The correct order of stability of $A l^{+}, A l^{+ 2}, A l^{+ 3}$ is:

A l^{+ 3} > A l^{+ 2} > A l^{+}

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A l^{+ 2} > A l^{+ 3} > A l^{+}

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A l^{+ 2} < A l^{+} > A l^{+ 3}

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A l^{+ 3} > A l^{+} > A l^{+ 2}

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Solution

The correct option is D $A l^{+ 3} > A l^{+} > A l^{+ 2}$
The electronic configuration of Al is:
$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{1}$

So, it tends to donate 3 valence electrons in order to achieve a stable configuration. Hence, it is most stable.

$A l^{+}$ contains completely filled $3 s$ orbital. Hence it is more stable than $A l^{+ 2}$ in which $3 s$ orbital is half-filled.

The correct order of stability is $A l^{+ 3} > A l^{+} > A l^{+ 2}$ .

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Similar questions

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