Question

Find the lewis acid strength of \[BB{r_3},BC{l_3},B{F_3}\] is in the order.
A.\[BB{r_3} < BC{l_3} < B{F_3}\]
B.\[BC{l_3} < B{F_3} < BB{r_3}\]
C.\[B{F_3} < BC{l_3} < BB{r_3}\]
D.\[BB{r_3} < B{F_3} < BC{l_3}\]

Answer 1

Hint: In order to answer this question we must know about the back bonding of each compound and their lewis acid structure. We know lewis acids are chemical entities that have empty orbital and property of the accept pair of the electron from lewis base.

Complete answer:
In, \[BB{r_3}\] Bromine \[\left( {Br} \right)\] is in Group \[7\] having \[7\] no of valance electron. Whereas boron \[\left( B \right)\] has\[6\] valance electron. So, lewis structure of \[BB{r_3}\] having \[24\]valance electron. The lewis acid structure of \[BB{r_3}\] very similar to \[BC{l_3}\] and \[B{F_3}\]. In \[B{F_3}\] boron has \[2p\]orbital vacancy and fluorine has filled unused \[2p\]orbital. To form a \[p\pi - p\pi \]bond, fluorine transfer these two electrons to \[2p\]orbital of boron.
Now we arrange the above compound according to their decreasing order of back bonding.
\[B{F_3} < BC{l_3} < BB{r_3}\]
As we know, stronger back bonding tends to weaker tendency to act as a lewis acid

Therefore option C. \[B{F_3} < BC{l_3} < BB{r_3}\] is correct.

Note:
Lewis acid is a substance that accepts a pair of electrons to form a covalent bond. A Lewis base is a substance that provides a pair of electrons to form a covalent bond Example of lewis acids: \[S{O_{3,}}AlC{l_3},B{F_{3,}}{K^ + },{H^ + },C{O_3}\]. Example of lewis bases: \[{H^ - },{F^ - },CO,{H_2}O,N{H_3}\]