Question

An alternative definition of electronegativity is Electronegativity= constant (I.E — E.A.) where I.E. is the ionization energy and E.A. is the electron affinity using the sign conventions of this book. Use data in Chapter 12 to calculate the (I.E. - E.A.) term for F, Cl, Br, and I. Do these values show the same trend as the electronegativity values given in this chapter? The first ionization energies of the halogens are 1678, 1255, 1138, and 1007 kJ/mol, respectively.

Answer 1

Answer:

The electronengativity values of given elements is as follows.

Fluorine - 4

Chlorine -3

Bromine - 2.9

Iodine- 2.5

Explanation:

Electronegativity  =consant (I.E-E.A)

The electron affinity and ionization energy values of the given elements is as follows.

(In attachment)

First we have to find the value of constant by using the fluorine atom to whom the electronengativity taken as "4".

Fluorine:

$4=constant[1678-(-327.8)]$

$Constant=0.0019942168$

By using this constant values we can find electronegatvity values of remaining elements.

Chlorine:

$Electronegativity=0.0019942168[1255+348.7]=3.1980\sim 3$

Therefore, electronegativity of chlorine is 3.

Bromine:

$Electronegativity=0.0019942168[1138+324.5]=2.91\sim 2.9$

Therefore, electronegativity of bromine is 2.9.

Iodine:

$Electronegativity=0.0019942168[1007+295.7]=2.59\sim 2.5$

Therefore, electronegativity of iodine is 2.5.