Calculate the electronegativity of fluorine from the following data:
`E_(H-H) = 104.4 kcal mol^(-1), E_(F-F) = 36.6 kcal mol^(-1)`
`E_(H-F) = 134.3 kcal mol^(-1), chi_(H) = 2.1`

Explanation

To calculate the electronegativity of fluorine (X_F) using the given bond energies and the electronegativity of hydrogen (X_H), we will use the formula for electronegativity difference:

\[ X_A - X_B = 0.208 \times \left( E_{AB} - \frac{1}{2}(E_{AA} + E_{BB}) \right)^{1/2} \]

Where:
- \( E_{AB} \) is the bond energy of the AB bond (H-F bond in this case).
- \( E_{AA} \) is the bond energy of the AA bond (H-H bond).
- \( E_{BB} \) is the bond energy of the BB bond (F-F bond).
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