JEE Chemistry The Solid State MCQs Set 10

Answer: (b) NaWO₃

Explanation:
1. W at 8 corners: 8 × 1/8 = 1
2. O at 12 edge centres: 12 × 1/4 = 3
3. Na at the body centre: 1
4. Formula = NaWO₃. Answer: (b)

Teacher's Note:
1. A corner atom is shared by 8 cells, an edge atom by 4 cells and a body-centre atom belongs wholly to the cell.
2. Count the atoms of each kind per unit cell.

Answer: (d) 2 and 4

Explanation:
1. bcc: 8 × 1/8 + 1 = 2 atoms (Na)
2. fcc: 8 × 1/8 + 6 × 1/2 = 4 atoms (Mg). Answer: (d)

Teacher's Note:
1. bcc has 2 atoms and fcc has 4 atoms per unit cell.
2. Corner atoms count 1/8 each and face atoms 1/2 each.

Answer: (c) \( \frac{\sqrt3 a}{2} \)

Explanation:
1. In CsCl the ions touch along the body diagonal
2. The body diagonal \( = \sqrt3 a = 2(r_+ + r_-) \)
3. Interionic distance \( r_+ + r_- = \frac{\sqrt3a}{2} \). Answer: (c)

Teacher's Note:
1. CsCl has a simple cubic arrangement with the other ion at the body centre.
2. The two ions touch along the body diagonal.

Answer: (d) X₅Y₄Z₈

Explanation:
1. Complete cell: X = 4, Y = 4 (octahedral voids), Z = 8 (tetrahedral voids)
2. One body diagonal contains 2 corner X atoms (2 × 1/8 = 1/4 of X), the body-centre Y (1) and two tetrahedral Z sites (2)
3. Remaining: X = 4 − 1/4 = 15/4, Y = 3, Z = 6
4. Ratio 15/4 : 3 : 6 = 5 : 4 : 8, so the formula is X₅Y₄Z₈. Answer: (d)

Teacher's Note:
1. A body diagonal passes through two corners, the centre and two tetrahedral voids.
2. Remove those particles and simplify the ratio.

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

Explanation:
1. In the tetragonal system two edges are equal and the third differs, with all angles 90°
2. Option (a) is cubic, (c) is orthorhombic and (d) is hexagonal. Answer: (b)

Teacher's Note:
1. Learn the seven crystal systems by their axial lengths and angles.
2. Tetragonal is a stretched cube.

Answer: (a) AB

Explanation:
1. A: 8 × 1/8 = 1 per cell
2. B: 1 at the body centre
3. Formula AB. Answer: (a)

Teacher's Note:
1. Corner atoms contribute one-eighth each.

Answer: (a) 5.783 Å

Explanation:
1. \( \rho = \frac{ZM}{N_Aa^3} \) with Z = 4 and M = 99
2. \( a^3 = \frac{4\times99}{6.022\times10^{23}\times3.4} = 1.934\times10^{-22}\ \mathrm{cm^3} \)
3. \( a = 5.783\times10^{-8} \) cm = 5.783 Å. Answer: (a)

Teacher's Note:
1. An fcc unit cell contains 4 formula units.
2. Convert cm to angstrom by multiplying by 10^8.

Answer: (d) \( \frac\pi6 \)

Explanation:
1. One atom per cell with \( r = \frac a2 \)
2. Packing fraction \( = \frac{\frac43\pi\left(\frac a2\right)^3}{a^3} = \frac\pi6 \approx 0.52 \). Answer: (d)

Teacher's Note:
1. In simple cubic packing the atoms touch along the cell edge.
2. The packing efficiency is 52.4 percent.

Answer: (d) cubic close packing

Explanation:
1. The ABAB... sequence is hexagonal close packing
2. The ABCABC... sequence is cubic close packing (fcc). Answer: (d)

Teacher's Note:
1. Cubic close packing and fcc are the same arrangement.

Answer: (b) Schottky defect

Explanation:
1. Equal numbers of cations and anions are missing from the lattice
2. Nothing is displaced to an interstitial site
3. This is a Schottky defect (the density decreases). Answer: (b)

Teacher's Note:
1. A Schottky defect is a pair of vacancies, one cation and one anion.
2. A Frenkel defect would show a displaced ion in an interstitial site.

Answer: (a) 346.4 pm

Explanation:
1. The ions touch along the body diagonal
2. Interionic distance \( = \frac{\sqrt3}{2}a = 0.866\times400 = 346.4 \) pm. Answer: (a)

Teacher's Note:
1. The nearest cation-anion distance is half the body diagonal.

Answer: (d) S

Explanation:
1. In a close-packed array of N spheres there are N octahedral voids and 2N tetrahedral voids
2. So S spheres give S octahedral sites. Answer: (d)

Teacher's Note:
1. Octahedral voids equal the number of spheres.

Answer: (b) \( 10^{23} \)

Explanation:
1. \( M = \frac{\rho a^3N_A}{Z} = \frac{6\times(2\times10^{-8})^3\times6.022\times10^{23}}{2} \approx 14.4 \) g/mol
2. Moles in 2.4 g \( = \frac{2.4}{14.4} \approx 0.166 \)
3. Atoms \( = 0.166\times6.022\times10^{23} \approx 10^{23} \). Answer: (b)

Teacher's Note:
1. bcc has Z = 2.
2. Atoms in a sample = moles times Avogadro's number.

Answer: (a) A₄B₅O₁₀

Explanation:
1. For N oxide ions there are 2N tetrahedral voids and N octahedral voids
2. A = (1/5)(2N) = 0.4N and B = (1/2)(N) = 0.5N
3. A : B : O = 0.4 : 0.5 : 1 = 4 : 5 : 10
4. Formula A₄B₅O₁₀. Answer: (a)

Teacher's Note:
1. Count voids relative to the number of lattice (oxide) ions.
2. Reduce the ratio to whole numbers.

Answer: (a) \( 2.50\times10^{21} \)

Explanation:
1. Formula units in 1 g \( = \frac{1}{60}\times6.022\times10^{23} = 1.0\times10^{22} \)
2. An NaCl-type cell has 4 formula units
3. Unit cells \( = \frac{1.0\times10^{22}}{4} = 2.5\times10^{21} \). Answer: (a)

Teacher's Note:
1. An NaCl-type cell contains four AB units.
2. Convert mass to formula units first.