Answer
a) In $SO_{4}^{2-}$ ion, oxidation state of $S = +6$
b) In $SO_{3}^{2-}$ ion, oxidation state of $S = +4$
c) In $HSO_{3}^{-}$ ion, oxidation state of $S = +4$
d) In $HSO_{4}^{-}$ ion, oxidation state of $ S = +6$
Work Step by Step
For polyatomic ion, algebraic sum of oxidation state of each element will be equal to charge of the ion.
a) In $SO_{4}^{2-}$, oxidation state of O = -2. Let's say oxidation state of S = $x$
Hence, $x + (-2)\times4 = -2$ by solving we get $x = +6$
So, oxidation state of $ S = +6$
b) In $SO_{3}^{2-}$, oxidation state of O = -2. Let's say oxidation state of S = $x$
Hence, $x + (-2)\times3 = -2$ by solving we get $x = +4$
So, oxidation state of $ S = +4$
c) In $HSO_{3}^{-}$, oxidation state of O = -2 and H = +1. Let's say oxidation state of S = $x$
Hence, $(+1) + x + (-2)\times3 = -1$ by solving we get $x = +4$
So, oxidation state of $ S = +4$
d) In $HSO_{4}^{-}$, oxidation state of O = -2 and H = +1. Let's say oxidation state of S = $x$
Hence, $(+1) + x + (-2)\times4 = -1$ by solving we get $x = +6$
So, oxidation state of $ S = +6$
