Question

Oxidation state of N in \[NO_4^ + \] ​ is:

Answer 1

Hint: We need to know that there are two types of reactions that take place in the reduction process which is oxidation and reduction. In the case of oxidation reaction, the electrons are removed from the substance. But in the case of the reduction process, the electrons are added to a substance. In the oxidation process, the oxygen is gained by the substance, and in the reduction process, the oxygen is lost by the substance.

Complete answer:
Let's consider the oxidation state of N be x. The compound \[NO_4^ + \] contains one positive charge and the oxidation state of oxygen is equal to \[ - 2\]. Hence,
\[x + 4 \times \left( { - 2} \right) = + 1\]
\[x - 8 = + 1\]
Therefore, \[x = + 9\]
Thus, the oxidation state of N in \[NO_4^ + \] ​ is equal to \[ + 9\].
The oxidizing agent is also known as oxidant. And the oxidizing agent is a species which has the ability to oxidize other substances. Which means, there is an increase in the oxidation state of the species by removing the electrons. Oxygen, hydrogen peroxide and halogens are the common oxidizing agents.
The oxidizing agent will try to exist in their highest oxidation state due to their strong tendency to accept the electron which undergoes the reduction. If the substance has stronger electron affinity, it has the greater oxidizing power.

Note:
We need to know that the compound with weaker electron affinity should be a weaker oxidizing agent. Because, it does not have a high capacity to attract the electrons. . During a chemical reaction, the oxidizing agent will remove the electrons from the species. And the oxidizing agent will undergo the reduction in a redox reaction. Fluorine is an example of the strongest oxidizing agent. Because it is highly electronegative in nature.