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As shown on the Periodic Table, Carbon has a Ground State of [He] 2s22p2 as represented by this orbital diagram.

Ground State

6C

 [He] 2s2 2p2

Carbon

 [He] 

The molecular geometry of CH4 (methane) is tetrahedral.

sp3 Hybridization: Methane Gas

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CH4

H

Now that we have 4 uniform, singly occupied orbitals, we can bring in H to make Methane.

Hybridization State

Tetrahedral

sp3 in 3D

C

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Notice there are only two singly occupied orbitals for bonding. Only CH2 can be formed at this stage.  CH2 does not exist in nature, so we know something else must happen for CH4 to be formed.

 [He] 

 [He] 2s2 2p2

CH4 bonds should all be equal, but remember that s bonds are stronger than p bonds, so nature has to do something to make the orbitals uniform. That something is hybridization.

 [He]

Promotion State

 [He] 2s2 2p2

To create equal bond strength, we mix different orbitals and create one set of uniform hybrid orbitals.

6C

As shown on the Periodic Table, Hydrogen has a Ground State of 1s1

1H

Hydrogen

sp3

p, 1

 [He]

p, 2

p, 3

s

Hybrid Orbital

As shown on the Periodic Table, Carbon has a Ground State of [He] 2s22p2   as represented by this orbital diagram.

3

4

To make CH4, we need 4 singly occupied orbitals to bond.

1

2

CH4 bonds should all be equal, but remember that s bonds are stronger than p bonds, so nature has to do something to make the orbitals uniform. That something is hybridization.  

Note that 3d10 is not included in the orbital diagram because it is a pseudo noble gas core and therefore does not take part in the chemical reaction.  Treat a pseudo noble gas core like the electrons inside the brackets with the symbol of the noble gas.

Now that we have 4 uniform, singly occupied orbitals, we can bring in H to make Methane.

As shown on the Periodic Table, Bromine has a Ground State of [Ar] 4s23d104p5 as represented by this orbital diagram.

An electron is promoted to create four singly occupied orbitals for bonding.

[He]

To make CH4, we need 4 singly occupied orbitals to bond.