Question

Ionization potential and electron affinity of fluorine are $17.42$ and $3.412eV$ respectively. The electronegativity of fluorine on Mulliken scale and Pauling scale is $x$ and $y$ respectively, then $\frac{(x+y)}{0.4}$ is:

Answer 1

According to Mulliken's Scale:
$\text{Electronegativity}=x=\frac{I.E+E.A}{2}=\frac{17.42+3.412}{2}=10.416$

Thus, the value of $x$ is $10.416$.

According to Pauling's Scale:
$\text{Electronegativity}=y=\frac{I.E.+E.A.}{2\times 2.8}=\frac{17.42+3.412}{5.6}=3.72$
Thus, value of $y$ is $3.72$
$\therefore$ $\frac{x+y}{0.4}=\frac{10.416+3.72}{0.4}=35.34$