Question

Sum of first three ionization energies of $Al$ is $53.0 \,eV$ $atom ^{-1}$ and the sum of first two ionization energies of $Na$ is $52.2 \,eV$ $atom^{-1}$.. 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. $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{0} 3 p^{0}$, whereas $Na ( II )$ has non-inert gas configuration i.e. $1 s^{2} 2 s^{2} 2 p^{5}$ and hence $Al ($ III $)$ is more stable than $Na ( II )$.