The formula Ni098 O1.00 shows that Ni : O = 0.98 : 1.00 = 98 : 100.
Thus, if there are 100 O-atoms, then Ni atoms = 98 charge on 100 O2′ ions = 100 x (-2) = - 200 suppose Ni atoms as Ni2+ = x
Then Ni atoms as Ni3+ = 98 - x
Total charge on Ni2+ and Ni3+
= (+2) x + (+3) (98 - x) = 294 - x
As metal oxide is neutral, total charge on cations = total charge on anions
294 - x = 200
x = 94
∴ % of Ni as Ni2+ = gx100 98
= 96%
% of Ni as Ni3+ = 1000 - 96 = 4.