98Ni atoms are associated with 100 O-atoms
Out of 98Ni atoms, suppose Ni present as Ni2+=x
Then Ni present as Ni3+=98−x
Total charge on xNi2+ and (98−x)Ni3+ + should be equal to charge on 100O2− ions
Hence, x×2+(98−x)×3=100×2
or 2x+294−3x=200
or x=94 ∴ Fraction of Ni present as Ni2+=9894×100=96%
Fraction of Ni present as Ni3+=984×100=4%