Question

The pair of electrons in the given carbanion $C{H_3}C \equiv {C^ - }$ , is present in which of the following orbitals:
$A.\,\,2p$
$B.\,\,s{p^3}$
$C.\,\,s{p^2}$
$D.\,\,sp$

Answer 1

:The hybridization concept is used in organic chemistry to explain chemical bonding. This theory is very useful to explain the covalent bonds in organic compounds. In the hybridization, the number of $\sigma \,$ -bond, $\pi$ -bond and number of electrons are calculated present in the carbanion along with the charge present on it.
Complete step-by-step answer: The hybridization is an intermixing of atomic orbitals of different shapes and has nearly same energy to give the same number of hybrid orbitals of the same shape, equal energy and orientation such that there will be less repulsion between the hybrid orbitals.
A carbanion is an anion in which carbon is trivalent (forms three bonds) and bears a formal negative charge. The negative charge of a carbanion is localized in any $sp$ hybridized orbital on carbon depending on their availability of lone pair. As a consequence, to its hybridization carbanions are assumed to be trigonal pyramidal, bent, and linear geometries, respectively.
The pair of electrons in the $C{H_3}C \equiv {C^ - }$ is present at $\equiv C$ carbon.
This carbon has $1\,\sigma \,$-bond and $2\,\pi$ -bonds and $11$ lone pair of electrons.
Now, we need to calculate the hybridization by using the hybridisation formula, i.e.,
$Hybridization\, = \,\dfrac{{Number{\text{ }}of{\text{ }}\sigma {\text{ }} - {\text{ }}electrons}}{2}$
${\text{ = }}\dfrac{{{\text{2 + 2 (negative ion)}}}}{{\text{2}}}$
$= \,2\, = \,sp$
An atom with a triple bond and one lone pair has two regions of electron density. Therefore, its hybridization would be $sp$(one orbital per region of electron density) in which both the electrons are being shared in two different orbitals. In this triple bond carbanion, the pair of electrons are being shared in the $s$ and $p$ orbital giving it a linear-shaped geometry of $sp$ hybridization. Even when there are two different bonds associated, it is $sp$ hybridized.

Therefore, the correct option is option $D.\,\,sp$.

Note: Carbanion possesses an unshared pair of electrons thus it is an electron-rich species. The negatively charged carbon is trivalent. The triple bond in carbanion consists of one $\sigma$ bond and two $\pi$ bonds. The sigma bond between the carbon atoms forms overlapping of $sp$ hybrid orbitals from each carbon atom.