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Q.

The compound having sp, sp2 and sp3 hybridized orbitals of carbon atoms in 1 : 3 : 2 ratio is

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a

CH3−CH=CH−CN

b

CH3−HC=C=CH−CH3

c

CH3−C≡C−CH=CH2

d

CH3−C≡C−CH=CH−CH3

answer is B.

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Detailed Solution

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CH3−CH=CH−CN has 1 sp, 2 sp2 & 1 sp2 carbons. hence 2 sp, 6 sp2 & 4 sp3 hybrid orbitals of carbons

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