Question

When a methyl radical is formed from \(\mathrm{CH}_{3} \mathrm{Cl}\), select the correct statement: (1) bond angle of \(109^{\circ} 28^{\prime}\) is retained (2) number of sigma bonds is three (3) carbon undergoes geometric change from tetrahedral to planar (4) hybridization changes sp' to sp² (a) 2,3 and 4 (b) 1,3 and 4 (c) 2 and 4 (d) 3 and 4

Short answer

The correct choice is (a) 2, 3, and 4.

Step-by-step solution

Evaluate the Bond Angle

When a methyl radical is formed from \(\mathrm{CH}_{3}\mathrm{Cl}\), it loses a chlorine atom, which affects the geometry of the molecule. Originally, \(\mathrm{CH}_{3}\mathrm{Cl}\) has a tetrahedral geometry with bond angles of approximately \(109^{\circ} 28^{\prime}\). However, for a methyl radical, the geometry becomes trigonal planar, and the bond angles become \(120^{\circ}\) due to the different hybridization and presence of an unpaired electron.

Count the Number of Sigma Bonds

In \(\mathrm{CH}_{3}\mathrm{Cl}\), carbon forms four sigma bonds with three hydrogen atoms and one chlorine atom. When the chlorine leaves, the methyl radical \(\cdot \mathrm{CH}_{3}\) is formed, which has three sigma bonds between carbon and the three hydrogen atoms.

Determine the Geometric Change

Initially, \(\mathrm{CH}_{3}\mathrm{Cl}\) has a tetrahedral shape. Removing one chlorine atom results in the geometry changing from tetrahedral to planar for the methyl radical, as it is now formed by three sigma bonds with one unpaired electron.

Consider the Change in Hybridization

In \(\mathrm{CH}_{3}\mathrm{Cl}\), carbon is \(sp^3\) hybridized due to forming four sigma bonds in a tetrahedral geometry. After losing chlorine, the methyl radical \(\cdot \mathrm{CH}_{3}\) has an \(sp^2\) hybridization with a planar geometry, since it forms only three sigma bonds and contains a lone electron in the unhybridized \(p\) orbital.

Select the Correct Statements

Based on the evaluation: statement 1 is incorrect (bond angle changes), statement 2 is correct (three sigma bonds remain), statement 3 is correct (geometry changes from tetrahedral to planar), and statement 4 is correct (hybridization changes from \(sp^3\) to \(sp^2\)). Thus, the correct choice is (a) 2, 3, and 4.

Key concepts

Hybridization

When it comes to understanding the methyl radical, hybridization is key. Hybridization describes how atomic orbitals mix to form new hybrid orbitals that make the chemical bonds in a molecule. Initially, in \(\text{CH}_3\text{Cl}\), carbon is \(sp^3\) hybridized, forming four sigma bonds in a tetrahedral arrangement. However, when a chlorine atom is removed to form the methyl radical \(\cdot\text{CH}_3\), things change. The carbon atom transitions to an \(sp^2\) hybridization. This means that instead of four hybrid orbitals, we have three \(sp^2\) orbitals and one unhybridized \(p\) orbital.\BREAKLINE|\ This unpaired electron lives in the unhybridized \(p\) orbital. The transition to \(sp^2\) means that carbon now forms three sigma bonds instead of four, which directly leads to a change in geometry, as we'll discuss next.

Sigma Bonds

Sigma bonds are one of the primary types of covalent bonds established between atoms in molecules. In the context of the methyl radical, let's understand their significance. Originally, in \(\text{CH}_3\text{Cl}\), carbon forms four sigma bonds due to its \(sp^3\) hybridization: three with hydrogen atoms and one with a chlorine atom.\BREAKLINE|\ Upon losing the chlorine atom, the structure simplifies to form the methyl radical \(\cdot\text{CH}_3\), which then has only three sigma bonds linking the carbon atom with the three hydrogen atoms; the unshared electron is in the unhybridized \(p\) orbital. These sigma bonds determine the overall shape and stability of the resulting radical.

Molecular Geometry

Molecular geometry is vital to understanding how molecules behave and react. Initially, \(\text{CH}_3\text{Cl}\) has a tetrahedral shape, characterized by bond angles of approximately \(109^{\circ} 28'\). However, when the chlorine atom is removed, converting the molecule into a methyl radical, the geometry shifts. Now, the structure is trigonal planar.\BREAKLINE|\ This change occurs because the carbon atom only has three \(sp^2\) hybridized orbitals available for bonding, positioning the hydrogen atoms in one plane. The presence of the lone electron in a \(p\) orbital allows this planar structure, with bond angles close to \(120^{\circ}\).\BREAKLINE|\ This geometry change significantly impacts how the methyl radical engages in further chemical reactions.

Bond Angle Changes

Bond angles provide insight into the spatial arrangement of atoms within molecules, which are influenced by the type of hybridization at play. Initially, \(\text{CH}_3\text{Cl}\) exhibits bond angles typical of a tetrahedral structure, around \(109^{\circ} 28'\), due to its \(sp^3\) hybridization.\BREAKLINE|\ However, the transition to a methyl radical results in the angles opening up to \(120^{\circ}\).\BREAKLINE|\ This increase occurs because the electron pairs move further apart as a result of the switch to \(sp^2\) hybridization and the planar molecular geometry it causes. Understanding these changes is essential for predicting how molecules will interact in various chemical contexts.