Why boron trihydride is weaker lewis acid than boron tribromide and boron tri-iodide?

There is no back-bonding in case of boron trihydride so it is a stronger lewis acid than boron trichloride ( even though chlorine atom is more electronegative than hydrogen atom).


Then **why boron trihydride is weaker lewis acid than boron tribromide and boron tri-iodide?**




Predicting the acidic strengths of $\ce{BX\_3}$ using back-bonding is *one plausible* theory. This article1 hints us to use bond-strength data to predict the same:



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> The weaker Lewis acid strength of $\ce{BF3}$ compared to $\ce{BCl3}$ with respect to bases such as $\ce{NMe3}$ is explained in terms of the ligand close-packing (LCP) model. The halogen ligands remain close-packed during the formation of the complex leading to an increase in the $\pu{B−Hal}$ bond length. Because a $\ce{B−F}$ bond is stronger than a $\ce{B−Cl}$ bond, more energy is required to stretch a $\ce{B−F}$ bond than a $\ce{B−Cl}$ bond. Hence $\ce{BF3}$ is a weaker acid than $\ce{BCl3}$.
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Pulling up the bond strength data, I get:


\begin{array}{c|c}
\mathbf{Bond} & \mathbf{Strength (KJ/mole) } \\\hline
\text{B-H} & \text{389}\\
\text{B-F} & \text{613} \\
\text{B-Cl} & \text{456}\\
\text{B-Br} & \text{377}\\
\text{B-I} & \text{?}\end{array}


Building up on these two pieces of information, it can be said that since $\ce{B-H}$ bond is stronger than $\ce{B-Br}$ and $\ce{B-I}$ (see below), $\ce{BH3}$ is a weaker acid than $\ce{BBr3}$ and $\ce{BI3}$.


I'm afraid I couldn't find the data for $\ce{B-I}$ bond, but since we can compare the bond-strengths of boron trifluorides on the basis of back-bonding theory, we can be sure that the strength of $\ce{B-I}$ bond will be $\lt \pu{377 kJ mol^-1}$.


**References**


1. Ligand Close-Packing and the Lewis Acidity of BF3 and BCl3
Bryan D. Rowsell, Ronald J. Gillespie, and George L. Heard
Inorganic Chemistry **1999** 38 (21), 4659-4662
DOI: [10.1021/ic990713m](https://pubs.acs.org/doi/abs/10.1021/ic990713m)


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