In a reaction, `4` mole of electrons are transferred to `1` mole of `HNO_(3)`, the possible product obtained due to reduction is:

Options

A. `0.5` mole of `N_(2)`
B. `0.5` mole of `N_(2)O`
C. `1` mole of `NO_(2)`
D. `1` mole of `NH_(3)`

Explanation

To solve the question regarding the reduction of HNO3 when 4 moles of electrons are transferred, we can follow these steps:

### Step 1: Determine the oxidation state of nitrogen in HNO3.
In HNO3, nitrogen (N) has an oxidation state of +5. This can be calculated based on the overall charge of the molecule and the known oxidation states of hydrogen (+1) and oxygen (-2).

### Step 2: Identify the possible products of nitrogen reduction.
When nitrogen is reduced, it can take on several oxidation states. The common oxidation states for nitrogen include:
- 0 (in N2)
- +1 (in N2O)
- +2 (in NO)
- +3 (in NO2)
- -3 (in NH3)

### Step 3: Calculate the change in oxidation state.
Since we are transferring 4 moles of electrons, we need to determine how far the nitrogen can be reduced. Starting from +5 and gaining 4 electrons, the oxidation state of nitrogen would change as follows:
- +5 (in HNO3) + 4 (electrons) = +1

### Step 4: Identify the product formed.
The oxidation state of +1 corresponds to nitrous oxide (N2O). Therefore, when 4 moles of electrons are added to 1 mole of HNO3, the nitrogen is reduced to an oxidation state of +1, resulting in the formation of nitrous oxide.

### Step 5: Write the balanced equation for the reaction.
To balance the reaction, we consider that 1 mole of HNO3 will produce 0.5 moles of N2O when reduced by 4 moles of electrons:
\[ 2 \text{HNO}_3 + 4 \text{e}^- \rightarrow \text{N}_2\text{O} + \text{other products} \]

### Conclusion:
The possible product obtained due to the reduction of 1 mole of HNO3 with 4 moles of electrons is **0.5 moles of N2O** (nitrous oxide).

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