JEE Chemistry The Solid State MCQs Set 06

Practice MCQs for JEE Chemistry The Solid State

Review structured MCQ sets for JEE Chemistry The Solid State. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Access The Solid State Questions and Solutions

Navigate directly to the 50 objective questions for The Solid State using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.

Multiple Choice Questions

Question 1: \( a \neq b \neq c,\ \alpha=\gamma=90^\circ \neq \beta \) represents which crystal system?
(a) Tetragonal
(b) Orthorhombic
(c) Monoclinic
(d) Triclinic
Show Answer & Explanation

Answer: (c) Monoclinic

Explanation:
1. In the monoclinic system all three edges are unequal and two angles are 90° while one angle (β) differs from 90°.
2. Hence the system is monoclinic.

Teacher's Note:
1. Learn the seven systems as a table of edges and angles.
2. Monoclinic: a ≠ b ≠ c, α = γ = 90°, β ≠ 90°.

Question 2: The most unsymmetrical crystal system is
(a) Cubic
(b) Hexagonal
(c) Triclinic
(d) Orthorhombic
Show Answer & Explanation

Answer: (c) Triclinic

Explanation:
1. Triclinic has a ≠ b ≠ c and α ≠ β ≠ γ ≠ 90°, i.e. no equalities at all.
2. So it has the least symmetry.

Teacher's Note:
1. Cubic is the most symmetrical, triclinic the least.
2. Count how many edges or angles are forced to be equal to judge symmetry.

Question 3: Bravais lattices are of
(a) 10 types
(b) 8 types
(c) 7 types
(d) 14 types
Show Answer & Explanation

Answer: (d) 14 types

Explanation:
1. The seven crystal systems together allow 14 distinct space lattices, called Bravais lattices.

Teacher's Note:
1. Cubic has 3 (simple, bcc, fcc), tetragonal 2, orthorhombic 4, monoclinic 2, and hexagonal, rhombohedral and triclinic 1 each.
2. Total = 3 + 2 + 4 + 2 + 1 + 1 + 1 = 14.

Question 4: The crystal system of a compound with unit cell dimensions a = 0.387, b = 0.387 and c = 0.504 nm and α = β = 90°, γ = 120° is
(a) Cubic
(b) Hexagonal
(c) Orthorhombic
(d) Rhombohedral
Show Answer & Explanation

Answer: (b) Hexagonal

Explanation:
1. Two edges are equal (a = b) and the third is different (c).
2. Two angles are 90° and the remaining angle γ is 120°, which is the hexagonal condition.

Teacher's Note:
1. Hexagonal: a = b ≠ c, α = β = 90°, γ = 120°.
2. Tetragonal also has a = b ≠ c but all angles are 90°.

Question 5: In a simple cubic cell, each point on a corner is shared by
(a) 2 unit cells
(b) 1 unit cell
(c) 8 unit cells
(d) 4 unit cells
Show Answer & Explanation

Answer: (c) 8 unit cells

Explanation:
1. Eight cubes meet at every corner of a cubic lattice.
2. So a corner atom contributes 1/8 to each cell.

Teacher's Note:
1. Corner = 1/8, face = 1/2, edge centre = 1/4, body centre = 1.
2. These contributions are used in every counting problem.

Question 6: In a face centred cubic cell, an atom at the face contributes to the unit cell
(a) 1 part
(b) 1/2 part
(c) 1/4 part
(d) 1/8 part
Show Answer & Explanation

Answer: (b) 1/2 part

Explanation:
1. A face is shared between two adjacent unit cells.
2. So each face-centred atom contributes 1/2.

Teacher's Note:
1. An fcc cell has 6 faces, so face atoms contribute 6 × 1/2 = 3.
2. Add 8 × 1/8 = 1 from corners to get 4 atoms per cell.

Question 7: In a body centred cubic unit cell, an atom at the body centre is shared by
(a) 1 unit cell
(b) 2 unit cells
(c) 3 unit cells
(d) 4 unit cells
Show Answer & Explanation

Answer: (a) 1 unit cell

Explanation:
1. The body-centre atom lies wholly inside the cell and is not shared.

Teacher's Note:
1. It contributes a full atom to the cell.
2. bcc has 8 × 1/8 + 1 = 2 atoms per cell.

Question 8: Which type of cubic lattice has the maximum number of atoms per unit cell?
(a) Simple cubic
(b) Body centred cubic
(c) Face centred cubic
(d) All have the same
Show Answer & Explanation

Answer: (c) Face centred cubic

Explanation:
1. Simple cubic: 8 × 1/8 = 1. bcc: 1 + 1 = 2. fcc: 1 + 3 = 4.
2. So fcc has the most atoms per cell.

Teacher's Note:
1. Remember 1, 2, 4 for sc, bcc, fcc.
2. The number of atoms per cell is called Z.

Question 9: The number of atoms in a unit cell of a monoatomic element of simple cubic, body centred cubic and face centred cubic lattice respectively is
(a) 4, 9 and 14
(b) 1, 2 and 4
(c) 4, 5 and 6
(d) 2, 3 and 5
Show Answer & Explanation

Answer: (b) 1, 2 and 4

Explanation:
1. Simple cubic = 8 × 1/8 = 1.
2. bcc = 8 × 1/8 + 1 = 2.
3. fcc = 8 × 1/8 + 6 × 1/2 = 4.

Teacher's Note:
1. Do not count lattice points naively (8, 9, 14); atoms shared between cells are only fractionally counted.
2. The answer 1, 2, 4 is used in density formulas.

Question 10: Which one of the following is a primitive unit cell?
(a) Simple cubic
(b) BCC
(c) FCC
(d) bcc and fcc both
Show Answer & Explanation

Answer: (a) Simple cubic

Explanation:
1. A primitive unit cell has lattice points only at the corners.
2. Only the simple cubic cell fits; bcc and fcc have extra points inside or on faces.

Teacher's Note:
1. Primitive cells have Z = 1.
2. Centred cells (bcc, fcc) are non-primitive.

Question 11: In a body centred cubic unit cell, a metal atom at the centre of the cell is surrounded by how many other metal atoms?
(a) 8
(b) 6
(c) 12
(d) 4
Show Answer & Explanation

Answer: (a) 8

Explanation:
1. The central atom touches the eight corner atoms along the body diagonals.

Teacher's Note:
1. Coordination number of bcc = 8.
2. For sc it is 6 and for fcc/hcp it is 12.

Question 12: A compound is formed by elements A and B. It crystallises in the cubic structure when atoms A are at the corners of the cube and atoms B are at the centre of the body. The simplest formula of the compound is
(a) AB
(b) AB2
(c) A2B
(d) AB4
Show Answer & Explanation

Answer: (a) AB

Explanation:
1. A: 8 × 1/8 = 1.
2. B: 1 at the body centre.
3. Ratio A : B = 1 : 1, so the formula is AB.

Teacher's Note:
1. Always convert positions to atoms per cell before writing the formula.
2. This is the CsCl-type arrangement.

Question 13: A compound formed by elements A and B has a cubic structure in which A atoms are at the corners of the cube and B atoms are at the face centres. The formula of the compound is
(a) A4B3
(b) A2B
(c) AB3
(d) A2B3
Show Answer & Explanation

Answer: (c) AB3

Explanation:
1. A: 8 × 1/8 = 1.
2. B: 6 × 1/2 = 3.
3. Formula = AB3.

Teacher's Note:
1. Corners give 1 atom and the six faces give 3 atoms.
2. Check the answer against likely oxidation states if given.

Question 14: Sodium metal crystallises in a BCC lattice with cell edge a = 4.29 Å. What is the radius of a sodium atom?
(a) 1.86 Å
(b) 1.90 Å
(c) 18.3 Å
(d) 1.12 Å
Show Answer & Explanation

Answer: (a) 1.86 Å

Explanation:
1. In bcc the atoms touch along the body diagonal: \( \sqrt{3}a = 4r \).
2. \( r = \dfrac{\sqrt{3}\times 4.29}{4} = 1.86 \) Å.

Teacher's Note:
1. bcc: r = √3a/4; fcc: r = a/(2√2); sc: r = a/2.
2. Always identify the touching direction first.

Question 15: An element has BCC structure with 12.08 × 1023 unit cells. The number of atoms in these cells is
(a) 12.08 × 1023
(b) 24.16 × 1023
(c) 48.38 × 1023
(d) 12.08 × 1022
Show Answer & Explanation

Answer: (b) 24.16 × 1023

Explanation:
1. A bcc cell has 2 atoms.
2. Atoms = 2 × 12.08 × 1023 = 24.16 × 1023.

Teacher's Note:
1. Number of atoms = Z × number of unit cells.
2. bcc has Z = 2.

Question 16: A metal has BCC structure and the edge length of its unit cell is 3.04 Å. The volume of the unit cell in cm3 will be
(a) 1.6 × 10−21 cm3
(b) 2.81 × 10−23 cm3
(c) 6.02 × 10−23 cm3
(d) 6.6 × 10−24 cm3
Show Answer & Explanation

Answer: (b) 2.81 × 10−23 cm3

Explanation:
1. a = 3.04 Å = 3.04 × 10−8 cm.
2. V = a3 = (3.04 × 10−8)3 = 2.81 × 10−23 cm3.

Teacher's Note:
1. 1 Å = 10−8 cm and 1 pm = 10−10 cm.
2. Cube the edge length, not the radius.

Question 17: An element having BCC geometry has atomic mass 50. The density of the unit cell, if its edge length is 290 pm, is
(a) 6.81 g cm−3
(b) 3.40 g cm−3
(c) 13.62 g cm−3
(d) 1.23 g cm−3
Show Answer & Explanation

Answer: (a) 6.81 g cm−3

Explanation:
1. \( \rho = \dfrac{ZM}{N_A a^3} \) with Z = 2.
2. \( a^3 = (2.9\times10^{-8})^3 = 2.439\times10^{-23} \) cm3.
3. \( \rho = \dfrac{2\times50}{6.022\times10^{23}\times2.439\times10^{-23}} = 6.81 \) g cm−3.

Teacher's Note:
1. Density of a unit cell = (Z × M)/(NA × a3).
2. Convert the edge to cm before cubing.

Question 18: A metal (atomic mass = 50) has a body centred cubic crystal structure. The density of the metal is 5.96 g cm−3. The volume of the unit cell is
(a) 13.9 × 10−24 cm3
(b) 27.8 × 10−24 cm3
(c) 6.95 × 10−24 cm3
(d) 55.6 × 10−24 cm3
Show Answer & Explanation

Answer: (b) 27.8 × 10−24 cm3

Explanation:
1. Mass of one cell = 2 × 50 / 6.022 × 1023 = 1.66 × 10−22 g.
2. V = mass/density = 1.66 × 10−22 / 5.96 = 2.78 × 10−23 cm3 = 27.8 × 10−24 cm3.

Teacher's Note:
1. Rearrange ρ = ZM/(NAV) to V = ZM/(ρNA).
2. Check that your power of ten matches an option.

Question 19: An element crystallises in BCC structure. The edge length of its unit cell is 288 pm. If the density of the crystal is 7.2 g cm−3, what is the atomic mass of the element?
(a) 51.8
(b) 103.6
(c) 25.9
(d) 207.2
Show Answer & Explanation

Answer: (a) 51.8

Explanation:
1. \( M = \dfrac{\rho a^3 N_A}{Z} \), a = 2.88 × 10−8 cm, Z = 2.
2. \( a^3 = 2.389\times10^{-23} \) cm3.
3. \( M = \dfrac{7.2\times2.389\times10^{-23}\times6.022\times10^{23}}{2} = 51.8 \) g mol−1.

Teacher's Note:
1. Option (b) comes from forgetting to divide by Z.
2. The element is vanadium (M about 51).

Question 20: An element of density 6.8 g cm−3 occurs in BCC structure with cell edge 290 pm. The number of atoms present in 200 g of the element is
(a) 2.4 × 1042
(b) 1.2 × 1042
(c) 1.2 × 1024
(d) 2.4 × 1024
Show Answer & Explanation

Answer: (d) 2.4 × 1024

Explanation:
1. Volume of a cell = (2.9 × 10−8)3 = 2.439 × 10−23 cm3.
2. Mass of one cell = 6.8 × 2.439 × 10−23 = 1.658 × 10−22 g, containing 2 atoms.
3. Atoms in 200 g = (200 / 1.658 × 10−22) × 2 = 2.4 × 1024.

Teacher's Note:
1. Find the mass of one cell first, then scale up.
2. Atoms in 200 g = number of cells × Z.

Question 21: An element A crystallises in fcc structure. 200 g of this element has 4.12 × 1024 atoms. The density of A is 7.2 g cm−3. The edge length of the unit cell is
(a) 26.97 × 10−24 cm
(b) 299.9 pm
(c) 5 × 10−12 cm
(d) 2.99 cm
Show Answer & Explanation

Answer: (b) 299.9 pm

Explanation:
1. Mass of one atom = 200 / 4.12 × 1024 = 4.854 × 10−23 g.
2. Mass of one fcc cell = 4 × 4.854 × 10−23 = 1.942 × 10−22 g.
3. V = 1.942 × 10−22 / 7.2 = 2.697 × 10−23 cm3, so a = 2.998 × 10−8 cm = about 299.9 pm.

Teacher's Note:
1. Option (a) is the volume, not the edge length.
2. Take a cube root only at the last step.

Question 22: The density of Li is 0.53 g cm−3 and its edge length is 3.5 Å. The number of Li atoms in a unit cell is (NA = 6.023 × 1023, M = 6.94 g mol−1)
(a) 2
(b) 8
(c) 4
(d) 6
Show Answer & Explanation

Answer: (a) 2

Explanation:
1. \( Z = \dfrac{\rho a^3 N_A}{M} \).
2. a3 = (3.5 × 10−8)3 = 4.2875 × 10−23 cm3.
3. Z = 0.53 × 4.2875 × 10−23 × 6.023 × 1023 / 6.94 = 1.97, about 2 (bcc).

Teacher's Note:
1. Z must come out as a whole number, so round to the nearest integer.
2. Z = 2 means bcc (lithium is bcc).

Question 23: CsBr crystallises in a body centred cubic lattice. The unit cell length is 436.6 pm. Given that the atomic masses of Cs = 133 and Br = 80 and Avogadro number = 6.023 × 1023 mol−1, the density of CsBr is
(a) 42.5 g cm−3
(b) 0.425 g cm−3
(c) 8.25 g cm−3
(d) 4.25 g cm−3
Show Answer & Explanation

Answer: (d) 4.25 g cm−3

Explanation:
1. CsBr has the CsCl-type structure: a cell holds 1 Cs and 1 Br atom, i.e. one formula unit (Z = 1, M = 133 + 80 = 213).
2. a = 4.366 × 10−8 cm, a3 = 8.323 × 10−23 cm3.
3. \( \rho = \dfrac{1\times213}{6.023\times10^{23}\times8.323\times10^{-23}} = 4.25 \) g cm−3.

Teacher's Note:
1. In CsCl-type crystals there is one formula unit (one Cs and one Br) per cell, so use Z = 1 with the formula mass.
2. Using Z = 2 with the formula mass would wrongly double the density (option c).

Question 24: A metal crystallises in the face centred cubic lattice. The edge of the unit cell is 408 pm. The diameter of the metal atom is
(a) 144 pm
(b) 204 pm
(c) 288 pm
(d) 408 pm
Show Answer & Explanation

Answer: (c) 288 pm

Explanation:
1. In fcc, atoms touch along the face diagonal: \( \sqrt{2}a = 4r \).
2. \( r = \dfrac{408}{2\sqrt{2}} = 144.2 \) pm.
3. Diameter = 2r = 288 pm.

Teacher's Note:
1. The question asks for diameter, not radius.
2. Diameter in fcc = a/√2.

Question 25: A metal has an fcc lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g cm−3. The molar mass of the metal is
(a) 20 g mol−1
(b) 40 g mol−1
(c) 30 g mol−1
(d) 27 g mol−1
Show Answer & Explanation

Answer: (d) 27 g mol−1

Explanation:
1. \( M = \dfrac{\rho a^3 N_A}{Z} \), Z = 4.
2. a3 = (4.04 × 10−8)3 = 6.594 × 10−23 cm3.
3. M = 2.72 × 6.594 × 10−23 × 6.022 × 1023 / 4 = 27 g mol−1 (aluminium).

Teacher's Note:
1. fcc means Z = 4.
2. Always keep units in g and cm.

Free study material for Chemistry

Download Chapter MCQs: JEE Chemistry The Solid State

JEE Chemistry The Solid State Objective Test Questions

Review structured objective questions for JEE Chemistry The Solid State. Built according to official JEE guidelines, these MCQ sets support daily revision and core concept reinforcement.

NCERT-Aligned Objective Questions and Solutions

Built using the official NCERT book for JEE, these Chemistry objective sets provide reliable academic guidance. Pair your practice with our recommended NCERT solutions to master optimal problem-solving approaches.

Next Steps in Your Exam Preparation

Explore our broader library of printable assignments, chapter notes, and mock tests designed to support continuous revision and secure higher marks in JEE assessments.

FAQs

Where can I access latest JEE Chemistry The Solid State MCQs Set 06?

You can get most exhaustive JEE Chemistry The Solid State MCQs Set 06 for free on StudiesToday.com. These MCQs for JEE Chemistry are updated for the 2026-27 academic session as per JEE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Chemistry JEE material?

Yes, our JEE Chemistry The Solid State MCQs Set 06 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the JEE paper is now competency-based.

How do practicing Chemistry MCQs help in scoring full marks in JEE exams?

By solving our JEE Chemistry The Solid State MCQs Set 06, JEE students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Chemistry.

Do you provide answers and explanations for JEE Chemistry The Solid State MCQs Set 06?

Yes, Chemistry MCQs for JEE have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused JEE exams.

Can I practice these Chemistry JEE MCQs online?

Yes, you can also access online interactive tests for JEE Chemistry The Solid State MCQs Set 06 on StudiesToday.com as they provide instant answers and score to help you track your progress in Chemistry.