Question

What is the hybridization of each carbon atom in acetonitrile?

Answer 1

Hint :We know that the hybridization can be defined as the mixing of two atomic orbitals which have the same energy levels and by this mixing they provide a degenerated new type of orbitals and the mixing is based on quantum mechanics.

Complete Step By Step Answer:
A region of electron density is just one, double, or triple bond and lone try of electrons. The variety of regions of electron density can provide you with the steric number of the atom that successively can provide you with its hybridization. During this case, the left atom is enclosed by four regions of lepton density as a result of its secured to four completely different atoms, i.e. the three hydrogen atoms and therefore the right atom. The steric variety are going to be up to four, which means that the left carbon is $ s{{p}^{3}} $ hybridized, i.e. it uses one $ s $ orbital and $ 3\text{ }p $ orbitals to create four $ s{{p}^{3}} $ hybrid orbitals. The proper carbon is enclosed by a pair of regions of lepton density as a result of its secured to two completely different atoms, i.e. the gas atom and therefore the left atom. during this case, the steric variety are going to be up to a pair of, which means that the proper carbon is $ ~sp $ hybridized, i.e. it uses one $ s $ orbital and one p orbital to create $ 2\text{ }sp $ hybrid orbitals. Consequently, the left carbon can have $ 109.5 $ bond angles and therefore the right carbon can have $ 180 $ bond angles. Therefore, its $ s{{p}^{3}} $ hybridization.
 $ {{H}_{3}}C-C\equiv N: $
The total number of valence electrons present in a molecule of acetonitrile will be equal to $ 16 $ because we have;
 $ 2\times 4{{e}^{-}}=8{{e}^{-}}~ $ from two atoms of carbon.
 $ 3\times 1{{e}^{-}}~=3{{e}^{-}}~ $ from two atoms of hydrogen.
 $ 1\times 5{{e}^{-}}~=5{{e}^{-}} $ from two atoms of nitrogen.
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.
 $ \left( 4\times 2{{e}^{-}}\text{ }+\text{ }1\times 6{{e}^{-}} \right)=14{{e}^{-}} $ . Thus, the remaining two 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.

Additional Information:
Always remember that the formula we used is only applicable for covalent compounds, not for coordination compounds because in coordination compounds we use crystal field theory to explain the structure of them and VSEPR theory is not applicable over there.

Note :
Remember that the VSEPR theory accounts for the presence of lone pairs and its impact on the structure of molecules. It considers that the presence of lone pair distorts the structure because lone pair repels a bond pair more than a bond pair repels a bond pair and that readjustment within the molecule takes place to minimize the repulsion and that alters the bond angle.