Question

The halides of group $IIIA$ behave as Lewis acid. The acceptor ability is maximum for halides of
A. $Tl$
B. $Ga$
C. $Al$
D. $B$

Answer 1

We know that Lewis acids are acids which can accept electrons. Similarly Lewis bases are bases which can donate electrons. Lewis acids have deficiency of electrons. The atomic number of boron is five and the atomic number of aluminium is thirteen. The electronic configuration of boron is $1{s^2},2{s^2}2{p^1}$ and the electronic configuration of aluminium is $1{s^2},2{s^2}2{p^6},3{s^2},3{p^1}$

Complete step by step solution:
We already know that the electronic configuration of Boron is $1{s^2},2{s^2}2{p^1}$. Boron do not have $d$ orbitals for bonding. We can observe that Boron has three valence electrons for bonding. Boron can form covalent bonds by using only $s$ and $p$ orbital. The orbitals present are $2s,2{p_x},2{p_y},2{p_z}$. We can see that it is electron deficient in nature. So boron is a great electron acceptor as compared to other group $IIIA$ elements. So the halide of boron has the maximum electron accepting tendency.
So from the above explanation it is clear to us that the correct answer of the given question is option: D.

Hence, the correct answer is option D.

Additional information:
The $IIIA$ group is in the P Block of the modern periodic table. It is in P Block because the last electron of the elements has been added to a $p$ orbital. When we go down in a group of the periodic table the atomic radius increases. The ionisation enthalpy decreases when we go down the group.

Note: Always remember that the halide of boron has the maximum ability of accepting electrons because of its electronic configuration. Boron do not have any $d$ orbital. Lewis acids are compounds which can accept electrons and Lewis bases are the compounds which can donate electrons.