A transition metal M can exists in two oxidation states +2 and +3. It forms an oxide whose experimental formula is given by ${\mathrm{M}}_{\mathrm{x}}\mathrm{O}$ where $x<1$. Then the ratio of metal ions in +3 state to those in +2 state is given by :

Given that the metal oxide's experimental formula is ${\mathrm{M}}_{\mathrm{x}}\mathrm{O}$.

Two oxidation states of the transition metal $\mathrm{M}$are possible.

$\begin{array}{ccc} & M^{2+}& M^{3+}\\ \text{ Let }& y& (x-y)\end{array}$

with the use of charge balancing.

$\begin{array}{r}2y+3(x-y)-2=0\\ \Rightarrow -y+3x-2=0\\ \Rightarrow y=3x-2\end{array}$

Metal ion to metal ratio,

$$\begin{gathered} \left(\frac{M^{3+}}{M^{2+}}\right)=\frac{x-y}{y} \\ =\frac{x-3x+2}{3x-2} \\ =\frac{2(1-x)}{3x-2} \end{gathered}$$