Question

Boron is unable to form ${\text{B}}{{\text{F}}_{\text{6}}}^{{\text{3 - }}}$ ions due to:
(A) Non-availability of d-orbitals.
(B) Small size of boron atoms.
(C) Non-metallic nature.
(D) Less reactivity towards the hydrogen.

Answer 1

In the above question, it is asked that why boron is unable to form ${\text{B}}{{\text{F}}_{\text{6}}}^{{\text{3 - }}}$ . The answer to this question lies in the maximum covalency of the atom. Boron is the first member of Group 13 and hence, it does not have any d-orbitals and hence, it cannot extend its octet.

Complete answer:
We know that the atomic number of boron is 5. So, it does not contain any d-orbitals.
The maximum valency for elements which does not have d orbital is equal to the number of unpaired electrons present in the valence shell of s and p orbital. The covalency is determined by the number of electrons shared while forming a chemical compound.
The maximum covalency of the first elements of each group is 4. This is because of small atomic size, absence of d-orbitals and high electronegativity.Since, boron is the first element of the Group 13, therefore, the maximum covalency of boron is 4. It is not possible for boron to extent its covalency. In ${\text{B}}{{\text{F}}_{\text{6}}}^{{\text{3 - }}}$ , boron is attached with 6 fluorine atom which makes its unstable.
Hence, we can conclude that boron is unable to form ${\text{B}}{{\text{F}}_{\text{6}}}^{{\text{3 - }}}$ ions due to non-availability of d orbitals.
So, option A is the correct option.

Note:
-Boron is a mineral that is found in food and the environment. People take boron supplements as medicine.
-Boron is used for building strong bones and treating osteoarthritis.
-Boron was once also being used as a food preservative.