Step 1: Understanding the Concept:
The stability of hydrides depends heavily on their bonding characteristics. Elements in the second period form different types of hydrides: ionic, covalent, and electron-deficient.
Step 2: Detailed Explanation:
Let's look at the nature of each given hydride:
HF (Hydrogen Fluoride): A highly stable covalent compound due to the extremely strong and polar H-F bond.
NH$_3$ (Ammonia): A very stable, electron-precise covalent hydride with strong N-H bonds.
LiH (Lithium Hydride): A stable, saline (ionic) hydride featuring strong electrostatic attractions between Li$^+$ and H$^-$ ions in its crystal lattice.
BeH$_2$ (Beryllium Hydride): Beryllium is a small atom with high ionization energy, leading it to form covalent rather than ionic bonds. However, it lacks enough electrons to form a complete octet, making BeH$_2$ an {electron-deficient} compound. It exists as a complex polymeric structure with 3-center 2-electron (3c-2e) hydrogen bridges.
Because of its electron-deficient and polymeric nature, BeH$_2$ is thermodynamically the least stable of the group. It is notoriously difficult to synthesize directly from its elements and decomposes relatively easily compared to the stable octets/lattices in the other choices.
Step 3: Final Answer:
BeH$_2$ is the least stable hydride.
A group 15 element forms \( d\pi - d\pi \) bond with transition metals. It also forms a hydride, which is the strongest base among the hydrides of other group members that form \( d\pi - d\pi \) bonds. The atomic number of the element is …….