Question

Hybridisation of ${{C}_{2}}$ and ${{C}_{3}}$ of ${{H}_{3}}C-CH=C=CH-C{{H}_{3}}$ are
(A)- $sp,s{{p}^{3}}$
(B)- $s{{p}^{2}},sp$
(C)- $s{{p}^{2}},s{{p}^{2}}$
(D)- $sp,sp$

Answer 1

The hybridisation is obtained by the number of sigma bonds formed with the adjacent atoms. And the unpaired orbitals with unpaired electrons overlap sideways to form pi-bonds in the compound.

Complete step by step answer:
The given compound is symmetric in nature, such that C-1 and C-5, C-2 and C-4 have the same number of groups attached to them. The central carbon C-3 atom forming double-bond with both the adjacent carbons (C-2 and C-4) atoms. The electronic configuration of the carbon atom being $1{{s}^{2}}2{{s}^{2}}2{{p}^{2}}$. In its excited state it has a total four unpaired electrons in the 2s and 2p orbital. Thus, it forms four bonds with adjacent atoms to complete its octet. Since there is a presence of double bond, so both the pi- and the sigma- bonds are present in the given compound. In the C-2 carbon atom, forming two sigma bonds with a methyl group and hydrogen atom. And one pi- and one sigma-bond with C-3 carbon. Then, in order to form the three sigma bonds, the two 2p-orbitals (${{p}_{x}},{{p}_{y}}$ ) and the 2s-orbital undergo hybridisation to form three $s{{p}^{2}}$ hybrid orbitals. And the ${{p}_{z}}$ perpendicular to the plane forms the pi-bond with adjacent C-3 carbon atoms through sideways overlapping. Similarly, the C-3 carbon atom, forming two sigma bonds with the adjacent carbon atoms (C-2 and C-4) and two pi-bonds with them. Then, to form the two sigma bonds, one 2p-orbital (${{p}_{z}}$) and the 2s-orbital undergo hybridisation to form two $sp$ hybrid orbitals. And the two 2p-orbitals (${{p}_{x}},{{p}_{y}}$) , perpendicular to the internuclear axis overlap sideways with the adjacent carbon atoms (C-2 and C-4) forming two pi-bonds.

Therefore, the Hybridisation of ${{C}_{2}}$ and ${{C}_{3}}$ of ${{H}_{3}}C-CH=C=CH-C{{H}_{3}}$ are option (B)- $s{{p}^{2}},sp$ respectively.

Note: During the hybridisation, the hybrid orbitals formed have similar energy. Also, in the sp hybrid orbitals the s-character is greater than the $s{{p}^{2}}$ hybrid orbital.