Question

In the equation
$ CrO^{2-}_{4}+SO^{2-}_{3} \to Cr(OH)^{-}_{4} + SO^{2-}_{4} $
the oxidation number of $ Cr $ changes from

• A. $ + 6 $ to $ +4 $
• B. $ + 6 $ to $ + 3 $
• C. $ + 8 $ to $ + 4 $
• D. $ + 4 $ to $ + 3 $

Answer 1

Answer: (B)

The oxidation number of hydrogen is $+1$ in all of its compounds except in metallic hydrides and the oxidation number of oxygen is $-2$ in all of its compounds except peroxides, superoxide and $OF_{2}$
$CrO_{4}^{2-}+SO_{3}^{2-} \rightarrow Cr \left(OH\right)_{4}^{-}+SO_{4}^{2-}$
Let the oxidation number of $Cr$ is $x$ is $CrO_{4}^{2-}$
$x + 4(−2) = − 2$
$x= 6$
and in $Cr(OH)_{4}^{-}$ the oxidation number of $Cr$ is $y$
$y + 4(−2) + 4(1) = − 1$
$y − 8 + 4 = − 1$
$y = 3$
Hence, oxidation number of $Cr$ changes from $+6$ to $+3$