Explanation

(b)
${CrO}_{4}^{2 -} + {SO}_{3}^{2 -} \longrightarrow {Cr(OH)}_{4}^{-} + {SO}_{4}^{2 -}$
Let the oxidation number of Cr is $x$ in ${CrO}_{4}^{2 -}$
$x + 4( - 2) = \ - 2$
$x =$ 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.