Problem 166 - A Complete Resource Book in Chemistry for JEE Main

Question

Underlined carbon is sp \(^{3}\) hybridized in (a) \(\mathrm{CH}_{3} \mathrm{CH}=\underline{\mathrm{CH}}_{2}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}\) (c) \(\mathrm{CH}_{3} \mathrm{CONH}_{2}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CN}\)

Step-by-Step Solution

Answer

Option (b) is sp3 hybridized.

Step 1: Understand the Concept of Hybridization

Hybridization involves the mixing of atomic orbitals to form new hybrid orbitals, which can form sigma bonds. Carbon typically undergoes sp, sp2, or sp3 hybridization depending on its bonding and molecular geometry. An sp3 hybridized carbon atom forms four sigma bonds and has a tetrahedral geometry.

Step 2: Analyze Option (a)

In the molecule \(\mathrm{CH}_{3} \mathrm{CH}=\underline{\mathrm{CH}}_{2}\), the underlined carbon is part of a double bond (\(\mathrm{CH}_2\)) which means it has an sp2 hybridization (one \(\sigma\) bond and one \(\pi\) bond). Hence, the underlined carbon is not sp3 hybridized.

Step 3: Analyze Option (b)

In the molecule \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}\), the underlined carbon is in the \(\mathrm{NH}_2\) group. The carbon in question (\(\mathrm{CH}_2\)) forms four single bonds with no double or triple bonds, resulting in a sp3 hybridization. Thus, the underlined carbon is sp3 hybridized.

Step 4: Analyze Option (c)

In the molecule \(\mathrm{CH}_{3} \mathrm{CONH}_{2}\), the underlined carbon in the amide group is part of a double bond in the carbonyl group (C=O). This carbon atom has sp2 hybridization because it forms one sigma and one pi bond.

Step 5: Analyze Option (d)

In the molecule \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CN}\), the underlined carbon is part of a triple bond found in alkynes, which suggests it is sp hybridized with two pi bonds and one sigma bond. Therefore, it is not sp3 hybridized.

Key Concepts

sp3 Hybridization

Hybridization is a concept in chemistry where atomic orbitals combine to form new, hybridized orbitals. These hybrid orbitals are crucial because they can form stronger bonds with neighboring atoms. The sp3 hybridization is when one s orbital mixes with three p orbitals, resulting in four equivalent sp3 hybrid orbitals.
This configuration is common in carbon atoms that are saturated, meaning every bond around the carbon is a single bond. A molecule like methane, CH4, is a classic example, where the carbon forms four sigma bonds with hydrogen.

• Each bond is composed of an electron pair shared between the carbon and the hydrogen.
• There are no pi bonds, which are typically found in double or triple bonds.

This sp3 hybridization results in a three-dimensional shape, ensuring the bonds are as far apart as possible from each other.

Sigma Bonds

Sigma (σ) bonds are the strongest type of covalent chemical bond. They occur when two atomic orbitals overlap directly between the nuclei of two atoms. In sp3 hybridization, every bond formed is a sigma bond. This is because the sp3 orbitals overlap end-to-end with orbitals from another atom.
This overlap allows electrons to be shared in a way that stabilizes the molecules. You can visualize them as cylindrical tubes hugging straight between two bonded atoms.

• Sigma bonds can rotate without breaking, allowing molecules significant freedom in movement.
• These bonds are critical to the structure of organic molecules, giving them stability and rigidity.

In simple hydrocarbons, sigma bonds form the backbone which defines the molecule's general shape and structure.

Tetrahedral Geometry

The geometry of an sp3 hybridized atom is tetrahedral. This means the atom is at the center of a shape resembling a triangular pyramid, with bonds projecting outwards towards the corners. The bond angles in a perfect tetrahedron are about 109.5 degrees.
In the case of carbon, each of the four sp3 hybrid orbitals forms a single sigma bond with other atoms, such as hydrogen in methane (CH4).

• This geometry maximizes the distance between electron pairs, which minimizes electron repulsion, thus stabilizing the molecule.
• Tetrahedral geometry is pivotal in determining the molecule's properties, influencing things like polarity and reactivity.

Understanding tetrahedral geometry is essential for envisioning how molecules interact in three dimensions.