Sum of first three ionisation energies of Al is 53-03 eV atom^(-1) and the sum of first two ionisation energies of Na is 52 20 eV atom^(-1) Out of Al(II) and Na(II)

Sum of first three ionisation energies of Al is 53-03 eV atom^(-1) and

1.	Sum of first three ionisation energies of Al is 53-03 eV atom^(-1) and the sum of first two ionisation energies of Na is 52 20 eV atom^(-1) Out of Al(II) and Na(II)
Answer» NA(II) is more stable than AL(III) Al(III) is more stable than Na(II) Both are equally unstable Both are equally stable Solution :I.E. is not only criteria for the stability of an OXIDATION state.