Solid State NEET PYQs (2005–2026) with Solutions

10 solved Solid State PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2023)

   NEET 2023HardVoids in Crystal Structure

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

   1. AOption A: 5
   2. BOption B: 4
   3. COption C: 3
   4. DOption D: 2

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   Correct answer: (A) 5

   Explanation

   If B atoms form a close packed structure containing n atoms, then the number of tetrahedral voids is 2n. Since A occupies 1/3 of tetrahedral voids, number of A atoms = 2n/3. Thus, ratio A : B = 2n/3 : n = 2 : 3. Therefore, the formula is A₂B₃ and x + y = 2 + 3 = 5.

2. Question 2 (NEET 2023)

   NEET 2023MediumUnit Cell

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

   1. AOption A: 1/2
   2. BOption B: 1/3
   3. COption C: 1/4
   4. DOption D: 1/12

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   Correct answer: (C) 1/4

   Explanation

   In an fcc unit cell, an edge-centred octahedral void is shared among four adjacent unit cells. Therefore, the fraction of one edge-centred octahedral void belonging to a single unit cell is 1/4.

3. Question 3 (NEET 2022)

   NEET 2022HardPoint 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:

   1. AOption A: (A) is not correct but (R) is correct
   2. BOption B: Both (A) and (R) are correct and (R) is the correct explanation of (A)
   3. COption C: Both (A) and (R) are correct but (R) is not the correct explanation of (A)
   4. DOption D: (A) is correct but (R) is not correct

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   Correct answer: (C) Both (A) and (R) are correct but (R) is not the correct explanation of (A)

   Explanation

   Assertion refers to Schottky defect where missing cations and anions maintain neutrality. Reason correctly describes Frenkel defect. Both are true, but Reason does not explain Assertion. Hence option (C) is correct.

4. Question 4 (NEET 2022)

   NEET 2022HardDensity of Unit Cell

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

   1. AOption A: 65 u
   2. BOption B: 63.1 u
   3. COption C: 31.55 u
   4. DOption D: 60 u

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   Correct answer: (B) 63.1 u

   Explanation

   Using: ρ = ZM/(a³Nₐ) For fcc, Z = 4 Substituting values: M ≈ 63.1 u Hence option (B) is correct.

5. Question 5 (NEET 2021)

   NEET 2021EasyVoids in Crystal Lattice

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

   1. AOption A: 12, 6
   2. BOption B: 8, 4
   3. COption C: 6, 12
   4. DOption D: 2, 1

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   Correct answer: (A) 12, 6

   Explanation

   For N close-packed atoms: - Number of octahedral voids = N - Number of tetrahedral voids = 2N A hexagonal primitive unit cell contains 6 atoms. Therefore: - Octahedral voids = 6 - Tetrahedral voids = 12 Hence the correct answer is 12, 6.

6. Question 6 (NEET 2021)

   NEET 2021EasyBravais Lattices

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

   1. AOption A: 3
   2. BOption B: 7
   3. COption C: 5
   4. DOption D: 2

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   Correct answer: (A) 3

   Explanation

   Body centred unit cells exist in cubic, tetragonal and orthorhombic crystal systems. Therefore total number = 3.

7. Question 7 (NEET 2020)

   NEET 2020EasyBody Centered Cubic Unit Cell

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

   1. AOption A: $\frac{\sqrt{3}}{4} \times 288\ \text{pm}$
   2. BOption B: $\frac{\sqrt{2}}{4} \times 288\ \text{pm}$
   3. COption C: $\frac{4}{\sqrt{3}} \times 288\ \text{pm}$
   4. DOption D: $\frac{4}{\sqrt{2}} \times 288\ \text{pm}$

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   Correct answer: (A) $\frac{\sqrt{3}}{4} \times 288\ \text{pm}$

   Explanation

   For a body centered cubic (bcc) structure: $$4r = \sqrt{3}a$$ where: - $$r$$ = atomic radius - $$a$$ = edge length of the cube Given: $$a = 288\ \text{pm}$$ Therefore: $$r = \frac{\sqrt{3}}{4}a$$ $$r = \frac{\sqrt{3}}{4} \times 288\ \text{pm}$$ Hence, option (A) is correct.

8. Question 8 (NEET 2018)

   NEET 2018HardDensity of crystal structures

   Iron exhibits bcc structure at room temperature. Above 900°C, it transforms to fcc structure. The ratio of density of iron at room temperature to that at 900°C (assuming molar mass and atomic radius of iron remains constant with temperature) is

   1. AOption A: √3 / √2
   2. BOption B: 4√3 / 3√2
   3. COption C: 1 / 2
   4. DOption D: 3√3 / 4√2

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   Correct answer: (D) 3√3 / 4√2

   Explanation

   Density: $$\rho = \frac{ZM}{N_A a^3}$$ For BCC: $$Z=2,\quad a=\frac{4r}{\sqrt3}$$ For FCC: $$Z=4,\quad a=2\sqrt2 r$$ Therefore, $$\frac{\rho_{bcc}}{\rho_{fcc}} = \frac{2/(4r/\sqrt3)^3}{4/(2\sqrt2 r)^3}$$ $$= \frac{3\sqrt3}{4\sqrt2}$$ Hence, option (D) is correct.

9. Question 9 (NEET 2017)

   NEET 2017MediumCrystal Defects and Electrical Properties of Solids

   Which is the incorrect statement?

   1. AOption A: Frenkel defect is favoured in those ionic compounds in which sizes of cation and anions are almost equal
   2. BOption B: FeO₀.₉₈ has non-stoichiometric metal deficiency defect
   3. COption C: Density decreases in case of crystals with Schottky defect
   4. DOption D: NaCl(s) is insulator, silicon is semiconductor, silver is conductor, quartz is piezoelectric crystal

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   Correct answer: (A) Frenkel defect is favoured in those ionic compounds in which sizes of cation and anions are almost equal, (B) FeO₀.₉₈ has non-stoichiometric metal deficiency defect

   Explanation

   Statement (1) is incorrect because Frenkel defect occurs when cations are much smaller than anions. Statement (2) is also incorrect because non-stoichiometric iron oxide usually exists as Fe₀.₉₃–Fe₀.₉₆O due to metal deficiency defects, not specifically Fe₀.₉₈O.

10. Question 10 (NEET 2016)

    NEET 2016MediumCrystal Structures and Coordination Number

    In calcium fluoride, having the fluorite structure, the coordination numbers for calcium ion $(Ca^{2+})$ and fluoride ion $(F^-)$ are:

    1. AOption A: 4 and 2
    2. BOption B: 6 and 6
    3. COption C: 8 and 4
    4. DOption D: 4 and 8

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    Correct answer: (C) 8 and 4

    Explanation

    In fluorite structure of $CaF_2$: - $Ca^{2+}$ ions form FCC arrangement. - $F^-$ ions occupy all tetrahedral voids. Thus: - Coordination number of $Ca^{2+}$ is $8$. - Coordination number of $F^-$ is $4$. Hence option (C) is correct.