Given formula of nickel oxide = Ni_0.98O_1.00

Therefore, Ni:O = 0.98:1.00

Total charge on O^-2 ion = 100 × (-2) = -200

Now, let the number of Ni²⁺ ions = x

Therefore, number of Ni⁺³ = 98- x

Now, since compound is neutral

Therefore, no. of Ni²⁺ ions + no. of Ni³⁺ ions +no. of O^-2 ions = 0

x (-2) + (98 - x) × (-3) + (-200) = 0

-2x – 294 +3x – 200 =0

x-94 = 0

x = 94

Therefore, the number of Ni²⁺ ions = 94

Number of ni³⁺ ions = 98- 94 = 4

Now, fraction of nickel exists as Ni²⁺ = 0.959

Fraction of nickel exists as Ni ³⁺ ions = 0.041