Experimentally it was found that a metal oxide has formula `M_(0.98)O`. Metal M, present as `M^(2+) and M^(3+)` in its oxide. Fraction of the metal which exists as `M^(3+)` would be

Options

A. `5.08%`
B. `7.01%`
C. `4.08%`
D. `6.05%`

Answer

C

Explanation

The formula `M_(0.98)O` Shows that if there are 100 O-atoms present as `O^(2-)` ions then there are 98 M atoms present as `M^(2+)` and `M^(3+)` . Suppose `M^(3+)=x` then `M^(2+)=98-x`. As the compoumd as a whole in neutral.
Total charge on `M^(3+)` and
`M^(2+)=` total charge on `100O^(2-)` ions
`x(+3)+(98-x)(+2)=100xx2`
`3x+196-2x=200`
or x=4
`:.` % of `M^(3+)=(4)/(98)xx100=4.08%`