Question

The oxidation state of sulphur in ${{\text{S}}_{\text{2}}}{{\text{O}}_{\text{6}}}^{{\text{2 - }}}$ is:

Answer 1

Hint: 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 answer:
The formula of ion given to us is ${{\text{S}}_{\text{2}}}{{\text{O}}_{\text{6}}}^{{\text{2 - }}}$. Here combining atoms are sulphur and oxygen. The negative charge on ions indicates it is an anion.
Calculate the oxidation state of sulphur using the oxidation number rules as follows:
As per the oxidation number oxidation number of oxygen is always -2 except in peroxide. In peroxides oxidation number of oxygen is -1.
An ion ${{\text{S}}_{\text{2}}}{{\text{O}}_{\text{6}}}^{{\text{2 - }}}$ is dithionate ion. Here the oxidation state of oxygen is -2.
So, calculate the oxidation state of sulphur as follows:

(Number of sulphur atom) (Oxidation state of sulphur) + (Number of oxygen atom) (Oxidation state of oxygen) = Charge on the ion
In the formula of ${{\text{S}}_{\text{2}}}{{\text{O}}_{\text{6}}}^{{\text{2 - }}}$ there are 2 sulphur atoms and 6 oxygen atoms and charge on an ion is -2.
Thus,
= (2) (Oxidation state of sulphur) + (6) (-2) = -2
Oxidation state of sulphur = +5

Hence, the oxidation state of sulphur in ${{\text{S}}_{\text{2}}}{{\text{O}}_{\text{6}}}^{{\text{2 - }}}$ is +5.

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.