JEE Chemistry The Solid State MCQs Set 07

Answer: (a) 12

Explanation:
1. Each sphere touches 6 in its own layer, 3 in the layer above and 3 in the layer below = 12.

Teacher's Note:
1. hcp and ccp both have CN 12.
2. Packing efficiency of both is 74%.

Answer: (a) hexagonal close packing (hcp) and cubic close packing (ccp)

Explanation:
1. ABAB repeats after two layers and gives hcp.
2. ABCABC repeats after three layers and gives ccp (fcc).

Teacher's Note:
1. The third layer decides the pattern: over the first layer gives hcp, over the second voids gives ccp.
2. ccp is the same as fcc.

Answer: (c) 68%

Explanation:
1. \( \text{Packing efficiency} = \dfrac{2\times\frac{4}{3}\pi r^3}{a^3} \) with \( a = \dfrac{4r}{\sqrt{3}} \).
2. This gives \( \dfrac{\sqrt{3}\pi}{8} = 0.68 \), i.e. 68%.

Teacher's Note:
1. sc 52.4%, bcc 68%, fcc/hcp 74%.
2. Void space in bcc = 32%.

Answer: (a) 32%

Explanation:
1. Packing efficiency of bcc = 68%.
2. Vacant space = 100 − 68 = 32%.

Teacher's Note:
1. The void percentage is just 100 minus the packing efficiency.
2. For fcc the void space is 26%.

Answer: (a) 74%

Explanation:
1. Both are closest packings with packing efficiency \( \dfrac{\pi}{3\sqrt{2}} = 0.74 \).

Teacher's Note:
1. 74% is the maximum possible packing for identical spheres.
2. The remaining 26% is void space.

Answer: (a) 26%

Explanation:
1. Packing efficiency = 74%, so empty space = 100 − 74 = 26%.

Teacher's Note:
1. Empty space = 100 − packing efficiency.
2. Do not confuse this with bcc (32%).

Answer: (c) bcc

Explanation:
1. hcp and ccp (= fcc) are closest packings with CN 12.
2. bcc has CN 8 and only 68% packing, so it is not a close packing.

Teacher's Note:
1. Close packing means CN 12 and 74% efficiency.
2. bcc and simple cubic are not close-packed.

Answer: (b) Face centred

Explanation:
1. Packing efficiency: sc 52.4%, bcc 68%, fcc 74%.
2. So fcc is the most closely packed.

Teacher's Note:
1. Higher CN and higher packing efficiency go together.
2. fcc has CN 12.

Answer: (b) Mg

Explanation:
1. Al, Cu and Ni are fcc (ccp).
2. Mg crystallises in hcp.

Teacher's Note:
1. Common hcp metals: Mg, Zn, Cd, Be, Ti.
2. Common fcc metals: Al, Cu, Ag, Au, Ni, Pb.

Answer: (a) Helium

Explanation:
1. Solid helium has an hcp structure.
2. Ne, Ar, Kr and Xe are ccp.

Teacher's Note:
1. He is the exception in the noble gas group.

Answer: (c) BCC

Explanation:
1. CN 12 is fcc/hcp, CN 8 is bcc and CN 6 is simple cubic.
2. So the lattice is bcc.

Teacher's Note:
1. Coordination numbers: sc 6, bcc 8, fcc/hcp 12.
2. This is a quick way to identify a lattice.

Answer: (c) the void is surrounded tetrahedrally by four spheres

Explanation:
1. A tetrahedral void is the hole at the centre of four touching spheres arranged at the corners of a tetrahedron.

Teacher's Note:
1. Octahedral voids are surrounded by six spheres.
2. The name refers to the arrangement of surrounding spheres.

Answer: (b) Half

Explanation:
1. For N spheres there are N octahedral and 2N tetrahedral voids.
2. So octahedral : tetrahedral = 1 : 2, i.e. half.

Teacher's Note:
1. Voids per atom: octahedral 1, tetrahedral 2.
2. In fcc: 4 octahedral and 8 tetrahedral voids per cell.

Answer: (b) 6

Explanation:
1. An octahedral hole is surrounded by six spheres.

Teacher's Note:
1. Tetrahedral hole: CN 4; octahedral hole: CN 6.

Answer: (c) Larger

Explanation:
1. Octahedral void radius = 0.414R.
2. Tetrahedral void radius = 0.225R.
3. So the octahedral void is larger.

Teacher's Note:
1. Larger voids accommodate bigger cations.
2. Radius ratios: tetrahedral 0.225 to 0.414, octahedral 0.414 to 0.732.

Answer: (b) 2

Explanation:
1. In a close-packed lattice, N atoms generate 2N tetrahedral voids.

Teacher's Note:
1. In fcc there are 8 tetrahedral voids for 4 atoms.
2. Octahedral voids per atom = 1.

Answer: (c) P₂Q

Explanation:
1. Per 4 Q atoms in ccp there are 8 tetrahedral voids, all filled by P.
2. P : Q = 8 : 4 = 2 : 1, so the formula is P₂Q.

Teacher's Note:
1. Antifluorite-type: anions ccp, cations in all tetrahedral voids (e.g. Na₂O).
2. Always scale to Z atoms of the lattice.

Answer: (b) 2Z

Explanation:
1. ABCABC is ccp, with tetrahedral voids twice the number of atoms.
2. So voids = 2Z.

Teacher's Note:
1. For fcc, Z = 4 and tetrahedral voids = 8.
2. Octahedral voids = Z.

Answer: (a) \( \dfrac{\pi}{6} \)

Explanation:
1. In sc, 2r = a and Z = 1.
2. \( \text{Fraction} = \dfrac{\frac{4}{3}\pi r^3}{(2r)^3} = \dfrac{\pi}{6} = 0.524 \).

Teacher's Note:
1. π/6 for sc, π√3/8 for bcc, π/(3√2) for fcc.
2. These three fractions must be memorised.

Answer: (a) 32%

Explanation:
1. Packing efficiency of bcc is 68%.
2. Free space = 100 − 68 = 32%.

Teacher's Note:
1. Free space is the same as void space.
2. bcc has more free space than fcc (26%).

Answer: (d) \( \dfrac12 a : \dfrac{\sqrt3}{4} a : \dfrac{1}{2\sqrt2} a \)

Explanation:
1. sc: r = a/2.
2. bcc: r = √3a/4.
3. fcc: r = a/(2√2).

Teacher's Note:
1. Each radius comes from the direction along which the atoms touch (edge, body diagonal, face diagonal).
2. Option (b) uses half the body diagonal, which is the distance between ions in a CsCl-type cell.

Answer: (c) 151.8 pm

Explanation:
1. bcc: \( r = \dfrac{\sqrt{3}a}{4} = \dfrac{1.732\times351}{4} = 151.98 \) pm, about 152 pm.

Teacher's Note:
1. Use r = √3a/4 for bcc.
2. Check: 4r should be the body diagonal √3a.

Answer: (b) 128

Explanation:
1. fcc: \( r = \dfrac{a}{2\sqrt{2}} = \dfrac{361}{2.828} = 127.6 \) pm, about 128 pm.

Teacher's Note:
1. In fcc, 4r = √2a.
2. Compare with bcc where 4r = √3a.

Answer: (c) 1

Explanation:
1. In ccp there are 4 atoms and 4 octahedral voids per cell (1 at the body centre and 12 edge centres × 1/4 = 3).
2. So 1 octahedral void per atom.

Teacher's Note:
1. Tetrahedral: 2 per atom; octahedral: 1 per atom.

Answer: (b) 352 pm

Explanation:
1. 530 kg m⁻³ = 0.530 g cm⁻³.
2. \( a^3 = \dfrac{2\times6.94}{0.530\times6.02\times10^{23}} = 4.35\times10^{-23} \) cm³.
3. a = 3.52 × 10⁻⁸ cm = 352 pm.

Teacher's Note:
1. 1 kg m⁻³ = 10⁻³ g cm⁻³.
2. Converting density units first avoids a factor of 1000 error.