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Question: Assertion: Boron can only form \({{B}}{{{F}}_4}^ - \), whereas aluminium forms \({{Al}}{{{F}}_6}^{...

Assertion:
Boron can only form BF4−{{B}}{{{F}}_4}^ - , whereas aluminium forms AlF63−{{Al}}{{{F}}_6}^{3 - }
Reason:
The first member of a group of elements in the s and p blocks shows anomalous behavior.
A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion.
C. Assertion is correct but Reason is incorrect.
D. Both Assertion and Reason are incorrect.

Explanation

Solution

Boron and aluminium lies in the group 1313. They show an observable trend, i.e. they have a maximum covalency of +3 + 3. Generally, the halides of these elements act as Lewis acids. This is due to their deficiency in the electrons.

Complete step by step answer:
As we know that the first element of s and p block elements have no d orbital. Thus they have a maximum covalency of +4 + 4. They cannot expand their valence shell for the accommodation of electron pairs, but other members can.
The electronic configuration of boron is 1s22s22p11{{{s}}^2}2{{{s}}^2}2{{{p}}^1} while that of aluminium is [Ne]3s23p1\left[ {{{Ne}}} \right]3{{{s}}^2}3{{{p}}^1}.
Aluminium has d orbitals, thus it can expand their valence shell. So we can say that the maximum covalency of aluminium is +6 + 6 and thus it can bond with six fluorine atoms. But in boron atoms, the d orbitals are absent, but it has one s-orbital and three p-orbitals. So the maximum covalency of boron is 44, thus cannot bond with six fluorine atoms. It can only bond with four fluorine atoms. Hence boron forms BF4−{{B}}{{{F}}_4}^ - and aluminium forms AlF63−{{Al}}{{{F}}_6}^{3 - }due to the unavailability of d orbitals.
So both Assertion and Reason are correct, but Reason is not the correct explanation for Assertion.

Hence the correct option is A.

Note: Halides of other elements in the group 1313undergo dimerization by hydrogen bridging. Halogen in halogen bridged molecules donates electrons to these elements. In AlF63−{{Al}}{{{F}}_6}^{3 - }, electrons need one s orbital, three p-orbitals and two d-orbitals.