2
$\begingroup$

In Concise Inorganic Chemistry by JD Lee (4th edition; adapted by Sudarshan Guha), page 77 under the topic "Back Bonding" it is given:

The $\mathrm{2p}\pi$-$\mathrm{2p}\pi$ back bonding in $\ce{B(OH)3}$ and $\ce{B(OCH3)3}$ explains why $\ce{B-O-H}$ and $\ce{B-O-C}$ angles are $112°$ and $113°$, respectively.

Based on the above statement I arrived at the following:

Due to back bonding in $\ce{B(OH)3}$ and $\ce{B(OCH3)3}$ there is a partial double bond character in the bond between boron and oxygen in both the compounds. The oxygen in both compounds is $\mathrm{sp}^3$ hybridised and ideal bond angle is $109°28'$. But due to the partial double bond character and as we know the repulsion between a double bond and single bond is greater than between two single bonds the bond angles $\ce{B-O-H}$ and $\ce{B-O-C}$ are larger than the ideal $109°28'$.

Kindly tell whether my understanding is correct or not. If not, kindly give an explanation for the observed bond angles and their deviation from ideality.

$\endgroup$

Your Answer

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

Browse other questions tagged or ask your own question.