In rock salt (4 formula units per cell) the Na+ and Cl- ions touch along the cell edge. With r(Na+)=100 pm, r(Cl-)=181 pm and pi=3.14, find the percentage of the unit cell volume occupied by the ions (nearest integer).

JEE Advanced levelNumerical answerDifficulty 3 of 3ss037

In rock salt (4 formula units per cell) the $Na X^{+}$ and $Cl X^{-}$ ions touch along the cell edge. With $r (Na X^{+}) = 100 pm$ , $r (Cl X^{-}) = 181 pm$ and $π = 3.14$ , find the percentage of the unit cell volume occupied by the ions (nearest integer).

Numerical answer type, answer in %, tolerance 0.5.

Edge

a = 2 (100 + 181) = 562 pm

. Ion volume per cell

= 4 \times \frac{4}{3} π (10 0^{3} + 18 1^{3}) = 4 \times \frac{4}{3} \times 3.14 \times 6929741 = 1.1605 e + 08 p m^{3}

. Cell volume

= 56 2^{3} = 1.7750 e + 08 p m^{3}

. Fraction

= 65.38% \approx 65%