If the ionization enthalpy and electron gain | vedclass

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(A) The electronegativity $(X)$ on the Pauling scale is related to the sum of ionization enthalpy $(I.E.)$ and electron gain enthalpy $(E.A.)$ by the

Vedclass · Accepted Answer

$2.8$

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(A) The electronegativity $(X)$ on the Pauling scale is related to the sum of ionization enthalpy $(I.E.)$ and electron gain enthalpy $(E.A.)$ by the formula: $X = \frac{I.E. + E.A.}{540}$ (where energies are in $kJ \ mol^{-1}$).Given: $I.E. = 275 \ kcal \ mol^{-1}$ and $E.A. = 86 \ kcal \ mol^{-1}$.Sum $= 275 + 86 = 361 \ kcal \ mol^{-1}$.Convert to $kJ \ mol^{-1}$ using $1 \ kcal = 4.184 \ kJ$: $361 	imes 4.184 = 1510.42 \ kJ \ mol^{-1}$.Electronegativity $= \frac{1510.42}{540} \approx 2.797 \approx 2.8$.

If the ionization enthalpy and electron gain enthalpy of an element are $275$ and $86 \ kcal \ mol^{-1}$ respectively,then the electronegativity of the element on the Pauling scale is

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