For there action, M^x+ + MnO₄^– → MO₃^– + Mn²⁺ + 1/2O₂, if one mole of MnO₄^– oxidises 1.67 moles of M^x+ to MO₃^–, then the value of 'x' in the reaction is

$$\begin{gathered} {\mathrm{M}}^{\mathrm{x}+} + \stackrel{+7}{\mathrm{Mn}}{\mathrm{O}}_4^{-}\to \stackrel{+5}{\mathrm{M}}{\mathrm{O}}_3^{-}+ {\mathrm{Mn}}^{2+} + 1/2 {\mathrm{O}}_2 \\ 1.67\mathrm{moles} 1\mathrm{mole} \end{gathered}$$

By cross multiply the change in oxidation states of reactants

$$\begin{gathered} {\mathrm{M}}^{\mathrm{x}+} + \stackrel{+7}{\mathrm{Mn}}{\mathrm{O}}_4^{-}\to \stackrel{+5}{\mathrm{M}}{\mathrm{O}}_3^{-}+ {\mathrm{Mn}}^{2+} + 1/2 {\mathrm{O}}_2 \\ 5\mathrm{moles} ? \end{gathered}$$

No. of moles of ${\mathrm{MnO}}_4^{-}$ = Change in oxidation state of M=$\frac{5}{1.67}\approx 3$

Hence, x=2