In 1931, Pauling defined the electronegativity of an atom as the tendency of the atom to attract electrons to itself when combined in a compound. The implication is that when a covalent bond is formed, the electrons used for bonding need not be shared equally by both atoms. If the bonding electrons spend more time around one atom, that atom will have a `del-`charge, and consequently the other atom will have a `del+` charge. In the extreme case, where the bonding electrons are round one atom all of the time, the bond is ionic. Pauling and others have attempted to relate the electronegativity difference between two atoms to the amount of ionic character in the bond between them.
Mulliken
In 1934, Mulliken suggested an alternative approach to electronegativity based on the ionization energy and electron affinity of an atom. Consider two atoms A and B. If an electron is transferred from A to B, forming ions `A^(+)` and `B^(-)`, then the energy change is the ionization energy of atom `A(I_(A))` minus the electron affinity of atom `B(E_(B))`, that is `I_(A)-E_(B)`. Alternatively, if the electron was transferred the other way to give `B^(+)` and `A^(-)` ions, then the energy changed would be `I_(B) - I_(A)`. If `A^(+)` and `B^(-)` are actually formed, then this process require less energy , and
`(I_(A) - E_(B)) lt (I_(B) - E_(A))`
Rerranging `(I_(A)+E_(A)) lt (I_(A)-E_(B))`
Now with respect to electronegativity for the same change, `E.N_(A) lt E.N_(B)`.
Thus Mulliken suggested that electronegativity is proportional to `I.E + E.A` and could be regarded as the average of the ionization energy and the electron affinity of an atom.
Electronegativity `=((I+E))/(2)`
Mulliken used I and E values measured in electron volts, and the values were about 2.8 times alrger than the Pauling values. It is to be noted that `I.E`. and `E.A` values are defined for singular gaseous atoms.
For a reaction `A(g)+B(g) rar AB(S)` the enthalpy change for the recation will be [assuming `AB(s)` to be an ionic compound] If `I.E_(A)+E.A_(A) lt I.E_(B) +E.A_(B)`

Options

A. `DeltaH_(I.E.A) +DeltaH_(eg_(A))+DeltaH_(I.E_(AB(s)))`
B. `DeltaH_(eg_(A)) +DeltaH_(I.E_(B)) +DeltaH_(L.E_(AB(s)))`
C. `I.E_(A)+E.A_(B)+DeltaH_(L.E_(AB_(s)))`
D. `I.E_(B)+E.A_(B)+Delta_(L.E_(AB(s)))`

Answer

C

Explanation

As `I.E_(A)+E.A_(A) lt I.E_(B)+E.A_(B)` the compound `AB(s)`, will be `A^(+)B^(-)(s)` and not `A^(-)B^(+)(s)`.

From Hess's Law
`DeltaH = DeltaH_(I.E.) +DeltaH_(eg)+DeltaH_(I.E.(s))`