Question

What is the oxidation state of Sn in $SnSO_4$?

Answer 1

+2

Answer 2

Let the oxidation state of Sn be $x$.

For the sulfate ion, $\text{SO}_4^{2-}$:

• Oxidation state of S = $+6$
• Each O = $-2$ and there are 4 O atoms: $4(-2) = -8$
• Sum for sulfate: $+6 - 8 = -2$

Since the compound is neutral:

$$x + (-2) = 0 \quad \Longrightarrow \quad x = +2$$