The bond energies of H-H, X-X and H-X are `104 K.cal, 38 K.cal` and `138 K.Cal` respectively the electron egativity of 'X' is `[sqrt(67) = 8.18]`

Options

A. `3.0`
B. `3.5`
C. `3.8`
D. `1.7`

Answer

To find the electronegativity of element 'X', we will use the bond energies provided and the formula derived from experimental data. Here is a step-by-step solution:

### Step 1: Understand the given bond energies
We have the following bond energies:
- H-H bond energy = 104 K.cal
- X-X bond energy = 38 K.cal
- H-X bond energy = 138 K.cal

### Step 2: Calculate the change in enthalpy (ΔH)
Using the formula for bond dissociation energy:
\[ \Delta H = \text{Bond energy of H-X} - \frac{1}{2} \times \text{Bond energy of H-H} + \text{Bond energy of X-X} \]
Substituting the values:
\[ \Delta H = 138 - \frac{1}{2} \times 104 + 38 \]
Calculating:
\[ \Delta H = 138 - 52 + 38 = 124 \text{ K.cal} \]

### Step 3: Use the electronegativity formula
The formula relating the electronegativity of X (denoted as \( \chi_X \)) and hydrogen (denoted as \( \chi_H \)) is:
\[ \chi_X - \chi_H = 0.208 \times \sqrt{\Delta H} \]
Given that the electronegativity of hydrogen (\( \chi_H \)) is approximately 2.1, we can substitute the values:
\[ \chi_X - 2.1 = 0.208 \times \sqrt{124} \]

### Step 4: Calculate \( \sqrt{124} \)
Calculating \( \sqrt{124} \):
\[ \sqrt{124} \approx 11.14 \]

### Step 5: Substitute back to find \( \chi_X \)
Now substituting \( \sqrt{124} \) back into the equation:
\[ \chi_X - 2.1 = 0.208 \times 11.14 \]
Calculating the right side:
\[ \chi_X - 2.1 \approx 2.32 \]
Thus:
\[ \chi_X \approx 2.1 + 2.32 = 4.42 \]

### Step 6: Finalize the answer
The electronegativity of element 'X' is approximately 4.42.