For the reaction, $M^{x +} + {M n O}_{4} {}^{-}⟶ {M O}_{3} {}^{-}+ {M n}^{2 +} + 1 / 2 O_{2}$ , if one mole of ${M n O}_{4}^{-}$ oxidizes $1.67$ moles of $M^{x +}$ to ${M O}_{3}^{-}$ , then the value of ' $x$ ' in the reaction is

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$5$

$3$

$2$

$1$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

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

The oxidation state of $M$ in ${M O}_{3}^{-}$ is $+ 5$ . The net change in electrons/oxidation state is $5 - x$

$M^{x +} ⟶ {M O}_{3} {}^{-}1$ mole conversion involves $5 - x$ moles of electrons

The oxidation state of $M n$ in ${M n O}_{4}^{-}$ is $+ 7$ and in ${M n}^{2 +}$ is $+ 2$ . The net change in electrons/oxidation state is $+ 5$

${M n O}_{4}^{-} ⟶ {M n}^{2 +} 1$ mole conversion involves $5$ mole electron

Given that one mole of ${M n O}_{4}^{-}$ oxidizes $1.67$ moles of $M^{x}$ to ${M O}_{3} {}^{-}\cdot 1.67$ mole can be written as $5 / 3$

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

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

$5 \ 5$ gets cancelled

$5 - x / 3 = 1$

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

Hence, the correct option is $C$ .

Watch 3-min video & get full concept clarity