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The unit cell edge length of NaF crystal...

The unit cell edge length of NaF crystal is `4.634Å`. If the ionic radius of `Na^(+)` ion is 95 pm, what is the ionic radius of `F^(-)` ion, assuming that anion-anion contact and face centred cubic lattice?

Text Solution

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Forn the face centred cubic lattice,
`r(Na^(+))+r(F^(-))=(a)/(2)=(4.634Å)/(2)=2.317Å`
`=(2.317xx100)"pm"=231.7"pm"`
`r(F^(-))=231.7-r(Na^(+))=231.7-95=136.7"pm"`
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