# NEET Chemistry The Solid State Questions Solved

Compute the percentage void space per unit volume of unit cell in zinc fluoride structure.

(A) 15.03%

(B) 22.18%

(C) 18.23%

(D) 25.07%

(D). Anions occupy fcc positions and half of the tetrahedral holes are occupied by cations. Since there are four anions and 8 tetrahedral holes per unit cell, the fraction of volume occupied by spheres per unit volume of the unit cell is

$=\frac{4×\left(\frac{4}{3}{\mathrm{\pi r}}_{\mathrm{a}}^{3}\right)+\frac{1}{2}×8×\left(\frac{4}{3}{\mathrm{\pi r}}_{c}^{3}\right)}{16\sqrt{2}{r}_{a}^{3}}=\frac{\mathrm{\pi }}{3\sqrt{2}}\left\{1+{\left(\frac{{r}_{c}}{{r}_{a}}\right)}^{3}\right\}$

for tetrahedral holes, $\frac{{r}_{c}}{{r}_{a}}=0.225$

$=\frac{\mathrm{\pi }}{3\sqrt{2}}\left\{1+{\left(0.225\right)}^{3}\right\}=0.7493$

$\therefore$  Void volume = 1 – 0.7493 = 0.2507/unit volume of unit cell.
% void space = 25.07%

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