JEE Chemistry States Of Matter MCQs Set 21

Answer: (b) AB₂O₄

Explanation:
1. Let the number of B atoms (ccp) be N = 4 per cell. Octahedral voids = N, tetrahedral voids = 2N.
2. A occupies half the octahedral voids: N/2 = 2; oxygen fills all tetrahedral voids: 2N = 8.
3. A : B : O = 2 : 4 : 8 = 1 : 2 : 4, so the formula is AB₂O₄.

Teacher's Note:
1. In ccp there is one octahedral and two tetrahedral voids per atom.
2. Reduce the ratio to its simplest whole numbers.

Answer: (b) 90%

Explanation:
1. For solid 1 (A touching along the body diagonal): 4r_A = √3 a₁, so a₁ = 4r_A/√3.
2. a₂ = 1.5a₁ = 6r_A/√3 = 2√3 r_A.
3. Atoms in solid 2: one A and one B. Volume = (4/3)π(r_A³ + 8r_A³) = 12πr_A³; a₂³ = 24√3 r_A³.
4. Packing efficiency = 12π/(24√3) = π/(2√3) = 0.907, about 90%.

Teacher's Note:
1. Find the edge of solid 1 from the touching condition along the body diagonal.
2. Count one A and one B in the cell of solid 2.

Answer: (a) Xe

Explanation:
1. From p(V − b) = RT, pV/RT = 1 + pb/RT, so Z = 1 + (b/RT)p.
2. The plot of Z versus p is a straight line with slope b/RT.
3. The steepest slope belongs to the gas with the largest b, which is Xe.

Teacher's Note:
1. The slope of Z against p is b/RT at fixed temperature.
2. Larger b means a larger molecular size and a steeper line.

Answer: (a) 2.0 pm

Explanation:
1. Moles = 10 × 10⁻³ L × 10⁻³ M = 10⁻⁵ mol, so molecules = 10⁻⁵ × 6 × 10²³ = 6 × 10¹⁸.
2. Area per molecule = 0.24/(6 × 10¹⁸) = 4 × 10⁻²⁰ cm².
3. The cube's face area is a², so a = 2 × 10⁻¹⁰ cm = 2 × 10⁻¹² m = 2 pm.

Teacher's Note:
1. Divide the total area by the number of molecules.
2. The face of the cube is the area covered per molecule.

Answer: (c) N₂ (300 K), O₂ (400 K), H₂ (300 K)

Explanation:
1. v_mp = √(2RT/M) is proportional to √(T/M).
2. √(T/M): H₂ at 300 K = √150 = 12.2; N₂ at 300 K = √(300/28) = 3.3; O₂ at 400 K = √(400/32) = 3.5.
3. Increasing order: N₂ (300 K) < O₂ (400 K) < H₂ (300 K), i.e. I, II, III.

Teacher's Note:
1. Compare T/M for each gas; a higher value gives a higher peak speed.
2. Light gases move much faster even at lower temperature.

Answer: (b) 1 : 2 : 4

Explanation:
1. Simple cubic: 8 × 1/8 = 1 atom.
2. bcc: 8 × 1/8 + 1 = 2 atoms.
3. fcc: 8 × 1/8 + 6 × 1/2 = 4 atoms. The ratio is 1 : 2 : 4.

Teacher's Note:
1. Corner atoms are shared by 8 cells and face atoms by 2 cells.
2. Memorise Z = 1, 2, 4 for sc, bcc, fcc.

Answer: (d) 422/x³

Explanation:
1. Z = 4 for fcc and a = x × 10⁻⁸ cm.
2. d = ZM/(N_A a³) = 4 × 63.55 / (6.023 × 10²³ × x³ × 10⁻²⁴).
3. d = 254.2/(0.6023 x³) = 422/x³ g cm⁻³.

Teacher's Note:
1. Use d = ZM/(N_A a³) with the edge in cm.
2. Convert Å to cm by multiplying by 10⁻⁸.

Answer: (d) Triclinic

Explanation:
1. Hexagonal: a = b ≠ c, α = β = 90°, γ = 120°.
2. Monoclinic: a ≠ b ≠ c, α = γ = 90°, β ≠ 90°. Tetragonal: a = b ≠ c, all angles 90°.
3. Only the triclinic cell has a ≠ b ≠ c and α ≠ β ≠ γ ≠ 90°.

Teacher's Note:
1. Triclinic is the least symmetric crystal system.
2. Compare the edge and angle conditions of each system.

Answer: (c) hcp lattice B, 1/3 tetrahedral voids A

Explanation:
1. Take 6 atoms in the hcp lattice, giving 12 tetrahedral voids.
2. If B forms the lattice (6 atoms), A must be 4 atoms for the ratio A : B = 2 : 3.
3. Fraction of tetrahedral voids occupied by A = 4/12 = 1/3.

Teacher's Note:
1. Let the larger count of atoms form the lattice and place the other in voids.
2. hcp has 2 tetrahedral voids per atom.

Answer: (d) 0.067 a

Explanation:
1. In bcc, R = √3 a/4.
2. At the middle of an edge the small sphere of radius r touches the two corner atoms along the edge: a/2 = R + r.
3. r = a/2 − √3a/4 = a(2 − √3)/4 = 0.067a.

Teacher's Note:
1. The interstitial sphere touches the two corner atoms along the edge.
2. Half the edge equals the sum of the two radii.

Answer: (b) 2p_A = 3p_B

Explanation:
1. Z = pV/nRT, so p = ZnRT/V.
2. p_A/p_B = (Z_A/Z_B)(V_B/V_A) = 3 × 1/2 = 3/2.
3. Therefore 2p_A = 3p_B.

Teacher's Note:
1. For equal n and T, pressure is proportional to Z/V.
2. Substitute the given ratios directly.

Answer: (b) 500 K

Explanation:
1. Pressure and volume are constant, so nT is constant.
2. n₁T₁ = n₂T₂ with n₂ = (3/5)n₁.
3. T₂ = 300 × 5/3 = 500 K.

Teacher's Note:
1. In an open vessel the pressure stays equal to the atmosphere.
2. Express the final answer in kelvin, as in option (b).

Answer: (c) Frenkel defect

Explanation:
1. In a Frenkel defect a cation leaves its lattice site and occupies an interstitial position.
2. Schottky and vacancy defects involve missing ions, and metal deficiency involves a variable oxidation state.

Teacher's Note:
1. Frenkel defect is also called a dislocation defect.
2. It is common when the cation is much smaller than the anion, as in AgBr.

Answer: (b) 2p_i (T₂/(T₁ + T₂))

Explanation:
1. Total moles are unchanged: 2p_iV/RT₁ = p_fV/RT₁ + p_fV/RT₂.
2. 2p_i/T₁ = p_f(T₂ + T₁)/(T₁T₂).
3. p_f = 2p_i T₂/(T₁ + T₂).

Teacher's Note:
1. The pressure is equal in both bulbs once connected.
2. Conserve the total number of moles.

Answer: (a) 1.86 Å

Explanation:
1. In bcc the atoms touch along the body diagonal: 4r = √3 a.
2. r = √3 × 4.29/4 = 1.86 Å.

Teacher's Note:
1. The body diagonal of a cube is √3 a.
2. It carries two corner radii and one full atom diameter, i.e. 4r.

Answer: (b) Z = 1 − a/(VRT)

Explanation:
1. At low pressure V is large, so b can be neglected relative to V: (p + a/V²)V = RT.
2. pV + a/V = RT, hence pV/RT = 1 − a/(VRT).
3. Z = 1 − a/(VRT).

Teacher's Note:
1. At low pressure attractions dominate and Z is less than 1.
2. At high pressure b dominates and Z = 1 + pb/RT.

Answer: (b) Silicon

Explanation:
1. Silicon has a giant covalent network (diamond-like structure).
2. Iodine, sulphur and white phosphorus form molecular crystals held by weak van der Waals forces.

Teacher's Note:
1. Network solids include diamond, silicon, SiC and quartz.
2. Molecular solids consist of discrete molecules.

Answer: (b) 4.08%

Explanation:
1. Take 1 mol of oxide: 0.98 mol of M and 1 mol of O²⁻. Let x mol be M³⁺, so M²⁺ = 0.98 − x.
2. Charge balance: 2(0.98 − x) + 3x = 2, so x = 0.04.
3. Fraction of M³⁺ = 0.04/0.98 × 100 = 4.08%.

Teacher's Note:
1. Apply electroneutrality to the non-stoichiometric oxide.
2. The metal deficiency is compensated by M³⁺ ions.