1. Identify the elements

We are given:

• $A = \mathrm{Ar}$
• $B = \mathrm{Br}$
• $C = \mathrm{F}$
• $D = \mathrm{S}$

We need the correct order of electron gain enthalpy.

2. Recall the trend of electron gain enthalpy

Electron gain enthalpy is the enthalpy change when an electron is added to a gaseous atom:

$X(g) + e^- \to X^-(g)$

• More negative electron gain enthalpy means greater tendency to accept an electron.
• Noble gases have positive or least favorable electron gain enthalpy because their valence shell is already complete.

3. Compare the given elements

(i) Argon, $\mathrm{Ar}$

• Noble gas with completely filled shell.
• Adding an electron is highly unfavorable.
• So $\mathrm{Ar}$ has the highest electron gain enthalpy value (most positive / least negative).

(ii) Sulfur, $\mathrm{S}$

• Chalcogen, tends to gain electron.
• Electron gain enthalpy is negative.

(iii) Bromine, $\mathrm{Br}$

• Halogen, electron gain enthalpy is strongly negative.

(iv) Fluorine, $\mathrm{F}$

• Also a halogen, but due to very small size, the incoming electron experiences greater electron-electron repulsion in compact $2p$ orbitals.
• Therefore, although highly negative, its electron gain enthalpy is less negative than chlorine and also less negative than bromine.

Thus among $\mathrm{F}$ and $\mathrm{Br}$:

$\mathrm{Br} < \mathrm{F}$

if we compare in the usual numerical sense of enthalpy values (because Br is more negative).

4. Overall order

Using actual periodic behavior:

• $\mathrm{Ar}$ is highest.
• $\mathrm{S}$ is less favorable than halogens but more favorable than noble gas.
• $\mathrm{Br}$ has more negative electron gain enthalpy than $\mathrm{F}$.

Hence the order is:

$\mathrm{Ar} > \mathrm{S} > \mathrm{F} > \mathrm{Br}$

Now substituting the labels:

• $A = \mathrm{Ar}$
• $D = \mathrm{S}$
• $C = \mathrm{F}$
• $B = \mathrm{Br}$

So:

$A > D > C > B$

5. Match with the options

Option A is:

$A > D > C > B$

So the correct option is A.

6. Compare with stored correct answer

Stored correct answer: B

But option B says:

$A > D > B > C$

That would imply:

$\mathrm{Ar} > \mathrm{S} > \mathrm{Br} > \mathrm{F}$

This is incorrect because bromine has more negative electron gain enthalpy than fluorine, so in the usual greater-than ordering of numerical enthalpy values we must write:

$\mathrm{F} > \mathrm{Br}$

not $\mathrm{Br} > \mathrm{F}$.

Therefore, the stored answer appears to be incorrect.