The hybridization of the positively charged c | vedclass

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(D) The positively charged carbon (carbocation) is bonded to three other atoms and has no lone pairs. The number of hybrid orbitals = $3 	ext{ (bond

Vedclass · Accepted Answer

$sp^3, sp^2$

Vedclass · Answer

(D) The positively charged carbon (carbocation) is bonded to three other atoms and has no lone pairs. The number of hybrid orbitals = $3 	ext{ (bond pairs)} + 0 	ext{ (lone pairs)} = 3$,which corresponds to $sp^2$ hybridization.The negatively charged carbon (carbanion) is bonded to three other atoms and has one lone pair. The number of hybrid orbitals = $3 	ext{ (bond pairs)} + 1 	ext{ (lone pair)} = 4$,which corresponds to $sp^3$ hybridization.Therefore,the hybridization is $sp^2$ and $sp^3$ respectively.

The hybridization of the positively charged carbon and the negatively charged carbon in the following structures are respectively:

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(Note: The image shows the structures of these carbanions.)

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