Give the oxidation number of carbon in the following:

$$\begin{gathered} \left(a\right)\mathit{C}{\mathit{F}}_{\mathbf{2}}\mathit{C}{\mathit{l}}_{\mathbf{2}} \\ \left(b\right)\mathit{N}{\mathit{a}}_{\mathbf{2}}{\mathit{C}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{4}} \\ \left(c\right)\mathit{H}\mathit{C}{\mathit{O}}_{\mathbf{3}}^{\mathbf{-}} \\ \left(d\right){\mathit{C}}_{\mathbf{2}}{\mathit{H}}_{\mathbf{6}} \end{gathered}$$

Oxidation state of Fluorine is -1.

Oxidation state of Chlorine is -1.

Let, oxidation state of carbon in CF₂Cl₂is x.

No you can write,

$$\begin{gathered} x+2\left(-1\right)+2\left(-1\right)=0 \\ x=+4 \end{gathered}$$

Hence, oxidation state of carbon in CF₂Cl₂ is +4.

Oxidation state of Sodium is +1.

Oxidation state of Oxygen is -2.

Let, oxidation state of carbon in Na₂C₂O₄is y.

No you can write,

$$\begin{gathered} 2\left(+1\right)+2\left(y\right)+2\left(-2\right)=0 \\ 2y=+6 \\ y=+3 \end{gathered}$$

Hence, oxidation state of carbon in Na₂C₂O₄is +3.

Oxidation state of Hydrogen is +1.

Oxidation state of Oxygen is -2.

Let, oxidation state of carbon in HCO₃^-is z.

No you can write,

$$\begin{gathered} \left(+1\right)+z+3\left(-2\right)=-1 \\ z=+4 \end{gathered}$$

Hence, oxidation state of carbon in HCO₃^- is +4.

Oxidation state of Hydrogen is +1.

Let, oxidation state of carbon in C₂H₆is k.

No you can write,

$$\begin{gathered} 2k+6\left(+1\right)=0 \\ 2k=-6 \\ k=-3 \end{gathered}$$

Hence, oxidation state of carbon in C₂H₆ is -3.