Name: Hybridisation of Atomic Orbitals
Uploaded: 2019-07-19
Duration: 3 min 28 s
Description: During hybridisation, orbitals having comparable energies, mix and form hybrid orbitals of equivalent energies. This video explains the concept with the help...

Question

Explain the formation of

{BF}_{3}

molecule using hybridisation.

Accepted Answer

The formation of ${BF}_{3}$ molecule using hybridisation is as follows:

${}_{5}B$ has an electronic configuration $1 s^{2} 2 s^{2} 2 {p_{x}}^{1}$ . As it has one unpaired electron $(2 {p_{x}}^{1})$ it should form only one covalent bond to give $B - F$ molecule. We get practically ${BF}_{3}$ molecule. The explanation is as follows:

(i) Boron first undergoes excitation to get electronic $1 s^{2} 2 s^{2} 2 {p_{x}}^{1} 2 {p_{y}}^{1}$ .

(ii) As it forms three identical $B - F$ bonds in ${BF}_{3}$ , it is suggested that an exciting boron atom undergoes hybridisation. There is an intermixing of $2 s, 2 p_{x}, 2 p_{y}$ orbitals and their redistribution into three identical orbitals called ${sp}^{2}$ hybrid orbitals. For three ${sp}^{2}$ orbitals to get separated to have minimum repulsion the angle between any two orbitals is $120 °$ at the central atom and each ${sp}^{2}$ orbital gets one electron. Now three fluorine atoms overlap their $2 p_{z}$ orbitals containing unpaired electrons $(F_{9,} 1 s^{2} 2 s^{2} 2 {p_{x}}^{2} 2 {p_{y}}^{2} 2 {p_{z}}^{1})$ the three ${sp}^{2}$ orbitals of boron contain unpaired electrons to form three ${σsp}^{2} - p$ bonds.

Explain the formation of ${BF}_{3}$ molecule using hybridisation.

Important Questions on Chemical Bonding

Learn and Prepare for any exam you want !

Explain the formation of BF3 molecule using hybridisation.

Important Questions on Chemical Bonding

Learn and Prepare for any exam you want !

Explain the formation of ${BF}_{3}$ molecule using hybridisation.