1. What is 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 value means greater tendency to accept an electron.

2. Compare $\mathrm{Cl}$ and $\mathrm{F}$

Although fluorine is more electronegative, its electron gain enthalpy is less negative than that of chlorine.

This is because the incoming electron enters the very small and compact $2p$ subshell in fluorine, where electron-electron repulsion is high.

Hence:

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

So chlorine has more negative electron gain enthalpy than fluorine.

3. Compare $\mathrm{Te}$ and $\mathrm{Po}$

Te and Po are in group 16. In general, electron gain enthalpy becomes less negative down a group.

Thus moving from Te to Po:

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

So tellurium has more negative electron gain enthalpy than polonium.

4. Now compare halogens with chalcogens

Halogens generally have much more negative electron gain enthalpy than group 16 elements, because they need only one electron to attain noble gas configuration.

Therefore:

$\mathrm{Cl},\mathrm{F} \text{ have more negative electron gain enthalpies than } \mathrm{Te},\mathrm{Po}$

So the increasing order (from less negative to more negative) should place Po and Te first, then F and Cl.

Thus:

$\mathrm{Po} < \mathrm{Te} < \mathrm{F} < \mathrm{Cl}$

5. Check options

• A: $\mathrm{F < Cl < Te < Po}$ ❌ halogens cannot be less negative than Te and Po
• B: $\mathrm{Po < Te < F < Cl}$ ✅ correct
• C: $\mathrm{Te < Po < Cl < F}$ ❌ wrong order for Te and Po, and also for Cl and F
• D: $\mathrm{Cl < F < Te < Po}$ ❌ reverse trend

6. Final answer

The correct order of electron gain enthalpies is:

$\boxed{\mathrm{Po < Te < F < Cl}}$

So the correct option is B.