Question

In the equation
$Cr{O}_4^{2-}+S{O}_3^{2-}\to Cr(OH{)}_4^{-}+S{O}_4^{2-}$
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 $O{F}_2$
$Cr{O}_4^{2-}+S{O}_3^{2-}\to Cr{\left(OH\right)}_4^{-}+S{O}_4^{2-}$
Let the oxidation number of $Cr$ is $x$ is $Cr{O}_4^{2-}$
$x+4(-2)=-2$
$x=6$
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 $+3$