Question

In $$\text{H}{\text{N}}_{\text{3}}$$, oxidation number of nitrogen is
A) -3
B) -1
C) -0.33
D) 0.5

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 fraction, zero, negative, or positive whole number. Use the oxidation number rules to calculate the oxidation state of sulphur.

Complete answer:
The formula of a compound given to us is $$\text{H}{\text{N}}_{\text{3}}$$. Here combining atoms are nitrogen and hydrogen. It is a neutral compound as there is no charge on it.
Calculate the oxidation state of nitrogen using the oxidation number rules as follows:
As per the oxidation number rules, 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.
Now, calculate the oxidation number of nitrogen as follows:
(Number of nitrogen atom) (Oxidation number of nitrogen) + (Number of H atom)(oxidation number of H atom)= 0
In the formula of nitrogen there are 3 nitrogen atoms and one hydrogen atom.
Thus,
(3) (Oxidation state of nitrogen) + (1) (+1) = 0
Oxidation state of nitrogen = -0.33
Hence, the oxidation number of nitrogen in $$\text{H}{\text{N}}_{\text{3}}$$ is -0.33.

Thus, the correct option is (C).

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 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 reduction the oxidation number of an atom decreases.