Question

For the reaction
$M ^{ x ^{+}}+ MnO _4^{-} \longrightarrow MO _3^{-}+ Mn ^{2+}+\frac{1}{2} O _2$. If one mole of $MnO _4^{-}$oxidises $1^* 67$ moles of $M ^{ x ^{+}}$to $MO _3^{-}$, then the value of $x$ in the reaction is

• A. 5
• B. 3
• C. 2
• D. 1

Answer 1

Answer: (C)

$\overset{\text{ +7}}{MnO }O^{-}_4 + 5e^{-} \rightarrow Mn^{2+}$
Since 1 mole of $MnO_4^{-}$ accepts $5$ moles of electrons, therefore, $5$ moles of electrons are lost by $1.67$ moles of $M^{x+}$
$\therefore $ I Mole of $M^{x+}$ will lose electrons
$= 5/1-67 = 3 $ moles (approx.)
Since $M^{x+}$ changes to $MO_3^{-}$
(where O.N. of $M = + 5)$ by accepting $3$ electrons
$\therefore \, x = + 5 —3 = + 2$.