Experiment shows that Nickel oxide has the formula `Ni_(0.96) O_(1.00)`.What fraction of Nickel exists as of `Ni^(2+)` and `Ni^(3+)` ions ?

Explanation

Formula is `Ni_(0.96)O_(1.00)`.
So the ration of Ni=O = 96.00
So if there are 100 atom of oxygen, as atoms of Ni
Let the number of atoms of `Ni^(+2) = x`
The number of atoms of `Ni^(+3) =96-x`
Charge on Ni = charge on O
So that oxygen has charge = 2
`3(96 -x) + 2x =2 ( 100)`
` 288 -3x + 2x = 200`
`-x =-88`
`x =88`
Percentage of `Ni^(+2) =` ( atom of `Ni^(+2) //` total number of atoms of Ni 100. )
`= 100 . ( 88//96)`
= 91.6%
Percentage of `Ni^(+3) = 100- Ni^(+2)`
`= 100 - 91.6`
=8.4%