Question

Calculate the electronegativity (eV) of chlorine from the bond energy of $Cl-F$ bond (61 kcal mol${}^{-1}$), $F-F$ bond ($38$ kcal mol${}^{-1}$) and $Cl-Cl$ bond ($58$ kcal mol${}^{-1}$) and electronegativity of fluorine ($4.0$ eV).
Based on Pauling's scale :
$(EN{)}_F-(EN{)}_{Cl}=k\sqrt{\Delta }=0.208\sqrt{\Delta }$
where, $\Delta$ is resonance energy and k is the conversion factor (which is $0.208$ for converting kcal into eV).

• A. $3.22$
• B. $1.34$
• C. $2.40$
• D. None of these

Answer 1

The correct option is A $3.22$

$\Delta =(BE{)}_{Cl-F}-\sqrt{\left(BE{)}_{Cl-Cl}\right(BE{)}_{F-F}}$
$=61-\sqrt{58\times 38}$
$=61-46.95=14.05kcal$
$(EN{)}_{Cl}=(EN{)}_F-0.208\sqrt{\Delta }$
$=4.0-0.208\sqrt{14.05}$
$=4.0-0.78=3.22eV$