Analysis shows that nickel oxide has the formula `Ni_(0.00)`. What fractions of nickel exist as `Ni^(2+)` and `Ni^(3+)` ions?

Explanation

The formula `Ni_(0).98^(O)` shows that
Ni : O = 0.98 : 1 = 98 : 100
Then, if there are 100 , `O^(2-)` atoms , then Ni atoms = 98
Charge on 100 `O^(-2)` ions = `100 xx (–2) = –200`
Suppose Ni atoms as `Ni^(2+)` = x
Then Ni atoms as `Ni^(3)+ = 98 – x`
Total charge of `Ni^(2+) & Ni^(3+)` = (+2) x + (+3) (98 – x) = 2x + 294 – 3x = 294 – x `Ni_(O.98)` O is neutral,
294 – x = 200 of x = 294 – 200 = 94
Magnitude of charge = | –200 | = 200
% of Ni as `Ni^(2+)= (94)/(98)xx100 = 95.92 %` and % Ni as `Ni^(3+)` = 100 - 95.92 = 4.08 %