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

Explanation

To solve the problem of determining the possible product obtained from the reduction of 1 mole of HNO₃ with the transfer of 4 moles of electrons, we can follow these steps:

### Step 1: Determine the oxidation state of nitrogen in HNO₃
In nitric acid (HNO₃), the oxidation state of nitrogen can be calculated as follows:
- Let the oxidation state of nitrogen be \( x \).
- The equation for the oxidation state is:
\[ x + 1 + 3(-2) = 0 \]
Simplifying this gives:
\[ x + 1 - 6 = 0 \implies x - 5 = 0 \implies x = +5 \]

### Step 2: Understand the effect of electron transfer
Since 4 moles of electrons are being added to HNO₃, the oxidation state of nitrogen will decrease. The initial oxidation state of nitrogen is +5, and adding 4 electrons will change the oxidation state:
\[ \text{New oxidation state} = +5 - 4 = +1 \]

### Step 3: Identify the product with nitrogen in the +1 oxidation state
Now we need to find a compound where nitrogen has an oxidation state of +1. We can analyze potential products:
- **N₂**: The oxidation state of nitrogen is 0 (not possible).
- **N₂O**: The oxidation state of nitrogen can be calculated as follows:
\[ 2x + (-2) = 0 \implies 2x = 2 \implies x = +1 \quad (\text{possible}) \]
- **NO**: The oxidation state of nitrogen is +2 (not possible).
- **NH₃**: The oxidation state of nitrogen is -3 (not possible).

### Step 4: Conclusion
The only compound where nitrogen has an oxidation state of +1 is N₂O. Therefore, the possible product obtained due to the reduction of HNO₃ with 4 moles of electrons is:
\[ \text{Product} = N₂O \]

### Final Answer
The possible product obtained due to the reduction of 1 mole of HNO₃ with 4 moles of electrons is **N₂O**.

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