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This question already has an answer here:

What are hybridisation states of each carbon atom in the following compounds?

  • $\ce{CH2=C=O}$
  • $\ce{CH3CH=CH2}$
  • $\ce{(CH3)2CO}$
  • $\ce{CH2=CHCN}$
  • $\ce{C6H6}$

How do I find the hybridisation of the orbitals of the carbon atom in such compounds?

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marked as duplicate by Martin - マーチン Oct 15 '18 at 9:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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My method for finding the hybridization of the orbitals of an atom in a molecule:

  • Look at the molecular structure (geometry), apply the following:
    1. Tetrahedral: sp3
    2. Trigonal planar: sp2
    3. Linear: sp
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For hydrocarbons, if the carbon atom forms only σ bonds, the orbitals of it will be sp3 hybridised. If the carbon atom forms a π bond, its orbitals are sp2 hybridised. If the carbon atom forms two π bonds, its orbitals are sp hybridised.

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    $\begingroup$ Seeing as three of the five examples asked about are not hydrocarbons, this is not particularly helpful. You also don't explain why this is the case, which would be more useful in order to provide a general answer. $\endgroup$ – bon Aug 5 '15 at 17:34
  • $\begingroup$ I wanted to say organic compounds. As others have suggested usual ways to identify the hybridisation, I gave a simple method to find it. ☺ $\endgroup$ – Rajini Aug 5 '15 at 17:36
  • $\begingroup$ Please note that atoms cannot be hybridised, only orbitals. It's common to refer to atoms to be hybridised, but technically this is not correct. $\endgroup$ – Martin - マーチン Oct 15 '18 at 9:20
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Just apply the bond method which is the simplest and most effective. Count the number of bond pairs of central atoms excluding pi-bonds and the number Of lone pairs and match the corresponding hybridisation.

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    $\begingroup$ Welcome to the site. Could you just be more elaborate, just? This answer would benefit from better descriptions, just, and some examples, just. $\endgroup$ – M.A.R. Aug 5 '15 at 16:55

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