Question

Which of the following, oxidation number of carbon is -1?
(A) $C{H}_4$
(B) $C_6{H}_6$
(C) $CC{l}_4$
(D) $HCl$

Answer 1

Hint: We can calculate the oxidation number:
The oxidation number of a free element is always 0.
The oxidation number of a monatomic ion equals the charge of the ion.
The oxidation number of H is +1, but it is -1 in when combined with less electronegative elements.
The oxidation number of O in compounds is usually -2, but it is -1 in peroxides.

Complete answer:
We have been provided with four compounds: $C{H}_4$, $C_6{H}_6$, $CC{l}_4$and $HCl$.
We need to calculate the oxidation number of all these compounds and evaluate whose oxidation number is -1.
 The oxidation number of carbon in methane $C{H}_4$ is –4, while its oxidation number in carbon dioxide is +4. The oxidation number of hydrogen in all compounds is +1. The oxidation number of oxygen in its elemental form is 0, while its value in carbon dioxide and water is –2.
The oxidation state of carbon in benzene is -1 and the oxidation state of hydrogen in benzene is +1. In a C-H bond (this bond is nonpolar covalent), the H is treated as if it has an oxidation state of +1. This means that every C-H bond will decrease the oxidation state of carbon by 1.
In $CC{l}_4$ the oxidation number for chlorine is -1 and for carbon it is +4.
The oxidation number of the compound HCl is zero. In a compound, hydrogen has an oxidation number of +1. Since the sum of the oxidation numbers of hydrogen and chlorine must equal zero, the oxidation number of chlorine must be -1.
so, only in benzene the oxidation number of carbon is -1.
So, from this we can say that the oxidation state of carbon in benzene $(C_6{H}_6)$ is -1.

Therefore, we can conclude that option (B) is correct.

Note:
Oxidation numbers are important to calculate as they are used by chemists to keep track of electrons within a compound. We can use guidelines to assign oxidation numbers to atoms in a compound. Changes in oxidation state during a reaction tell us that there is a transfer of electrons.