Question -
Answer -
The formula of nickeloxide is Ni0.98O1.00.
Therefore, the ratio ofthe number of Ni atoms to the number of O atoms,
Ni : O = 0.98 : 1.00 = 98 : 100
Now, total charge on 100O2− ions = 100 × (−2)
= −200
Let the number of Ni2+ ions be x.
So, the number of Ni3+ ions is 98 − x.
Now, total charge on Ni2+ ions = x(+2)
= +2x
And, total charge on Ni3+ ions = (98 − x)(+3)
= 294 − 3x
Since, the compound isneutral, we can write:
2x + (294 −3x) + (−200) = 0
⇒ −x + 94 = 0
⇒ x = 94
Therefore, number of Ni2+ ions = 94
And, number of Ni3+ ions = 98 − 94 = 4
Hence, fraction of nickelthat exists as Ni2+
= 0.959
And, fraction of nickelthat exists as 
= 0.041
Alternatively, fractionof nickel that exists as Ni3+ = 1 − 0.959
= 0.041