An imaginary lattice is formed as given
`Cl_(2) rarr 2Cl" "DeltaH = x_(1) " "Cl rarr Cl^(+) + e^(-) " "DeltaH = x_(2)`
`Cl + e^(-) rarr Cl^(-) " "Delta H = x_(3)" "Cl^(+) + Cl^(-) rarr Cl^(+)Cl^(-) " "Delta H = x_(4)`
Thus , enthalpy change when `Cl^(+)Cl^(-)` is formed is:

Options

A. `(+(x_1)/2 + x_2 + x_3 + x_4)`
B. `(-(x_1)/2 + x_2 + x_3 + x_4)`
C. `(x_1 + x_2 - x_3 - x_4)`
D. `-(x_1 + x_2 - x_3 - x_4)`

Explanation

To find the enthalpy change when \( \text{Cl}^+ \text{Cl}^- \) is formed, we will analyze the given reactions step by step and calculate the overall enthalpy change.

### Step-by-Step Solution:

1. **Identify the Reactions**:
- The reactions provided are:
1. \( \text{Cl}_2 \rightarrow 2\text{Cl} \) with \( \Delta H = x_1 \)
2. \( \text{Cl} \rightarrow \text{Cl}^+ + e^- \) with \( \Delta H = x_2 \)
3. \( \text{Cl} + e^- \rightarrow \text{Cl}^- \) with \( \Delta H = x_3 \)
4. \( \text{Cl}^+ + \text{Cl}^- \rightarrow \text{Cl}^+ \text{Cl}^- \) with \( \Delta H = x_4 \)

2. **Determine the Sign of Each Enthalpy Change**:
- **For \( x_1 \)**: The reaction involves breaking \( \text{Cl}_2 \) into two chlorine atoms. This requires energy (endothermic), so \( x_1 \) is positive.
- **For \( x_2 \)**: The formation of \( \text{Cl}^+ \) from \( \text{Cl} \) requires ionization energy (endothermic), so \( x_2 \) is positive.
- **For \( x_3 \)**: The formation of \( \text{Cl}^- \) from \( \text{Cl} \) involves gaining an electron, which releases energy (exothermic), so \( x_3 \) is negative.
- **For \( x_4 \)**: The formation of the ionic compound \( \text{Cl}^+ \text{Cl}^- \) is an exothermic process, so \( x_4 \) is negative.

3. **Write the Overall Enthalpy Change**:
- The overall enthalpy change \( \Delta H \) for the formation of \( \text{Cl}^+ \text{Cl}^- \) can be calculated as follows:
\[ \Delta H = x_1 + x_2 + x_3 + x_4 \]
- Substituting the signs:
\[ \Delta H = x_1 + x_2 - x_3 - x_4 \]

4. **Conclusion**:
- The enthalpy change when \( \text{Cl}^+ \text{Cl}^- \) is formed is given by:
\[ \Delta H = x_1 + x_2 - x_3 - x_4 \]

### Final Answer:
Thus, the enthalpy change when \( \text{Cl}^+ \text{Cl}^- \) is formed is \( x_1 + x_2 - x_3 - x_4 \).

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