In which one of the following molecules strongest back donation of an electron pair from halide to boron is expected?

We have four boron trihalides: $$\text{BF}_3,\; \text{BCl}_3,\; \text{BBr}_3,\; \text{BI}_3$$. In every one of these molecules the central boron atom possesses only six valence electrons, so it is electron-deficient and can accept electron density. Each halogen atom, on the other hand, has three lone pairs and can donate one of those lone pairs into the empty $$2p$$ orbital of boron. This donation of a filled orbital on the halogen into an empty orbital on boron is called back bonding (or $$p\;(\pi)\!-\!p\;(\pi)$$ back donation when both orbitals are $$p$$ orbitals).

The strength of such back donation depends primarily on the efficiency of orbital overlap. The guiding principle is:

$$\text{Better size and energy match } \Longrightarrow \text{stronger overlap } \Longrightarrow \text{stronger back bonding}.$$

For boron, the relevant empty orbital is the $$2p$$ orbital. Therefore the ideal donor would also have a $$2p$$ orbital, because orbitals of the same principal quantum number $$n$$ (and therefore similar size and energy) overlap most effectively. Let us compare each halogen:

• In $$\text{F}$$ the lone pair resides in a $$2p$$ orbital (same principal quantum number as boron’s $$2p$$).
• In $$\text{Cl}$$ the lone pair is in a $$3p$$ orbital (larger and higher in energy than $$2p$$).
• In $$\text{Br}$$ the lone pair is in a $$4p$$ orbital (even larger and higher in energy).
• In $$\text{I}$$ the lone pair is in a $$5p$$ orbital (largest mismatch of all).

Because the radial size and energy separation between $$2p$$ and $$2p$$ orbitals are minimal, the overlap in $$\text{BF}_3$$ is far superior to that in the other trihalides. As the halogen becomes heavier (from F to I) the difference in size and energy between its lone-pair orbital and boron’s empty $$2p$$ orbital keeps increasing, so the overlap - and hence the back donation - keeps weakening.

Therefore the order of back-bond strength is

$$\text{BF}_3 \; > \; \text{BCl}_3 \; > \; \text{BBr}_3 \; > \; \text{BI}_3.$$

So the molecule with the strongest halide → boron back donation is $$\text{BF}_3$$, which corresponds to option 4.

Hence, the correct answer is Option 4.