Analysis shows that nickel oxide has the formula `Ni_(0.98)O_(1.00)`. What fractions of nickel exist as `Ni^(2+)andNi^(3+)` ions.

Explanation

Suppose number of Ni atoms as `Ni^(2+)` ions per formula unit = x
`therefore" "` Number of Ni atoms as `Ni^(3+)` ions per formula unit = 0.98-x
Total charge of `Ni^(2+)andNi^(3+)` ions per formula unit
`=x xx(+2)+(0.98-x)(+3)`
Total charge of `O^(2-)` ions per formula unit = (-2)`xx`1 = -2
Total cationic charge + total anionic charge = 0
`therefore" "+2x+2.94-3x-2=0or,x=0.94`
`therefore" "` % of Ni atoms as `Ni^(2+)` ions = `0.94/0.98xx100=95.9%` and % of Ni atom as `Ni^(3+)` ions = `((0.98-0.94)/0.98)xx100=4.1%`