A) \[CrO_{4}^{2-}\to C{{r}_{2}}O_{7}^{2-}\]
B) \[Cr\to CrC{{l}_{3}}\]
C) \[Na\to N{{a}^{+}}\]
D) \[2{{S}_{2}}O_{3}^{2-}\to {{S}_{4}}O_{6}^{2-}\]
Correct Answer: A
Solution :
[a] \[CrO_{4}^{2-}\to C{{r}_{2}}O_{7}^{2-}\] O.N of Cr does not change. It remains +6 on both sides. Hence there is no oxidation or reduction.
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