Question

A solid compound $XY$ has $NaCl$ structure. If the radius of the cation $\left(X^{+}\right)$ is $100$ pm, the radius (in pm) of the anion $\left(Y^{-}\right)$ will be:

• A. $275.1$
• B. $322.5$
• C. $241.5$
• D. $165.7$

Answer 1

The correct option is C $241.5$

Radius ratio of $NaCl$ like crystal $=\frac{r^{+}}{r^{-}}=0.414$

Substituting the values of the radius of the cation in this expression, we get

$r^{-}=\frac{100}{0.414}=241.5$ pm

Thus, the radius of the anion $\left(Y^{-}\right)$ will be $241.5$ pm.