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

Options

A. 0.9592
B. 0.0408
C. 0.0605
D. 0.9395

Explanation

The formula `M_(0).98^(O)` shows that
M : O= 0.98 : 1 = 98 : 100
Then if there are 100 , ` O^(2-)` atoms then M atoms = 98
Charge on 100 `O^(2-) "ions "= 100 xx (–2) = –200 `
Suppose M atoms as `M^(2+)` = x
Then Ni atoms as `M^(3+)` = 98-x
Total charge of `M^(2+)& M^(3+) = (+2)x + (+3) ( 98-x) = 2x+294 - 3x==294 -x`
`M_(O.98) ` O is neutral,
294 -x = 200 of x = 294 -200 = 94
Magnitude of charge = |-200 | = 200
% of as `M^(2+) = (94)/(98)xx100 = 95.92 % ` and % of M as `M^(3+) = 100 - 95.92= 4.08 % `