First, recall that electron-gain enthalpy $\Delta_{\mathrm{eg}}H$ is often quoted as a (negative) numerical value. A “more negative” electron-gain enthalpy means that an atom gains an electron more favorably.

1. Known Trends in the Periodic Table

Halogens (Group 17):

Although one might expect the electron-gain enthalpy to become steadily less negative as we go down the group (due to increasing size and shielding), there is an anomaly at the top:

$ \text{Most negative: } \mathrm{Cl} > \mathrm{F} > \mathrm{Br} > \mathrm{I} > \mathrm{At} \quad(\text{in terms of magnitude of negativity}). $

Numerically (since these values are negative), that ordering translates to:

$ \Delta_{\mathrm{eg}}H(\mathrm{Cl}) < \Delta_{\mathrm{eg}}H(\mathrm{F}) < \Delta_{\mathrm{eg}}H(\mathrm{Br}) < \Delta_{\mathrm{eg}}H(\mathrm{I}) < \Delta_{\mathrm{eg}}H(\mathrm{At}). $

The smaller (more negative) number on a number line corresponds to a larger (more exothermic) electron-gain enthalpy.

Chalcogens (Group 16):

A similar pattern holds, with an anomaly near oxygen. Going down from S to Se to Te to Po, the electron-gain enthalpy becomes less negative:

$ \text{Most negative: } \mathrm{S} > \mathrm{Se} > \mathrm{Te} > \mathrm{Po} \quad(\text{in magnitude}). $

Numerically (all negative values), that means:

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

2. Checking Each Option

We interpret each statement as a comparison of numerical values (remembering they are negative).

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

Iodine’s electron-gain enthalpy is more negative than astatine’s.

Numerically, $\Delta_{\mathrm{eg}}H(\mathrm{I})$ is indeed a “smaller” (i.e. more negative) number than $\Delta_{\mathrm{eg}}H(\mathrm{At})$.

So this statement is correct.

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

Tellurium (Te) is above polonium (Po) in Group 16, so Te typically has a more negative electron-gain enthalpy than Po.

Numerically, that means $\Delta_{\mathrm{eg}}H(\mathrm{Te})$ is smaller (more negative) than $\Delta_{\mathrm{eg}}H(\mathrm{Po})$.

This statement is correct.

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

Among halogens, $\mathrm{Cl}$ actually has the most negative $\Delta_{\mathrm{eg}}H$. Fluorine’s value, while also negative, is a bit less so (because of small-orbital repulsion in F).

Numerically, that means $\Delta_{\mathrm{eg}}H(\mathrm{Cl})$ is indeed a smaller (more negative) number than $\Delta_{\mathrm{eg}}H(\mathrm{F})$.

So this statement is correct.

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

In Group 16, sulfur (S) has a more negative electron-gain enthalpy than selenium (Se).

Numerically, that implies

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

not the other way around.

The statement says $\Delta_{\mathrm{eg}}H(\mathrm{Se}) < \Delta_{\mathrm{eg}}H(\mathrm{S})$, which would mean Se is more negative than S—but that is not correct.

Hence, (D) is the incorrect statement.