A) \[M{{n}_{2}}{{O}_{3}}\]
B) \[Mn{{O}_{2}}\]
C) \[Mn{{O}_{4}}\]
D) \[MnO_{4}^{2-}\]
Correct Answer: B
Solution :
\[\overset{+2}{\mathop{MnS{{O}_{4}}}}\,\xrightarrow[{}]{{}}\overset{+4}{\mathop{Mn{{O}_{2}}}}\,\] In this conversion, there is loss of two electrons, hence equivalent weight of\[MnS{{O}_{4}}\]is the half of its molecular weight.
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