Hybridisation In Organic Molecules

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In a molecule of propyne, the two carbon atoms joined by the triple bond adopt which hybridisation state to accommodate that bonding?

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The hybridisation of a carbon atom is decided by the number of sigma bonds and lone pairs around it, which together equal the steric number. In propyne, CH_3-C#CH, the two triple-bonded carbons each form one sigma bond within the triple bond plus one additional sigma bond to a neighbouring group, giving a steric number of two. A steric number of two corresponds to sp hybridisation, leaving two unhybridised p orbitals on each carbon to form the two pi bonds of the triple bond and producing a linear arrangement. The choice sp^2 is wrong because that applies to carbons with a steric number of three, such as those in a double bond. The choice sp^3 describes carbons with four sigma bonds, like the methyl carbon, not the triple-bonded ones. The choice sp^3d is irrelevant for carbon, which cannot expand its octet. This follows the NCERT correlation between sigma-bond count and hybridisation. Plausibility check: a linear, 180-degree triple-bond geometry is only consistent with sp hybridisation.

Hybridisation In Organic Molecules

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