Question

If a small amount of $KMn{O}_4$ is added to conc.$H_{2}S{O}_4$, a green coloured solution with oxidation number of $Mn$ as $+7$ is formed. Probable compound is :

• A. $Mn{O}_{3}HS{O}_4$
• B. $K_{2}Mn{O}_4$
• C. $Mn{O}_{2}HS{O}_4$
• D. $Mn{O}_2$

Answer 1

The correct option is A $Mn{O}_{3}HS{O}_4$

Let $x$ be the oxidation state of the $Mn$.
$\underset{\uparrow x}{Mn}\underset{\uparrow -6-1=0}{O_3}HS{O}_4$
$x=+7$
$K_{2}Mn{O}_4$
$+2+x-8=0;x=+6$
$Mn{O}_{2}HS{O}_4$
$x-4-1=0;x=5$
$Mn{O}_2$
$x=4$
Therefore, the compound is $Mn{O}_{3}HS{O}_4$.