Question

Calculate the oxidation number of sulphur, chromium and nitrogen in ${\mathrm{H}}_2{\mathrm{SO}}_5,{\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-}\mathrm{and}{\mathrm{NO}}_3^{-}$respectively.

Answer 1

Oxidation state: The oxidation state, also known as the oxidation number in chemistry, is the hypothetical charge of an atom if all of its bonds to other atoms were entirely ionic.

Part 1: Calculating the oxidation state of Sulphur ($\mathrm{S}$) in ${\mathrm{H}}_2{\mathrm{SO}}_5$

• Let $\mathrm{x}$ be the oxidation number of Sulphur

$$\begin{gathered} 2(+1)+\mathrm{x}+5(-2)=0 \\ \mathrm{x}=+8 \end{gathered}$$

• Sulphur can not have $+8$ oxidation state because $\mathrm{S}$ has only six valance electrons.
• This fallacy is removed by calculating oxidation of $\mathrm{S}$ by chemical bonding method.
• In ${\mathrm{H}}_2{\mathrm{SO}}_5$ two Oxygen ($\mathrm{O}$) atoms are in $-1$ oxidation state.
• Let us assume the oxidation number of Sulphur is $\mathrm{x}$.

$$\begin{gathered} 2(+1)+\mathrm{x}+3(-2)+2(-1)=0 \\ +2+\mathrm{x}-6-2=0 \\ \mathrm{x}=+6 \end{gathered}$$

Part 2: Calculating the oxidation state of Chromium ($\mathrm{Cr}$) in ${\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-}$

• Let $\mathrm{x}$ be the oxidation number of Chromium in Dichromate ion
  $$\begin{gathered} 2\mathrm{x}+7(-2)=-2 \\ \mathrm{x}=+6 \end{gathered}$$
• Hence the oxidation number of Chromium in Dichromate ion is $+6$.

Part 3: Calculating the oxidation state of Nitrogen ($\mathrm{N}$) in ${\mathrm{NO}}_3^{-}$

• Let $\mathrm{x}$ be the oxidation number of Nitrogen in the Nitrate ion (${\mathrm{NO}}_3^{-}$).
• $\mathrm{x}+3(-2)=-1$
• From the structure:

• $$\begin{gathered} \mathrm{x}+1(-1)+1(-2)+1(-2)=0 \\ \mathrm{x}=+5 \end{gathered}$$
• Thus, the oxidation number of Nitrogen in the Nitrate ion (${\mathrm{NO}}_3^{-}$) is $+5$