Question

The anion which cannot be formed by boron is:
A. ${\text{B}}{{\text{F}}_6}^{3 - }$
B. ${\text{B}}{{\text{H}}_4}^ -$
C. ${\text{B}}{\left( {{\text{OH}}} \right)_4}^ -$
D. ${\text{B}}{{\text{O}}_2}^ -$

Answer 1

The number of electron pairs an atom share with other atoms is termed as covalency. The compound formation depends upon its maximum covalency. It is the ability of an element to form a maximum number of covalent bonds with other elements to form stable electronic configuration.

Complete step by step answer:
Boron atom has three valence electrons. It has the ability to form triple bonds because of its valency. It has a total of five electrons. Thus its electronic configuration is $1{{\text{s}}^2}2{{\text{s}}^2}2{{\text{p}}^1}$. It has only two shells. It does not have ${\text{d}}$ orbitals. Covalency is the number of orbitals in a valence shell. The outermost shell contains one $2{\text{s}}$ orbitals and three $2{\text{p}}$ orbitals. Thus its maximum covalency is four.
Therefore, we can tell that boron has three valence electrons with maximum covalency four. Thus it can form bonds with a maximum of four atoms. In ${\text{B}}{{\text{H}}_4}^ -$ and ${\text{B}}{\left( {{\text{OH}}} \right)_4}^ -$ four atoms are bonded with boron. In ${\text{B}}{{\text{O}}_2}^ -$, two oxygen atoms are bonded with boron. But in ${\text{B}}{{\text{F}}_6}^{3 - }$, six fluorine atoms are there, which is not possible since it cannot extend its orbital because it has no ${\text{d}}$ orbital.
Hence ${\text{B}}{{\text{F}}_6}^{3 - }$ is the anion which cannot be formed by boron.

So, the correct answer is Option A .

Note:
Fluorine has a total of nine electrons. Its electronic configuration is $1{{\text{s}}^2}2{{\text{s}}^2}2{{\text{p}}^5}$. When a boron atom is excited, the electron in $2{\text{s}}$ orbital gets excited to $2{\text{p}}$ orbital. Therefore it has a total of three electrons in ${\text{s}}$ and ${\text{p}}$ orbital. So only three fluorine atoms can be accommodated. It has no empty ${\text{d}}$ orbitals.