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We have X atoms of A. If all the atoms gain one electron each, the energy released is a eV. UY atoms of A lose one electron each, then the energy absorbed is b eV. Theo the electronegativity of A on the Mulliken scale will be:

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$\frac{a + b}{2}$

$\frac{1}{2} (a + b) \times N_{a}$

$\frac{1}{2} (\frac{a}{X} + \frac{b}{Y}) \times N_{a}$

$\frac{1}{2} (\frac{a}{X} + \frac{b}{Y})$

answer is B.

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Detailed Solution

For A. electron affinity per atom $= \frac{a}{X} e v$

For A. ionization energy per atom $= \frac{b}{Y} e v$

E.N. (Mulliken) $= \frac{1}{2} (\frac{a}{X} + \frac{b}{Y})$

Hence, option B is correct.

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