Question
The equivalent weight of $$MnS{O_4}$$ is half of its molecular weight when it is converted to :
A.
$$M{n_2}{O_3}$$
B.
$$Mn{O_2}$$
C.
$$MnO_4^ - $$
D.
$$MnO_4^{2 - }$$
Answer :
$$Mn{O_2}$$
Solution :
For equivalent weight of $$MnS{O_4}$$ to be half of its molecular weight, change in oxidation state must be equal to 2. It is possible only when oxidation state of $$Mn$$ in product is $$+ 4.$$ Since oxidation state of $$Mn$$ in
$$\eqalign{
& MnS{O_4}\,{\text{is}}\,{\text{ + }}\,{\text{2}}{\text{.}}\,\,{\text{So,}}\,Mn{O_2}\,{\text{is}}\,{\text{correct}}\,{\text{answer}}{\text{.}} \cr
& {\text{In}}\,Mn{O_2},O.S.\,{\text{of}}\,Mn\, = + 4 \cr
& \therefore {\text{Change}}\,{\text{in}}\,O.S.\,{\text{of}}\,Mn = + 4 - \left( { + 2} \right) = + 2 \cr} $$