Question

A binary solid $\left(A^{+}{B}^{-}\right)$ has a rock salt structure. If the edge length is $400$ pm and radius of cation $\left(A^{+}\right)$ is 75 pm, the radius (in pm) of anion $\left(B^{-}\right)$ will be:

• A. $100$
• B. $125$
• C. $250$
• D. $325$

Answer 1

Answer: (B)

$125$

Given that the binary solid $\left(A^{+}{B}^{-}\right)$ has rock salt structure (NaCl type structure).
Distance between $N{a}^{+}$ and $C{l}^{-}=\frac{1}{2}\times \text{edge length}=\frac{1}{2}\times 400=200$ pm
As radius of cation $=75$ pm, radius of anion $=200-75=125$ pm