Q.KF crystallizes in fcc structure like sodium chloride. Calculate the distance between $K^+$ and $F^-$ in KF. (Given: density of KF is $2.48\ \text{g cm}^{-3}$)

Step 1. KF has the NaCl-type (rock-salt) structure, so $z=4$ formula units per unit cell, and $M = M(K) + M(F) = 39 + 19 = 58\ \text{g mol}^{-1}$.

Step 2. From $\rho = \dfrac{zM}{a^3 N_A}$, solve for $a^3$: $a^3 = \dfrac{zM}{\rho N_A} = \dfrac{4\times58}{2.48\times6.023\times10^{23}} = \dfrac{232}{1.494\times10^{24}} = 1.553\times10^{-22}\ \text{cm}^3$.

Step 3. Taking the cube root: $a = \left(1.553\times10^{-22}\right)^{1/3} \approx 5.376\times10^{-8}\ \text{cm} = 537.6$ pm. …

Find the edge length a from the density formula (z=4, M=58 for KF), then halve it to get the nearest-neighbour c …

• Forgetting that in the NaCl-type structure the ion-ion distance is a/2, NOT the full edge length a -- the cation and anion sub-lattices interpenetrate so t …