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. 0.0701
B. 0.0408
C. 0.0605
D. 0.0508

Explanation

The formula M0.98O 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 H atoms as `M^(3) = 98 - `x
Total charge of `M^(2+) & M^(3+) = (+2) x `+ (+3) (98 – x) = 2x + 294 – 3x = 294 – x MO.`98^(O)` is neutral,
294 - x = 200 `rArr ` x = 294 - 200 =94
Magnitude of charge = | - 200 | = 200
% of M as `M^(2+) = (94)/(98) xx `100 = 95.92% and % of M as `M^(3+)` = 100 - 95.92 = 4.08 %