Extra Questions: Class 12 Chemistry Chapter 1 The Solid State

Genuinely harder, HOTS-level practice for Class 12 Chemistry Chapter 1 (The Solid State), going beyond the standard exercise. These Class 12 Chemistry Chapter 1 important questions are handy for last-minute exam practice.

1. Q1 (Multi-step density problem). A metal crystallises in a bcc lattice with a unit cell edge of 300 pm. If the density of the metal is 5.96 g/cm³, calculate its atomic mass, and identify the coordination number and packing efficiency of this structure.
   Solution: Z=2 for bcc. M = d·a³·N_A/Z = 5.96×(300×10⁻¹₀)³×6.022×10²³/2 ≈ 48.4 g/mol (close to titanium, 47.9 g/mol). Coordination number for bcc = 8; packing efficiency = 68%.
2. Q2 (Assertion-Reason). Assertion (A): Frenkel defects do not change the density of an ionic solid. Reason (R): In a Frenkel defect, no ion actually leaves the crystal — it only shifts to an interstitial site.
   Solution: Both A and R are true, and R correctly explains A — since the displaced ion stays within the same crystal (just at a different site), the total mass and volume of the crystal are unchanged, so density is unaffected. Answer: Both true, R is the correct explanation of A.
3. Q3 (Conceptual synthesis). Explain why ZnS shows Frenkel defects but NaCl shows Schottky defects, connecting your answer to the radius ratio of the two ions in each compound.
   Solution: Frenkel defects occur when there is a large size difference between the cation and anion, allowing the smaller ion to fit into an interstitial space; Schottky defects occur when the two ions are of comparable size and high coordination number, so no ion is small enough to occupy an interstitial gap without severe strain. In ZnS, Zn²⁺ (74 pm) is much smaller than S²⁻ (184 pm) — a large radius-ratio mismatch that favours Frenkel defects. In NaCl, Na⁺ (102 pm) and Cl⁻ (181 pm) are closer in size with a stable 6:6 coordination, favouring Schottky defects instead.
4. Q4 (Application). A compound forms hcp packing of anions with cations occupying all the tetrahedral voids. What is the simplest formula of the compound, and give one real example.
   Solution: In close packing, the number of tetrahedral voids is twice the number of packing spheres. If all tetrahedral voids are filled and there are n anions, there are 2n cations, giving the ratio cation:anion = 2:1, formula M₂X. A real example following this pattern is Na₂O (antifluorite structure).

More on This Chapter

NCERT Solutions | Revision Notes | Class 12 Chemistry Book

Frequently Asked Questions

Are these questions relevant if my board doesn’t examine this chapter?
Yes — these test the underlying concepts (defects, packing, density calculations) that reappear in later physical chemistry chapters and in competitive exams.

Written by Satish

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