Question

Explain geometry of methane molecule on the basis of Hybridization.

Answer 1

Formation of methane (CH₄) molecule on the basis of sp³ hybridization:

1. Methane molecule (CH₄) has one carbon atom and four hydrogen atoms.
2. The ground state electronic configuration of C (Z = 6) is 1s² 2s² \[\ce{2p^1_{{x}}}\] \[\ce{2p^1_{{y}}}\] \[\ce{2p^0_{{z}}}\].
   Electronic configuration of carbon:
   

   Ground state:	1s	2s	2p
   	↑↓	↑↓	↑	↑

   Excited state:	1s	2s	2p
   	↑↓	↑	↑	↑	↑

   sp3 Hybrid orbitals:	↑↓	↑	↑	↑	↑
   	1s	(four sp3 hybrid orbitals)

3. In order to form four equivalent bonds with hydrogen, the 2s and 2p orbitals of C-atom undergo sp³ hybridization.
4. One electron from the 2s orbital of the carbon atom is excited to the 2p_z orbital. Then the four orbitals 2s, p_x, p_y, and p_z mix and recast to form four new sp³ hybrid orbitals having the same shape and equal energy. They are maximum apart and have tetrahedral geometry with an H–C–H bond angle of 109°28'. Each hybrid orbital contains one unpaired electron.
5. Each of these sp³ hybrid orbitals with one electron overlaps axially with the 1s orbital of the hydrogen atom to form one C–H sigma bond. Thus, in CH₄ molecule, there are four C–H bonds formed by the sp³–s overlap.
   Diagram: