In the reaction :
`3CuO+2NH_3 rarr N_2+3H_2O+3Cu`
the change of `NH_3` to `N_2` involve

Options

A. Loss of 6 electrons per mol of `N_2`
B. Loss of 3 electrons per mol of `N_2`
C. Gain of 6 electrons per mol of `N_2`
D. Gain of 3 electrons per mol of `N_2`

Answer

To determine the change of \( NH_3 \) to \( N_2 \) in the reaction \( 3CuO + 2NH_3 \rightarrow N_2 + 3H_2O + 3Cu \), we need to analyze the oxidation states of nitrogen in both \( NH_3 \) and \( N_2 \).

### Step-by-Step Solution:

1. **Identify the oxidation states**:
- In \( NH_3 \) (ammonia), the oxidation state of nitrogen (N) is \( -3 \).
- In \( N_2 \) (molecular nitrogen), the oxidation state of nitrogen is \( 0 \).

2. **Determine the change in oxidation state**:
- The change in oxidation state for nitrogen going from \( NH_3 \) to \( N_2 \) is:
\[ \text{Change} = 0 - (-3) = 3 \]
- This indicates that each nitrogen atom in \( NH_3 \) is losing 3 electrons.

3. **Account for the number of nitrogen atoms**:
- The balanced equation shows that 2 moles of \( NH_3 \) produce 1 mole of \( N_2 \). Since \( N_2 \) contains 2 nitrogen atoms, we need to consider the total loss of electrons for both nitrogen atoms.
- Therefore, the total loss of electrons for the conversion of 2 moles of \( NH_3 \) to 1 mole of \( N_2 \) is:
\[ \text{Total electrons lost} = 2 \times 3 = 6 \text{ electrons} \]

4. **Conclusion**:
- The change of \( NH_3 \) to \( N_2 \) involves the loss of **6 electrons**.

### Final Answer:
The change of \( NH_3 \) to \( N_2 \) involves the loss of **6 electrons**.

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