Question

The shapes and hybridisation of $B{F_3}$ and \[B{H_4}^ - \] respectively are:
A) $B{F_3}$ - Trigonal, \[s{p^2}\] hybridisation;
\[B{H_4}^ - \]- Square planar, \[s{p^3}\] hybridisation
B) $B{F_3}$ - Triangular, \[s{p^3}d\] hybridisation;
\[B{H_4}^ - \]- Square planar, \[s{p^3}d\] hybridisation
C) $B{F_3}$ - Trigonal, \[s{p^2}\] hybridisation;
\[B{H_4}^ - \]- Tetrahedral, \[s{p^3}\] hybridisation
D) $B{F_3}$ - Tetrahedral, \[s{p^3}\] hybridisation;
\[B{H_4}^ - \]- Tetrahedral, \[s{p^3}\] hybridisation

Answer 1

Hint: We find hybridisation of each central atom by calculating lone pairs and bond pairs on the central atom.
According to Valence shell electron pair repulsion theory, depending on bond pairs and lone pairs we can determine hybridisation, shapes and structures.
Number of hybrid orbitals = lone pair + sigma bond pair
Depending on hybrid orbitals we can determine hybridisation.

Complete step by step answer:
Firstly we have to understand the concept of hybridisation.
Hybridization is the concept of mixing atomic orbitals into new hybrid orbitals (with different energies and shapes, than the component atomic orbitals) to form chemical bonds using valence bond theory and Valence shell Electron pair repulsion theory.
This concept of hybridisation gives us shape structure, bond angles and many such factors.
To know hybridisation, first we need to calculate hybrid orbitals involved.
Number of hybrid orbitals = lone pair + sigma bond pair
Let us take an example of $B{F_3}$ , we have 3 valence electrons in Boron(B), and all 3 have been bonded as sigma bonds with Fluorine(F).
So, Hybrid orbitals = Number of sigma bonds = 3.
Thus 1 s-orbital and 2 p-orbitals are involved in Hybridisation and we can name this hybridisation as \[s{p^2}\] and as we have only 3 bond pairs, so shape will be trigonal structure, so that minimum repulsion happens in bond pairs.
Now, Let us take another example of \[B{H_4}^ - \] , we have 3 valence electrons in Boron(B) and 1 negative charge adds one electron to boron giving rise to 1 negative charge on compound, so total valence electron on Boron is 4 now, and all 4 have been bonded as sigma bond with Hydrogen(H).
So, Hybrid orbitals = Number of sigma bonds = 4.
Thus 1 s-orbital and 3 p-orbitals are involved in Hybridisation and we can name this hybridisation as \[s{p^3}\] and as we have only 4 bond pairs, so shape will be tetrahedral structure, so that minimum repulsion happens in bond pairs.
The structures of $B{F_3}$ and \[B{H_4}^ - \] can be shown below respectively:

Thus the final correct answer is option (C).

Note: In determination of shapes and structures of molecules, take care of arrangement of bond pairs and lone pairs, so that the repulsion is least. Even though, in both molecules, Boron (B) is a central atom, but they can have different hybridisation and structure. It is not necessary that all compounds of the same central atom will have the same hybridisation, it depends on atoms attached to it, lone pairs and charges on molecules.