Question

Calculate the electronegativity of chlorine from the given data: 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 is $4.0$.

• A. $2.33$
• B. $3.22$
• C. $3.33$
• D. $2.67$

Answer 1

The correct option is B $3.22$

According to Pauling's Scale:
$\Delta$ = ${\chi }_F-{\chi }_{Cl}$ = $0.208\times \sqrt{E_{Cl-F}-\sqrt{(E_{F-F}\times E_{Cl-Cl})}}$
where, $E_{Cl-F}$ = Bond energy of $Cl–F$ bond $\left(kcal/mol\right)$
$E_{F-F}$ = Bond energy of $F–F$ bond $\left(kcal/mol\right)$
$E_{Cl-Cl}$ = Bond energy of $Cl–Cl$ bond $\left(kcal/mol\right)$
${\chi }_F$ = Electronegativity of $F$
${\chi }_{Cl}$ = Electronegativity of $Cl$
${\chi }_F-{\chi }_{Cl}$ = $0.208\times \sqrt{61-\sqrt{58\times 38)}}$
${\chi }_F-{\chi }_{Cl}$ = $0.208\times \sqrt{14.05}$
${\chi }_{Cl}$ = ${\chi }_F-0.208\times \sqrt{14.05}$
${\chi }_{Cl}$ = $4.0-0.78$
${\chi }_{Cl}$ = $3.22$