A metal exists as an oxide with formula $M_{0.96} \mathrm{O}$. Metal, $M$ can exist as $M^{2+}$ and $M^{3+}$ in its oxide $M_{0.96} \mathrm{O}$. The percentage of $M^{3+}$ in the oxide is nearly

We are given metal oxide with formula $M_{0.96} \mathrm{O}$. Here, $M$ can exist as $M^{2+}$ as well as $M^{3+}$. Here, for every 100 oxide ions there are $96 M$ ions. Let out of this $96 M$ atoms, $M^{2+}$ exists $=x$ and $M^{3+}$ will be $=(96-x)$. Total positive charge on $96 M$-atoms $=2 \times x+3 \times(96-x)$ Total negative charge on $100 \mathrm{O}$-atoms $=100 \times 2$ $=200$ As per electrical neutrality concept $\begin{aligned} 2 x+3(96-x) &=200 \\ 2 x+288-3 x &=200 \\ -x &=200-288 \\ -x &=-88 \text { or } x=88 \\ 96-88 &=8 \end{aligned}$ So, the ratio of $M^{3+}$ will be $\frac{8}{96} \times 100=8.3 \%$