Question

The anion which can not be formed by boron is:
A. $B{F_6}^{ - 3}$
B. $B{H_4}^{ - 1}$
C. $B{\left( {OH} \right)_4}^{ - 1}$
D. $B{O_2}^{ - 1}$

Answer 1

Hint: Boron is the metalloid present in the group $13$ in period $2$ in p-block. Its atomic number is $5$ and the electronic configuration of boron is $1{s^2}2{s^2}2{p^1}$. We can solve this problem by explaining about the property of boron.

Complete step by step answer:
Boron$\left( B \right)$is the element having one electron in p-orbital in its ground state hence it is placed in the p-block. For completion of its octet, it forms a covalent bond because the removal of electrons from its shell needs more ionization energies, so it is difficult to form an oxidation state of $ + 3$.
Therefore, for a stable structure of its compound, boron undergoes hybridisation and forms more stable and same energy level orbitals which take part in the bond formation. For hybridisation, one electron from the $2s$ orbital promoted into the $2p$ orbital, then $s\& p$ orbital mixed to form $s{p^2}$ and $s{p^3}$ hybrid orbital and these orbital participate in bond formation.
Boron has five electrons only so it expands its orbital by hybridisation but by hybridisation it has only six electrons. After hybridisation it is only capable of having six electrons around itself so it forms anions by which it completes its octet having $8$ electrons around itself.
In the given question, $B{F_6}^{ - 3}$ anion is not formed by the boron because boron does not have d-orbital. Due to absence of $2d$ orbitals, boron can not expand its orbital beyond $8$. This anion has a $12$ valence electron around boron.
The anion which can not be formed by boron is $B{F_6}^{ - 3}$.
So, the correct answer is “Option A”.

Note: Boron has only five electrons hence it rarely follows the octet rule. All the elements want to complete its octet therefore the forms bond with each other. Boron forms the covalent bond mainly.