Question

The ionisation enthalpies of atoms A and B are $200$ and $300kcalmo{l}^{-l}$ respectively. The electron gain enthalpies of these atoms are $80.0$ and $85.0kcalmo{l}^{-l}$ respectively. Which of the atoms has higher electronegativity?

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

Answer 1

The correct option is B B

$\text{Electronegativity}=\frac{\text{Ionisation energy + Electron affinity}}{2}$
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,
$X_A$ = $\frac{200+80}{2\times 62.5}$ = 2.24

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