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
$$\begin{gathered} A.1.42eV \\ B.1.89eV \\ C.2.67eV \\ D.3.22eV \end{gathered}$$

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.
$$(EN{)}_F-(EN{)}_{Cl}=k\sqrt[]{\mathrm{\Delta }}=0.208\sqrt[]{\mathrm{\Delta }}$$ ………………..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
$$\mathrm{\Delta }=(BE{)}_{Cl-F}-\sqrt{{(BE)}_{Cl-Cl}{(BE)}_{F-F}}$$ …………equation 2

 Now, we will substitute the given band energy of Cl –F, F-F, and Cl-Cl in the modified equation.
$$=61-\sqrt{58\times 38}$$
Now we can solve the calculation as
$$=61-46.95$$
= 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
$$(EN{)}_F-(EN{)}_{Cl}=k\sqrt[]{\mathrm{\Delta }}=0.208\sqrt[]{\mathrm{\Delta }}$$
Again we have to do small modification with respect to $(EN{)}_F$
$$(EN{)}_F=0.208\sqrt[]{\mathrm{\Delta }}-(EN{)}_{Cl}$$
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.