Question

The state of hybridization of B in \[{\rm{BC}}{{\rm{l}}_{\rm{3}}}\]
A) \[sp\]
B) \[s{p^2}\]
C) \[s{p^3}\]
D) \[s{p^2}{d^2}\]

Answer 1

Hint: In the process of hybridization, the atomic orbitals comprising similar energies undergo intermixing to give a new group of orbitals possessing different shapes and energies. Here, a formula is to be used, to check hybridization.

Formula Used:The formula is,
The formula of hybridization is,
\[H = \dfrac{{V + X - C + A}}{2}\]
Here, V stands for the count of valence electrons, X stands for count of monovalent atoms of the compound, C stands for charge of cation and A is for the charge of anion.
The H value of 4 is for the \[s{p^3}\] hybridization, the H value of 3 is for a \[s{p^2}\] hybridized molecule. And the H value of 2 is for a \[sp\] hybridized molecule.

Complete step by step solution:We know, Boron has an atomic number of 5, so it has valence electrons of 3. And the count of monovalent atoms bonded to the boron atom is \[{\rm{BC}}{{\rm{l}}_{\rm{3}}}\]is 3. So,
\[H = \dfrac{{3 + 3}}{2} = 3\]
So, the hybridization of Boron atom in the molecule of \[{\rm{BC}}{{\rm{l}}_{\rm{3}}}\]is \[s{p^2}\].

Therefore, option B is right.

Note: Always remember that, there is another method to check the hybridization of a compound. We have to count the number of groups that surround the central atom of the compound. The surrounding groups are atoms that form bonds with the atom and the count of lone pairs. If the surrounding group count is 4, the hybridization is \[s{p^3}\]. The count of 3 is for the \[s{p^2}\]hybridized molecule. And the count of 2 indicates a sp hybridized molecule.