Question

Apart from tetrahedral geometry, another possible geometry of ${{C}}{{{H}}_{{4}}}$ could be square planar with the four H atoms at the corners of the square and the C atom at its centre. Why ${{C}}{{{H}}_{{4}}}$ is not a square planar?

Answer 1

Hint: Geometry, shape and hybridization of a molecule can be calculated by using some rules i.e. first write the Lewis structure then calculate the number of sigma bonds and lone pair then determine the steric number (steric number is composed of number of lone pairs and sigma bonds) and then finally assign the hybridization and shape of the molecule.

Complete step by step answer:
Firstly, we will learn about the concept of hybridization, which can be defined as the mixing of orbitals of the same energy to give new degenerate orbitals.
Electronic configuration of carbon in the ground state can be represented as:
${{C}}\;\;{{:}}\;\;{{1}}{{{s}}^{{2}}}\;{{2}}{{{s}}^{{2}}}\;{{2p}}_{{x}}^{{1}}\;{{2p}}_{{y}}^{{1}}$
Electronic configuration of carbon in the excited state can be represented as:
${{C}}\;\;{{:}}\;\;{{1}}{{{s}}^{{2}}}\;{{2}}{{{s}}^1}\;{{2p}}_{{x}}^{{1}}\;{{2p}}_{{y}}^{{1}}\;{{2p}}_{{z}}^{{1}}$
In ${{C}}{{{H}}_{{4}}}$, the central atom carbon contains four valence electrons and each electron is shared by hydrogen atoms which results in four sigma bonds and no lone pairs. Based on the number of sigma bonds and lone pairs, the steric number will be four. Hence, ${{C}}{{{H}}_{{4}}}$ molecule has ${{s}}{{{p}}^3}$ hybridization and the geometry will be tetrahedral.
For square planar, ${{ds}}{{{p}}^{{2}}}$ hybridization is required, which is not possible due to the absence of d-orbital. The bond angle of square planar geometry has bond angle of $90^\circ $ and is lesser than tetrahedral bond angle which is $109.5^\circ $. Due to the repulsive forces in square planar geometry makes it less stable than tetrahedral.
Therefore, the geometry of ${{C}}{{{H}}_{{4}}}$ molecule is tetrahedral and not square planar.

Note: We need to remember that the shape does not count lone pairs but the shape is an outcome of a lone pair. For example, in ${{N}}{{{H}}_3}$ molecule the structure will be tetrahedral but due to the presence of lone pair the shape of ammonia molecule is comes out to be trigonal pyramidal. If there is no lone pair then the shape and the geometry of the molecule will be the same.