Which of the following forms the most stable gaseous negative ion?

The correct option is Cl.

To understand why chlorine ($\text{Cl}$) forms the most stable gaseous negative ion among the halogens ($\text{F}$, $\text{Cl}$, $\text{Br}$, and $\text{I}$), we must examine the concept of electron affinity.
Electron affinity is the energy change that occurs when an electron is added to a neutral gaseous atom to form a negative ion (anion). A more negative electron affinity value indicates a more release of energy, representing a more stable negative ion.

Typically, electron affinity decreases (becomes less negative) as we move down a group in the periodic table because the atomic size increases, placing the incoming electron further from the attractive force of the nucleus.
However, there is a well-known exception between fluorine ($\text{F}$) and chlorine ($\text{Cl}$).

Fluorine is a very small atom with its outer electrons residing in the relatively compact 2p subshell.
Due to this extremely small size and high electron density, when an extra electron is added to a fluorine atom, it experiences strong inter-electronic repulsion from the electrons already present in the small valence shell.

In contrast, chlorine has a larger atomic size with its valence electrons in the roomier 3p subshell.
The incoming electron experiences much less inter-electronic repulsion when entering the larger valence shell of chlorine.
As a result, chlorine releases more energy upon accepting an electron compared to fluorine, giving it the most negative electron gain enthalpy (electron affinity) of all the elements in the periodic table.

Therefore, chlorine ($\text{Cl}$) forms the most stable gaseous negative ion (${\text{Cl}}^{-}$).