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Calculate the coordination number of \(\mathrm{Na}^{+}\)in \(\mathrm{NaCl}\) crystal, given radius of \(\mathrm{Na}^{+}\)and \(\mathrm{Cl}^{-}\)are \(95 \mathrm{pm}\) and \(181 \mathrm{pm}\) respectively.
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6
For \(\mathrm{NaCl}\) crystal
Radius ratio, \(\frac{r_{\mathrm{Na}^{+}}}{r_{\mathrm{Cl}^{-}}}=\frac{95}{181}=0.5248\), which is, in the range of \(0.414-0.732\).
So, coordination number of \(\mathrm{Na}^{+}\)in \(\mathrm{NaCl}\) crystal \(=6\)
Radius ratio, \(\frac{r_{\mathrm{Na}^{+}}}{r_{\mathrm{Cl}^{-}}}=\frac{95}{181}=0.5248\), which is, in the range of \(0.414-0.732\).
So, coordination number of \(\mathrm{Na}^{+}\)in \(\mathrm{NaCl}\) crystal \(=6\)
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