Question

An element has AiH₁ 1314 and AegH₁ = -141 (KJmol-¹) respectively. The E.N of the element on Mullickan scale is:​

Answer 1

Answer:

According to Mulliken, electronegativity of an atom is average of ionization energy and electron affinity (in eV).

n  

m

​

=  

2

IE+EA

​

 

If ionisation energy and electron affinity are in kcal mol  

−1

.

n=  

125

IE+EA

​

=  

125

275+86

​

=2.88

Explanation: