Problem 12 - General Chemistry: Principles and Modern Applications

Question

Propose a hybridization scheme to account for bonds formed by the central carbon atom in each of the following molecules: (a) hydrogen cyanide, HCN; (b) methyl alcohol, \(\mathrm{CH}_{3} \mathrm{OH} ;\) (c) acetone, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}\) (d) carbamic acid,

Step-by-Step Solution

Answer

Hybridizations for each molecule are: HCN - sp hybridization, CH3OH - sp3 hybridization, (CH3)2CO - sp2 hybridization, NH2COOH - sp2 hybridization.

Step 1: Determine Hybridization for Hydrogen Cyanide (HCN)

In HCN, the central carbon atom is bonded to a hydrogen atom and a nitrogen atom. Therefore, it is forming two sigma bonds which will use two of the carbon atom's valence electrons. This suggests that carbon's 2s and one of the 2p orbitals will hybridize into two sp hybrid orbitals.

Step 2: Determine Hybridization for Methyl Alcohol (CH3OH)

In CH3OH, the central carbon atom is bonded to three hydrogen atoms and one oxygen atom. The carbon atom therefore forms four sigma bonds using all its four valence electrons. This indicates that carbon's 2s and all the three 2p orbitals will hybridize into four sp3 hybrid orbitals.

Step 3: Determine Hybridization for Acetone (CH3COCH3)

In CH3COCH3 or (CH3)2CO, the central carbon atom bonded to two other carbon atoms and one oxygen atom. This results in three sigma bonds with other atoms and one pi bond with the oxygen atom, therefore using all its four valence electrons. Hence, it can be said that carbon's 2s and two of the 2p orbitals will hybridize into three sp2 hybrid orbitals.

Step 4: Determine Hybridization for Carbamic Acid (NH2COOH)

In NH2COOH, the central carbon atom is bonded to one nitrogen atom, one oxygen atom and one hydroxyl group. This results in three sigma bonds with other atoms and one pi bond with one of the oxygen atoms, using all its four valence electrons. As such, it can be inferred that carbon's 2s and two of the 2p orbitals will hybridize into three sp2 hybrid orbitals.

Key Concepts

Sigma Bonds

Sigma bonds (\( \sigma \) bonds) are one of the most basic types of chemical bonds. They occur when two atomic orbitals overlap head-to-head. This direct overlap allows the bonding electrons to be shared across a region centered along the bond axis.

**Properties of Sigma Bonds:**

• They are single covalent bonds and can occur with s or p atomic orbitals.
• They have higher bond strength compared to pi bonds when they are in isolation.
• Once a sigma bond is formed, it restricts rotation around the bond axis.

In the molecule methyl alcohol (\( \mathrm{CH}_3\mathrm{OH} \)), the carbon forms four sigma bonds — three with hydrogen atoms and one with an oxygen atom. Understanding sigma bonds is essential for visualizing the core scaffold of many organic molecules.

sp3 Hybridization

The concept of sp3 hybridization is instrumental in understanding the geometry of carbon-containing molecules. When a carbon atom undergoes sp3 hybridization, it mixes one s orbital with three p orbitals, creating four equivalent sp3 hybrid orbitals.

**Characteristics of sp3 Hybridization:**

• Molecules with sp3 hybridized carbon typically have a tetrahedral geometry.
• The bond angles in a perfect tetrahedron are approximately 109.5 degrees.
• An example of a molecule with sp3 hybridization is methane (\( \mathrm{CH}_{4} \)) where carbon forms four sigma bonds with hydrogen atoms.

For methyl alcohol (\( \mathrm{CH}_{3} \mathrm{OH} \)), the carbon atom adopts an sp3 hybridization since it forms four sigma bonds, resulting in tetrahedral geometry.

Pi Bonds

Pi bonds (\( \pi \) bonds) are formed when two lobes of an orbital on one atom overlap side by side with two lobes of an orbital on another atom. This typically occurs in the presence of p orbitals.

**Features of Pi Bonds:**

• They are often found in double or triple bonds, complementing a sigma bond.
• Pi bonds result in electron density above and below the bond axis.
• Due to their nature, pi bonds restrict the rotation around the double or triple bond axis.

For example, in acetone (\( \left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO} \)), the central carbon forms one pi bond with the oxygen and two additional sigma bonds. Understanding pi bonds helps in determining the rigidity and unsaturation level in organic molecules.