Electron gain enthalpy with negative sign of fluorine is less than that of chlorine due to :

The electron gain enthalpy with a negative sign represents the energy released when an electron is added to a neutral gaseous atom. A more negative value indicates greater energy release, meaning the atom has a higher tendency to gain an electron. The question states that fluorine has a less negative electron gain enthalpy than chlorine, meaning chlorine releases more energy upon gaining an electron compared to fluorine.

To understand why, consider the atomic structures of fluorine and chlorine. Fluorine (atomic number 9) is in period 2, group 17, and has the electron configuration $$1s^2 2s^2 2p^5$$. Chlorine (atomic number 17) is in period 3, group 17, with the configuration $$1s^2 2s^2 2p^6 3s^2 3p^5$$. Generally, electron gain enthalpy becomes more negative down a group due to increasing atomic size. However, fluorine is an exception because its electron gain enthalpy is less negative than chlorine's.

The primary reason is the smaller atomic size of fluorine. Fluorine's valence electrons are in the second shell (n=2), resulting in a compact atomic radius. When an extra electron is added to form the fluoride ion ($$F^-$$), it enters the already crowded 2p subshell. This causes significant electron-electron repulsion because the high electron density is confined in a small space. The repulsion reduces the energy released during electron addition, making the electron gain enthalpy less negative.

In contrast, chlorine has a larger atomic size because its valence electrons are in the third shell (n=3). When an electron is added to form the chloride ion ($$Cl^-$$), it occupies the more spacious 3p subshell. The electron density is spread over a larger volume, reducing electron-electron repulsion. Consequently, more energy is released, making chlorine's electron gain enthalpy more negative than fluorine's.

Now, evaluating the options:

Option A suggests high ionization enthalpy of fluorine. Ionization enthalpy is the energy required to remove an electron, which is unrelated to electron gain enthalpy (energy change when adding an electron). Fluorine does have high ionization enthalpy, but this does not explain the electron gain enthalpy trend.

Option B claims a smaller size for chlorine, but chlorine is larger than fluorine due to an additional electron shell. This is incorrect.

Option C states the smaller size of fluorine, which aligns with our reasoning. The compact size leads to high interelectronic repulsion upon electron addition, reducing the energy released.

Option D mentions a bigger size of fluorine's 2p orbital. However, the 2p orbital is smaller than chlorine's 3p orbital because it is closer to the nucleus (n=2 vs. n=3). Thus, fluorine's 2p orbital is not bigger; it is smaller.

Therefore, the correct reason is the smaller size of the fluorine atom.

Hence, the correct answer is Option C.