High School

Calculate the electronegativity of fluorine from the following data:

- Bond energy of H-H bond: \( E_{H-H} = 104.2 \, \text{kcal mol}^{-1} \)
- Bond energy of F-F bond: \( E_{F-F} = 36.6 \, \text{kcal mol}^{-1} \)
- Bond energy of H-F bond: \( E_{H-F} = 134.6 \, \text{kcal mol}^{-1} \)
- Electronegativity of Hydrogen (\( X_H \)) = 2.1

Choose the correct electronegativity of fluorine:

A. 3.87
B. 2.76
C. -3.87
D. -2.76

Answer :

The electronegativity of fluorine is 3.87. Option A is the correct answer.

To calculate the electronegativity of fluorine, we can use the concept of bond energies and the Pauling scale.

Given:

- Bond energy of H-H bond (Eₕ₋ₕ) = 104.2 kcal/mol

- Bond energy of F-F bond (E₋) = 36.6 kcal/mol

- Bond energy of H-F bond (Eₕ₋) = 134.6 kcal/mol

- Electronegativity of hydrogen (Xₕ) = 2.1

1. Calculate the electronegativity of fluorine (X₋) using the bond energies:

X₋ - Xₕ = √(Eₕ₋ - Eₕ₋ₕ) - √(E₋ - Eₕ₋ₕ)

X₋ - 2.1 = √(134.6 - 104.2) - √(36.6 - 104.2)

X₋ - 2.1 = √30.4 - √(-67.6)

X₋ - 2.1 = 5.51 - 8.22

X₋ - 2.1 = -2.71

X₋ = -2.71 + 2.1

X₋ = 3.87

Therefore, the electronegativity of fluorine is 3.87.

The correct answer is option A.

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