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Chemical Bonding And Molecular Structure

Question
CBSEENCH11008419

In the hydrocarbon

straight C with 6 below straight H subscript 3 space minus straight C with 5 below straight H equals straight C with 4 below straight H minus straight C with 3 below straight H subscript 2 minus straight C with 2 below identical to straight C with 1 below straight H

The state of hybridization of carbon 1, 3, and 5 are in the following sequence

  • sp2, sp, sp3

  • sp, sp3, sp2

  • sp, sp2, sp3

  • sp3, sp2, sp

Solution

B.

sp, sp3, sp2

negative straight C minus straight C minus sp cubed
minus straight C equals straight C minus sp squared
minus straight C space identical to space straight C minus space sp
equals straight C equals straight C equals sp

straight H subscript 3 straight C with 6 below and sp cubed on top space minus space straight C with 5 below and sp squared on top straight H space equals straight C with 4 below and sp squared on top straight H space minus space straight C with 3 below and sp cubed on top straight H subscript 2 minus straight C with 2 below and sp on top identical to straight C with 1 below and sp on top straight H
Hence the state of hybridization of carbon 1,3 and 5 are sp, sp3, sp2 respectively.