0.5 moles of gaseousnon-metallic X^(-) anions (having positive electron affinity) requires 806.4 kJ energy to get completely converted into gaseous X^(+) ions. Calculate Pauling's electronegativity of the elements X. Use Avogardo's no = 6 xx 10^(23) and 1eV = 1.6 xx 10^(-19)J. [Use the fact that, Pauling's electronegativity =("Mulliken's electronegativity")/(2.8) and Mulliken's electronegativity =("Ionisation energy+ Electron affinity")/(2)]

0.5 moles of gaseousnon-metallic X^(-) anions (having positive electro

1.	0.5 moles of gaseousnon-metallic X^(-) anions (having positive electron affinity) requires 806.4 kJ energy to get completely converted into gaseous X^(+) ions. Calculate Pauling's electronegativity of the elements X. Use Avogardo's no = 6 xx 10^(23) and 1eV = 1.6 xx 10^(-19)J. [Use the fact that, Pauling's electronegativity =("Mulliken's electronegativity")/(2.8) and Mulliken's electronegativity =("Ionisation energy+ Electron affinity")/(2)]
Answer» SOLUTION :Let `X rarr X^(+)e^(-)` `:. IE` (energy absorbed) = aeV PER atom and `X +e^(-) rarr X^(-)` `:. EA` (energy RELEASED) =- beV per atom Now `(N)/(2)` of `X^(-)` will lose two elecctrons to give `(N)/(2)X^(+)` `:. X^(-) rarr X +E^(-)` `:.` energy absorbed `= + beV` per atom. `X rarr X^(+) +e^(-)` `:.` energy absorbed `=+ aeV` per atom. `:. a xx (N)/(2) + b xx (N)/(2) = (806.4xx 1000)/(1.6 xx 10^(-19)) eV` or `(a+b) = (806.4 xx 1000 xx 2)/(1.6 xx 10^(-19) xx 6 xx 10^(23)) = (806.4 xx 2)/(1.6 xx 60) = 16.8` `xhi_(m) = (IE +EA)/(2) = (16.8)/(2) = 8.4` `:.` Mulliken's electronegativity = 8.4 Pauling's electronegativity `= (8.4)/(2.8) = 3.0`