Question

Boron cannot form which one of the following anions?
A. $B{{F}_{6}}^{-3}$
B. $B{{H}_{4}}^{-}$
C. $B{{\left( OH \right)}_{4}}^{-}$
D. $B{{O}_{3}}^{-}$

Answer 1

To solve this question, we must know the valency of boron as well as the period number of the element as the valency will tell us about the number of the electrons present in the outermost shell and the period number will tell us about the number of shells present in the atom in total.

Complete step by step solution:
-Valency is the number of the electrons present in the outermost shell of an atom which is also called the valence shell of an atom. Those electrons are important to be studied as they are the most loosely held electrons of the atoms.
-Boron is an atom with the atomic number 5. It comes right after Li and Be.
-If we have a look at the periodic table, we can find that B is present in period 2 and group 3 of the periodic table
-As the atomic number of B is 5, the number of electrons in B is also 5 since we know,
Atomic no. = No. of proton = No. of electron
-Now according to Hund’s rule, the 5 electrons of Boron are filled in their orbitals as $1{{s}^{2}}2{{s}^{2}}2{{p}^{1}}$
-Thus, we see that there are 2 electrons in the 1st orbital, 3 electrons in the 2nd orbital which is also the last shell of Boron. Thus there can only be 2 orbitals present in boron named s and p orbitals.
So the electronic configuration of Boron is (2,3).
-Now, we know that every atom tries to reach the electronic configuration of the noble gases to become stable like them. The configuration of noble gases is of the form $n{{s}^{2}}n{{p}^{6}}$ . This configuration is reached either by sharing of electrons or losing of electrons.
-Boron can lose its electrons to achieve its noble configuration of He. To form anions, it needs to gain electrons from the radicals it attaches itself to. This is done to try and attain 8 electrons in the valence shell.
-Looking at the options, we can see that the last three options require boron to form 4 bonds and attain 6 electrons in its outermost shell. The first option would lead to the entering of 12 electrons in the valence shell but it is not possible as it needs 6 bonds which is not possible in boron due to absence of the d-orbital.
-As there are no 2d-orbitals present, boron cannot extend the number of electrons entering it forming more bonds. So the compound $B{{F}_{6}}^{-3}$ cannot be formed.

Thus the correct option is A.

Note: There are many compounds which do not fulfill the octet rule and yet they exist in nature and in a very stable form. They can have more than 8 electrons and are called hypervalent compounds. They can have less than 8 electrons and are called hypervalent compounds. There also consist compounds with odd electrons like NO, NOCl.