The correct order of increasing size is:

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

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

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

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

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Solution

The correct option is D $A l^{+ 3} < M g^{+ 2} < N a^{+} < F^{-}$
The isoelectronic species radius increases with lower atomic charge. Thus correct order is
${A l}^{3 +} < {M g}^{2 +} < N a < F^{-}$

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Similar questions

$N a^{+} . M g^{2 +}, A l^{3 +}$ and $S i^{4 +}$ are isoelectronic. The order of their ionic size is

N a^{+}

M g^{2 +}

A l^{3 +}

S i^{4 +}

N^3- , O^2- , F^- , Mg^2+ , Na^+ , Al^3+ Arrange in the increasing order of ionic radii.

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

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

N^{3 -}, O^{2 -}, F^{-}, N e

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

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