Experimentally it was found that a metal oxide has formula M0.98O. Metal M, is present as M2+ and M3+ in its oxide. Fraction of the metal which exists as M3+ would be

oxygen is in the oxidation state of -2, In order for M0.98 to be neutral, its total oxidation state must be +2.Let x represent the fraction of M3+.M2+ fraction will then be (0.98-x).When the compound becomes neutal,3x+2(0.98-x)=23x+1.96-2x=2x=2-1.96x=0.04Therefore, the correct answer is 0.04

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Experimentally it was found that a metal oxide has formula M_0.98O. Metal M, is present as M²⁺ and M³⁺ in its oxide. Fraction of the metal which exists as M³⁺ would be

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answer is 0.04.

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Detailed Solution

oxygen is in the oxidation state of -2, In order for M_0.98 to be neutral, its total oxidation state must be +2.
Let x represent the fraction of M³⁺.
M²⁺ fraction will then be (0.98-x).
When the compound becomes neutal,
$\begin{array}{l} 3 x + 2 (0.98 - x) = 2 \\ 3 x + 1.96 - 2 x = 2 \\ x = 2 - 1.96 \\ x = 0.04 \end{array}$