Which of the following ionic species has the greatest proton affinity to form stable compound?

\(NH_2^-\)

The correct answer is option 2. \(NH_2^-\).

Proton affinity refers to the ability of an ion or a molecule to attract and bind a proton (H⁺) to form a stable compound. It's often related to the basicity of the species.

We can assess their proton affinities based on their basicity:

Hydrosulfide ion (\(HS^-\)): It has a lone pair of electrons and can accept a proton to form hydrogen sulfide (\(H_2S\)). It's a relatively strong base but weaker than other species in the list.

Amide ion (\(NH_2^-\)): It also has a lone pair of electrons and can accept a proton to form ammonia (\(NH_3\)). It's a stronger base than hydrosulfide ion.

Fluoride ion (\(F^-\)): It's highly electronegative and can stabilize a negative charge well. However, its basicity is weaker compared to the amide ion and hydrosulfide ion.

Iodide ion (\(I^-\)): It's the largest and least electronegative ion among the options. It has the weakest ability to stabilize a negative charge, but its basicity is stronger than that of fluoride ion.

Based on these considerations, the species with the greatest proton affinity to form a stable compound would be the one that has the strongest basicity. In this case, it's: 2. \(NH_2^-\) (Amide ion)