Question

Oxidation number of ${\text{S}}$ in ${({\text{C}}{{\text{H}}_{\text{3}}}{\text{)}}_{\text{2}}}{\text{S}}$ is:
A) 0
B) +1
C) +2
D) +3

Answer 1

Oxidation state is the number of electrons that the atom of an element loses or gains during chemical bond formation. The oxidation number can be zero, negative, or positive. Use the oxidation number rules to calculate the oxidation state of sulphur.

Complete step by step answer:
The formula of a compound given to us is ${({\text{C}}{{\text{H}}_{\text{3}}}{\text{)}}_{\text{2}}}{\text{S}}$. Here combining atoms are carbon, sulphur and hydrogen. It is a neutral compound as there is no charge on it.
Calculate the oxidation state of sulphur using the oxidation number rules as follows:
As per the oxidation number rules, oxidation number of H is always +1 except in metal hydride it is -1.
So, for a given compound oxidation number of ${\text{H}}$ is +1.
In the given compound ${({\text{C}}{{\text{H}}_{\text{3}}}{\text{)}}_{\text{2}}}{\text{S}}$ each carbon atom is bonded to 3 H atom and 1 sulphur atom. So the oxidation number of carbon is - 4 because carbon is more electronegative than sulphur.
Now, calculate the oxidation number of sulphur as follows:
(Number of sulphur atom) (Oxidation number of sulphur) + (Number of carbon atom) (Oxidation number of carbon) + (Number of H atom)(oxidation number of H atom)= 0
In the formula of ${({\text{C}}{{\text{H}}_{\text{3}}}{\text{)}}_{\text{2}}}{\text{S}}$ there is 1 sulphur atom, 6 hydrogen atoms and 2 carbon atoms.
Thus,
= (Oxidation state of sulphur) (1) + (2) (-4) + (6) (+1) = 0
= Oxidation state of sulphur = 8 - 6
Oxidation state of sulphur = 2
Hence, the oxidation number of sulphur in ${({\text{C}}{{\text{H}}_{\text{3}}}{\text{)}}_{\text{2}}}{\text{S}}$ is +2.

Thus, the correct option is (C) +2.

Note: Oxidation state is the only apparent charge which represents the positive or negative character of the atom. It is necessary to denote the charge of oxidation state even if it is positive. Oxidation is the loss of electrons while reduction is the gain of electrons. During oxidation, the oxidation number of an atom increases while during reduction the oxidation number of an atom decreases.