Question

In a reaction, 4 moles of electrons are transferred to one mole of HNO3 when acted as an oxidant. The possible reduction product is?
A.(1/2) mole of ${ N }_{ 2 }$
B.(1/2) mole of ${ N }_{ 2 }O$
C.1 mole of ${ NO }_{ 2 }$
D.1 mole of ${ NH }_{ 3 }$

Answer 1

Hint: To get the correct answer you just need to find the oxidation number of nitrogen both in nitric acid and the product it forms. Then identify the change where 4 electrons are involved. This will be your answer.

Complete step by step answer:
First, let’s find the oxidation state of nitrogen in ${ HNO }_{ 3 }$,
1+x+3(-2) = 0
x = +5
Hence, the oxidation state of nitrogen in ${ HNO }_{ 3 }$ is +5.
Using the same method we can calculate the oxidation state of other molecules that are in the given options. In ${ N }_{ 2 }$, ${ N }_{ 2 }O$, ${ NO }_{ 2 }$​ , and ${ NH }_{ 3 }$​ the oxidation state of nitrogen is 0, +1, +4, and -3 respectively.

Now, we will look at each option one by one,
-Option (A) In ${ N }_{ 2 }$, the oxidation state of both nitrogens is 0. So, it needs 5 moles of electrons per N atom, which makes 10 moles of electrons for the formation of 1 mol ${ N }_{ 2 }$ molecule as a product. Hence 5 moles of electrons for (1/2) mole of ${ N }_{ 2 }$. This answer is incorrect.
-Option (B) In ${ N }_{ 2 }O$, the oxidation state of both nitrogens is +1. So, it needs 4 moles of electrons per N atom, which makes 8 moles of electrons for the formation of 1 mol ${ N }_{ 2 }O$ molecule as a product. Hence 4 moles of electrons for (1/2) mole of ${ N }_{ 2 }O$. This answer is correct.
-Option (C) In ${ NO }_{ 2 }$, the oxidation state of nitrogen is +4. So, it needs 1 mole of electrons per N atom for the formation of 1 mol ${ NO }_{ 2 }$ molecule as a product. Hence, this answer is incorrect.
-Option (D) In ${ NH }_{ 3 }$, the oxidation state of nitrogen is -3. So, it needs 8 moles of electrons per N atom for the formation of 1 mol ${ NH }_{ 3 }$ molecule as a product. Hence, this answer is incorrect.

Therefore, we can conclude that the correct answer to this question is option B.

Note: We should know that in nitric acid, nitrogen has an already maximum oxidation state +5 and it can increase further due to the absence of d orbitals in nitrogen. Hence it can act as an oxidizing agent only.