Question

What is the oxidation number of the sulfur atoms in ${\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3$?

Answer 1

Oxidation number: Oxidation number denotes the oxidation state of an element in a compound ascertained according to a set of rules formulated on the basis that electron pair in a covalent bond belongs entirely to a more electronegative element.

Oxidation number of the sulfur atoms in ${\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3$

• From the arbitrary law of computation of oxidation state $\mathrm{S}$-atom which is not linked with oxygen will possess -2 oxidation state.
• So, the oxidation number of accepted sulfur marked by number 1 is $-2$.
• The oxidation state of oxygen is $-2$.
• The oxidation state of sodium is $+1$.
• Consider the oxidation number of sulfur marked by number 2 is $x$.
• $$\begin{gathered} 2\times \left(+1\right)+3\times (-2)+x-2=0 \\ \left(\text{For Na}\right)\left(\text{For O}\right)\left({\text{for S}}_{\text{1}}\right) \\ +2-6+x=2 \\ x=+6 \end{gathered}$$
• The oxidation state of sulfur marked by number 1 is -2, and the oxidation state of other sulfur marked by number 2 is $+6$.