Question

The correct Lewis acid order for boron halides is:
A)${\text{BB}}{{\text{r}}_{\text{3}}}{\text{ > BC}}{{\text{l}}_{\text{3}}}{\text{ > B}}{{\text{I}}_{\text{3}}}{\text{ > B}}{{\text{F}}_{\text{3}}}$
B)${\text{B}}{{\text{I}}_{\text{3}}}{\text{ > B}}{{\text{F}}_{\text{3}}}{\text{ > BB}}{{\text{r}}_{\text{3}}}{\text{ > BC}}{{\text{l}}_{\text{3}}}$
C)${\text{B}}{{\text{F}}_{\text{3}}}{\text{ > BC}}{{\text{l}}_{\text{3}}}{\text{ > BB}}{{\text{r}}_{\text{3}}}{\text{ > B}}{{\text{I}}_{\text{3}}}$
D)${\text{B}}{{\text{I}}_{\text{3}}}{\text{ > BB}}{{\text{r}}_{\text{3}}}{\text{ > BC}}{{\text{l}}_{\text{3}}}{\text{ > B}}{{\text{F}}_{\text{3}}}$

Answer 1

Hint:To solve this question, it is required to have knowledge about Lewis acidity and back bonding. Lewis acidity of a compound is defined as its tendency to accept electrons. It is prominent in compounds which have vacant orbitals to accept electrons. Back bonding occurs when there is a transfer of electrons from the electron rich species to anti-bonding pi orbital of the other species attached to it. It occurs when there is a lone pair on the Lewis base and a vacant p-orbital on the Lewis acid species.

Complete step by step answer:
Boron is an electron deficient species. Its electronic configuration is ${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{1}}}$. It donates the three electrons in its valence shell to each halogen atom to form ${\text{B}}{{\text{X}}_{\text{3}}}$ . This means that boron has three vacant p-orbitals where it can accept new electrons and thus acts as Lewis acid.
In the case of ${\text{B}}{{\text{F}}_{\text{3}}}$ the small size of boron and fluorine means that the atoms are really close to each other, where fluorine is the electron rich species with lone pairs and boron is the electron deficient species with an empty p-orbital. As a result, the fluorine atom donates a lone pair to the anti-bonding p-orbital of boron. So, the ${\text{B - F}}$ bond acquires partial double bond character and also contains a filled p-orbital. Thus, it does not act as a Lewis acid due to the absence of vacant p-orbital and inability to accept extra electrons.
In ${\text{BC}}{{\text{l}}_3}$ this effect is not so prominent as in ${\text{B}}{{\text{F}}_{\text{3}}}$ due to the increase in size in chlorine atom. The ${\text{2p\pi - 2p\pi }}$ back-bonding effect is more prominent than ${\text{2p\pi - 3p\pi }}$ back bonding effect. Due to the absence of back-bonding, ${\text{BC}}{{\text{l}}_3}$ shows Lewis acid character better than ${\text{B}}{{\text{F}}_{\text{3}}}$ . Similarly, this trend is followed for the other halogens as well with Lewis acidic character increasing down the group as back-bonding becomes less and less prominent. Thus, the correct order of Lewis acid character of boron halides will be : ${\text{B}}{{\text{I}}_{\text{3}}}{\text{ > BB}}{{\text{r}}_{\text{3}}}{\text{ > BC}}{{\text{l}}_{\text{3}}}{\text{ > B}}{{\text{F}}_{\text{3}}}$
$\therefore $ The correct option is option D, i.e. ${\text{B}}{{\text{I}}_{\text{3}}}{\text{ > BB}}{{\text{r}}_{\text{3}}}{\text{ > BC}}{{\text{l}}_{\text{3}}}{\text{ > B}}{{\text{F}}_{\text{3}}}$.

Note: In the questions involving halogens, we generally apply the concept of electronegativity. In this case, if we did so then fluorine being most electronegative would pull electron density towards itself and boron fluoride would be a better Lewis acid. But, the small size of fluorine atom and the small size of boron atoms causes back-bonding to occur and instead causes boron fluoride to have the least Lewis acidic character.