• question_answer The equivalent mass of $MnS{{O}_{4}}$ is half of its molecular mass when it is converted to A) $M{{n}_{2}}{{O}_{3}}$                B) $Mn{{O}_{2}}$ C) $Mn{{O}_{4}}$                               D) $MnO_{4}^{2-}$

The equivalent mass will be half of molecular mass if change in oxidation number is 2. $\overset{+2}{\mathop{MnS{{O}_{4}},}}\,\overset{+3}{\mathop{M{{n}_{2}}{{O}_{3}},}}\,\overset{+4}{\mathop{Mn{{O}_{2}},}}\,\overset{+8}{\mathop{Mn{{O}_{4}},}}\,\overset{+6}{\mathop{MnO_{4}^{2-}}}\,$ $\overset{+2}{\mathop{MnS{{O}_{4}}}}\,\xrightarrow[{}]{{}}\overset{+4}{\mathop{Mn{{O}_{2}}}}\,$