Question

The radius of $N{a}^{+}$ ions is $95$ pm and that of $C{l}^{-}$ ions is $181$pm. Predict whether the Co-ordination number of $N{a}^{+}$ ion is $6$ or $4$.

Answer 1

Given:-

Radius of ${Na}^{+}$ ion $\left(r_{{Na}^{+}}\right)=95pm$

Radius of ${Cl}^{-}$ ion $\left(r_{{Cl}^{-}}\right)=181pm$

Therefore,

$\frac{r_{{Na}^{+}}}{r_{{Cl}^{-}}}=\frac{95}{181}=0.525$

$\therefore$ Coordination no. $=6(\because 0.414<\frac{r^{+}}{r^{-}}<0.732)$

Hence the coordination no. of ${Na}^{+}$ ion is $6$.