JEE Chemistry The Solid State MCQs Set 09

Answer: (c) A is true, but R is false.

Explanation:
1. Assertion is true: metals conduct due to mobile electrons.
2. Reason is false: Schottky defects are vacancies in ionic crystals and have nothing to do with metallic conduction.

Teacher's Note:
1. Metallic conduction arises from delocalised electrons.
2. Schottky defect is a feature of ionic solids.

Answer: (d) A is false, but R is true.

Explanation:
1. Graphite is hexagonal, so the assertion is false.
2. Tetragonal has all three angles 90°; γ = 120° belongs to hexagonal, so the reason is also false.

Teacher's Note:
1. Tetragonal: a = b ≠ c, all angles 90°.
2. Hexagonal: a = b ≠ c, γ = 120°.

Answer: (d) A is false, but R is true.

Explanation:
1. bcc has 68% and hcp has 74% packing efficiency, so the assertion is false.
2. The reason is false too: bcc has Z = 2 and hcp Z = 6 (per hexagonal cell) with different arrangements.

Teacher's Note:
1. hcp and fcc share 74%.
2. bcc is less densely packed.

Answer: (c) A is true, but R is false.

Explanation:
1. Assertion is true: N atoms give 2N tetrahedral and N octahedral voids.
2. Reason is false as stated: tetrahedral void radius is 0.225R and octahedral is 0.414R, so the ratio is about 0.54, not a simple one-half rule, and it is not the cause of the 2 : 1 count.

Teacher's Note:
1. The 2 : 1 ratio comes from geometry of layer stacking, not from void size.
2. Tetrahedral voids are smaller than octahedral voids.

Answer: (d) A is false, but R is true.

Explanation:
1. CN of Cs⁺ in CsCl is 8, so the assertion is false.
2. Cl⁻ ions are at the corners of a simple cube (Cs⁺ is at the centre), so the reason is false.

Teacher's Note:
1. CsCl is 8 : 8 coordinated.
2. Do not call CsCl a bcc crystal; the lattice is simple cubic with a two-ion basis.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Both statements are true.
2. The name antifluorite follows from the reversed positions, but the reason merely gives a name; it does not explain why Na₂O has that structure.

Teacher's Note:
1. A reason that only restates the name is not an explanation.
2. Use option B for such cases.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Both are true.
2. Fewer Fe²⁺ sites are occupied, and the missing positive charge is made up by oxidising some Fe²⁺ to Fe³⁺, which explains the deficiency.

Teacher's Note:
1. Fe_0.95O is a typical non-stoichiometric formula.
2. Metal-deficient oxides are p-type semiconductors.

Answer: (d) The fraction of the total volume occupied by the atoms in a primitive cell is 0.48

Explanation:
1. Statements 1 to 3 are correct.
2. A primitive (simple cubic) cell has packing fraction π/6 = 0.52, not 0.48.

Teacher's Note:
1. Diamond has 8 corners × 1/8 + 6 faces × 1/2 + 4 inside = 8 atoms.
2. Packing fraction of sc is 0.524.

Answer: (d) Two axial angles are the same and one is different

Explanation:
1. Rhombohedral has a = b = c and α = β = γ ≠ 90°.
2. So the statement that one angle differs is wrong.

Teacher's Note:
1. Rhombohedral is cubic skewed along the body diagonal.
2. Statements (a), (b) and (c) are all consistent with this.

Answer: (d) 241.5 pm

Explanation:
1. For an octahedral void the minimum ratio is 0.414.
2. r⁻ = 100/0.414 = 241.5 pm.

Teacher's Note:
1. The question uses the limiting ratio for the NaCl structure.
2. r⁻ = r⁺/0.414.

Answer: (a) 127 pm

Explanation:
1. Four atoms per cell means fcc.
2. \( r = \dfrac{a}{2\sqrt{2}} = \dfrac{361}{2.828} = 127.6 \) pm.

Teacher's Note:
1. Z = 4 identifies fcc.
2. Copper has this edge length.

Answer: (a) M₃N₂

Explanation:
1. ccp: 4 M atoms, 8 tetrahedral voids.
2. N = 8/3.
3. M : N = 4 : 8/3 = 3 : 2, so M₃N₂.

Teacher's Note:
1. Scale to 4 lattice atoms and multiply by the fraction of voids filled.
2. Check the ratio by clearing fractions.

Answer: (d) \( \dfrac{\sqrt3}{2}a \)

Explanation:
1. The body diagonal of a cube is √3a.
2. The body centre to a corner is half of it, √3a/2.

Teacher's Note:
1. This is the same distance as r⁺ + r⁻ in CsCl.
2. For bcc metals it equals 2r.

Answer: (b) B₃A₂

Explanation:
1. hcp has 2 atoms and 4 tetrahedral voids per cell.
2. A = 4/3 per 2 B.
3. B : A = 2 : 4/3 = 3 : 2, formula B₃A₂.

Teacher's Note:
1. hcp: 2 lattice atoms per cell and 2 tetrahedral voids per atom.
2. Clear fractions to get whole numbers.

Answer: (a) X₂Y

Explanation:
1. hcp: 2 X and 4 tetrahedral voids per cell.
2. Y = 1/4 × 4 = 1.
3. Formula X₂Y.

Teacher's Note:
1. Number of tetrahedral voids = 2 × number of atoms.
2. Then multiply by the occupied fraction.

Answer: (a) \( \dfrac{16}{\sqrt3} \) Å

Explanation:
1. bcc: \( \sqrt{3}a = 4r = 16 \) Å.
2. \( a = \dfrac{16}{\sqrt{3}} \) Å.

Teacher's Note:
1. bcc edge a = 4r/√3.
2. fcc edge a = 2√2 r.

Answer: (d) HCP ≈ fcc > bcc > simple cubic

Explanation:
1. hcp and fcc: 74%, bcc: 68%, simple cubic: 52.4%.

Teacher's Note:
1. hcp and fcc are equal in packing efficiency.
2. Write the values down in decreasing order.

Answer: (c) 8, 6

Explanation:
1. A cube has 8 corners and 6 faces.
2. So there are 8 positions for X and 6 for Y (contributing 1 and 3 atoms per cell, formula XY₃).

Teacher's Note:
1. The question asks for positions, not atoms per cell.
2. Contribution per cell would be 8 × 1/8 = 1 and 6 × 1/2 = 3.

Answer: (c) \( \dfrac{3\sqrt3}{4\sqrt2} \)

Explanation:
1. bcc: Z = 2, \( a = \dfrac{4r}{\sqrt3} \); fcc: Z = 4, \( a = 2\sqrt2 r \).
2. \( \rho \propto Z/a^3 \). Ratio \( = \dfrac{2/(4r/\sqrt3)^3}{4/(2\sqrt2 r)^3} = \dfrac{3\sqrt3}{4\sqrt2} \approx 0.92 \).

Teacher's Note:
1. Compute density ratio from Z/a³ with a expressed in r.
2. Ratio below 1 means bcc is less dense than fcc.

Answer: (a) unpaired electron occupying an anionic vacancy

Explanation:
1. When an alkali halide is heated in alkali metal vapour, anions leave their sites and electrons occupy the vacancies.
2. These trapped electrons absorb visible light and impart colour.

Teacher's Note:
1. F comes from the German word for colour.
2. F-centres are metal-excess defects.

Answer: (b) On heating it turns yellow

Explanation:
1. On heating, ZnO loses oxygen and gets excess Zn²⁺ in interstitial sites with free electrons; this colour centre makes it yellow.
2. It turns white again on cooling.

Teacher's Note:
1. ZnO is a metal-excess (interstitial) defect solid.
2. It does not sublime.

Answer: (a) SO₂ (solid)

Explanation:
1. Solid SO₂ consists of discrete SO₂ molecules held by weak intermolecular forces.
2. SiC and graphite are covalent network solids; ZnS is ionic.

Teacher's Note:
1. Molecular solids are soft with low melting points.
2. Dry ice and solid SO₂ are examples.