Ionisation potential and electron affinity of fluorine are `17.42` and `3.45 eV` respectively. Calculate the electronegativity of fluorine.

Explanation

`chi("Mulliken")=(17.42+3.45)/(2)=10.35 chi_("Pulling")= (chi("Mulliken"))/(2.8)=(10.435)/(2.8)=3.726`
Alternately : When I.E. and E.A are given in eV then we can directly use I.E. E.A (Pauling) `=(I.E+E.A)/(5.6)`
i.e, `chi("Pauling")=(17.42+3.45)/(5.6)=3.726`