Hybridisation of C2 and C3 of H3C−CH=C=CH−CH3 are (A)- sp,sp3 (B)- sp2,sp (C)- sp2,sp2 (D)- sp,sp

Amswer: (b)

Explanation

The presented compound is symmetric in nature, with the same number of groups connected to C-1 and C-5, C-2 and C-4. The centre carbon atom (C-3) forms a double bond with two surrounding carbon atoms (C-2 and C-4) in this structure. The carbon atom’s electrical configuration is 1s22s22p2. It possesses four unpaired electrons in the 2s and 2p orbitals in its excited state. To complete its octet, it creates four bonds with nearby atoms. Both the pi- and sigma-bonds are present in the given compound due to the existence of a double bond. Forming two sigma bonds with a methyl group and a hydrogen atom in the C-2 carbon atom. With C-3 carbon, there is one pi-bond and one sigma-bond.

The two 2p-orbitals (px,py) and the 2s-orbital are then hybridised to generate three sp2 hybrid orbitals, which form the three sigma bonds. Through sideways overlapping, the pz perpendicular to the plane creates the pi-bond with neighbouring C-3 carbon atoms. Similarly, the C-3 carbon atom forms two sigma bonds and two pi-bonds with the nearby carbon atoms (C-2 and C-4). The two sigma bonds are then formed by hybridising one 2p-orbital (pz) with the 2s-orbital to generate two sp hybrid orbitals. And two pi-bonds are formed when two 2p-orbitals (px,py) perpendicular to the internuclear axis overlap sideways with nearby carbon atoms (C-2 and C-4).

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