Question

Which is the correct order for ionic radii?

• A. $L{i}^{+}<A{l}^{3+}<M{g}^{2+}<K^{+}$
• B. $A{l}^{3+}<M{g}^{2+}<K^{+}<L{i}^{+}$
• C. $M{g}^{2+}<A{l}^{3+}<K^{+}<L{i}^{+}$
• D. None of the above

Answer 1

The correct option is A

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

$L{i}^{+}\Rightarrow 1s^2$

$A{l}^{3+}\Rightarrow 1s^{2}2s^{2}2p^6$

$M{g}^{2+}\Rightarrow 1s^{2}2s^{2}2p^6$

$K^{+}\Rightarrow 1s^{2}2s^{2}2p^{6}3s^{2}3p^6$

Principal quantum number ($n$) is maximum in $K^{+}$, therefore it is maximum in size.

Down the group, atomic radius increases because the principal quantum number ($n$) increases. And the valence electrons are farther from the nucleus.

$A{l}^{3+}$ and $M{g}^{2+}$ are isoelectronic.

$\therefore$ $A{l}^{3+}<M{g}^{2+}$

Higher the ionic charge on the atom, more will be the ionic radius. Now, in option (a), the correct order will be $Mg2+>Li+>Be2+$

Because $Mg2+$ has principal quantum number ($n$) = $3$ and $B{e}^{2+}$ gas $n$ = $2$

Among $Li2+$ and $Be2+$, $L{i}^{+}>B{e}^{2+}$ because along a period from left to right, atomic radius increases.

Now, in option (b)

$H^{+}<L{i}^{+}<H^{-}$

For $H^{+}$, there are no electrons, so no atomic radius. In the case of $L{i}^{+}$ and $H^{-}$, $L{i}^{+}<H^{-}$, because $n$ is least in $L{i}^{+}$, i.e., $n$ = $1$.

Therefore, it is the smallest in size. So, the correct order is $L{i}^{+}<A{l}^{3+}<M{g}^{2+}<K^{+}$