Question

Among boron halides, which is the strongest Lewis acid?
A. ${\text{B}}{{\text{F}}_{\text{3}}}$
B. ${\text{B}}{{\text{I}}_{\text{3}}}$
C. ${\text{BC}}{{\text{l}}_{\text{3}}}$
D. ${\text{BB}}{{\text{r}}_{\text{3}}}$

Answer 1

We know that the compounds that have empty orbitals which help them accept electrons from Lewis bases are known as Lewis acids. The strength of the Lewis acid is affected by the bond between boron and the halogen.

Complete step by step solution:
In boron halides, the central atom is boron. Boron has three valence electrons. The p orbitals of halides are not completely filled.
When boron halides are formed, the three halogen atoms share one electron each and get bonded to the central boron atom by forming three bonds. As three bonds are formed, the boron atom now has six electrons. The boron atom needs two electrons more to complete its octet.
The halogen atom then passes two electrons to the vacant p-orbital of boron and a ${\text{p }}\pi - {\text{p }}\pi$ bond is formed between the halogen atom and central boron atom. As a result, the electron deficiency of boron decreases and the Lewis acid character of the boron trihalide decreases.
-> In ${\text{B}}{{\text{F}}_{\text{3}}}$, the boron and fluorine atoms are small in size and are close to each other. When the fluorine atom donates electrons to the boron atom, the ${\text{B}} - {\text{F}}$ bond acquires partial double bond character and contains filled p-orbital. Thus, ${\text{B}}{{\text{F}}_{\text{3}}}$ does not act as a Lewis acid due to the absence of vacant p-orbital and inability to accept extra electrons.
->In ${\text{BC}}{{\text{l}}_{\text{3}}}$, the chlorine atom is larger in size as compared to the fluorine atom. As a result, the back bonding effect is less prominent than that in ${\text{B}}{{\text{F}}_{\text{3}}}$. Thus, ${\text{BC}}{{\text{l}}_{\text{3}}}$ is a stronger Lewis acid than ${\text{B}}{{\text{F}}_{\text{3}}}$.
->In ${\text{BB}}{{\text{r}}_{\text{3}}}$, the bromine atom is larger in size than the chlorine atom. Thus, ${\text{BB}}{{\text{r}}_{\text{3}}}$ is a stronger Lewis acid than ${\text{BC}}{{\text{l}}_{\text{3}}}$.
-> In ${\text{B}}{{\text{I}}_{\text{3}}}$, the iodine atom is larger in size than the bromine atom. Thus, ${\text{B}}{{\text{I}}_{\text{3}}}$ is a stronger Lewis acid than ${\text{BB}}{{\text{r}}_{\text{3}}}$.
Thus, the strongest Lewis acid is ${\text{B}}{{\text{I}}_{\text{3}}}$.

**Thus, the correct option is (B) ${\text{B}}{{\text{I}}_{\text{3}}}$.

Note: **
The Lewis acid character increases as we move down the group. This is because down the group the back-bonding becomes less prominent. Thus, the acidic nature increases down the group.