Question

Boron can form only $\text{B}{{\text{F}}_{\text{4}}}^{\text{-}}$ . But aluminium can form $\text{Al}{{\text{F}}_{6}}^{\text{-}}$ . Explain.

Answer 1

Boron belongs to the second period of the periodic table while aluminium belongs to the third group. The elements in the second group have the second shell as the valence shell which contains two subshells while the elements of the third period having the third shell as the valence shell have three subshells.

Complete step by step solution:
Both the elements, born and aluminium belong to the same group and thus have the same valence shell configuration. The general electronic configuration of the elements of group 13 is $\text{n}{{\text{s}}^{\text{2}}}\text{n}{{\text{p}}^{\text{3}}}$ and for boron the valence shell is the second shell while for aluminium it is the third shell. In the second shell, there being two subshells, they can accommodate a total of eight electrons.
The third shell being having higher capacity than the second shell due to the presence of the d-subshell, it can accommodate eighteen electrons including all the subshells. So the boron atoms accept an extra pair of electrons from only one fluoride anion to form the fluoroborate anion to satisfy its octet configuration, while aluminium can accommodate electron pairs from three fluoride anions to form the aluminium hexafluoride anion and those extra electron pairs could be accommodated in the 3d orbitals of the aluminium atom.
Hence, Boron can form only $\text{B}{{\text{F}}_{\text{4}}}^{\text{-}}$ . But aluminium can form $\text{Al}{{\text{F}}_{6}}^{\text{-}}$ as well.

Note:
Just like aluminium, the other members of the third period can also accommodate extra electrons in the third shell to expand their octet. This leads to the formation of the silicates by silicon as well and the penta covalency of phosphorus.