Problem 125 - A Complete Resource Book in Chemistry for JEE Main

Question

When a methyl radical is formed from \(\mathrm{CH}_{3} \mathrm{Cl}\), select the correct statement: (1) bond angle of \(109^{\circ} 28\) ' is retained (2) number of sigma bonds is three (3) carbon undergoes geometric change from tetrahedral to planar (4) hybridization changes \(\mathrm{sp}^{3}\) to \(\mathrm{sp}^{2}\) (a) 2,3 and 4 (b) 1,3 and 4 (c) 2 and 4 (d) 3 and 4

Step-by-Step Solution

Answer

The correct statements are 3 and 4, so the answer is (d).

Step 1: Identify the initial state

The initial molecule is \( \mathrm{CH}_{3} \mathrm{Cl} \), in which carbon is tetrahedrally hybridized (\( \mathrm{sp}^3 \) hybridization) with a bond angle of approximately \(109^{\circ} 28'\).

Step 2: Analyze the formation of methyl radical

When a methyl radical is formed, one chlorine atom leaves, creating \( \mathrm{CH}_3^\bullet \) with one unpaired electron at carbon.

Step 3: Determine changes in hybridization

In \( \mathrm{CH}_3^\bullet \), the carbon atom changes its hybridization from \( \mathrm{sp}^3 \) to \( \mathrm{sp}^2 \). This is because the molecule becomes planar to allow equal distribution of charge around the newly formed radical.

Step 4: Assess geometric changes

The geometric structure of the methyl radical \( \mathrm{CH}_3^\bullet \) becomes planar (trigonal planar geometry) with a bond angle close to \(120^{\circ}\), which differs from the original tetrahedral structure.

Step 5: Count the number of sigma bonds

The methyl radical \( \mathrm{CH}_3^\bullet \) has three \( \sigma \)-bonds, corresponding to the bonds between carbon and the three hydrogen atoms.

Step 6: Select the correct statement(s)

Given these analyses, options (3) carbon undergoes geometric change from tetrahedral to planar, and (4) hybridization changes from \( \mathrm{sp}^{3} \) to \( \mathrm{sp}^{2} \) are correct. Thus, the correct choice is (d) 3 and 4.

Key Concepts

Hybridization

Hybridization is a concept used to explain the bonding and structure of molecules by combining atomic orbitals into new hybrid orbitals. These hybrid orbitals have different shapes and energies than the original atomic orbitals.

• In an \( \text{sp}^3 \) hybridized carbon, one \( s \) and three \( p \) orbitals mix to form four equivalent \( \text{sp}^3 \) hybrid orbitals.
• These orbitals are directed towards the corners of a tetrahedron, giving a tetrahedral geometry with bond angles of approximately \(109^{\circ} 28'\).

In the methyl radical \( \text{CH}_3^\bullet \), the hybridization of the carbon atom changes from \( \text{sp}^3 \) to \( \text{sp}^2 \). This transformation occurs because one of the \( \text{sp}^3 \) hybrid orbitals becomes an unpaired electron.
This leads to the formation of three \( \text{sp}^2 \) hybrid orbitals and one unhybridized \( p \) orbital, resulting in a planar structure with 120° bond angles, which we will discuss in detail below.

Tetrahedral Geometry

Tetrahedral geometry is a fundamental concept that involves four bonded groups arranged around a central atom. In the case of \( \text{CH}_3\text{Cl} \), the carbon atom with \( \text{sp}^3 \) hybridization exhibits this geometry.
Key features of tetrahedral geometry include:

• Four equivalent bonds symmetrically arranged in space, each forming the shape of a tetrahedron.
• Bond angles of about \(109^{\circ} 28'\).
• A three-dimensional structure where no two bonds are collinear, ensuring maximum separation of electron pairs, thus minimizing repulsion.

Upon the formation of a methyl radical \( \text{CH}_3^\bullet \), tetrahedral geometry is altered. The removal of a chlorine atom and the presence of an unpaired electron lead to a significant geometric transformation. This change accommodates the electron density more effectively in a two-dimensional plane, giving rise to what is known as trigonal planar geometry.

Trigonal Planar Geometry

Trigonal planar geometry is a key structural feature that emerges when a carbon atom in a molecule such as \( \text{CH}_3^\bullet \) adopts \( \text{sp}^2 \) hybridization. This configuration results when one of the \( sp^3 \) hybrid orbitals transforms into a p orbital, providing room for equal charge distribution around the carbon atom.

• In this geometry, the three \( \sigma \)-bonds between carbon and hydrogen lie in a single plane, each with bond angles of approximately \(120^{\circ}\).
• This planar arrangement allows for optimal overlap of the bonding electrons, ensuring stability of the methyl radical.
• The empty \( \text{p} \) orbital, which remains unhybridized, is oriented perpendicular to the plane of the molecule, ready to interact with other species.

Transitioning from tetrahedral to trigonal planar geometry represents a fundamental structural change in the behavior of the molecule. This shift leads to different chemical properties and reactivity, emphasizing the importance of understanding geometrical and hybridization states in organic chemistry.