If I and E are ionisation energy and electron affinity of an element in kJ mole^–1 electronegativity is given as

Given data: Ionization energy and electron affinity of an element

Concept used: Relation between Ionization energy and electron affinity

Detailed solution:  

We know 

$\mathrm{E}.\mathrm{N}.=\frac{\left(\mathrm{Ionization} \mathrm{Potential}+\mathrm{elctron} \mathrm{affinity}\right)}{2\times 2.8\times 96}$

Since, 1 ev/atom=96kJ/mol

$\mathrm{E}.\mathrm{N}.=\frac{\left(\mathrm{I}+\mathrm{E}\right)}{544}$