question_answer 25)

Apart from tetrahedral geometry, another possible geometry for \[C{{H}_{4}}\] is square planar with four H-atoms at the comers of the square and C-atom at its centre. Explain why \[C{{H}_{4}}\] is not square planar?

Answer:

The electronic configuration of carbon atom is \[1{{s}^{2}},2{{s}^{2}},2p_{x}^{1}2p_{y}^{1}2p_{z}^{0}\]. For tetravalency, 2.s-orbital is unpaired and electron is shifted to \[2{{p}_{z}}\] -orbital, i.e., four singly occupied orbitals undergo hybridization, i.e. \[s{{p}^{3}}\]-hybridization giving tetrahedral geometry. For square planar geometry, one d-orbital is required as \[ds{{p}^{2}}\]- hybridization is to take place which is not possible for carbon atom as d-orbital is not present.