Question

When zinc is treated with conc. $HN{O_3}$, an oxide of nitrogen is formed. The change in the oxidation state of nitrogen during this reaction is:
(A) +5 to 0
(B) +5 to +2
(C) +5 to +4
(D) +5 to +1

Answer 1

In order to find the change in the oxidation state of the nitrogen when zinc is reacted with concentrated nitric acid, we must first know what will be produced during the given reaction. From the product it will be easy to calculate the change in the oxidation state.

Complete answer:
Let us see the given question. In the reaction, the zinc is a metal which will react with the concentrated nitric acid, in order to give a nitrogen dioxide gas along with zinc nitrate and water molecules. Nitrogen dioxide can be used as a catalyst, rocket fuel, bleaching agent and as an explosive. The reaction between zinc and concentrated nitric acid is given below:
$Zn + 4HN{O_3}(conc) \to Zn{(N{O_3})_2} + 2N{O_2} + 2{H_2}O$
In the above given reaction, the oxidation state of nitrogen in conc. $HN{O_3}$ is +5. It will undergo a reduction reaction and then the oxidation state of nitrogen in $N{O_2}$ is +4.
Therefore, we can say that the change in the oxidation state of the nitrogen when zinc is reacted with concentrated nitric acid is from +5 to +4.

Hence the correct answer is option (C).

Note:
We have to remember that when zinc is treated with dilute nitric acid a very different product will be formed. In this reaction, the zinc treated with dilute nitric acid will form ${N_2}O$ instead of $N{O_2}$.