Discuss the shape of CH tripple bond CH on the basis of hybridization.

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Solution

sp Hybridisation in HC $\equiv$ CH

In sp hybridisation, one s (2s) and one p (2p_x) orbitals present in the valence shell of the excited carbon atom get hybridised to form two equivalent orbitals while two remaining two 2p-orbitals (2p_y and 2p_z) do not take part in hybridisation. One sp hybrid orbital of one carbon atom overlaps axially with sp hybrid orbital of the other carbon atom to form C-C sigma bond. The remaining hybridised orbital of each carbon atom overlaps axially with half-filled orbital of hydrogen forming sigma bonds. each of the two unhybridised orbitals of one carbon atom overlaps sidewise with the orbitals of the other carbon atom to form two pi-bonds.
So the triple bond between two carbon atoms is made up of one sigma and two pi bonds. Thus, in acetylene molecule:
C $\equiv$ C bond length = 120pm
C-H bond length = 106pm
H-C-C bond angle = 108 $°$
Hence acetylene molecule has linear shape.

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