Question

Calculate the electronegativity of chlorine from bond energy of Cl-F bond (61 kcal mol\(^{-1}\)), F-F (38 kcal mol\(^{-1}\)) and Cl-Cl bond (58 kcal mol\(^{-1}\)) and electronegativity of fluorine 4.0 eV

• A. 1.42 eV
• B. 1.89 eV
• C. 2.67 eV
• D. 3.22 eV

Hint:

Pauling scale: \(\chi_F = 4.0\), \(\chi_{Cl} = 3.16\) (standard), close to 3.22.

Answer 1

The Correct Option is D

To calculate the electronegativity of chlorine based on the given bond energies and the electronegativity of fluorine, we can use the concept introduced by Linus Pauling. According to Pauling's scale and the relationship between bond energy and electronegativity, the difference in electronegativity (\( \Delta \chi \)) between two atoms in a diatomic molecule can be estimated using the bond energies of the molecules involved.

The formula used is:

\(\Delta \chi = 0.208 \times \sqrt{E_{\text{AB}} - \frac{(E_{\text{AA}} + E_{\text{BB}})}{2}}\)

Where:

• \( E_{\text{AB}} \) is the bond energy of Cl-F bond = 61 kcal/mol
• \( E_{\text{AA}} \) is the bond energy of Cl-Cl bond = 58 kcal/mol
• \( E_{\text{BB}} \) is the bond energy of F-F bond = 38 kcal/mol.

Substituting these values into the formula, we get:

\(\Delta \chi = 0.208 \times \sqrt{61 - \frac{(58 + 38)}{2}}\)

\(\Delta \chi = 0.208 \times \sqrt{61 - 48}\)

\(\Delta \chi = 0.208 \times \sqrt{13}\)

\(\Delta \chi \approx 0.208 \times 3.60555 \approx 0.74995\)

Since the electronegativity of fluorine (\( \chi_F \)) is given as 4.0 eV, we can find the electronegativity of chlorine (\( \chi_{Cl} \)) using:

\(\chi_{Cl} = \chi_F - \Delta \chi\)

\(\chi_{Cl} = 4.0 - 0.75 = 3.25 eV\)

This solution approximates to the closest value provided in the options. Therefore, the correct answer considering significant figures and similar calculations, would be:

The approximated electronegativity of chlorine is 3.22 eV.