NEET The Solid State PYQs & Practice Questions

10 free NEET Chemistry questions on The Solid State, starting with previous year questions (2016–2023). Tap “Show answer” to check each one. No login needed.

Topics: Types of Solids · Crystal Systems and Unit Cells · Packing Efficiency · Defects in Solids · Electrical Properties · Coordination Number · Density of Unit Cell · Magnetic Properties · Voids and Close Packing · Unit Cell and Crystal Systems · Close Packing and Voids · Radius Ratio and Ionic Structures

1. Q1NEET 2023Close Packed Structures and Voids

   A compound is formed by two elements A and B. The element B forms a cubic close packed structure and atoms of A occupy 1/3 of the tetrahedral voids. If the formula of the compound is AxBy, then the value of x + y is:

   • A. 4
   • B. 3
   • C. 2
   • D. 5
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   Answer: C — 2

   If B forms a ccp lattice with N atoms, there are 2N tetrahedral voids. A occupies 1/3 of these voids, so number of A atoms = 2N/3. Ratio A:B = (2N/3):N = 2:3, giving formula A2B3, so x = 2, y = 3, and x + y = 5.

2. Q2NEET 2023Voids in FCC Lattice

   What fraction of one edge centred octahedral void lies in one unit cell of fcc?

   • A. 1/3
   • B. 1/4
   • C. 1/12
   • D. 1/2
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   Answer: A — 1/3

   An octahedral void located at the edge centre of an fcc unit cell is shared among 4 unit cells, so the fraction of that edge-centred octahedral void lying within one unit cell is 1/4.

3. Q3NEET 2022Density of Unit Cell

   Copper crystallises in fcc unit cell with cell edge length of 3.608 x 10^-8 cm. The density of copper is 8.92 g/cm^3. Calculate the atomic mass of copper.

   • A. 31.55 u
   • B. 60 u
   • C. 65 u
   • D. 63.1 u
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   Answer: C — 65 u

   Using density formula d = ZM/(Na x a^3), where Z=4 for fcc: 8.92 = (4 x M)/(6x10^23 x (3.608x10^-8)^3). Solving gives M = 62.84 ≈ 63 u, closest to 63.1 u.

4. Q4NEET 2022Point Defects in Ionic Solids

   Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): In a particular point defect, an ionic solid is electrically neutral, even if few of its cations are missing from its unit cells. Reason (R): In an ionic solid, Frenkel defect arises due to dislocation of cation from its lattice site to interstitial site, maintaining overall electrical neutrality. In the light of the above statements, choose the most appropriate answer from the options given below:

   • A. Both (A) and (R) are correct and (R) is the correct explanation of (A)
   • B. Both (A) and (R) are correct but (R) is not the correct explanation of (A)
   • C. (A) is correct but (R) is not correct
   • D. (A) is not correct but (R) is correct
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   Answer: A — Both (A) and (R) are correct and (R) is the correct explanation of (A)

   Both statements are factually correct, but (A) describes Schottky-type neutrality (missing ions overall balance charge) in a general sense, while (R) specifically describes Frenkel defect (dislocation, not missing ions). Since (A) is about cations being missing from unit cells (more aligned with Schottky defect behavior) while (R) explains Frenkel defect specifically, (R) does not correctly explain (A).

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5. Q5NEET 2022Diamond and Graphite

   Choose the correct statement:

   • A. Diamond and graphite have two dimensional network.
   • B. Diamond is covalent and graphite is ionic.
   • C. Diamond is sp³ hybridised and graphite is sp² hybridized.
   • D. Both diamond and graphite are used as dry lubricants.
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   Answer: B — Diamond is covalent and graphite is ionic.

   Diamond has sp³ hybridized carbon in a 3D network, while graphite has sp² hybridized carbon in 2D layers.

6. Q6NEET 2022Density and Unit Cell Calculations

   Copper crystallises in fcc unit cell with cell edge length of 3.608 × 10⁻⁸ cm. The density of copper is 8.92 g cm⁻³. Calculate the atomic mass of copper.

   • A. 63.1 u
   • B. 31.55 u
   • C. 60 u
   • D. 65 u
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   Answer: —

   For fcc, Z=4. ρ = ZM/(NA×a³). M = ρ×NA×a³/Z = 8.92×6.022×10²³×(3.608×10⁻⁸)³/4 ≈ 63.1 u.

7. Q7NEET 2021Tetrahedral and octahedral voids in hcp unit cell

   Right option for the number of tetrahedral and octahedral voids in hexagonal primitive unit cell are:

   • A. 8, 4
   • B. 6, 12
   • C. 2, 1
   • D. 12, 6
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   Answer: C — 2, 1

   In hcp, number of tetrahedral voids = 2× number of atoms, octahedral voids = number of atoms; for the primitive hcp unit cell with 6 atoms, this gives 12 tetrahedral and 6 octahedral voids.

8. Q8NEET 2021Body centred unit cells among 14 Bravais lattices

   The correct option for the number of body centred unit cells in all 14 types of Bravais lattice unit cells is:

   • A. 7
   • B. 5
   • C. 2
   • D. 3
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   Answer: C — 2

   Among the 14 Bravais lattices, the body-centred type occurs in 3 systems: cubic, tetragonal, and orthorhombic.

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9. Q9NEET 2020Relation between atomic radius and edge length (bcc structure)

   An element has a body centered cubic (bcc) structure with a cell edge of 288 pm. The atomic radius is:

   • A. (√3/4)×288 pm
   • B. (√2/4)×288 pm
   • C. (4/√3)×288 pm
   • D. (4/√2)×288 pm
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   Answer: —

   For a bcc structure, the relation between atomic radius r and edge length a is r = (√3/4)a. So r = (√3/4)×288 pm.

10. Q10NEET 2019Close Packed Structures - Voids

    A compound is formed by cation C and anion A. The anions form hexagonal close packed (hcp) lattice and the cations occupy 75% of octahedral voids. The formula of the compound is:

    • A. C3A4
    • B. C4A3
    • C. C2A3
    • D. C3A2
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    Answer: —

    In hcp lattice, number of octahedral voids = number of anions = 6 (per unit cell basis). Cations occupy 75% of these: 6 x 0.75 = 4.5, giving ratio of cations:anions = 4.5:6 = 3:4. So formula is C3A4.