Sum of first three ionization energies of $\mathrm{Al}$ is $53.0 \mathrm{eV}$ atom $^{-1}$ and the sum of…
Sum of first three ionization energies of $\mathrm{Al}$ is $53.0 \mathrm{eV}$ atom $^{-1}$ and the sum of first two ionization energies of $\mathrm{Na}$ is $52.2 \mathrm{eV}$ atom $^{-1}$. Out of Al(III) and $\mathrm{Na}(\mathrm{II})$
Na (II) is more stable than Al (III)
$\mathrm{Al}(\mathrm{III})$ is more stable than $\mathrm{Na}(\mathrm{II})$
Both are equally stable
Both are equally unstable
Solution
Ionization energy is not the only criteria for the stability of an oxidation state. Al (III) is more stable because it has inert gas configuration i.e. \(1 s^2 2 s^2 2 p^6 3 s^0 3 p^0\), whereas \(N a\) (II) has non-inert gas configuration i.e. \(1 s^2 2 s^2 2 p^5\) and hence \(A l\) (III) is more stable than \(N a\) (II).