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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)

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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

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Correct answer: (b) Al (III) is more stable than Na (II)

Explanation:

Ionization energy is not the only the criteria for the stability of an oxidation state.

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