Question

Oxidation number of $S$ in $(C{H}_3{)}_{2}S$ is

• A. $0$
• B. $+1$
• C. $-2$
• D. $+3$

Answer 1

Answer: (C)

$-2$

Let $x$ be the oxidation number of $S$.
The oxidation number of carbon is $+4$ and the oxidation number of hydrogen is $-1$. The sum of all the oxidation numbers is $0$ as the compound is neutral.
Hence, $2(+4+3(-1\left)\right)+x=0$ or, $x=-2$.
Thus, the oxidation state of sulphur is $-2$.