Reactions of boron trihalides

We need to first note that this is a Lewis acid-base reaction involving donation of a lone pair from ammonia to the boron atom. In particular, it is donated to the vacant 2p orbital on the boron atom. What it entails is a change in the geometry of the boron trihalide species. Previously, the molecular geometry about B is trigonal planar and after reaction, it changes to a tetrahedral geometry. Clearly, we have three energetic factors to consider: Firstly, the strength of the B-N bond formed. Secondly, the strength of B-Halogen $pi$ interactions. Thirdly, the energy involved in the change of geometry. Here, it seems that the most significant factor that greatly varies between the boron trihalides is the strength of the so-called " $pi$ back-bonding interactions".

In addition to the $sigma$ bonding between the boron atom and the halogen atoms, there is also additional $pi$ bonding interactions between them. Specifically, the vacant 2p orbital of boron can overlap with and interact with the filled p orbitals of the halogen atoms. Note that these orbitals are parallel to each other and are thus compatible to interact with each other. The strongest $pi$ bonds are formed between fluorine and boron while the weakest ones are the ones between iodine and boron. Recall that $pi$ bonds are strongly distance-dependent. As the $sigma$ bond between B and I is the weakest and also because I is the largest halogen here with the most diffuse orbitals, the distance between B and I would be the longest and hence, we can reasonably predict that B-I $pi$ bonding would be the weakest.

To change configuration from a planar one to a tetrahedral one, the B-Halogen $pi$ bonding would have to be significantly weakened. Note that in the tetrahedral configuration, the halogen's p orbitals and boron's 2p orbital are no longer parallel and interaction becomes significantly reduced. Since the $pi$ bonds between B and I are most easily broken, the configuration change would occur most easily for boron triiodide.