Answer
a) In compound NaCl, oxidation state of Na = +1, Cl = -1.
b) In compound $CaF_{2}$ , oxidation state of Ca = +2, F = -1.
c) In compound $SO_{2}$ , oxidation state of S = +4, O = -2.
d) In compound $H_{2}S$ , oxidation state of H = +1, S = -2.
Work Step by Step
Algebraic sum of oxidation state for a compound is zero. Now, let's find oxidation state of each atom.
a) In NaCl, Cl will have oxidation state of -1. So, Na will have the oxidation state of +1 so that sum of oxidation state = 0.
b) In $CaF_{2}$ , oxidation state of F = -1. Let's say, oxidation state of Ca = $x$.
Hence, $x + (-1)\times2 = 0$. By solving we get $x=+2$
Therefore, oxidation state of Ca = +2.
c) In $SO_{2}$, oxidation state of O = -2. Let's say oxidation state of S = $x$
Hence, $x + (-2) \times2 = 0$ By solving this equation we get $x = +4$
Therefore, oxidation state of S = +4.
d) In $H_{2}S$, oxidation state of H = +1. Let's say oxidation state of S = $x$
Hence, $(+1)\times2 + x = 0$ By solving this equation we get $x= -2$
So, oxidation state of S = -2.
