The density of diamond is at . The carbon-carbon bond distance (in pm) in diamond is — Solid State Chemistry Question

💡 Solution & Explanation

Diamond has 8 atoms per unit cell. $d = \frac{8 \times 12}{N_A \times a^3} = 2\sqrt{3} \implies a^3 = \frac{48}{6.0 \times 10^{23} \times \sqrt{3}} = \frac{80}{\sqrt{3}} \times 10^{-24} \text{ cm}^3$. The bond distance $l = \frac{\sqrt{3}a}{4} \implies a = \frac{4l}{\sqrt{3}}$. Thus, $a^3 = \frac{64l^3}{3\sqrt{3}}$. Equating volumes: $\frac{64l^3}{3\sqrt{3}} = \frac{80}{\sqrt{3}} \times 10^{-24} \implies l^3 = \frac{240}{64} \times 10^{-24} = 3.75 \times 10^{-24}$. Given $(1.55)^3 = 3.75$, $l = 1.55 \times 10^{-8} \text{ cm} = 155 \text{ pm}$. Therefore, correct answer is 0155.