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 $4$ .

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Solution

Given:-
Radius of ${N a}^{+}$ ion $(r_{{N a}^{+}}) = 95 p m$
Radius of ${C l}^{-}$ ion $(r_{{C l}^{-}}) = 181 p m$
Therefore,
$\frac{r_{{N a}^{+}}}{r_{{C l}^{-}}} = \frac{95}{181} = 0.525$
$∴$ Coordination no. $= 6 (∵ 0.414 < \frac{r^{+}}{r^{-}} < 0.732)$
Hence the coordination no. of ${N a}^{+}$ ion is $6$ .

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