Question

Ionization potential and electron affinity of fluorine are $17.42$ and $3.45eV$ respectively. The electronegativity of fluorine on Mulliken scale and Pauling scale is $x$ & $y$ respectively, then $\frac{(x+y)}{1.011}$is:

• A. 12
• B. 13
• C. 14
• D. 15

Answer 1

The correct option is C 14

According to Mulliken's Scale:
Electronegativity = $x$ = $\frac{I.E+E.A}{2}$ = $\frac{17.42+3.45}{2}$ = $10.435$

Thus, the value of $x$ is $10.435$.

According to Pauling's Scale:
Electronegativity = $y$ = $\frac{I.E.+E.A.}{2\times 2.8}$ = $\frac{17.42+3.45}{5.6}$ = $3.73$
Thus, Value of $y$ is $3.73$
$\therefore$ $\frac{x+y}{1.011}$ = $\frac{10.435+3.73}{1.011}$ = 14