Question

Calculate the electronegativity of chlorine from bond energy of $Cl - F$ bond $\left(61\, kcal\, mol ^{-1}\right)$ $F - F \left(38 \,kcal \,mol ^{-1}\right)$ 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

Answer 1

Answer: (D)

Based on Paulings scale
$(E N)_{F}=(E N)_{C l}=k \sqrt{\Delta}=0.208 \sqrt{\Delta}$
where, $\Delta=$ resonance energy
and $k=$ conversion factor which is $0.208$ for converting $k$ cal into $e V$.
$\Delta=(B E)_{C l-F}-\sqrt{(B E)_{C l-C l}(B E)_{F-F}}$
$=61-\sqrt{58 \times 38}$
$=61-46.95=14.05\, kcal$
$(E N)_{C l}=(E N)_{F}-0.208 \sqrt{\Delta}$
$=4.0-0.208 \sqrt{14.05}$
$=4.0-0.78=3.22 \,e V$