Question

Calculate the electronegativity of chlorine if the electronegativity of fluorine is 4.0.Given excess bond energy of Cl - F.bond is 14.68 kcal mol ^(-1)

Answer 1

Answer:

i think method

Explanation:

Δ=(BE)

Cl−F

−

(BE)

Cl−Cl

(BE)

F−F

=61−

58×38

=61−46.95=14.05kcal

(EN)

Cl

=(EN)

F

−0.208

Δ

=4.0−0.208

14.05

=4.0−0.78=3.22eV

Answer 2

Given:

The electronegativity of fluorine = 4.0.

Δ = 14.68 kcal/mol.

To find:

Electronegativity of chlorine.

Formula to be used:

$E.N_{Cl} =E.N_{F} -0.208\sqrt{}$Δ

The excess bond energy is multiplied to 0.208 to convert it to eV.

Step-by-step explanation:

Step 1 of 1

The excess bond energy, Δ = 14.68 kcal/mol.

$E.N_{F}=4.0$

The electronegativity of chlorine can thus be calculated as:

$E.N_{Cl} =E.N_{F} -0.208\sqrt{}$Δ

The excess bond energy is calculated by Δ$=(BE)_{Cl-F} -\sqrt{(BE)_{Cl-Cl} (BE)_{F-F} }$

So,

$E.N_{Cl} =4.0 -0.208\sqrt{14.68}\\\\E.N_{Cl} =4.0 -0.208*3.83\\\\E.N_{Cl} =4.0 -0.7966\\\\E.N_{Cl} =3.2033$

THE ELECTRONEGATIVITY OF CHLORINE IS 3.022 eV.