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NEET CHEMISTRYEasy

The molecule among the following that has the hybridization sp2,sp2,sp, and spsp^2, sp^2, sp, \text{ and } sp from left to right atoms is:

A

HC≡C−C≡HHC \equiv C - C \equiv H

B

CH2=CH−C≡CHCH_2 = CH - C \equiv CH

C

CH2=CH−CH=CH2CH_2 = CH - CH = CH_2

D

CH2−CH=CH−CH3CH_2 - CH = CH - CH_3

Step-by-Step Solution

  1. Hybridisation Rules: The hybridisation of a carbon atom can be determined by the number of sigma (σ\sigma) bonds and lone pairs attached to it [NCERT 11th, Ch 12, Sec 12.3; NCERT 11th, Ch 4, Sec 4.6].
  • Carbon bonded to 4 atoms (4 single bonds) →sp3\rightarrow sp^3.
  • Carbon bonded to 3 atoms (1 double bond) →sp2\rightarrow sp^2.
  • Carbon bonded to 2 atoms (1 triple bond or 2 double bonds) →sp\rightarrow sp.
  1. Analyze Option B (CH2=CH−C≡CHCH_2 = CH - C \equiv CH):
  • C1 (Left): Forms a double bond with C2 and two single bonds with H. Steric number = 3. Hybridisation = sp2sp^2.
  • C2: Forms a double bond with C1, a single bond with C3, and a single bond with H. Steric number = 3. Hybridisation = sp2sp^2.
  • C3: Forms a single bond with C2 and a triple bond with C4. Steric number = 2. Hybridisation = spsp.
  • C4 (Right): Forms a triple bond with C3 and a single bond with H. Steric number = 2. Hybridisation = spsp.
  • Sequence: sp2,sp2,sp,spsp^2, sp^2, sp, sp. This matches the required sequence.
  1. Analyze Other Options:
  • Option A (HC≡C−C≡HHC \equiv C - C \equiv H): All carbons are triple-bonded →sp,sp,sp,sp\rightarrow sp, sp, sp, sp.
  • Option C (CH2=CH−CH=CH2CH_2 = CH - CH = CH_2): All carbons are double-bonded →sp2,sp2,sp2,sp2\rightarrow sp^2, sp^2, sp^2, sp^2.
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