The correct order of ionic radius is:

N a^{+} > F^{- 1} > M g^{2 +} > O^{2 -} > A l^{3 +}

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O^{2 -} > F^{- 1} > N a^{+} > M g^{2 +} > A l^{3 +}

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A l^{3 +} > M g^{2 +} > N a^{+} > F^{-} > O^{2 -}

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N a^{+} > M g^{2 +} > A l^{3 +} > O^{2 -} > F^{-}

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Solution

The correct option is B $O^{2 -} > F^{- 1} > N a^{+} > M g^{2 +} > A l^{3 +}$
Higher the cationic charge smaller the radius, higher the anionic charge higher the radius for isoelectronic species.

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Similar questions

i) N e i i) O^{2 -} i i i) M g^{2 +} i v) F

v) A l^{3 +} v i) C l^{-} v i i) K v i i i) N a

O^{2 -}, F^{-}, N a^{+}, M g^{2 +}

A l^{3 +}

O^{2 -}, F^{-}, N a^{+}, M g^{2 +}

A l^{3 +}

O^{2 -}, F^{-}, N a^{+}, M g^{2 +}

A l^{3 +}

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