Define the relation among the $E N, I E a n d E . A .$

E N = \frac{I E \times E A}{2}

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E N = I E + E A

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2 E N = I E - E A

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E N = \frac{I E + E A}{2}

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E N = \frac{I E - E A}{2}

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Solution

The correct option is D $E N = \frac{I E + E A}{2}$
The correct relationship is $E N = \frac{I E + E A}{2}$
According to mulliken's scale, the electronegatiity is the average value of ionization potential and electron affinity of an atom.

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Similar questions

(IE = Ionisation energy in eV/atom, EA = Electron affinity in eV/atom, EN = Electronegativity in Mulliken scale)

A^{+} B^{-}

(IE = Ionisation energy in eV/atom, EA = Electron affinity in eV/atom, EN = Electronegativity in Mulliken scale)

A^{+} B^{-}

A^{-} B^{+}

A

B

A^{+} B^{-}

M g (g) + 2 F (g) \to M g^{2 +} (g) + 2 F^{-} (g)

(I E)_{1}

M g (g) = 737.7 k J m o l^{- 1}

(I E)_{2}

M g (g) = 1451 k J m o l^{- 1}

(E A)

F (g) = - 328 k J m o l^{- 1}

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