Which reaction involves neither oxidation nor reduction?

C r O_{4}^{2 -} \to C r_{2} O_{7}^{2 -}

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C r \to C r C l_{3}

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N a \to N a^{+}

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S_{2} O_{6}^{2 -} \to S_{4} O_{6}^{2 -}

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Solution

The correct option is A $C r O_{4}^{2 -} \to C r_{2} O_{7}^{2 -}$
Oxidation number of $C r$ in $C r O_{4}^{2 -} = x + 4 \times (- 2) = - 2$

$⟹ x = 6$

Oxidation number of $C r$ in $C r_{2} O_{7}^{2 -} = 2 x + 7 \times (- 2) = - 2$

$⟹ x = 6$

The oxidation number of Cr does not change. Hence, it undergoes neither oxidation nor reduction.

Hence, the correct option is $A$

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C O_{2}

C r O_{4}^{2 -}

C r_{2} O_{7}^{2 -}

C r O_{4}^{2 -} (a q) C O_{2} - - \to C r_{2} O_{7}^{2 -}

N a - e^{-} \to N a^{+}

S_{2} O {_{6}}^{2 -} (a q) + C r_{2} O {_{7}}^{2 -} (a q) \to S_{4} O {_{6}}^{2 -} (a q) + C r^{3 +} (a q)

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