The bond dissociation energies of `X_2, Y_2 and XY` are in the ratio of `1 : 0.5 : 1. DeltaH` for the formation of XY is -200 kJ mol^(-1)`. The bond dissociation energy of `X_2` will be

Options

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

Explanation

To find the bond dissociation energy of \( X_2 \), we can use the information given in the problem regarding the bond dissociation energies and the enthalpy change for the formation of \( XY \).

### Step 1: Define the bond dissociation energies

Let:
- The bond dissociation energy of \( X_2 \) be \( D(X_2) = x \)
- The bond dissociation energy of \( Y_2 \) be \( D(Y_2) = 0.5x \) (since the ratio is 1:0.5)
- The bond dissociation energy of \( XY \) be \( D(XY) = x \)

### Step 2: Write the formation reaction

The formation of \( XY \) from \( X_2 \) and \( Y_2 \) can be represented as:
\[ X_2 + Y_2 \rightarrow 2XY \]

### Step 3: Write the enthalpy change equation

The enthalpy change (\( \Delta H \)) for this reaction can be expressed in terms of the bond dissociation energies:
\[ \Delta H = [\text{Bonds broken}] - [\text{Bonds formed}] \]
Substituting the bond dissociation energies into the equation, we have:
\[ \Delta H = [D(X_2) + D(Y_2)] - [2D(XY)] \]
Substituting the values:
\[ \Delta H = [x + 0.5x] - [2x] \]
\[ \Delta H = 1.5x - 2x \]
\[ \Delta H = -0.5x \]

### Step 4: Set up the equation with the given value of \( \Delta H \)

We know that \( \Delta H \) for the formation of \( XY \) is -200 kJ/mol. Therefore, we can set up the equation:
\[ -0.5x = -200 \]

### Step 5: Solve for \( x \)

To find \( x \), we can rearrange the equation:
\[ 0.5x = 200 \]
\[ x = \frac{200}{0.5} = 400 \text{ kJ/mol} \]

### Conclusion

Thus, the bond dissociation energy of \( X_2 \) is \( 400 \text{ kJ/mol} \).