The correct order of radii is

N < Be < B

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F^{-} < O^{2 -} < N^{3 -}

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Na < Li < K

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F e^{3 +} < F e^{2 +} < F e^{4 +}

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Solution

The correct option is B $F^{-} < O^{2 -} < N^{3 -}$
Among isolectronic species, greater the negative charge, greater the ionic size, hence $F^{-} < O^{2 -} < N^{3 -}$

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Similar questions

The correct order of ionic radii of the species $N^{3 -}, O^{2 -}, N a^{+}$ and $F^{-}$ is:

N^{3 -}

O^{2 -}

N^{3 -}

N a^{+}

F^{-}

M g^{2 +}

O^{2 -}

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

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