Question

Calculate the electronegativity of chlorine from bond energy of $Cl-F$ bond $\left(61kcalmo{l}^{-1}\right)$ $F-F\left(38kcalmo{l}^{-1}\right)$ and $Cl-Cl$ bond $(58kcalmo{l}^{-1}$ ) and electronegativity of fluorine $4.0eV$

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

Answer 1

Answer: (D)

Based on Paulings scale
$(EN{)}_F=(EN{)}_{Cl}=k\sqrt{\Delta }=0.208\sqrt{\Delta }$
where, $\Delta =$ resonance energy
and $k=$ conversion factor which is $0.208$ for converting $k$ cal into $eV$.
$\Delta =(BE{)}_{Cl-F}-\sqrt{(BE{)}_{Cl-Cl}(BE{)}_{F-F}}$
$=61-\sqrt{58\times 38}$
$=61-46.95=14.05kcal$
$(EN{)}_{Cl}=(EN{)}_F-0.208\sqrt{\Delta }$
$=4.0-0.208\sqrt{14.05}$
$=4.0-0.78=3.22eV$