$ \Delta \mathrm{H}_{\mathrm{eg}}(\mathrm{Cl})=-349 \mathrm{~kJ} / \text { mole } $

$\Delta \mathrm{H}_{\mathrm{eg}}(\mathrm{F})=-333 \mathrm{~kJ} / \mathrm{mole} $

$ \Delta \mathrm{H}_{\mathrm{eg}}(\mathrm{I})=-296 \mathrm{~kJ} / \mathrm{mole} $

$ \Delta \mathrm{H}_{\mathrm{eg}}(\mathrm{Se})=-195 \mathrm{~kJ} / \mathrm{mole} $

$ \Delta \mathrm{H}_{\mathrm{eg}}(\mathrm{S})=-200 \mathrm{~kJ} / \text { mole } $

$ \Delta \mathrm{H}_{\mathrm{eg}}(\mathrm{Te})=-190 \mathrm{~kJ} / \mathrm{mole} $

$ \Delta \mathrm{H}_{\mathrm{eg}}(\mathrm{Po})=-174 \mathrm{~kJ} / \mathrm{mole} $

Electron gain enthalpy of $\mathrm{Se}$ is less negative than that of sulphur.

Option A: Δ_egH(I) < Δ_egH(At)

Iodine (I) is above Astatine (At) in Group 17 (Halogens), so ΔegH(I) should be more negative than ΔegH(At). This statement is correct.

Option B: Δ_egH(Te) < Δ_egH(Po)

Tellurium (Te) is above Polonium (Po) in Group 16 (Chalcogens), so ΔegH(Te) should be more negative than ΔegH(Po). This statement is correct.

Option C: Δ_egH(Cl) < Δ_egH(F)

Chlorine (Cl) is below Fluorine (F) in Group 17 (Halogens). However, the electron gain enthalpy of Chlorine is more negative than that of Fluorine. So, this statement is incorrect.

Option D: Δ_egH(Se) < Δ_egH(S)

Selenium (Se) is below Sulfur (S) in Group 16 (Chalcogens), so Δ_egH(Se) should be less negative (higher) than Δ_egH(S). This statement is correct.

Note :

Although the general trend across a period in the periodic table is for the electron gain enthalpy to become more negative, there is an exception between Fluorine (F) and Chlorine (Cl) in the halogen group (Group 17).

The electron gain enthalpy of Chlorine is more negative than that of Fluorine for the following reasons:

Small atomic size of Fluorine: Fluorine has the smallest atomic size among all the elements in the halogen group. As a result, the electron-electron repulsion in the 2p orbital of Fluorine is significant when an extra electron is added. This repulsion reduces the energy released during the electron gain process, leading to a less negative electron gain enthalpy value for Fluorine compared to Chlorine.
High effective nuclear charge in Fluorine: Fluorine has a higher effective nuclear charge than Chlorine, which means the positively charged nucleus attracts the negatively charged electrons more strongly in Fluorine. The high effective nuclear charge in Fluorine also contributes to the electron-electron repulsion in the 2p orbital, further reducing the energy released when gaining an electron.

Despite the general trend of electron gain enthalpy becoming more negative across a period, the combination of Fluorine's small atomic size and high effective nuclear charge results in Chlorine having a more negative electron gain enthalpy than Fluorine.