Concept: The ability of a compound to form dimers or polymers is related to the availability of vacant orbitals and the number of valence electrons, which dictates the type of bonding (e.g., 3-center-2-electron bonds) it can form to achieve stability.

Why (A) is correct:

Boron in BH₃ is sp² hybridized and has one vacant p-orbital. To achieve an octet, two BH₃ units dimerize to form B₂H₆ (diborane) through two 3-center-2-electron B-H-B bonds. Each boron atom in B₂H₆ is sp³ hybridized. This dimerization satisfies the electron deficiency of boron.

Beryllium in BeH₂ is sp hybridized and has two vacant p-orbitals. Each beryllium atom needs to form four bonds to achieve an octet. In the solid state, BeH₂ forms a polymeric structure where each Be atom is tetrahedrally coordinated by four hydrogen atoms, forming 3-center-2-electron Be-H-Be bonds. This extensive bonding is possible due to the presence of two vacant orbitals on each beryllium atom, allowing for the formation of multiple 3-center-2-electron bonds, leading to a polymeric structure. Dimerization is also possible, but the presence of two vacant orbitals allows for further extension into a polymer.

Option Analysis:

• A) BH₃ has one vacant orbital while BeH₂ has two vacant orbitals, so BeH₂ can extend bonding further. This is correct. The number of vacant orbitals directly correlates with the extent of electron-deficient bonding (like 3c-2e bonds) and thus the ability to form dimers or polymers.
• B) BH₃ forms only terminal B–H bonds, whereas BeH₂ forms only ionic bonds. This is incorrect. Both BH₃ (in B₂H₆) and BeH₂ form 3-center-2-electron bonds, which are covalent in nature. BeH₂ is predominantly covalent.
• C) BH₃ has higher electronegativity of boron compared to beryllium, preventing polymer formation. This is incorrect. Electronegativity differences primarily influence bond polarity, not necessarily the extent of polymerization in electron-deficient compounds. Both are electron-deficient.
• D) BeH₂ cannot form three-center two-electron bonds, while BH₃ can. This is incorrect. Both BeH₂ and BH₃ (in their aggregated forms) stabilize themselves by forming 3-center-2-electron bonds.

Correct Answer: (A)