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lf electron affinity of an element is 6 eV/atom and ionisation energy is 16.4 eV/atom then electronegativity of element on Pauling scale is :

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2.5

3.5

4.0

3.0

answer is C.

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Detailed Solution

E.N. on Mullikan scale

$= \frac{IE + EA}{2} (in eV) \Rightarrow \frac{6 + 16.4}{2} = \frac{22.4}{2} = 11.2 eV EN on Pauling scale = \frac{Mullikan scale}{2.8} = \frac{11.2}{2.8} = 4$

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