Answer
a) In the compound $CH_{4}$, oxidation state of $C = -4, H= +1$
b) In the compound $CH_{2}Cl_{2}$, oxidation state of $C = 0, H = +1, Cl = -1$.
c) In the compound $CuCl_{2}$, oxidation state of $Cu = +2, Cl = -1$.
d) In the compound $HI$, oxidation state of $H = +1, I = -1$
Work Step by Step
Algebraic sum of oxidation state for a compound is zero. Now, let's find oxidation state of each atom in compound.
a) In $CH_{4}$, oxidation state of H = +1, Let's say oxidation state of C = $x$
Hence, $x + (+1) \times4 = 0 $ By solving this equation we get x = -4.
Hence, oxidation state of $C = -4$
b) In $CH_{2}Cl_{2}$, oxidation state of H = +1, Cl = -1. Let's say oxidation state of C = $x$.
Hence, $x + (+1)\times2 + (-1)\times2 = 0$, By solving we get $x = 0$
So, oxidation state of $C =0$
c) In $CuCl_{2}$, oxidation state of Cl = -1, Let's say oxidation state of Cu = $x$
Hence, $x + (-1)\times2 = 0$ By solving we get $ x = +2$
So, oxidation state of $Cu = +2$.
d) In $HI$, oxidation state of $H = +1$. So, oxidation state of $I = -1$.
