For the reaction, $M^{x+}+{MnO}_4{}^{-}⟶{MO}_3{}^{-}+{Mn}^{2+}+1/2O_2$, if one mole of ${MnO}_4^{-}$oxidizes $1.67$ moles of $M^{x+}$ to ${MO}_3^{-}$, then the value of ' $x$ ' in the reaction is

Let us assume oxidation state of $M$ in $M^{x+}$ is $x$

The oxidation state of $M$ in ${MO}_3{ }^{-}$is $+5$. The net change in electrons/oxidation state is $5-x$

$M^{x+}⟶{MO}_3{}^{-}1$ mole conversion involves $5-x$ moles of electrons

The oxidation state of $Mn$ in ${MnO}_4^{-}$is $+7$ and in ${Mn}^{2+}$ is $+2$. The net change in electrons/oxidation state is $+5$

${MnO}_4^{-}⟶{Mn}^{2+}1$ mole conversion involves $5$ mole electron

Given that one mole of ${MnO}_4{ }^{-}$oxidizes $1.67$ moles of $M^x$ to ${MO}_3{}^{-}·1.67$ mole can be written as $5/3$

1 mole conversion $M^{x+}⟶{MO}_3{ }^{-}$involves $5-x$ moles of electrons, then for $5/3$ mole involves $5/3(5-x)$

Therefore, $\frac{5}{3}\times (5-x)=5$

$5\mathrm{\backslash }5$ gets cancelled

$5-x/3=1$

$5-x=3; x=5-3; x=2$

Hence, the correct option is $C$.