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.{\text{ }}1.42{\text{ }}eV \\\ B.{\text{ }}1.89{\text{ }}eV \\\ C.{\text{ }}2.67{\text{ }}eV \\\ D.{\text{ }}3.22{\text{ }}eV \\\$$

Answer 1

Hint- We can find the electronegativity of chlorine by using Pauling's scale equation. Therefore we have to calculate the resonance energy by using Cl –F, F-F , and Cl-Cl bonds initially. Then we substitute the resonance energy and given electronegativity of fluorine in Pauling’s equation to calculate the electronegativity of chlorine.
Formula used:
We used Pauling’s scale equation to calculate the electronegativity of chlorine.
${(EN)_F} - {(EN)_{Cl}} = k\sqrt[{}]{\Delta } = 0.208\sqrt[{}]{\Delta }$ ………………..equation 1
Where,
Δ = resonance energy
${(EN)_F}$ = Electronegativity of fluorine
${(EN)_{Cl}}$ = Electronegativity of chlorine
And $k$= conversation factor (which is 0.2080 of converting $kcal$ into$eV$).
Given:
The 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 the electronegativity of fluorine - 4.0 eV

Complete step by step answer:
Step 1
Initially, we have to find the resonance energy from Cl –F, F-F, and Cl-Cl bond. So we modify Pauling's equation with respect to Δ (resonance energy).
Therefore, we will get
$\Delta = {(BE)_{Cl - F}} - \sqrt {{{(BE)}_{Cl - Cl}}{{(BE)}_{F - F}}}$ …………equation 2

Now, we will substitute the given band energy of Cl –F, F-F, and Cl-Cl in the modified equation.
$= {\text{ }}61{\text{ }} - \sqrt {58 \times 38}$
Now we can solve the calculation as
$= {\text{ }}61{\text{ }}-{\text{ }}46.95$
= 14.05 $kcal$
Therefore, we have found the resonance energy is Δ = 14.05 $kcal$
Step 2
Now we have to calculate the electronegativity of chlorine using Pauling’s equation.
Here Pauling’s scale equation
${(EN)_F} - {(EN)_{Cl}} = k\sqrt[{}]{\Delta } = 0.208\sqrt[{}]{\Delta }$
Again we have to do small modification with respect to ${(EN)_F}$
${(EN)_F} = 0.208\sqrt[{}]{\Delta } - {(EN)_{Cl}}$
Let us substitute the resonance energy and electronegativity of fluorine.
$= 4.0 - 0.208\sqrt {14.05}$
$= 4.0 - (0.208 \times 3.7483)$
$= 4.0 - 0.7796$
$= 3.22{\text{ eV}}$

Therefore, we have found the electronegativity of choline is $= 3.22{\text{ eV}}$

Note: We must remember that Pauling came up with a slightly more futuristic equation for the relative electronegativity of two atoms in a molecule.
And also the Pauling scale is based on a pragmatic relationship between the electronegativity of bonded atoms and energy of a bond.