Question

Explain the hybridization in the ethane molecule. Draw the structure to demonstrate chemical bonding in it.

Answer 1

Hint: Hybridization of a molecule can also be calculated by a formula:
Hybridization = $${\displaystyle \frac{1}{2}}(V+M-C+A)$$

where, V = valence electrons,

M = monovalent atom linked to central atom,

C = charge on cation,

A = charge on anion.

After getting the numerical value,
2 = $$sp$$
3 = $$s{p}^2$$
4 = $$s{p}^3$$
5 = $$s{p}^{3}d$$
6 = $$s{p}^3{d}^2$$

Complete step by step solution:

First let’s draw the structure of ethane molecule.

From the structure we can see each carbon forms four bonds, three bonds with a Hydrogen atom and one with the adjacent carbon atom. To find the hybridisation of the molecule, first we will isolate both of the Carbon atoms and write down their electron configuration, which is 1s² 2s² 2p². From the electronic configuration we can see that only two unpaired electrons are available in orbital but we know that carbon can form four bonds, making it obvious that hybridization is required to make the four unpaired electrons available for this bonding. As a result, four sp³ hybrid orbitals are formed on each carbon.

Now, both of the carbons will have four bonds arranged in tetrahedral geometry. The carbon-carbon sigma bond is formed by overlapping of the sp³ hybrid orbital of the one carbon with the other carbon. which also has three Hydrogens bonded to it in the similar manner, while the six carbon-hydrogen sigma bonds are formed from overlaps between the sp³ orbitals on the two carbons and the 1s orbitals of six hydrogen atoms. And we get a molecule with a total of seven sigma bonds and eventually the ethane molecule.

Note: Because they are formed from the end-on-end overlap of two orbitals, sigma bonds are free to rotate. In this case of ethane molecules, the two methyl (CH₃) groups are free to rotate.