Question

The anion which cannot be formed by boron is:
(A) $BF_6^{3 - }$
(B) $BH_4^ -$
(C) $B(OH)_4^ -$
(D) $BO_2^ -$

Answer 1

By finding the number of valence electrons and also covalency of the Boron atom, we can determine which anion can be formed. For finding the covalency and valence electron, we must know the atomic number and electronic configuration of Boron. Boron has an atomic number of 5 and the electronic configuration is $1{s^2}2{s^2}2{p^1}$.

Complete answer:
Boron is a chemical element which is represented by the symbol B. It has an atomic number 5. It belongs to Group 13 and period 2. It has an electronic configuration of $1{s^2}2{s^2}2{p^1}$. Boron will belong to the p-block element. The number of valence electrons present in it is 3. Therefore, it forms three bonds due to its valency. From the electronic configuration we can say that Boron has only two subshells namely: s and p. It does not contain d subshell.
When we consider $BF_6^{3 - }$,
We can see that Boron atom has been attached to six fluorine atoms, therefore it might have a hybridisation of . But the Boron atom does not contain d subshell hence this hybridisation is not possible. Therefore, Boron will not form $BF_6^{3 - }$anion.
Boron atom has valency 3, but it has a covalency of 4. Covalency is the number of orbitals present in the valence shell, i.e. s subshell has one orbital and p subshell has three orbitals. The maximum covalency in Boron atom is 4. It can form a maximum of four bonds. Therefore, Boron can form $BH_4^ -$, $B(OH)_4^ -$and $BO_2^ -$anions.

Hence, the correct answer is option (A) $BF_6^{3 - }$

Additional information:
Characteristics of Boron:
- Boron is a pure crystalline substance.
- Boron is black in colour.
- It is a semiconductor of electricity.
- Boron is very brittle.
- Boron is usually present in many steels.
- In silicon and germanium, boron is added as a doping agent to improve the electrical conductivity.

Note: - Fluorine has an atomic number of 17 and its electronic configuration is given as $1{s^2}2{s^2}2{p^5}$. Hence the number of valence electrons in Fluorine is 7. Boron will be having its valence electron in the 2s and 2p orbital, i.e. two electrons in 2s and one electron in 2p. When the atom gets excited the electron from 2s will be excited to 2p. Total number of valence electrons in boron is three and hence it can only form three bonds with the fluorine atom.