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The type of hybridization of the carbon atoms from left to right in CH3 − CH = CH − CN is

  • A.\(sp^{3}, sp, sp, sp^{2}\)
  • B.\(sp^{3}, sp^{2}, sp, sp\)
  • C.\(sp^{3}, sp^{2}, sp^{2}, sp\)
  • D.\(sp^{3}, sp^{2}, sp^{2}, sp^{2}\)
  • E.\(sp^{3}, sp^{2}, sp, sp^{2}\)

Answer: C

The hybridisations are sp3, sp2, sp2, sp.

Count the electron domains on each carbon in CH3–CH=CH–C≡N:

  1. CH3: four single bonds, sp3.
  2. CH=: one double bond, sp2.
  3. =CH–: the other end of the double bond, sp2.
  4. C≡N: a triple bond, sp.

Options with sp on the second or third carbon treat a double-bonded carbon as triple-bonded, and option D gives the nitrile carbon sp2 instead of sp.

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