frac{r^+}{r^-}=\frac{180}{187}=0.96 \).
2. A ratio between 0.732 and 1.000 means coordination number 8, which is the CsCl-type structure.

Teacher's Note:
1. Radius ratio ranges: 0.155–0.225 (CN 3), 0.225–0.414 (CN 4), 0.414–0.732 (CN 6), 0.732–1 (CN 8).
2. Cations of almost the same size as anions give the highest coordination number.

Answer: (a) Na₂O

Explanation:
1. In antifluorite the anions form the fcc lattice and the cations occupy all the tetrahedral voids, so the formula is M₂X.
2. Na₂O fits this pattern: O²⁻ fcc, Na⁺ in all 8 tetrahedral voids. MgO has rock salt structure.

Teacher's Note:
1. Fluorite CaF₂: Ca²⁺ in fcc, F⁻ in all tetrahedral holes. Antifluorite reverses the roles.
2. Other antifluorite examples are Li₂O and K₂O.

Answer: (a) \( 1\times10^{-3}\ \text{mol}\% \)

Explanation:
1. Each Sr²⁺ replaces two Na⁺ ions to keep the crystal neutral, so one Na⁺ site becomes vacant.
2. One cation vacancy per Sr²⁺ ion, so vacancies = \( 10^{-3}\ \text{mol}\% \).

Teacher's Note:
1. Aliovalent doping creates vacancies: M²⁺ replaces two M⁺ and leaves one vacancy.
2. Count the number of dopant ions, not the number of ions replaced.

Answer: (c) Does not change

Explanation:
1. In a Frenkel defect an ion moves from its normal site to an interstitial site.
2. No ion leaves the crystal, so the mass and the volume remain the same and the density is unchanged.

Teacher's Note:
1. Compare Schottky defect: ions leave the crystal, so the density decreases.
2. Frenkel defect is a dislocation defect.

Answer: (a) Equal numbers of cation and anion vacancies

Explanation:
1. Schottky defect is a vacancy defect in which equal numbers of cations and anions are missing from the lattice to keep the crystal neutral.
2. Cation vacancies with interstitial cations describe Frenkel defect.

Teacher's Note:
1. Schottky: found in compounds with high coordination number and cations and anions of similar size (NaCl, KCl, CsCl).
2. Density decreases in Schottky defect.