Question

Calculate the oxidation number of sulphur in ${\mathrm{Na}}_2{\mathrm{S}}_4{\mathrm{O}}_6$.

Answer 1

• The chemical formula of the given ion is ${\mathrm{Na}}_2{\mathrm{S}}_4{\mathrm{O}}_6$.

• Let us assume the oxidation state of the sulphur ($\mathrm{S}$) is $\mathrm{x}$.
• There is a total of six oxygen ($\mathrm{O}$) atoms with oxidation state ($-2$), two sodium ($\mathrm{Na}$) atoms with oxidation state ($+1$) and the remaining four sulphur atoms.
• The total charge on the compound is

$$\begin{gathered} 4\times \left(\mathrm{x}\right)+2\times (+1)+6\times (-2)=0 \\ \Rightarrow 4\mathrm{x}+2–12=0 \\ \Rightarrow 4\mathrm{x}–10=0 \\ \Rightarrow 4\mathrm{x}=10 \\ \Rightarrow \mathrm{x}=+2.5 \end{gathered}$$

The oxidation state of the Sulphur in the given ion is $+2.5$.