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

Question

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 (A) 2.8 (B) 0.0 (C) 4.0 (D) 2.6

Tardigrade Admin · Accepted Answer

I.E. + E.A = 275 + 86 = 361 kcal mol-1 =361× 4.184=1510.42 kJ mol-1 electronegativity =(I . E . + E . A ./540) So, electronegativity =(1510.42/540)=2.797=2.8

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

2.8

0.0

4.0

2.6

I.E. + E.A = 275 + 86 = 361 kcal mol^-1
$= 361 \times 4.184 = 1510.42$ kJ mol^-1
electronegativity $= \frac{I . E . + E . A .}{540}$
So, electronegativity $= \frac{1510.42}{540} = 2.797 = 2.8$

Q. 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

2.8

0.0

4.0

2.6

I.E. + E.A = 275 + 86 = 361 kcal mol-1 =361×4.184=1510.42 kJ mol-1 electronegativity =540I.E.+E.A.​ So, electronegativity =5401510.42​=2.797=2.8

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

I.E. + E.A = 275 + 86 = 361 kcal mol^-1
$= 361 \times 4.184 = 1510.42$ kJ mol^-1
electronegativity $= \frac{I . E . + E . A .}{540}$
So, electronegativity $= \frac{1510.42}{540} = 2.797 = 2.8$