Meaning of electron gain enthalpy

Electron gain enthalpy is the enthalpy change when an electron is added to an isolated gaseous atom: $X(g) + e^- \rightarrow X^-(g)$

More negative electron gain enthalpy means greater tendency to accept an electron.

Elements involved

The elements are from group 16: $\text{O, S, Se, Te}$

General trend down the group

Usually, electron gain enthalpy becomes less negative down a group because atomic size increases and the incoming electron is added farther from the nucleus.

So a naive trend would be: $\text{O} > \text{S} > \text{Se} > \text{Te}$ in size of negativity? Actually, this must be checked carefully because oxygen is an exception.

Special case of oxygen

Oxygen is very small in size. The added electron enters the compact $2p$ orbital, where inter-electronic repulsion is high.

Because of this, oxygen has less negative electron gain enthalpy than sulfur.

Thus: $\Delta_{eg}H(\text{S}) < \Delta_{eg}H(\text{O})$ if written in actual signed values.

But the question says "($-$ ve value)", meaning we compare the magnitude of negative value. Therefore: $\text{S} > \text{O}$ in negative electron gain enthalpy magnitude.

Order among S, Se, Te

After sulfur, the usual down-group trend applies approximately: $\text{S} > \text{Se} > \text{Te}$ in magnitude of negative electron gain enthalpy.

Combining the trend

Therefore, the correct order of negative electron gain enthalpy is: $\boxed{\text{S} > \text{Se} > \text{Te} > \text{O}}$

Match with options

The printed options in the prompt appear corrupted for C and D, but the chemically correct order is: $\boxed{\text{O} < \text{S} > \text{Se} > \text{Te}}$ or equivalently, $\boxed{\text{S} > \text{Se} > \text{Te} > \text{O}}$

This corresponds to Option C as intended.

Comparison with stored answer

Stored correct answer = C.

My derived answer also corresponds to C.