In the reaction `A^(-n2)+xe^(-)rarrA^(-n1)` . Here, x will be :

Options

A. `n_(1)+n_(2)`
B. `n_(2)-n_(1)`
C. `n_(1)-n_(2)`
D. `n_(1)xxn_(2)`

Answer

To determine the value of \( x \) in the reaction \( A^{-n_2} + x e^{-} \rightarrow A^{-n_1} \), we need to analyze the change in charge from \( A^{-n_2} \) to \( A^{-n_1} \).

### Step-by-Step Solution:

1. **Identify the Change in Charge**:
- The initial charge on species \( A \) is \( -n_2 \).
- The final charge on species \( A \) is \( -n_1 \).
- The change in charge can be expressed as:
\[ \text{Change in charge} = -n_1 - (-n_2) = -n_1 + n_2 = n_2 - n_1 \]

2. **Determine the Number of Electrons Required**:
- Since electrons are negatively charged, adding electrons will decrease the overall charge of the species.
- To find \( x \), which represents the number of electrons added, we need to equal the change in charge to the number of electrons:
\[ x = n_2 - n_1 \]

3. **Final Expression for \( x \)**:
- Therefore, the number of electrons \( x \) required for the reaction can be expressed as:
\[ x = n_2 - n_1 \]

### Conclusion:
The value of \( x \) in the reaction \( A^{-n_2} + x e^{-} \rightarrow A^{-n_1} \) is given by:
\[ \boxed{n_2 - n_1} \]