Question

Ionization potential and electron affinity of fluorine are $17.42$ and $3.45$ eV respectively. The electronegativity of fluorine on Mulliken scale and Pauling scale is $x$ & $y$ ,then $\frac{(x+y)}{2.023}$ is:

Answer 1

(1) Mulliken's scale
Electronegativity $=\frac{\text{Ionisation potential + electron affinity}}{2}$
Electronegativity $=\frac{17.42+3.45}{2}=10.435=x$
(2) Pauling scale
Electronegativity $=\frac{\text{Ionisation potential + electron affinity}}{5.6}$
Electronegativity $=\frac{17.42+3.45}{5.6}=3.73=y$
$\frac{(x+y)}{2.023}=\frac{(10.435+3.73)}{2.023}=7$