2
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ Backbonding is ineffective in case of tribromides and triiodides since their 4p and 5p orbitals dont really overlap well with the 2p of boron. Morever they are electronegative and can make the boron more positive, and increase the tendency to accept a pair of electrons. $\endgroup$ – Pritt Balagopal May 17 '17 at 6:39
4
$\begingroup$

Predicting the acidic strengths of $\ce{BX_3}$ using back-bonding is one plausible theory. This article http://pubs.acs.org/doi/abs/10.1021/ic990713m hints us to use bond-strength data to predict the same:

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 $\ce{BCl3}$.

Pulling up the bond strength data, I get:

Bond |  Strength (KJ/mole) 
--------------------------
B-H  |  389 
B-F  |  613 
B-Cl |  456 
B-Br |  377
B-I  |   ?

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 $\lt \pu{377 kJ mol^-1}$.

$\endgroup$
  • $\begingroup$ Did you write $\ce {B}-\ce {Cl} $ where you meant $\ce {B}-\ce {Br} $? $\endgroup$ – Oscar Lanzi May 17 '17 at 13:03
  • 1
    $\begingroup$ @OscarLanzi Shame I answered so badly, edited the answer! $\endgroup$ – Berry Holmes May 17 '17 at 13:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.