Question

Explain the formation of BF₃ molecule with the help of Valence Bond theory?

Answer 1

Formation of BF₃ molecule :

1. ₅B has electronic configuration 1s² 2s² 2p_x . 

2. The excited electronic configuration of ₅B is 1s² 2s¹ 2p_x² 2p_y². 

3. As it forms three identical B-F bonds in BF₃ 

4. It is suggested that excited ‘B’ atom undergoes hybridisation. 

5. There is an intermixing of 2s, 2p_x , 2p_y orbitals and their redistribution into three identical orbitals called sp² hybrid orbitals. 

6. For three sp² orbitals to get separated to have minimum repulsion the angle between any two orbitals is 120° at the central atom and each sp² orbital gets one electron. 

7. Now three fluorine atoms overlap their 2p_z orbitals containing unpaired electrons (F₉ 1s² 2s² 2p_x² 2p_y² 2p_x²) the three sp² orbitals of ‘B’ that contain unpaired electrons to form three ssp^2-p bonds.