If the interionic distance between Na^{+} and | vedclass

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(A) The interionic distance $d$ is the sum of the ionic radii: $d = r_{Na^{+}} + r_{F^{-}} = 2.31 \ \mathring{A}$.Since $F^{-}$ is an anion and $Na^{+

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$0.95 \ \mathring{A}$ and $1.36 \ \mathring{A}$

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(A) The interionic distance $d$ is the sum of the ionic radii: $d = r_{Na^{+}} + r_{F^{-}} = 2.31 \ \mathring{A}$.Since $F^{-}$ is an anion and $Na^{+}$ is a cation,the radius of the anion is larger than the radius of the cation $(r_{F^{-}} > r_{Na^{+}})$.Based on periodic trends and experimental values,$r_{Na^{+}} \approx 0.95 \ \mathring{A}$ and $r_{F^{-}} \approx 1.36 \ \mathring{A}$.Summing these: $0.95 \ \mathring{A} + 1.36 \ \mathring{A} = 2.31 \ \mathring{A}$.

If the interionic distance between $Na^{+}$ and $F^{-}$ ions is $2.31 \ \mathring{A}$,then the radii of $Na^{+}$ and $F^{-}$ are:

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