JEE Chemistry The Solid State MCQs Set 26

Practice MCQs for JEE Chemistry The Solid State

Access targeted multiple-choice questions for The Solid State designed to align with the latest JEE academic syllabus for JEE Chemistry. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

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Multiple Choice Questions

Question 1: The unit cell with parameters α = β = γ = 90° and a = b ≠ c is:
(a) cubic
(b) triclinic
(c) hexagonal
(d) tetragonal
Show Answer & Explanation

Answer: (d) tetragonal

Explanation:
1. Three right angles with two equal edges and one different edge define the tetragonal system.
2. Answer: (d)

Teacher's Note:
1. Recall the geometry of the lattice first, then count atoms per unit cell. Check the result against the known limiting values.

Question 2: Which of the following crystal lattices has the minimum empty space?
(a) simple cubic
(b) body centred cubic
(c) face centred cubic
(d) all have equal empty space
Show Answer & Explanation

Answer: (c) face centred cubic

Explanation:
1. Packing efficiency: simple cubic 52.4%, bcc 68%, fcc 74%.
2. FCC is the most closely packed, so it has the least empty space.
3. Answer: (c)

Teacher's Note:
1. Recall the geometry of the lattice first, then count atoms per unit cell. Check the result against the known limiting values.

Question 3: Molybdenum (atomic weight 96 g mol−1) crystallises as a BCC crystal. If the density is 10.3 g/cm3, the radius of the Mo atom is:
(a) 111 pm
(b) 314 pm
(c) 135.96 pm
(d) none of these
Show Answer & Explanation

Answer: (c) 135.96 pm

Explanation:
1. a3 = ZM/(dNA) = 2 × 96/(10.3 × 6.022 × 1023) = 3.095 × 10−23 cm3, so a = 314 pm.
2. For bcc, r = √3a/4 = 1.732 × 314/4 = 136 pm.
3. Answer: (c)

Teacher's Note:
1. Recall the geometry of the lattice first, then count atoms per unit cell. Check the result against the known limiting values.

Question 4: The coordination numbers of cation and anion in the anti-fluorite structure (Na2O) are:
(a) 4 : 8
(b) 8 : 4
(c) 6 : 6
(d) 4 : 4
Show Answer & Explanation

Answer: (a) 4 : 8

Explanation:
1. In Na2O, O2− forms ccp and Na+ fills all tetrahedral voids.
2. Na+ has CN 4 and O2− has CN 8.
3. Answer: (a)

Teacher's Note:
1. Recall the geometry of the lattice first, then count atoms per unit cell. Check the result against the known limiting values.

Question 5: Select the correct statement for the CsCl crystal:
(a) Cs+ forms a simple cubic lattice and Cl− forms a simple cubic lattice
(b) Cl− occupies the body centre of Cs+
(c) Cs+ occupies the body centre of Cl−
(d) all are correct
Show Answer & Explanation

Answer: (d) all are correct

Explanation:
1. CsCl has two interpenetrating simple cubic lattices.
2. Each ion sits at the body centre of the cube formed by the other, so all statements are correct.
3. Answer: (d)

Teacher's Note:
1. Recall the geometry of the lattice first, then count atoms per unit cell. Check the result against the known limiting values.

Question 6: The radius of divalent cation A2+ is 94 pm and the radius of divalent anion B2− is 146 pm. The structure of compound AB is:
(a) rock salt
(b) zinc blende
(c) anti-fluorite
(d) CsCl type
Show Answer & Explanation

Answer: (a) rock salt

Explanation:
1. r+/r− = 94/146 = 0.644.
2. This lies in 0.414 to 0.732, the octahedral range, so the structure is rock salt.
3. Answer: (a)

Teacher's Note:
1. Recall the geometry of the lattice first, then count atoms per unit cell. Check the result against the known limiting values.

Question 7: The coordination number and the number of carbon atoms in a unit cell of diamond are:
(a) 4 and 4
(b) 4 and 8
(c) 8 and 4
(d) 6 and 4
Show Answer & Explanation

Answer: (b) 4 and 8

Explanation:
1. Each carbon in diamond is bonded tetrahedrally to 4 others.
2. The diamond cubic unit cell has 8 carbon atoms (fcc points plus half the tetrahedral voids).
3. Answer: (b)

Teacher's Note:
1. Recall the geometry of the lattice first, then count atoms per unit cell. Check the result against the known limiting values.

Question 8: An element has a BCC structure with edge length 288 pm. The density of the element is 7.2 g/cm3. How many atoms are present in 208 g of the element?
(a) 6 × 1023
(b) 2.4 × 1023
(c) 24 × 1023
(d) 12 × 1023
Show Answer & Explanation

Answer: (c) 24 × 1023

Explanation:
1. M = dNAa3/Z = 7.2 × 6.022 × 1023 × (2.88 × 10−8)3/2 = 51.8 g mol−1.
2. Moles in 208 g = 4.0, so atoms = 4 × 6.022 × 1023 = 24 × 1023.
3. Answer: (c)

Teacher's Note:
1. Recall the geometry of the lattice first, then count atoms per unit cell. Check the result against the known limiting values.

Question 9: A compound is formed by two elements M and N. Element N forms ccp and atoms of M occupy 1/3rd of the tetrahedral voids. The formula of the compound is:
(a) M3N2
(b) M2N3
(c) M3N
(d) MN3
Show Answer & Explanation

Answer: (b) M2N3

Explanation:
1. Per N atom there are 2 tetrahedral voids.
2. M = 2 × 1/3 = 2/3, so M : N = 2/3 : 1 = 2 : 3 and the formula is M2N3.
3. Answer: (b)

Teacher's Note:
1. Recall the geometry of the lattice first, then count atoms per unit cell. Check the result against the known limiting values.

Question 10: X-ray diffraction studies of an element show that it crystallises in an FCC unit cell with edge length 3.6 × 10−8 cm. In a separate experiment its density is 8.92 g/cm3. Calculate the atomic mass of the element.
(a) 156
(b) 63
(c) 40
(d) 23
Show Answer & Explanation

Answer: (b) 63

Explanation:
1. M = dNAa3/Z = 8.92 × 6.022 × 1023 × (3.6 × 10−8)3/4.
2. = 8.92 × 28.1/4 = 62.7 ≈ 63.
3. Answer: (b)

Teacher's Note:
1. Recall the geometry of the lattice first, then count atoms per unit cell. Check the result against the known limiting values.

Question 11: If r stands for the radius of the atom in simple cubic, body centred cubic and face centred cubic systems, the ratio of edge lengths of the cubes in these systems is respectively:
(a) (1/2)r : √3 r : (1/√2) r
(b) (1/2)r : (√3/2) r : (√2/2) r
(c) 2r : 4r/√3 : 4r/√2
(d) 2r : 4r/√2 : 4r/√3
Show Answer & Explanation

Answer: (c) 2r : 4r/√3 : 4r/√2

Explanation:
1. Simple cubic: a = 2r. bcc: √3a = 4r so a = 4r/√3. fcc: √2a = 4r so a = 4r/√2.
2. Answer: (c)

Teacher's Note:
1. Recall the geometry of the lattice first, then count atoms per unit cell. Check the result against the known limiting values.

Question 12: The fraction of total volume occupied by the atoms present in FCC is:
(a) π/6
(b) π/(3√2)
(c) π/(4√2)
(d) π/4
Show Answer & Explanation

Answer: (b) π/(3√2)

Explanation:
1. Z = 4 and a = 2√2 r.
2. Fraction = 4 × (4/3)πr3/(2√2 r)3 = π/(3√2) = 0.74.
3. Answer: (b)

Teacher's Note:
1. Recall the geometry of the lattice first, then count atoms per unit cell. Check the result against the known limiting values.

Question 13: CaS exists in a cubic closed packed structure of S2− in which Ca2+ occupy tetrahedral voids. What is the percentage of tetrahedral voids occupied by Ca2+?
(a) 25%
(b) 50%
(c) 75%
(d) 100%
Show Answer & Explanation

Answer: (b) 50%

Explanation:
1. Ca : S = 1 : 1.
2. Tetrahedral voids = 2 per S2−, so 1 of 2 is occupied, i.e. 50%.
3. Answer: (b)

Teacher's Note:
1. Recall the geometry of the lattice first, then count atoms per unit cell. Check the result against the known limiting values.

Question 14: The atomic radius of strontium (Sr) is 215 pm and it crystallises in a cubic closed packed structure. The edge length of the cube is:
(a) 430 pm
(b) 608.2 pm
(c) 496.5 pm
(d) none of these
Show Answer & Explanation

Answer: (b) 608.2 pm

Explanation:
1. For ccp (fcc), a = 2√2 r.
2. a = 2 × 1.414 × 215 = 608.2 pm.
3. Answer: (b)

Teacher's Note:
1. Recall the geometry of the lattice first, then count atoms per unit cell. Check the result against the known limiting values.

Question 15: Analysis shows that nickel oxide has the formula Ni0.98O. What fraction of nickel exists as Ni2+? (Ni is in +2 and +3 oxidation states)
(a) 0.94
(b) 0.959
(c) 0.98
(d) 0.02
Show Answer & Explanation

Answer: (b) 0.959

Explanation:
1. Let Ni2+ = x, so Ni3+ = 0.98 − x. 2x + 3(0.98 − x) = 2 gives x = 0.94.
2. Fraction = 0.94/0.98 = 0.959.
3. Answer: (b)

Teacher's Note:
1. Recall the geometry of the lattice first, then count atoms per unit cell. Check the result against the known limiting values.

Question 16: Which of the following is NOT an ionic solid?
(a) NaCl
(b) MgO
(c) CsCl
(d) SiC
Show Answer & Explanation

Answer: (d) SiC

Explanation:
1. SiC is a covalent network solid.
2. NaCl, MgO and CsCl are ionic.
3. Answer: (d)

Teacher's Note:
1. Recall the geometry of the lattice first, then count atoms per unit cell. Check the result against the known limiting values.

Question 17: Diamond has a packing efficiency of:
(a) 52.4%
(b) 34%
(c) 68%
(d) 74%
Show Answer & Explanation

Answer: (b) 34%

Explanation:
1. Z = 8 and r = √3a/8.
2. Efficiency = 8 × (4/3)πr3/a3 = π√3/16 = 0.34, i.e. 34%.
3. Answer: (b)

Teacher's Note:
1. Recall the geometry of the lattice first, then count atoms per unit cell. Check the result against the known limiting values.

Question 18: A solid compound PQ has the CsCl structure and the radius ratio r+/r− has its limiting value (0.732). If the radius of the cation is 120 pm, find the radius of the anion.
(a) 88 pm
(b) 140 pm
(c) 164 pm
(d) 120 pm
Show Answer & Explanation

Answer: (c) 164 pm

Explanation:
1. Limiting radius ratio for the cubic (8-coordinate) arrangement = 0.732.
2. r− = 120/0.732 = 164 pm.
3. Answer: (c)

Teacher's Note:
1. Recall the geometry of the lattice first, then count atoms per unit cell. Check the result against the known limiting values.

Question 19: The vacant space in a simple cubic lattice unit cell is:
(a) 32%
(b) 52%
(c) 48%
(d) 25%
Show Answer & Explanation

Answer: (c) 48%

Explanation:
1. Packing efficiency of simple cubic = π/6 = 52.4%.
2. Vacant space = 100 − 52 = 48%.
3. Answer: (c)

Teacher's Note:
1. Recall the geometry of the lattice first, then count atoms per unit cell. Check the result against the known limiting values.

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