1.

Sum of first three ionization energies of Al is \[53.0\text{ }eV\text{ }ato{{m}^{-1}}\] and the sum of first two ionization energies of Na is \[52.2\text{ }eV\text{ }ato{{m}^{-1}}.\] 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» C. Both are equally stable


Discussion

No Comment Found

Related MCQs