$C l_{2}$ disproportionate to $C l^{-}$ and species $A$ , such that the equivalent mass of $C l_{2}$ is $71$ . Thus, the oxidation number of chlorine in species $A$ is:

Open in App

Solution

The disproportion reaction is:
$C l_{2} \to C l^{-} + C l O^{-}$
0 -1 +1
So, the equivalent weight of $C l_{2}$ is $\frac{71}{1}$ .
The oxidation state of $C l$ in $C l O^{-}$ is $+ 1.$

Suggest Corrections

Similar questions

C l^{-}

C l_{2}

S_{2} O_{3}^{2 -} + {C l}_{2} ⟶ H {S O}_{4}^{-} + {C l}^{-}

[C o (N H_{3})_{4} (H_{2} O)_{2}] C l_{2}, [N i (H_{2} O)_{6}] C l_{2}, [C o (H_{2} O)_{5} C l] C l, [P t (N H_{3})_{2} C l_{2}], [P t (e n) C l_{2}], [P t (e n)_{2} C l_{2}], [P b C l_{2} B r I], [C o (N H_{3}) 4 C l_{2}], [C o (e n)_{2} B r_{2}]

C l_{2}

Cl O^{-}, {ClO}_{2}^{-}, {ClO}_{3}^{-} and {ClO}_{4}^{-}

Also, write the reaction for each of the species that disproportionate.

Join BYJU'S Learning Program