A) 2, 16, 5
B) 2, 5, 16
C) 16, 5, 2
D) 5, 16, 2
Correct Answer: B
Solution :
\[\operatorname{MnO}_{4}^{-} \xrightarrow[{}]{}\,\,M{{n}^{2+}}\] \[\operatorname{Mn}{{O}_{4}}+8{{H}^{+}}\,\xrightarrow[{}]{}\,\,M{{n}^{2\,+}}\,\,+\,\,4{{H}_{2}}O\] \[[\operatorname{Mn}O_{4}^{-}+8{{H}^{+}}+\,\,5{{e}^{\Theta }}\,\,\xrightarrow{{}}\,\,M{{n}^{2+}}+\,\,4\,{{H}_{2}}O]\times 2\]\[[{{\operatorname{C}}_{2}}O_{4}^{2-}\,\xrightarrow{{}}\,\,C{{O}_{2}}+2{{e}^{\Theta }}]\,\,\times \,\,5\] ... (ii) \[[MnO_{4}^{-}\,+\,\,8{{H}^{+}}\,\,+\,\,5{{e}^{\Theta }}\,\xrightarrow{{}}\,\,M{{n}^{2\,-}}\,+\,\,4{{H}_{2}}O]\,\,\times \,\,2\] Addition \[[{{C}_{2}}O_{4}^{2-}\,\,\xrightarrow{{}} C{{O}_{2}} +\,\,2{{e}^{\Theta }}]\,\,\times 5\] \[2MnO_{4}^{-}+16{{H}^{+}}\,\,+\,\,10{{e}^{\Theta }}\,\,+\,\,5{{C}_{2}}O_{4}^{2\,-}\] \[2\overset{2+}{\mathop{Mn}}\,\,\,+\,\,8{{H}_{2}}O\,\,+\,\,5C{{O}_{2}} +10{{e}^{\Theta }}\]
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