Question

KF has NaCl structure. The edge length of its unit cell has been found to be 537.6 pm. The distance between $K^{+}$ and $F^{-}$ in KF is

• A. 537.6 pm
• B. 1075.2 pm
• C. 268.5 pm
• D. None of these

Answer 1

Answer: (C)

For Nacl type crystals
$a=2\left({\left(\text{r}\right)}_{\text{cation}}\text{+}{\left(\text{r}\right)}_{\text{anion}}\right)$
$\Rightarrow {\text{r}}_{\text{cation}}\text{+}{\text{r}}_{\text{anion}}=\frac{a}{2}=\frac{537}{2}=268.5\text{pm}\text{.}$