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 $\left(58kcalmo{l}^{-1}\right)$ and electronegativity of fluorine $4.0eV$.

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

Answer 1

Answer: (D)

$3.22eV$

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