# What is the hybridization of each carbon atom in acetonitrile?

Jul 19, 2017

Here's what I got.

#### Explanation:

Start by drawing the Lewis structure of acetonitrile, $\text{CH"_3"CN}$.

The total number of valence electrons present in a molecule of acetonitrile will be equal to $16$ because you have

• $2 \times {\text{4 e"^(-) = "8 e}}^{-} \to$ from two atoms of carbon, $\text{C}$
• $3 \times {\text{1 e"^(-) = "3 e}}^{-} \to$ from three atoms of hydrogen, $\text{H}$
• $1 \times {\text{5 e"^(-) = "5 e}}^{-} \to$ from one atom of nitrogen, $\text{N}$

Now, the two carbon atoms will be bonded together via a single bond. One of the two carbon atoms will be bonded to the nitrogen atom via a triple bond and the other will be bonded to the three hydrogen atoms via single bonds.

This will account for

$4 \times {\text{2 e"^(-) + 1 xx "6 e"^(-) = "14 e}}^{-}$

The remaining $2$ valence electrons will be added on the nitrogen atom as a lone pair. In order to find the hybridization of the two carbon atoms, you must count the regions of electron density that surround the atoms.

A region of electron density is simply

• a single, double, or triple bond
• a lone pair of electrons

The number of regions of electron density will give you the steric number of the atom, which in turn will give you its hybridization.

In this case, the left carbon atom is surrounded by $4$ regions of electron density because it is bonded to four different atoms, i.e. the $3$ hydrogen atoms and the right carbon atom.

The steric number will be equal to $4$, which implies that the left carbon is $s {p}^{3}$ hybridized, i.e. it uses one $s$ orbital and three $p$ orbitals to form four $s {p}^{3}$ hybrid orbitals. The right carbon is surrounded by $2$ regions of electron density because it is bonded to two different atoms, i.e. the nitrogen atom and the left carbon atom.

In this case, the steric number will be equal to $2$, which implies that the right carbon is $s p$ hybridized, i.e. it uses one $s$ orbital and one $p$ orbital to form two $s p$ hybrid orbitals. Consequently, the left carbon will have ${109.5}^{\circ}$ bond angles and the right carbon will have ${180}^{\circ}$ bond angles. 