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 were 100 O-atoms present as `O^(2-)` ions, then there 98 M atoms present as `M^(2+)` and `M^(3+)` . Suppose `M^(3+) =x` , then `M^(2+)` =98-x . As the compound as a whole is neutral, Total charge on `M^(3+)` and `M^(2+)` =Total charge on 100 `O^(2-)` ions
x (+3) + (98-x) (+2)=100 x 2 or 3x + 196 - 2x =200 or x=4
`therefore` % of `M^(3+)=4/98xx100`=4.08 %