The formula of nickel oxide is Ni_0.98O_1.00.

Therefore, the ratio of the number of Ni atoms to the number of O atoms, Ni : O = 0.98 : 1.00 = 98 : 100

Now, total charge on 100 O^2- 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 Ni²⁺ ions = (98 - x)(+3)

= 294 - 3x

Since, the compound is neutral, 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 nickel that exists as Ni²⁺ = \(\frac{94}{98}\)

= 0.959

And, fraction of nickel that exists as Ni³⁺ =  \(\frac{4}{98}\)

= 0.041

Alternatively, fraction of nickel that exists as Ni³⁺ = 1 - 0.959

= 0.041