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$\text{N}{{\text{H}}_{3}}$is oxidised to NO by ${{\text{O}}_{2}}$ in basic medium. Number of equivalents of $\text{N}{{\text{H}}_{3}}$oxidised by 1 mole of ${{\text{O}}_{2}}$ is:
A. 4
B. 5
C. 6
D. 7

Last updated date: 13th Jun 2024
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Hint: Oxidised reagents are those reagents which itself gets reduced but oxidises the other molecule. Whereas reducing agents are those reagents which itself gets oxidised but reduces the other molecule.

Complete step by step answer:
- Firstly, we have to write the equation between ammonia and oxygen in which the ammonia will reduce into nitrogen oxide:
$\text{4N}{{\text{H}}_{3}}\text{ + 5}{{\text{O}}_{2}}\text{ }\to \text{ 4NO + 6}{{\text{H}}_{2}}\text{O}$
- As we can see that the oxidation state of nitrogen changes from -3 to +2 so it undergoes oxidation reaction.
- But there are a total of 4 moles of ammonia and nitrogen oxide so the difference between the oxidation state of both is:
$\begin{align}& \left( 3\text{ }\cdot \text{ 4} \right)\text{ - }\left( \text{2 }\cdot \text{ 4} \right)\text{ = 12 - 8} \\ & \text{= 4} \\ \end{align}$
- So, the n-factor is 4.
- Now, we have to calculate the number of equivalents of ammonia so let's understand the definition of it i.e.
The number of equivalents is an amount of the electron or ions that can be transferred in a chemical reaction.
- So, here a total of 4 electrons are transferred from the reactant side to the product side.
Therefore, option A. is the correct answer.

Note: The formula to find equivalent mass is the ratio of the molar mass of the substance and acidity or basicity of n-factor. The single elements have an oxidation state of zero according to the rules to calculate the oxidation state.