Arrange the following ions in order of their increasing radii.
$L i^{+}, M g^{2 +}, K^{+}, A l^{3 +}$ :

A l^{3 +} < M g^{2 +} < L i^{+} < K^{+}

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M g^{2 +} < A l^{3 +} < L i^{+} < K^{+}

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A l^{3 +} < M g^{2 +} < K^{+} < L i^{+}

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None of these

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Solution

The correct option is A $A l^{3 +} < M g^{2 +} < L i^{+} < K^{+}$
$A l^{3 +} < M g^{2 +} < L i^{+} < K^{+}$
$A l^{3 +}$ and $M g^{2 +}$ are iso-electronic ions.
In iso-electronic species, as the number of protons (atomic number) goes on increasing, size goes on decreasing due to stronger attraction on the electrons : $K^{+}$ is larger than $L i^{+}$ because on moving down the group, size decreases.

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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 +}

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

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

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

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

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