Question

Find the oxidation number of carbon in${\text{CHC}}{{\text{l}}_{\text{3}}}$ is:
A.+1
B.+2
C.+4
D. -2

Answer 1

Hint: Oxidation state is the number of electrons that the atom of an element loses or gains during chemical bond formation. The oxidation number can be zero, negative, or positive whole number. Use the oxidation number rules to calculate the oxidation state of carbon.

Step by step answer: The formula of a compound given to us is ${\text{CHC}}{{\text{l}}_{\text{3}}}$. Here combining atoms are carbon, hydrogen and chlorine. It is a neutral compound as there is no charge on it.
Calculate the oxidation state of carbon using the oxidation number rules as follows:
As per the oxidation number rules, the oxidation number of H is always +1 except in metal hydride it is -1.
So, for a given compound oxidation number of ${\text{H}}$ is +1.
As per the rules, the oxidation number of chlorine is -1.
Now, calculate the oxidation number of carbon as follows:
(Number of carbon atom) (Oxidation number of carbon) + (Number of hydrogen atom) (Oxidation number of hydrogen) + (Number of chlorine atom)(oxidation number of chlorine atom)= 0
In the formula of ${\text{CHC}}{{\text{l}}_{\text{3}}}$ there is 1 carbon atom, 1 hydrogen atoms and 3 chlorine atoms.
Thus,
(1) (Oxidation state of carbon) + (1) (+1) + (3) (-1) = 0
Oxidation state of carbon = -2
Hence, the oxidation number of carbon in ${\text{CHC}}{{\text{l}}_{\text{3}}}$ is -2.
Thus, the correct option is (D) -2

Note: Oxidation state is the only apparent charge which represents the positive or negative character of the atom. It is necessary to denote the charge of the oxidation state even if it is positive. Oxidation is the loss of electrons while reduction is the gain of electrons. During oxidation, the oxidation number of an atom increases while during the reduction oxidation number of an atom decreases.