Answer
a) In $CO_{3}^{2-}$ ion, oxidation state of $C = +4, O=-2$.
b) In $OH^{-}$ ion, oxidation state of $O = -2, H = +1$
c) In $NO_{3}^{-}$ ion, oxidation state of $N = +5, O = -2$
d) In $NO_{2}^{-}$ ion, oxidation state of $ N = +3, O = -2$
Work Step by Step
For a polyatomic ion, algebraic sum of oxidation state will be equal to the charge of the ion.
a) In $CO_{3}^{2-}$ , oxidation state of O = -2. Let's say oxidation state of $C = x$
Hence, $x + (-2)\times3 = -2$ by solving we get $x = +4$
So, oxidation state of $C = +4$
b) In $OH^{-}$, oxidation state of O = -2 and H = +1 so that algebraic sum is equal to -1.
c) In $NO_{3}^{-}$, oxidation state of O = -2, let's say oxidation of N = $x$
Hence, $x + (-2)\times3 = -1$, by solving we get $x = +5$
So, oxidation state of $N = +5$
d) In $NO_{2}^{-}$, oxidation state of O = -2, let's say oxidation of N = $x$
Hence, $x + (-2)\times2= -1$, by solving we get $x = +3$
So, oxidation state of $N = +3$