MCQ
From left to right, $sp^2, sp^2, sp, sp$ hybridisation is present in
- ✓$C{H_2} = CH - C \equiv N$
- B$CH_2=C=CH-CH_3$
- C$HC \equiv C - C \equiv CH$
- D$CH \equiv C - CH = C{H_2}$
✓
Answer
Correct option: A.
$C{H_2} = CH - C \equiv N$
a
Option $A$ is correct if the combination is, $sp ^2- sp ^2- sp$, because from left first two carbon atoms have one empty $p$ orbital due to $\pi$ bond, thus $sp ^2$ hybridisation. But last carbon atom has a triple bond that means two $p$ orbitals are empty, thus $sp$ hybridisation.
Option $A$ is correct if the combination is, $sp ^2- sp ^2- sp$, because from left first two carbon atoms have one empty $p$ orbital due to $\pi$ bond, thus $sp ^2$ hybridisation. But last carbon atom has a triple bond that means two $p$ orbitals are empty, thus $sp$ hybridisation.
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