JEE Chemistry The Solid State MCQs Set 08

Answer: (c) 0.225 to 0.414

Explanation:
1. Tetrahedral (CN 4) is stable for r⁺/r⁻ between 0.225 and 0.414.

Teacher's Note:
1. Triangular 0.155–0.225, tetrahedral 0.225–0.414, octahedral 0.414–0.732, cubic 0.732–1.
2. Remember the series 0.155, 0.225, 0.414, 0.732.

Answer: (c) 0.414

Explanation:
1. Octahedral coordination (CN 6) exists for radius ratio 0.414 to 0.732.
2. Its lower limit is 0.414.

Teacher's Note:
1. Below 0.414 the cation is too small and fits tetrahedral holes.
2. Above 0.732 the cation gets CN 8.

Answer: (c) 6

Explanation:
1. Radius ratio 0.414–0.732 corresponds to octahedral geometry, CN 6.

Teacher's Note:
1. Link radius ratio ranges to CN: 0.225–0.414 gives 4; 0.414–0.732 gives 6; 0.732–1 gives 8.

Answer: (d) 0.52

Explanation:
1. r(Na⁺) = 95 pm, r(Cl⁻) = 181 pm.
2. Ratio = 95/181 = 0.52, which is in the octahedral range.

Teacher's Note:
1. Hence Na⁺ occupies octahedral holes of the Cl⁻ lattice.
2. CN = 6 : 6.

Answer: (d) All halides of alkali metals have rock-salt type structure

Explanation:
1. Statements 1 to 3 are correct for NaCl.
2. CsCl, CsBr and CsI are not rock-salt type (they are CsCl type), so statement 4 is incorrect.

Teacher's Note:
1. Most alkali halides are rock salt type; Cs halides are the exception.
2. LiCl to RbI mostly adopt NaCl structure.

Answer: (c) Octahedral

Explanation:
1. Na⁺ ions fill all the octahedral voids of the ccp Cl⁻ lattice.

Teacher's Note:
1. 4 octahedral voids per cell hold 4 Na⁺ ions.
2. This matches 4 Cl⁻ ions per cell.

Answer: (c) Corners as well as centres of the faces of the cube

Explanation:
1. Cl⁻ ions form an fcc lattice: corners and face centres.

Teacher's Note:
1. Na⁺ occupies edge centres and the body centre.
2. Both ions form interpenetrating fcc lattices.

Answer: (c) 4

Explanation:
1. Cl⁻: 8 × 1/8 + 6 × 1/2 = 4.
2. Na⁺: 12 × 1/4 + 1 = 4.
3. So 4 formula units.

Teacher's Note:
1. Z = 4 for rock-salt structure.
2. This value is used in density problems.

Answer: (d) Vacant

Explanation:
1. Na⁺ fills only the octahedral voids.
2. The 8 tetrahedral voids stay empty.

Teacher's Note:
1. Rock salt: octahedral voids full, tetrahedral voids empty.
2. Zinc blende has the opposite: half the tetrahedral voids filled.

Answer: (b) 6

Explanation:
1. MgO has the rock salt structure, so each Mg²⁺ has 6 oxide neighbours.

Teacher's Note:
1. Rock salt gives 6 : 6 coordination.
2. Other examples: KCl, AgCl (ionic form), CaO.

Answer: (c) 8 : 8

Explanation:
1. Each Cs⁺ is at the body centre of a cube of 8 Cl⁻ ions, and vice versa.

Teacher's Note:
1. CsCl is 8 : 8 with 1 formula unit per cell.
2. The Cl⁻ ions form a simple cubic arrangement.

Answer: (a) one CsCl unit

Explanation:
1. Cl⁻: 8 × 1/8 = 1; Cs⁺ at the body centre = 1.
2. So one formula unit per cell.

Teacher's Note:
1. Z = 1 for CsCl.
2. Contrast with Z = 4 for NaCl.

Answer: (a) by application of high pressure

Explanation:
1. Under high pressure NaCl changes from CN 6 to the closer-packed CN 8 (CsCl type).
2. (Conversely, CsCl changes to NaCl type on heating above about 760 K.)

Teacher's Note:
1. Pressure favours the denser structure with higher coordination number.
2. Heating favours lower coordination.

Answer: (d) 8

Explanation:
1. CsCl-type structure has CN 8 for both ions.

Teacher's Note:
1. Other CsCl-type compounds: CsBr, CsI, NH₄Cl.
2. Radius ratio about 0.732 to 1.

Answer: (a) 4 : 4

Explanation:
1. In zinc blende, S²⁻ forms ccp and Zn²⁺ fills half of the tetrahedral voids.
2. Each ion has 4 neighbours.

Teacher's Note:
1. 4 : 4 coordination is typical of ZnS, CuCl, AgI.
2. Z = 4 formula units per cell.

Answer: (d) 8 : 4

Explanation:
1. Ca²⁺ forms fcc; F⁻ fills all 8 tetrahedral voids.
2. Each Ca²⁺ has 8 F⁻ neighbours; each F⁻ has 4 Ca²⁺ neighbours.

Teacher's Note:
1. CN ratio is the inverse of the ion ratio 1 : 2.
2. The order is cation first, then anion.

Answer: (b) 4

Explanation:
1. Ca²⁺ in fcc: 4 per cell.
2. F⁻: 8 in tetrahedral voids.
3. So 4 CaF₂ units.

Teacher's Note:
1. Z = 4 for fluorite.
2. Antifluorite (e.g. Na₂O) also has Z = 4.

Answer: (d) All of these

Explanation:
1. ZnS, CuCl and AgI all adopt the zinc blende (4 : 4) structure.

Teacher's Note:
1. Zinc blende has Z = 4.
2. Wurtzite (also ZnS) is another 4 : 4 structure.

Answer: (c) interchanging the positions of positive and negative ions in the lattice

Explanation:
1. In antifluorite the anions occupy the fcc positions and the cations fill all the tetrahedral voids, the reverse of CaF₂.

Teacher's Note:
1. Examples: Li₂O, Na₂O, K₂O.
2. Their CN is 4 (cation) and 8 (anion).

Answer: (c) AB₂

Explanation:
1. B forms ccp: 4 B per cell with 8 tetrahedral holes.
2. 25% of 8 = 2 A.
3. A : B = 2 : 4 = 1 : 2, formula AB₂.

Teacher's Note:
1. Always find the lattice atoms per cell first, then the filled holes.
2. Half the tetrahedral voids would give AB (zinc blende).

Answer: (b) 6

Explanation:
1. Radius ratio = 95/181 = 0.525, which lies between 0.414 and 0.732.
2. So the CN is 6.

Teacher's Note:
1. Use the radius ratio rule to predict structure.
2. Exceptions exist but this is the standard method.

Answer: (b) NaCl type

Explanation:
1. Radius ratio = 1.46/2.16 = 0.676.
2. It lies in 0.414–0.732, so CN is 6 and the structure is NaCl type.

Teacher's Note:
1. Radius ratio above 0.732 would give CsCl type.
2. Ratio below 0.414 would give ZnS type.

Answer: (b) 125 pm

Explanation:
1. In rock salt, a = 2(r⁺ + r⁻).
2. 400 = 2(75 + r⁻) gives r⁻ = 125 pm.

Teacher's Note:
1. Along the edge: anion–cation–anion touching means a = 2r⁺ + 2r⁻.
2. For CsCl-type, √3a = 2(r⁺ + r⁻).

Answer: (a) 200 pm

Explanation:
1. In a CsCl-type cell the ions touch along the body diagonal: \( r^+ + r^- = \dfrac{\sqrt{3}a}{2} \).
2. \( a = \dfrac{2\times173}{\sqrt{3}} = 199.8 \) pm, about 200 pm.

Teacher's Note:
1. Closest approach means the cation–anion distance.
2. Do not confuse it with the anion–anion distance, which equals a.

Answer: (a) 2a pm

Explanation:
1. In rock salt the Na⁺–Cl⁻ distance is half the edge.
2. So the edge = 2a pm.

Teacher's Note:
1. Edge = 2(r⁺ + r⁻) for rock salt.
2. The nearest-neighbour distance is a/2.