Meaning of electron gain enthalpy

Electron gain enthalpy, $\Delta_{\mathrm{eg}}H$, is the enthalpy change when an electron is added to an isolated gaseous atom: $X(g) + e^- \to X^-(g)$

More negative $\Delta_{\mathrm{eg}}H$ means greater tendency to accept an electron.

General trend

Across a period, electron gain enthalpy generally becomes more negative.

Down a group, it generally becomes less negative due to increase in size and shielding.

However, there are important exceptions:

• $\mathrm{Cl}$ has more negative electron gain enthalpy than $\mathrm{F}$.
• $\mathrm{S}$ has more negative electron gain enthalpy than $\mathrm{O}$. Similar size/crowding effects are relevant in p-block comparisons.

Check each option

Option A

$\Delta_{\mathrm{eg}}H(\mathrm{I}) < \Delta_{\mathrm{eg}}H(\mathrm{At})$

I and At are in group 17. Down the group, electron gain enthalpy becomes less negative.

So iodine should have more negative electron gain enthalpy than astatine: $\Delta_{\mathrm{eg}}H(\mathrm{I}) < \Delta_{\mathrm{eg}}H(\mathrm{At})$

Hence, A is correct.

Option B

$\Delta_{\mathrm{eg}}H(\mathrm{Te}) < \Delta_{\mathrm{eg}}H(\mathrm{Po})$

Te and Po are in group 16. Down the group, electron gain enthalpy becomes less negative.

Therefore Te should have more negative value than Po: $\Delta_{\mathrm{eg}}H(\mathrm{Te}) < \Delta_{\mathrm{eg}}H(\mathrm{Po})$

Hence, B is correct.

Option C

$\Delta_{\mathrm{eg}}H(\mathrm{Cl}) < \Delta_{\mathrm{eg}}H(\mathrm{F})$

This is the well-known exception in halogens. Because the incoming electron enters the compact $2p$ orbital in F, interelectronic repulsion is higher. In Cl, electron addition occurs in the larger $3p$ orbital, so more energy is released.

Thus: $\Delta_{\mathrm{eg}}H(\mathrm{Cl}) < \Delta_{\mathrm{eg}}H(\mathrm{F})$

Hence, C is correct.

Option D

$\Delta_{\mathrm{eg}}H(\mathrm{Se}) < \Delta_{\mathrm{eg}}H(\mathrm{S})$

S and Se are in group 16. Due to the small size of sulfur, adding an electron to S causes relatively greater repulsion than in Se. Therefore sulfur is an exception-like case here, and selenium can have a slightly more negative electron gain enthalpy than sulfur.

So the correct relation is: $\Delta_{\mathrm{eg}}H(\mathrm{Se}) < \Delta_{\mathrm{eg}}H(\mathrm{S})$

Therefore, D is also correct, not incorrect.

Conclusion

Since A, B, C, and D are all correct relations, there is no incorrect option among the given choices.

In particular, option D is actually true because $\Delta_{\mathrm{eg}}H(\mathrm{Se})$ is slightly more negative than $\Delta_{\mathrm{eg}}H(\mathrm{S})$.

Comparison with stored answer

The stored answer is D, but from known periodic trends and exceptions, D is not incorrect. So I disagree with the stored answer.