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NEET CHEMISTRYEasy

In the given hydrocarbon, the type/state of hybridization of carbons 1, 3, and 5 respectively are:

A

sp2, sp, sp3

B

sp, sp3, sp2

C

sp, sp2, sp3

D

sp3, sp2, sp

Step-by-Step Solution

  1. Hybridisation Rules: The state of hybridisation of carbon depends on the number of sigma (σ\sigma ) and \pi (π\pi ) bonds attached to it [NCERT 11th, Ch 12, Sec 12.3; Source 62, 163, 165, 166].
  • sp3 hybridised: Carbon bonded to 4 atoms via single bonds (4 σ\sigma bonds).
  • sp2 hybridised: Carbon bonded to 3 atoms (1 double bond, 3 σ\sigma + 1 π\pi ).
  • sp hybridised: Carbon bonded to 2 atoms (1 triple bond or 2 double bonds, 2 σ\sigma + 2 π\pi ).
  1. Structure Analysis (Reconstructed): Although the image is missing, the correct option (sp, sp3, sp2) corresponds to a hydrocarbon chain numbered such that:
  • C1 is sp: It must be part of a triple bond (e.g., HC≡C−HC≡C-).
  • C3 is sp3: It must be a saturated carbon (e.g., −CH2−-CH_2-).
  • C5 is sp2: It must be part of a double bond (e.g., −CH=-CH=).
  • A likely structure fitting this description is Hex-4-en-1-yne (HC1≡C2−C3H2−C4H=C5H−C6H3HC^1≡C^2-C^3H_2-C^4H=C^5H-C^6H_3).
  1. Assignment:
  • Carbon 1 (HC≡HC≡): Triple bond →→ sp.
  • Carbon 3 (−CH2−-CH_2-): Single bonds →→ sp3.
  • Carbon 5 (=CH−=CH-): Double bond →→ sp2.
  • This matches Option 2.
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