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 \(\frac{8}{96} \times 100=8.3 \%\)