Question

The sums of first and second ionization energies and those of third and fourth ionization energies (in MJ mol${}^{-1}$) of nickel and platinum are :

	$(IE{)}_1+(IE{)}_2$	$(IE{)}_3+(IE{)}_4$
Ni	2.49	8.80
Pt	2.66	6.70

Based on this information, the most common oxidation states of $Ni$ and $Pt$ are respectively:

• A. $+2,+2$
• B. $+4,+2$
• C. $+2,+4$
• D. $+4,+4$

Answer 1

The correct option is C $+2,+4$

For $Ni$, $(IE{)}_1+(IE{)}_2<<(IE{)}_3+(IE{)}_4$.

Thus, the difference is very large.

Hence, the stable oxidation of Ni is $+2$.

For $Pt$ $(IE{)}_1+(IE{)}_2<(IE{)}_3+(IE{)}_4$.

Thus, the difference is small.

Hence, the stable oxidation of $Ni$ is $+4.$