Question

The ionic radii of $A^{+}$ and $B^{-}$ ions are $0.98\times {10}^{-10}$ m and $1.81\times {10}^{-10}$ m. The coordination number of each ion in $AB$ is

• A. $8$
• B. $2$
• C. $6$
• D. $4$

Answer 1

The correct option is C $6$

Radius ratio, $\frac{r_{+}}{r_{-}}=\frac{0.98\times {10}^{-10}}{1.81\times {10}^{-10}}=0.541$
It lies in the range of $0.414$ to $0.732$. Hence, coordination number of each ion will be $6$ as the compound will have $NaCl$ type structure i.e. octahedral arrangement.