A) \[{{I}^{-}}\]
B) \[CO_{3}^{2-}\]
C) \[{{S}^{2-}}\]
D) \[NO_{2}^{-}\]
Correct Answer: B
Solution :
| Except \[CO_{3}^{2-}\] all other ions \[{{I}^{-}},{{S}^{2-}},NO_{2}^{-},\] are oxidised by acidified \[KMn{{O}_{4}}\]forming \[{{I}_{2}},S,NO_{3}^{-}\] respectively. |
| \[2MnO_{4}^{-}+10I+16{{H}^{+}}\to 2M{{n}^{2+}}5{{I}_{2}}+8{{H}_{2}}O.\] |
| \[{{S}^{2-}}\]is oxidised to S in presence of \[KMn{{O}_{4}}\] |
| \[2KMn{{O}_{4}}+3{{H}_{2}}S{{O}_{4}}+5{{H}_{2}}S\to \]\[{{K}_{2}}S{{O}_{4}}+2MnS{{O}_{4}}+5S+8{{H}_{2}}O.\] |
| Nitrites are oxidised to nitrates |
| \[2KMn{{O}_{4}}+5KN{{O}_{2}}+3{{H}_{2}}S{{O}_{4}}\to \]\[{{K}_{2}}S{{O}_{4}}+2MnS{{O}_{4}}+5KN{{O}_{3}}+3{{H}_{2}}O.\] |
| \[C{{O}_{3}}^{2-}\]is not oxidised by \[{{\operatorname{KMnO}}_{4}}\]because in \[C{{O}_{3}}^{2-}\] C is already in its highest +4 oxidation state. Thus \[C{{O}_{3}}^{2-}\] does not decolourise \[KMn{{O}_{4}}\] solution. |
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