Concept: $\pi$-back bonding (or back donation) occurs when a filled orbital on one atom (usually a lone pair or a filled p-orbital) overlaps with an adjacent empty orbital (usually a vacant p-orbital or d-orbital) on another atom. This typically happens between a Lewis acid (electron pair acceptor) and a Lewis base (electron pair donor) where the central atom has an empty orbital and the surrounding atoms have lone pairs.

Why (B) BF${}_3$ is correct:

In BF${}_3$, boron (B) is the central atom and has an empty 2p orbital. Each fluorine (F) atom has three lone pairs in its 2p orbitals. The empty 2p orbital of boron can accept electron density from the filled 2p orbitals of the fluorine atoms, forming a $\pi$-bond. This delocalization of electrons stabilizes the molecule and reduces the electron deficiency of boron. This is a classic example of p$\pi$-p$\pi$ back bonding.

Option Analysis:

• A) AlCl${}_3$: Aluminum (Al) has an empty 3p orbital, and chlorine (Cl) has lone pairs in its 3p orbitals. While back bonding is possible, it is much weaker and less significant than in BF${}_3$ due to the larger size of Al and Cl atoms, leading to less effective orbital overlap (3p-3p overlap is less efficient than 2p-2p overlap). AlCl${}_3$ primarily exists as a dimer (Al${}_2$Cl${}_6$) in the solid and vapor phases, where Al achieves an octet through coordinate bonds.
• C) SF${}_4$: Sulfur (S) is the central atom and has lone pairs and expanded octet. Fluorine has lone pairs. Sulfur has vacant 3d orbitals, and fluorine has 2p orbitals. While d$\pi$-p$\pi$ bonding can occur in some sulfur compounds, SF${}_4$ is generally described by hypervalent bonding rather than significant back bonding from fluorine to sulfur. The primary bonding is covalent, and sulfur already has an octet (or expanded octet) with lone pairs.
• D) BH${}_3$: Boron (B) has an empty 2p orbital, but hydrogen (H) does not have any lone pairs or filled p-orbitals to donate electrons for back bonding. Therefore, back bonding is not possible in BH${}_3$. BH${}_3$ exists as a dimer (B${}_2$H${}_6$) where boron achieves an octet through three-center two-electron bonds.

Correct Answer: (B)