Question

Sum of first three ionization energies of $Al$ is $53.0eV$ $ato{m}^{-1}$ and the sum of first two ionization energies of $Na$ is $52.2eV$ $ato{m}^{-1}$.. Out of Al(III) and Na(II)

• A. Na (II) is more stable than Al (III)
• B. Al (III) is more stable than Na (II)
• C. Both are equally stable
• D. Both are equally unstable

Answer 1

Answer: (B)

lonization 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. $1s^{2}2s^{2}2p^{6}3s^{0}3p^0$, whereas $Na(II)$ has non-inert gas configuration i.e. $1s^{2}2s^{2}2p^5$ and hence $Al($ III $)$ is more stable than $Na(II)$.