If the bond dissociation energies of `XY`,`X_(2)` and `Y_(2)` are in the ratio of `1:1:0.5` and `DeltaH_(f)` for the formation of `Xy` is `-200 KJ//mol`. The bond dissociation energy of `X_(2)` will be `:-`

Options

A. 800 KJ `"mol"^(-1)`
B. 100 KJ `"mol"^(-1)`
C. 200 KJ `"mol"^(-1)`
D. 400 KJ `"mol"^(-1)`

Explanation

To solve the problem, we need to find the bond dissociation energy of \( X_2 \) given the bond dissociation energies of \( XY \), \( X_2 \), and \( Y_2 \) in the ratio of \( 1:1:0.5 \) and the formation enthalpy \( \Delta H_f \) for \( XY \) as \( -200 \, \text{kJ/mol} \).

### Step-by-Step Solution:

1. **Define Bond Dissociation Energies:**
Let the bond dissociation energy of \( XY \) be \( x \). Then, according to the given ratio:
- Bond dissociation energy of \( X_2 = x \)
- Bond dissociation energy of \( Y_2 = 0.5x \)

2. **Write the Reaction for Formation of \( XY \):**
The formation of \( XY \) can be represented as:
\[ \frac{1}{2} X_2 + \frac{1}{2} Y_2 \rightarrow XY \]

3. **Apply Hess's Law:**
According to Hess's law, the enthalpy change for the reaction can be expressed as:
\[ \Delta H = \text{(Bond energies of reactants)} - \text{(Bond energies of products)} \]
For our reaction:
\[ \Delta H_f = \left( \frac{1}{2} \times x + \frac{1}{2} \times 0.5x \right) - x \]

4. **Substitute the Known Value of \( \Delta H_f \):**
We know \( \Delta H_f = -200 \, \text{kJ/mol} \):
\[ -200 = \left( \frac{1}{2} x + \frac{1}{4} x \right) - x \]

5. **Simplify the Equation:**
Combine the terms:
\[ -200 = \left( \frac{3}{4} x \right) - x \]
This simplifies to:
\[ -200 = -\frac{1}{4} x \]

6. **Solve for \( x \):**
Multiply both sides by -4:
\[ 800 = x \]

7. **Conclusion:**
The bond dissociation energy of \( X_2 \) is \( 800 \, \text{kJ/mol} \).

### Final Answer:
The bond dissociation energy of \( X_2 \) is \( 800 \, \text{kJ/mol} \).