Is the given order correct according to the ionic radii?
${A l}^{3 +} < {M g}^{2 +} < N a^{+} < F^{-}$

True

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False

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Solution

The correct option is A True
All these ions have 10 electrons in their shell, therefore, these are $i s o - e l e c t r o n i c$ species.

The more the positive charge, the smaller is the ionic radius.

Therefore, the order of ionic radii will be:

$A l^{3 +} < M g^{2 +} < N a^{+} < F^{-}$ (increasing order)

Hence, it is the correct order.

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Similar questions

K^{+}, N a^{+}, A l^{3 +} and M g^{2 +}

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

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

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

A l^{3 +}

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