Question

The ionisation enthalpies of atoms A and B are 400 and 300 $kcalmo{l}^{-l}$ respectively. The electron gain enthalpies of these atoms are 80.0 and 85.0 $kcalmo{l}^{-l}$ respectively. Which of the atoms has a higher electronegativity?

• A. A
• B. B
• C. Both have same electronegativity
• D. Can't be predicted

Answer 1

The correct option is A A

$\text{Electronegativity}=\frac{\text{Ionisation energy + Electron affinity}}{2}$
Since units are in $kcalmo{l}^{-1}$, we need to convert it. The formula is applicable when I.E and E.A are given in eV/atom. So,
$X_A$ = $\frac{400+80}{2\times 62.5}$ = 3.84

$X_B$ = $\frac{300+85}{2\times 62.5}$ = 3.08
Hence, electronegativity of A is greater than that of B.