98 Ni atoms are associated with 100 O atoms.
Out of 98 Ni atoms, suppose Ni present as Ni2+ = x
Then Ni present as Ni3+ = 98 – x
Total charge on x Ni2+ and (98 – x) Ni3+ should be equal to total charge on 100 O2– ions.
Hence, 2 × x + 3 × (98 – x) = 100 × 2 or 2x + 294 – 3x = 200
So, x = 94
Fraction of Ni present as Ni2+ = 94 / 98 x 100 = 96