Question

The decreasing order of power of boron halides to act as Lewis acid is:
$A.$ $B{F_3} > BC{l_3} > BB{r_3}$
$B.$ $BB{r_3} > BC{l_3} > B{F_3}$
$C.$ \[BC{l_3} > B{F_3} > BB{r_3}\]
$D.$ \[BC{l_3} > BB{r_3} > B{F_3}\]

Answer 1

Hint:We know that the compounds which have empty orbital which help them accept electrons from Lewis bases are known as Lewis acid. Generally the power of boron halides to act as Lewis acid affected by the bond between boron and the halogen attach to it.

Complete step by step answer:
Since, we know in boron halides the central atom is boron $\left( B \right)$ . We know Boron has three valence electrons that are the p orbital of boron halides that are not completely filled so they can act as Lewis acid.
When the formation of boron halides takes place, the three halogen atoms present around the boron would share one electron each to get bonded to the boron atom by forming three chemical bonds. As three bonds are formed the boron atom has six electrons, three its own and three from halogens. Thus, the boron atom required two electrons to complete its octet.
As we know the strength of Lewis acid depends on the bond present in them, in $B{F_3}$ the boron and fluorine atoms are small in size and very close to each other. When the fluorine atom donates its electron to the boron atom, the $B - F$ bond acquires a partial double bond character and contains filled p-orbitals. Therefore, $B{F_3}$ molecules do not act as Lewis acid.
In $BC{l_3}$ we know the chlorine atom is larger than fluorine as a result, the back bonding effect is less prominent that in $B{F_3}$ and in $BB{r_3}$ the bromine atom is larger in size than chlorine so it will act as stronger Lewis acid than $BC{l_{3.}}$
So the correct order of decreasing order is $BB{r_3} > BC{l_3} > B{F_3}$ .so the correct option is $B.$

Note:
 It is to be noted that in boron halides as we move down the group the Lewis acid strength increases because down the group the back-bonding becomes less prominent due to increase in size of halogen.