Chapter 11, Problem 11E.1P

(a)

Interpretation Introduction

Interpretation:

The point group of ${\text{CH}}_{\text{3}}\text{Cl}$ has to be stated.

Concept introduction:

A symmetry operation is defined as an action on an object to reproduce an arrangement that is identical to its original spatial arrangement.  The group of symmetry operations of which at least one point is kept fixed is called point group.  The symmetry operations can be identity, rotation, reflection, inversion and improper rotation.

Answer to Problem 11E.1P

The point group of ${\text{CH}}_{\text{3}}\text{Cl}$ is ${\text{C}}_{\text{3v}}$.

Explanation of Solution

The symmetry elements in ${\text{CH}}_{\text{3}}\text{Cl}$ are shown below.

Figure 1

The structure of ${\text{CH}}_{\text{3}}\text{Cl}$ is tetrahedral with three same and one different group attached to the central atom.  It has a threefold axis.  The molecule has three vertical planes.

Therefore, the point group of ${\text{CH}}_{\text{3}}\text{Cl}$ is ${\text{C}}_{\text{3v}}$.

(b)

Interpretation Introduction

Interpretation:

The number of normal modes of vibrations in ${\text{CH}}_{\text{3}}\text{Cl}$ has to be calculated.

Concept introduction:

The complex vibrations exhibit by the polyatomic molecule is known as normal modes of vibrations.  The vibrational modes of a molecule are IR or Raman active.  If a molecule has centre of symmetry, then the modes which are IR-active will be Raman inactive and the modes that are IR-inactive will be Raman active.  The total number of vibrational degrees of freedom for nonlinear molecule is represented by $3N-6$.

Answer to Problem 11E.1P

The number of normal modes of vibration in ${\text{CH}}_{\text{3}}\text{Cl}$ is $9$.

Explanation of Solution

The total number of atoms in the molecule (${\text{CH}}_{\text{3}}\text{Cl}$) is $5$.

The total number of vibrational modes of a nonlinear molecule is given by the formula below.

  $\text{Number}\text{of}\text{vibrational}\text{modes}=3N-6$        (1)

Where,

• $N$ is the total number of atoms in the molecule.

Substitute the value of $N$ in equation (1).

  $\begin{array}{c}\text{Number}\text{of}\text{vibrational}\text{modes}=3\left(5\right)-6\\ =15-6\\ =9\end{array}$

Therefore, the number of normal modes of vibration in ${\text{CH}}_{\text{3}}\text{Cl}$ is $9$.

(c)

Interpretation Introduction

Interpretation:

The symmetry species of the normal modes of vibration of ${\text{CH}}_{\text{3}}\text{Cl}$ has to be stated.

Concept introduction:

The characters of the irreducible representations of the given point group can be multiplied by each other.  The only condition is the characters of the same symmetry operations are multiplied together.  The multiplication of the characters is commutative.

The great orthogonality theorem for the reducible representation can be expressed as shown below.

  $a_{\Gamma }=\frac{1}{h}{\displaystyle \sum _{\begin{array}{l}\text{all}\text{classes}\\ \text{of}\text{point}\text{group}\end{array}}N\cdot {\chi }_{\Gamma }\cdot {\chi }_{\text{linear}\text{combo}}}$

Answer to Problem 11E.1P

The symmetry species of normal modes of vibrational for ${\text{CH}}_{\text{3}}\text{Cl}$ is $3{\text{A}}_1+4\text{E}$.

Explanation of Solution

The character table of $C_{\text{3v}}$ is shown below.

$C_{\text{3v}}$	$E$	$2C_{\text{3}}$	$3{\sigma }_{\text{v}}$
${\text{A}}_{\text{1}}$	$1$	$1$	$1$	$z$, $z^2$, $x^2+y^2$
${\text{A}}_{\text{2}}$	$1$	$1$	$-1$	$R_z$
$\text{E}$	$2$	$-1$	$0$	$\left(x,y\right)$, $\left(xy,x^2-y^2\right)\left(yz,zx\right)$, $\left(R_x,R_y\right)$

The irreducible representation of $C_{\text{3v}}$ for ${\text{CH}}_{\text{3}}\text{Cl}$ is shown below.

$C_{\text{3v}}$	$E$	$2C_{\text{3}}$	$3{\sigma }_{\text{v}}$
${\Gamma }_{\text{red}}$	$15$	$0$	$3$

The great orthogonality theorem for the reducible representation can be expressed as shown below.

  $a_{\Gamma }=\frac{1}{h}{\displaystyle \sum _{\begin{array}{l}\text{all}\text{classes}\\ \text{of}\text{point}\text{group}\end{array}}N\cdot {\chi }_{\Gamma }\cdot {\chi }_{\text{linear}\text{combo}}}$        (2)

Where,

• $a_{\Gamma }$ is the number of times the irreducible representation appears in a linear combination.
• $h$ is the order of the group.
• ${\chi }_{\Gamma }$ is the character of the class of the irreducible representation.
• ${\chi }_{\text{linear}\text{combo}}$ is the character of the class linear combination.
• $N$ is the number of symmetry operations.

The order of the group is $6$.

The great orthogonality theorem orthogonality of the irreducible representation of ${\text{A}}_{\text{1}}$, ${\text{A}}_{\text{2}}$, and $\text{E}$ is shown below.

Substitute the value of the order of the group, character of the class of the irreducible representation from character table of $C_{\text{3v}}$ point group, the character of the class linear combination and number of symmetry operations for ${\text{A}}_{\text{1}}$ in the equation (2).

  $\begin{array}{c}{a}_{{\text{A}}_1}=\frac{1}{6}\left[\left(1\cdot 1\cdot 15\right)+\left(2\cdot 1\cdot 0\right)+\left(3\cdot 1\cdot 3\right)\right]\\ =4\end{array}$

The number of times the irreducible representation for ${\text{A}}_{\text{1}}$ appears in a linear combination is $4$.

Similarly, for ${\text{A}}_{\text{2}}$, substitute the value of order of the group, character of the class of the irreducible representation from character table of $C_{\text{3v}}$ point group, character of the class linear combination and number of symmetry operations in the equation (2).

    $\begin{array}{c}{a}_{{\text{A}}_2}=\frac{1}{6}\left[\left(1\cdot 1\cdot 15\right)+\left(2\cdot 1\cdot 0\right)+\left(3\cdot -1\cdot 3\right)\right]\\ =1\end{array}$

The number of times the irreducible representation for ${\text{A}}_{\text{2}}$ appears in a linear combination is $1$.

Similarly, for $\text{E}$, substitute the value of the order of the group, character of the class of the irreducible representation from character table of $C_{\text{3v}}$ point group, the character of the class linear combination and number of symmetry operations in the equation (2).

    $\begin{array}{c}{a}_{\text{E}}=\frac{1}{6}\left[\left(1\cdot 2\cdot 15\right)+\left(2\cdot -1\cdot 0\right)+\left(3\cdot 0\cdot 3\right)\right]\\ =5\end{array}$

The number of times the irreducible representation for $\text{E}$ appears in a linear combination is $5$.

The displacement of the atoms of ${\text{CH}}_{\text{3}}\text{Cl}$ span ${\text{4A}}_{\text{1}}+{\text{A}}_{\text{2}}\text{+5E}$.

From the $C_{\text{3v}}$ character table, ${\text{A}}_{\text{1}}$ and $\text{E}$ are the symmetry species for translational motion while ${\text{A}}_{\text{2}}$ and $\text{E}$ are the symmetry species for rotational motion.

The symmetry species for the normal mode of vibration is calculated by the formula shown below.

    $\text{Vibrational}=\text{Displacement}\text{of}\text{atom}-\left(\text{rotational}+\text{translational}\right)$

Substitute the symmetry species rotational, translational and displacement of the atoms in the above expression.

    $\begin{array}{c}\text{Vibrational}={\text{4A}}_{\text{1}}+{\text{A}}_{\text{2}}\text{+5E}-\left({\text{A}}_{\text{1}}+\text{E}+{\text{A}}_{\text{2}}+\text{E}\right)\\ ={\text{4A}}_{\text{1}}+{\text{A}}_{\text{2}}\text{+5E}-{\text{A}}_{\text{1}}-\text{E}-{\text{A}}_{\text{2}}-\text{E}\\ =3{\text{A}}_1+4\text{E}\end{array}$

Therefore, the symmetry species of normal modes of vibrational for ${\text{CH}}_{\text{3}}\text{Cl}$ is $3{\text{A}}_1+4\text{E}$.

(d)

Interpretation Introduction

Interpretation:

The vibrational modes of ${\text{CH}}_{\text{3}}\text{Cl}$ which are infrared active have to be stated.

Concept introduction:

As mentioned in the concept introduction in part (b).

Answer to Problem 11E.1P

The vibrational modes of ${\text{CH}}_{\text{3}}\text{Cl}$ which are infrared active are $\text{E}$ and ${\text{A}}_{\text{1}}$.

Explanation of Solution

The character table of $C_{\text{3v}}$ is shown below.

$C_{\text{3v}}$	$E$	$2C_{\text{3}}$	$3{\sigma }_{\text{v}}$
${\text{A}}_{\text{1}}$	$1$	$1$	$1$	$z$, $z^2$, $x^2+y^2$
${\text{A}}_{\text{2}}$	$1$	$1$	$-1$
$\text{E}$	$2$	$-1$	$0$	$\left(x,y\right)$, $\left(xy,x^2-y^2\right)\left(yz,zx\right)$

The symmetry species of the normal modes of vibration of ${\text{CH}}_{\text{3}}\text{Cl}$ is $3{\text{A}}_1+4\text{E}$.

The symmetry species that contains $x$, $y$, $z$ are infrared active.

The irreducible representation $\text{E}$ and ${\text{A}}_{\text{1}}$ has $x$, $y$, and $z$.  Therefore, these are infrared active.

(e)

Interpretation Introduction

Interpretation:

The vibrational modes of ${\text{CH}}_{\text{3}}\text{Cl}$ which is Raman active have to be stated.

Concept introduction:

As mentioned in the concept introduction in part (b).

Answer to Problem 11E.1P

The vibrational modes of ${\text{CH}}_{\text{3}}\text{Cl}$ which are Raman active are $\text{E}$ and ${\text{A}}_{\text{1}}$.

Explanation of Solution

The character table of $C_{\text{3v}}$ is shown below.

$C_{\text{3v}}$	$E$	$2C_{\text{3}}$	$3{\sigma }_{\text{v}}$
${\text{A}}_{\text{1}}$	$1$	$1$	$1$	$z$, $z^2$, $x^2+y^2$
${\text{A}}_{\text{2}}$	$1$	$1$	$-1$
$\text{E}$	$2$	$-1$	$0$	$\left(x,y\right)$, $\left(xy,x^2-y^2\right)\left(yz,zx\right)$

The symmetry species of the normal modes of vibration of ${\text{CH}}_{\text{3}}\text{Cl}$ is $3{\text{A}}_1+4\text{E}$.

The symmetry species that contains quadratic form is Raman active.

The irreducible representation $\text{E}$ and ${\text{A}}_{\text{1}}$ has a quadratic form. Therefore, these are Raman active modes.