Analysis shows that nickel oxide has the formula `Ni_(0.96)` O. What fractions of the nickel exist as `Ni^(2+)` and `Ni^(3+)`.

Explanation

98 Ni atom are associated with 100 O atom. Out of 98 Ni atoms, suppose Ni present as `Ni^(2+) =x`
Then Ni present as `Ni^(3+)= 98-x`
Then Ni present as `Ni^(3+) = 98-x`
Total change on x `Ni^(2+)` and `(98-x) Ni^(3+)` should be equal to total change on `100 O^(2-)` Hence. `2 xx x + 3 xx (98-x) = 100 xx 2`
Or, `2x + 294-3x = 200`
`therefore x = 94`
`therefore` Fraction of Ni present as `Ni^(2+) = 94/98 xx 10 = 96%`
Fraction of `Ni^(3+)` present as `Ni^(3+) = 4/98 xx 100 = 4%`