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

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$\frac{I + E}{129}$

$\frac{I + E}{5.6}$

$\frac{I + E}{544}$

$\frac{I + E}{2}$

answer is D.

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Detailed Solution

Given data: Ionization energy and electron affinity of an element

Concept used: Relation between Ionization energy and electron affinity

Detailed solution:

We know

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

Since, 1 ev/atom=96kJ/mol

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

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