JEE Chemistry The Solid State MCQs Set 04

Answer: (b) NaWO₃

Explanation:
1. W: 8 corners \(\times\) 1/8 = 1.
2. O: 12 edge centres \(\times\) 1/4 = 3.
3. Na: 1 at the body centre.
4. Formula = NaWO\(_3\).

Teacher's Note:
1. Edge-centre atoms are shared by four cells, so each counts 1/4.
2. The totals 1 : 3 : 1 give the formula directly.

Answer: (c) 546 pm

Explanation:
1. Fluorite has Z = 4 formula units per cell, M = 78 g/mol.
2. \( a^3 = \dfrac{Z M}{\rho N_A} = \dfrac{4 \times 78}{3.18 \times 6.022\times10^{23}} = 1.629\times10^{-22}\ \text{cm}^3 \).
3. \( a = 5.46\times10^{-8} \) cm = 546 pm.

Teacher's Note:
1. Use \(a^3 = ZM/(\rho N_A)\) and take the cube root at the end.
2. Convert cm to pm by multiplying by 10\(^{10}\).

Answer: (d) Coordination number of each cation and anion is six

Explanation:
1. In NaCl each Na⁺ is surrounded by six Cl⁻ ions at the octahedral positions.
2. Each Cl⁻ is likewise surrounded by six Na⁺ ions.
3. So the coordination number of both ions is 6 (6 : 6 coordination).

Teacher's Note:
1. For a 1 : 1 compound the two coordination numbers must be equal.
2. Rock salt 6 : 6, zinc blende 4 : 4, caesium chloride 8 : 8.

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

Explanation:
1. In CaF₂ each Ca²⁺ has 8 F⁻ neighbours and each F⁻ has 4 Ca²⁺ neighbours (8 : 4).
2. In CsCl each ion has 8 neighbours of the opposite kind (8 : 8).

Teacher's Note:
1. The ratio of coordination numbers in CaF₂ must be 2 : 1 because the formula has twice as many anions as cations.
2. Anions occupy all tetrahedral voids, so each anion has four cation neighbours.

Answer: (c) it is formed by four spheres, the centres of which form a regular tetrahedron

Explanation:
1. A tetrahedral void sits between one sphere of one layer and three touching spheres of the adjacent layer.
2. The centres of these four spheres lie at the corners of a regular tetrahedron, which gives the void its name.

Teacher's Note:
1. Merely being bounded by four spheres is not enough; the geometry of the centres matters.
2. An octahedral void is surrounded by six spheres whose centres form a regular octahedron.

Answer: (d) vacant

Explanation:
1. In rock salt the Cl⁻ ions form a ccp lattice and the Na⁺ ions occupy all the octahedral voids.
2. The 8 tetrahedral voids per cell are not occupied by any ion.
3. They are vacant.

Teacher's Note:
1. Rock salt fills octahedral voids only.
2. Zinc blende and fluorite are the structures that use tetrahedral voids.

Answer: (d) 12

Explanation:
1. Take a face-centre atom: 4 corner atoms in its own face, 4 face-centre atoms in the cell below-and-around, and 4 in the neighbouring cell make 12.
2. Hence the coordination number of FCC (and of ccp/hcp) is 12.

Teacher's Note:
1. Coordination numbers: simple cubic 6, BCC 8, FCC and HCP 12.
2. Close packing always gives 12.

Answer: (c) C₂A₃

Explanation:
1. If there are N anions, there are N octahedral voids.
2. Cations = 2N/3.
3. C : A = 2/3 : 1 = 2 : 3, so the formula is C₂A₃ (as in corundum Al₂O₃).

Teacher's Note:
1. Octahedral voids equal the number of close-packed atoms.
2. Multiply by the fraction occupied, then clear fractions.

Answer: (a) X₂Y₄Z

Explanation:
1. X: 4 atoms per FCC cell.
2. Y: all 8 tetrahedral voids = 8.
3. Z: half of 4 octahedral voids = 2.
4. X : Y : Z = 4 : 8 : 2 = 2 : 4 : 1, so the formula is X₂Y₄Z.

Teacher's Note:
1. FCC cell: 4 atoms, 8 tetrahedral voids, 4 octahedral voids.
2. Always reduce the ratio to its simplest form.

Answer: (b) body-centred cubic

Explanation:
1. Coordination number 8 for each kind of atom means each atom sits at the centre of a cube formed by the other kind.
2. That is the arrangement of two interpenetrating simple cubic lattices, which is the body-centred cubic type (CsCl-like).

Teacher's Note:
1. A BCC-type cell with different atoms at corners and centre gives 8 : 8 coordination.
2. FCC would give 12 or 6.

Answer: (b) 190.68 pm

Explanation:
1. In the CsCl-type cell the ions touch along the body diagonal: \( \sqrt{3}\,a = 2(r_+ + r_-) \).
2. \( r_+ + r_- = \dfrac{1.732 \times 480}{2} = 415.7 \) pm.
3. \( r_+ = 415.7 - 225 \) = 190.7 pm.

Teacher's Note:
1. The body diagonal carries one cation diameter plus two half-anion diameters, i.e. 2r\(_+\) + 2r\(_-\).
2. Do not use the face-diagonal relation for this structure.

Answer: (a) 1 Cs⁺ and 1 Cl⁻

Explanation:
1. Cl⁻ ions at the eight corners contribute 8 \(\times\) 1/8 = 1.
2. One Cs⁺ sits wholly inside the cell.
3. So the cell has 1 Cs⁺ and 1 Cl⁻ (Z = 1).

Teacher's Note:
1. Z for CsCl is 1, unlike NaCl where Z = 4.
2. Mass of cell = (molar mass)/N_A.

Answer: (b) 0.225 r_a

Explanation:
1. Limiting radius ratios: triangular void 0.155, tetrahedral void 0.225, octahedral void 0.414, cubic void 0.732.
2. For a tetrahedral void the ratio is 0.225.

Teacher's Note:
1. Radius ratio increases with coordination number.
2. Memorise 0.155, 0.225, 0.414, 0.732 for coordination numbers 3, 4, 6, 8.

Answer: (c) ABBAABBA

Explanation:
1. In a close-packed stacking the layer above must sit in the depressions of the layer below, so two adjacent layers can never be identical.
2. ABBAABBA has the same layer B directly on top of B, which is not allowed.
3. The other three sequences have no identical neighbouring layers.

Teacher's Note:
1. Valid close packing needs each layer different from the layers immediately above and below it.
2. ABAB is hcp and ABCABC is ccp.

Answer: (b) 100 pm

Explanation:
1. ZnS-type has coordination number 4 for each ion, so the cation sits in a tetrahedral void and r\(_+\)/r\(_-\) = 0.225.
2. r\(_-\) = 22.5/0.225 = 100 pm.

Teacher's Note:
1. Use the limiting ratio 0.225 for the ideal tetrahedral fit.
2. Ratio of cation to anion radius is always below 1 for a stable ionic crystal.

Answer: (a) 335.1 pm

Explanation:
1. In the CsCl-type cell the oppositely charged ions touch along the body diagonal, which is half of \( \sqrt{3}\,a \) for one cation-anion contact.
2. Distance = \( \dfrac{\sqrt{3}}{2}a = 0.866 \times 387 \) = 335.2 pm.

Teacher's Note:
1. The cation-anion distance is half the body diagonal.
2. Compare: for NaCl type it is a/2.

Answer: (a) 77.07 pm

Explanation:
1. A carbon atom in an alternate tetrahedral void touches the nearest FCC atom, at a distance \( \dfrac{\sqrt{3}}{4}a \) from it.
2. This distance equals 2r.
3. \( r = \dfrac{\sqrt{3}\,a}{8} = \dfrac{1.732 \times 356}{8} \) = 77.07 pm.

Teacher's Note:
1. The C-C bond length is \(\sqrt{3}\)a/4 = 154 pm, which is twice the radius.
2. Do not give the bond length as the radius.

Answer: (d) CsCl

Explanation:
1. Schottky defect is found in ionic compounds of high coordination number in which cation and anion are of comparable size, such as NaCl, KCl, CsCl.
2. ZnS and AgCl show Frenkel defect (small cation, large size difference).
3. CaF₂ shows Frenkel (anti-Frenkel type) rather than Schottky.

Teacher's Note:
1. Schottky: comparable ion size, high coordination number.
2. Frenkel: very different sizes, low coordination number.

Answer: (a) 53 pm

Explanation:
1. The largest void in FCC is the octahedral void, with radius = a/2 − r, where r = \( a/(2\sqrt{2}) \).
2. r = 128 pm, so the void radius = 181 − 128 = 53 pm.
3. (Check: 0.414 \(\times\) 128 = 53 pm.)

Teacher's Note:
1. Octahedral voids are larger than tetrahedral voids.
2. The tetrahedral void in the same lattice would take only about 29 pm.

Answer: (c) 0.293

Explanation:
1. Along an edge the corner atoms cover 2r of the length a = \( 2\sqrt{2}\,r \).
2. Fraction covered = \( 2r/(2\sqrt{2}\,r) = 1/\sqrt{2} = 0.707 \).
3. Fraction not covered = 1 − 0.707 = 0.293.

Teacher's Note:
1. Along the edge of an FCC cell the atoms do not touch each other.
2. They touch along the face diagonal.