Question

Which of the following, oxidation number of carbon is -1?
(a) $\text{C}{{\text{H}}_{\text{4}}}$
(b) ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}$
(c) $\text{CC}{{\text{l}}_{4}}$
(d) $\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}$

Answer 1

Hint: Knowing the oxidation number gives us an idea about the number of electrons gained or lost during the chemical reactions. We need to examine each of the compounds to find the required oxidation number of carbon.

Step by step answer:
> The oxidation number of carbon in Methane ($\text{C}{{\text{H}}_{\text{4}}}$) is -4. This is because hydrogen has an oxidation number of +1 and $\text{C}{{\text{H}}_{\text{4}}}$ contains 4 hydrogen atoms. So, according to the following equation, we can see that to attain stability, the carbon atom has to have - 4 oxidation number.
$\begin{align}
& {{\text{C}}^{\text{x}}}{{\text{H}}_{4}}^{\text{+1}} \\
& \text{Let x be the oxidation number of C} \\
& x\text{ + 4}\times \text{1 = 0} \\
& \Rightarrow \text{ x = }-4 \\
& \Rightarrow \text{ x = }-4 \\
& \therefore \text{ x = }-4 \\
\end{align}$

> In the case of ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}$ or Benzene, the oxidation state of the hydrogen atom in each CH bond is treated as +1. This is because 6 carbon atoms and 6 hydrogen atoms satisfy the stable structure of benzene. Hence, for each hydrogen atom having oxidation number of +1, we will have each Carbon atom with an oxidation number of -1. To attain stability, every CH bond will decrease the oxidation number of carbon by 1. This theory can be proved with the equation given below.
$\begin{align}
& {{\text{C}}_{\text{6}}}^{\text{x}}{{\text{H}}_{\text{6}}}^{\text{+1}} \\
& \text{Let x be the oxidation number of C} \\
& 6x\text{ + 6}\times \text{1 = 0} \\
& \Rightarrow \text{ 6x = }-\text{6} \\
& \Rightarrow \text{ x = }-1 \\
& \therefore \text{ x = }-1 \\
\end{align}$

> The oxidation number of carbon in compound $\text{CC}{{\text{l}}_{4}}$ is + 4. This is because each chlorine atom in the compound carries -1 charge and therefore an oxidation number of -1. There are a total of 4 chlorine atoms in the compound, which makes the total oxidation number of chlorine alone as -4. Therefore, to attain stability, carbon needs to have an oxidation number of +4. This theory can be proved by the following algebraic equation.

$\begin{align}
& {{\text{C}}^{\text{x}}}\text{C}{{\text{l}}_{4}}^{-\text{1}} \\
& \text{Let x be the oxidation number of C} \\
& x\text{ + 4}\times (-\text{1) = 0} \\
& \Rightarrow \text{ x = +}4 \\
& \Rightarrow \text{ x = +}4 \\
& \therefore \text{ x = +}4 \\
\end{align}$

> In $\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}$, the hydrogen atoms have an oxidation number of +1 and the chlorine atoms have an oxidation number of -1. The number of atoms of hydrogen and chlorine is equal. Hence, cancelling the oxidation number and resulting 0. To attain stability, the carbon atom has to have an oxidation number of zero. This can be backed by the equation given below.
$\begin{align}
& {{\text{C}}^{\text{x}}}{{\text{H}}_{2}}^{\text{+1}}\text{C}{{\text{l}}_{2}}^{-1} \\
& \text{Let x be the oxidation number of C} \\
& x\text{ + 2}\times \text{1 + 2}\times \text{(}-1\text{) = 0} \\
& \Rightarrow \text{ x + 2 }-\text{ 2= 0} \\
& \Rightarrow \text{ x = 0} \\
& \therefore \text{ x = 0} \\
\end{align}$

Note: Losing an electron makes the compound positive and the element is known as cation. Gaining an electron makes the compound negative, and the element is known as an anion.