JEE Chemistry The Solid State MCQs Set 13

Chemistry Objective Questions and Answers: 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.

Download The Solid State MCQs with Answers

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: When Fe0.93O is heated in the presence of oxygen it converts to Fe₂O₃. How many of the following statements are correct? (A) The equivalent weight of Fe0.93O is molecular weight/0.79. (B) The numbers of moles of Fe²⁺ and Fe³⁺ in 1 mole of Fe0.93O are 0.79 and 0.14 respectively. (C) Fe0.93O is metal deficient with a lattice made of cubic close-packed O²⁻ ions. (D) The percentage composition of Fe²⁺ and Fe³⁺ in Fe0.93O is 85 % and 15 % respectively.
(a) 1
(b) 2
(c) 3
(d) 4
Show Answer & Explanation

Answer: (d) 4

Explanation:
1. Let Fe²⁺ = x and Fe³⁺ = y: x + y = 0.93 and 2x + 3y = 2 (charge balance), giving x = 0.79 and y = 0.14 (B true).
2. Only Fe²⁺ is oxidised to Fe³⁺, so 0.79 mol of electrons are lost per mole: equivalent weight = M/0.79 (A true).
3. The oxide ions form a ccp lattice with some Fe sites vacant (C true). % Fe²⁺ = 0.79/0.93 = 85 %, Fe³⁺ = 15 % (D true). All four are correct.

Teacher's Note:
1. Use charge neutrality to find the Fe²⁺ : Fe³⁺ ratio in non-stoichiometric oxides.
2. Metal deficiency defect: some cations are missing and some others are in a higher oxidation state.

Question 2: A cubic solid is made of two elements X and Y. Atoms of X are present on every alternate corner and one at the centre of the cube. Y is at two-thirds of the total faces (face centres). The empirical formula of the compound is
(a) X₂Y1.5
(b) X2.5Y
(c) XY2.5
(d) X1.5Y₂
Show Answer & Explanation

Answer: (d) X1.5Y₂

Explanation:
1. X: 4 alternate corners × 1/8 = 0.5, plus 1 at the centre = 1.5.
2. Y: two-thirds of 6 faces = 4 faces, each shared by 2 cells: 4 × 1/2 = 2.
3. Formula = X1.5Y₂.

Teacher's Note:
1. A corner atom contributes 1/8, a face-centre atom 1/2 and a body-centre atom 1 to the unit cell.
2. Count the atoms of each element, then write the simplest ratio.

Question 3: A metal M forms a hexagonal close-packed structure. The total number of voids in 0.02 mol of it is (NA = 6.02 × 10²³)
(a) 1.20 × 10²²
(b) 2.41 × 10²²
(c) 3.61 × 10²²
(d) 7.22 × 10²²
Show Answer & Explanation

Answer: (c) 3.61 × 10²²

Explanation:
1. In a close-packed structure each atom has 1 octahedral void and 2 tetrahedral voids, i.e. 3 voids per atom.
2. Number of atoms = 0.02 × 6.02 × 10²³ = 1.204 × 10²².
3. Total voids = 3 × 1.204 × 10²² = 3.61 × 10²² (that is 36 × 10²¹).

Teacher's Note:
1. Voids per atom: octahedral 1 and tetrahedral 2 in both ccp and hcp.
2. Convert moles to atoms with Avogadro's number first.

Question 4: Iron(II) oxide, FeO, crystallises in a cubic lattice with unit cell edge 5.0 Å. If the density is 4.0 g cm⁻³, the number of FeO units per unit cell is (Fe = 56, O = 16; NA = 6.0 × 10²³ mol⁻¹)
(a) 2
(b) 4
(c) 6
(d) 8
Show Answer & Explanation

Answer: (b) 4

Explanation:
1. Z = ρ a³ NA/M with a = 5.0 × 10⁻⁸ cm, so a³ = 1.25 × 10⁻²² cm³.
2. Z = 4.0 × 1.25 × 10⁻²² × 6.0 × 10²³/72 = 300/72 = 4.17.
3. Nearest integer: Z = 4.

Teacher's Note:
1. Convert the edge to cm before cubing it.
2. Z must be a whole number; round off to the nearest integer.

Question 5: A metal oxide sample has the formula M0.83O, where M exists in +2 and +3 oxidation states. The percentage of metal ions in the +2 state is nearest to
(a) 59 %
(b) 41 %
(c) 83 %
(d) 17 %
Show Answer & Explanation

Answer: (a) 59 %

Explanation:
1. Let M²⁺ = x and M³⁺ = y per formula unit: x + y = 0.83 and 2x + 3y = 2.
2. Solving: y = 2 − 1.66 = 0.34 and x = 0.49.
3. % M²⁺ = 0.49/0.83 × 100 ≈ 59 %.

Teacher's Note:
1. Use charge balance with the oxide ion charge of −2.
2. The cation deficiency is compensated by some ions being in the higher oxidation state.

Question 6: A metal M crystallises in two lattices: face-centred cubic (edge 2.0 Å) and body-centred cubic (edge 2.5 Å). The ratio of the densities of the fcc lattice to the bcc lattice is nearest to
(a) 2
(b) 1
(c) 8
(d) 4
Show Answer & Explanation

Answer: (d) 4

Explanation:
1. Density ρ = ZM/(NAa³), so the ratio ρfcc/ρbcc = (Zfcc/afcc³)/(Zbcc/abcc³).
2. Zfcc = 4 and Zbcc = 2: ratio = (4/8)/(2/15.625) = 0.5/0.128 = 3.9.
3. Nearest integer: 4.

Teacher's Note:
1. For the same metal M and NA cancel.
2. Z = 4 for fcc and Z = 2 for bcc.

Free study material for Chemistry

Practice MCQs for JEE Chemistry The Solid State

About The Solid State MCQs for JEE Chemistry

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.

How to Verify Your MCQ Answers

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.

Enhance Speed with Online MCQ Tests

Wrap up your chapter revision by testing your knowledge against standard question formats. Everything on our platform is provided free of charge.

FAQs

Where can I access latest JEE Chemistry The Solid State MCQs Set 13?

You can get most exhaustive JEE Chemistry The Solid State MCQs Set 13 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 13 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 13, 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 13?

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 13 on StudiesToday.com as they provide instant answers and score to help you track your progress in Chemistry.