Comprehensive Study Notes

Qualitative information about molecular wave functions and properties can often be obtained from the symmetry of the molecule. By the symmetry of a molecule, we mean the symmetry of the framework formed by the nuclei held fixed in their equilibrium positions. (Our starting point for molecular quantum mechanics will be the Born-Oppenheimer approximation, which regards the nuclei as fixed when solving for the electronic wave function; see Section 13.1.) The symmetry of a molecule can differ in different electronic states. For example, HCN is linear in its ground electronic state, but nonlinear in certain excited states. Unless otherwise specified, we shall be considering the symmetry of the ground electronic state.

Symmetry Elements and Operations

A symmetry operation is a transformation of a body such that the final position is physically indistinguishable from the initial position, and the distances between all pairs of points in the body are preserved. For example, consider the trigonal-planar molecule $\mathrm{BF}_{3}$ (Fig. 12.1a), where for convenience we have numbered the fluorine nuclei. If we rotate the molecule counterclockwise by $120^{\circ}$ about an axis through the boron nucleus and perpendicular to the plane of the molecule, the new position will be as in Fig. 12.1b. Since in reality the fluorine nuclei are physically indistinguishable from one another, we have carried out a symmetry operation. The axis about which we rotated the molecule is an example of a symmetry element. Symmetry elements and symmetry operations are related but different things, which are often confused. A symmetry element is a geometrical entity (point, line, or plane) with respect to which a symmetry operation is carried out.

We say that a body has an $\boldsymbol{n}$-fold axis of symmetry (also called an $n$-fold proper axis or an $n$-fold rotation axis) if rotation about this axis by $360 / n$ degrees (where $n$ is an

FIGURE 12.1 (a) The $\mathrm{BF}{3}$ molecule. (b) $\mathrm{BF}{3}$ after a $120^{\circ}$ rotation about the axis through B and perpendicular to the molecular plane.

(a)

(b)

integer) gives a configuration physically indistinguishable from the original position; $n$ is called the order of the axis. For example, $\mathrm{BF}{3}$ has a threefold axis of symmetry perpendicular to the molecular plane. The symbol for an $n$-fold rotation axis is $C{n}$. The threefold axis in $\mathrm{BF}{3}$ is a $C{3}$ axis. To denote the operation of counterclockwise rotation by $(360 / n)^{\circ}$, we use the symbol $\hat{C}{n}$. The "hat" distinguishes symmetry operations from symmetry elements. $\mathrm{BF}{3}$ has three more rotation axes; each $\mathrm{B}-\mathrm{F}$ bond is a twofold symmetry axis (Fig. 12.2).

A second kind of symmetry element is a plane of symmetry. A molecule has a plane of symmetry if reflection of all the nuclei through that plane gives a configuration physically indistinguishable from the original one. The symbol for a symmetry plane is $\sigma$ (lowercase sigma). (Spiegel is the German word for mirror.) The symbol for the operation of reflection is $\hat{\sigma} . \mathrm{BF}{3}$ has four symmetry planes. The plane of the molecule is a symmetry plane, since nuclei lying in a reflection plane do not move when a reflection is carried out. The plane passing through the B and $\mathrm{F}{1}$ nuclei and perpendicular to the plane of the molecule is a symmetry plane, since reflection in this plane merely interchanges $\mathrm{F}{2}$ and $\mathrm{F}{3}$. It might be thought that this reflection is the same symmetry operation as rotation by $180^{\circ}$ about the $C{2}$ axis passing through B and $\mathrm{F}{1}$, which also interchanges $\mathrm{F}{2}$ and $\mathrm{F}{3}$. This is not so. The reflection carries points lying above the plane of the molecule into points that also lie above the molecular plane, whereas the $\hat{C}{2}$ rotation carries points lying above the molecular plane into points below the molecular plane. Two symmetry operations are equal only when they represent the same transformation of three-dimensional space. The remaining two symmetry planes in $\mathrm{BF}{3}$ pass through $\mathrm{B}-\mathrm{F}{2}$ and $\mathrm{B}-\mathrm{F}{3}$ and are perpendicular to the molecular plane.

The third kind of symmetry element is a center of symmetry, symbolized by $i$ (no connection with $\sqrt{-1}$ ). A molecule has a center of symmetry if the operation of inverting all the nuclei through the center gives a configuration indistinguishable from the original one. If we set up a Cartesian coordinate system, the operation of inversion through the origin (symbolized by $\hat{i}$ ) carries a nucleus originally at $(x, y, z)$ to $(-x,-y,-z)$. Does $\mathrm{BF}_{3}$ have a center of symmetry? With the origin at the boron nucleus, inversion gives the result shown in Fig. 12.3. Since we get a configuration that is physically distinguishable from

FIGURE 12.4 Effect of inversion in $\mathrm{BF}_{6}$.

the original one, $\mathrm{BF}{3}$ does not have a center of symmetry. For $\mathrm{SF}{6}$, inversion through the sulfur nucleus is shown in Fig. 12.4, and it is clear that $\mathrm{SF}{6}$ has a center of symmetry. (An operation such as $\hat{i}$ or $\hat{C}{n}$ may or may not be a symmetry operation. Thus, $\hat{i}$ is a symmetry operation in $\mathrm{SF}{6}$ but not in $\mathrm{BF}{3}$.)

The fourth and final kind of symmetry element is an $\boldsymbol{n}$-fold rotation-reflection axis of symmetry (also called an improper axis or an alternating axis), symbolized by $S{n}$. A body has an $S{n}$ axis if rotation by $(360 / n)^{\circ}(n$ integral) about the axis, followed by reflection in a plane perpendicular to the axis, carries the body into a position physically indistinguishable from the original one. Clearly, if a body has a $C{n}$ axis and also has a plane of symmetry perpendicular to this axis, then the $C{n}$ axis is also an $S{n}$ axis. Thus the $C{3}$ axis in $\mathrm{BF}{3}$ is also an $S{3}$ axis. It is possible to have an $S{n}$ axis that is not a $C{n}$ axis. An example is $\mathrm{CH}{4}$. In Fig. 12.5 we have first carried out a $90^{\circ}$ proper rotation $\left(\hat{C}{4}\right)$ about what we assert is an $S{4}$ axis. As can be seen, this operation does not result in an equivalent configuration. When we follow the $\hat{C}{4}$ operation by reflection in the plane perpendicular to the axis and passing through the carbon atom, we do get a configuration indistinguishable from the one existing before we performed the rotation and reflection. Hence $\mathrm{CH}{4}$ has an $S{4}$ axis. The $S{4}$ axis is not a $C{4}$ axis, although it is a $C{2}$ axis. There are two other $S{4}$ axes in methane, each perpendicular to a pair of faces of the cube in which the tetrahedral molecule is inscribed.

The operation of counterclockwise rotation by $(360 / n)^{\circ}$ about an axis, followed by reflection in a plane perpendicular to the axis, is denoted by $\hat{S}{n}$. An $\hat{S}{1}$ operation is a $360^{\circ}$ rotation about an axis, followed by a reflection in a plane perpendicular to the axis. Since a $360^{\circ}$ rotation restores the body to its original position, an $\hat{S}{1}$ operation is the same as reflection in a plane; $\hat{S}{1}=\hat{\sigma}$. A plane of symmetry has an $S_{1}$ axis perpendicular to it.

Consider now the $\hat{S}{2}$ operation. Let the $S{2}$ axis be the $z$ axis (Fig. 12.6). Rotation by $180^{\circ}$ about the $S{2}$ axis changes the $x$ and $y$ coordinates of a point to $-x$ and $-y$, respectively, and leaves the $z$ coordinate unaffected. Reflection in the $x y$ plane then converts the $z$ coordinate to $-z$. The net effect of the $\hat{S}{2}$ operation is to bring a point originally at $(x, y, z)$ to $(-x,-y,-z)$, which amounts to an inversion through the origin: $\hat{S}{2}=\hat{i}$. Any axis passing through a center of symmetry is an $S{2}$ axis. Reflection in a plane and inversion are special cases of the $\hat{S}_{n}$ operation.

FIGURE 12.5 An $S{4}$ axis in $\mathrm{CH}{4}$.

FIGURE 12.6 The $\hat{S}_{2}$ operation.

The $\hat{S}{n}$ operation may seem an arbitrary kind of operation, but it must be included as one of the kinds of symmetry operations. For example, the transformation from the first to the third $\mathrm{CH}{4}$ configuration in Fig. 12.5 certainly meets the definition of a symmetry operation, but it is neither a proper rotation nor a reflection nor an inversion.

Performing a symmetry operation on a molecule gives a nuclear configuration that is physically indistinguishable from the original one. Hence the center of mass must have the same position in space before and after a symmetry operation. For the operation $\hat{C}{n}$, the only points that do not move are those on the $C{n}$ axis. Therefore, a $C{n}$ symmetry axis must pass through the molecular center of mass. Similarly, a center of symmetry must coincide with the center of mass; a plane of symmetry and an $S{n}$ axis of symmetry must pass through the center of mass. The center of mass is the common intersection of all the molecular symmetry elements.

In discussing the symmetry of a molecule, we often place it in a Cartesian coordinate system with the molecular center of mass at the origin. The rotational axis of highest order is made the $z$ axis. A plane of symmetry containing this axis is designated $\sigma{v}$ (for vertical); a plane of symmetry perpendicular to this axis is designated $\sigma{h}$ (for horizontal).

Products of Symmetry Operations

Symmetry operations are operators that transform three-dimensional space, and (as with any operators) we define the product of two such operators as meaning successive application of the operators, the operator on the right of the product being applied first. Clearly, the product of any two symmetry operations of a molecule must be a symmetry operation.

As an example, consider $\mathrm{BF}{3}$. The product of the $\hat{C}{3}$ operator with itself, $\hat{C}{3} \hat{C}{3}=\hat{C}{3}^{2}$, rotates the molecule $240^{\circ}$ counterclockwise. If we take $\hat{C}{3} \hat{C}{3} \hat{C}{3}=\hat{C}{3}^{3}$, we have a $360^{\circ}$ rotation, which restores the molecule to its original position. We define the identity operation $\hat{E}$ as the operation that does nothing to a body. We have $\hat{C}{3}^{3}=\hat{E}$. (The symbol comes from the German word Einheit, meaning unity.)

Now consider a molecule with a sixfold axis of symmetry, for example, $\mathrm{C}{6} \mathrm{H}{6}$. The operation $\hat{C}{6}$ is a $60^{\circ}$ rotation, and $\hat{C}{6}^{2}$ is a $120^{\circ}$ rotation; hence $\hat{C}{6}^{2}=\hat{C}{3}$. Also $\hat{C}{6}^{3}=\hat{C}{2}$. Therefore, a $C{6}$ symmetry axis is also a $C{3}$ and a $C_{2}$ axis.

Since two successive reflections in the same plane bring all nuclei back to their original positions, we have $\hat{\sigma}^{2}=\hat{E}$. Also, $\hat{i}^{2}=\hat{E}$. More generally, $\hat{\sigma}^{n}=\hat{E}, \hat{i}^{n}=\hat{E}$ for even $n$, while $\hat{\sigma}^{n}=\hat{\sigma}, \hat{i}^{n}=\hat{i}$ for odd $n$.

Do symmetry operators always commute? Consider $\mathrm{SF}{6}$. We examine the products of a $\hat{C}{4}$ rotation about the $z$ axis and a $\hat{C}{2}$ rotation about the $x$ axis. Figure 12.7 shows that $\hat{C}{4}(z) \hat{C}{2}(x) \neq \hat{C}{2}(x) \hat{C}{4}(z)$. Thus symmetry operations do not always commute. Note that we describe symmetry operations with respect to a fixed coordinate system that does not move with the molecule when we perform a symmetry operation. Thus the $C{2}(x)$ axis does not move when we perform the $\hat{C}_{4}(z)$ operation.

FIGURE 12.7 Products of two symmetry operations in $\mathrm{SF}{6}$.
Top: $\hat{C}{2}(x) \hat{C}{4}(z)$.
Bottom: $\hat{C}{4}(z) \hat{C}_{2}(x)$.

Symmetry and Dipole Moments

As an application of symmetry, we consider molecular dipole moments. Since a symmetry operation produces a configuration that is physically indistinguishable from the original one, the direction of the dipole-moment vector must remain unchanged after a symmetry operation. (This is a nonrigorous, unsophisticated argument.) Hence, if we have a $C_{n}$ axis

FIGURE 12.8 The symmetry elements of $\mathrm{H}{2} \mathrm{O}$. of symmetry, the dipole moment must lie along this axis. If we have two or more noncoincident symmetry axes, the molecule cannot have a dipole moment, since the dipole moment cannot lie on two different axes. $\mathrm{CH}{4}$, which has four noncoincident $C{3}$ axes, has no dipole moment. If there is a plane of symmetry, the dipole moment must lie in this plane. If there are several symmetry planes, the dipole moment must lie along the line of intersection of these planes. In $\mathrm{H}{2} \mathrm{O}$ the dipole moment lies on the $C_{2}$ axis, which is also the intersection of the two symmetry planes (Fig. 12.8). A molecule with a center of symmetry cannot have a dipole moment, since inversion reverses the direction of a vector. A monatomic molecule has a center of symmetry. Hence atoms do not have dipole moments. (There is one exception to this statement; see Prob. 14.4.) Thus we can use symmetry to discover whether a molecule has a dipole moment. In many cases symmetry also tells us along what line the dipole moment lies.

Symmetry and Optical Activity

Certain molecules rotate the plane of polarization of plane-polarized light that is passed through them. Experimental evidence and a quantum-mechanical treatment (Kauzmann, pp. 703-713) show that the optical rotary powers of two molecules that are mirror images of each other are equal in magnitude but opposite in sign. Hence, if a molecule is its own mirror image, it is optically inactive: $\alpha=-\alpha, 2 \alpha=0, \alpha=0$, where $\alpha$ is the optical rotary power. If a molecule is not superimposable on its mirror image, it may be optically active. If the conformation of the mirror image differs from that of the original molecule only by rotation about a bond with a low rotational barrier, then the molecule will not be optically active.

What is the connection between symmetry and optical activity? Consider the $\hat{S}{n}$ operation. It consists of a rotation $\left(\hat{C}{n}\right)$ and a reflection $(\hat{\sigma})$. The reflection part of the
$\hat{S}{n}$ operation converts the molecule to its mirror image, and if the $\hat{S}{n}$ operation is a symmetry operation for the molecule, then the $\hat{C}_{n}$ rotation will superimpose the molecule and its mirror image:

\(
\text { molecule } \xrightarrow{\hat{C}_{n}} \text { rotated molecule } \xrightarrow{\hat{\sigma}} \text { rotated mirror image }
\)

We conclude that a molecule with an $S{n}$ axis is optically inactive. If the molecule has no $S{n}$ axis, it may be optically active.

Since $\hat{S}{1}=\hat{\sigma}$ and $\hat{S}{2}=\hat{i}$, a molecule with either a plane or a center of symmetry is optically inactive. However, an $S_{n}$ axis of any order rules out optical activity.

A molecule can have a symmetry element and still be optically active. If a $C{n}$ axis is present and there is no $S{n}$ axis, the molecule can be optically active.

Symmetry Operations and Quantum Mechanics

What is the relation between the symmetry operations of a molecule and quantum mechanics? To classify the states of a quantum-mechanical system, we consider those operators that commute with the Hamiltonian operator and with each other. For example, we classified the states of many-electron atoms using the quantum numbers $L, S, J$, and $M{J}$, which correspond to the operators $\hat{L}^{2}, \hat{S}^{2}, \hat{J}^{2}$, and $\hat{J}{z}$, all of which commute with one other and with the Hamiltonian (omitting spin-orbit interaction). The symmetry operations discussed in this chapter act on points in three-dimensional space, transforming each point to a corresponding point. All the quantum-mechanical operators we have discussed act on functions, transforming each function to a corresponding function. Corresponding to each symmetry operation $\hat{R}$, we define an operator $\hat{O}_{R}$ that acts on functions in the following manner. Let $\hat{R}$ bring a point originally at $(x, y, z)$ to the location ( $x^{\prime}, y^{\prime}, z^{\prime}$ ):

\(
\begin{equation}
\hat{R}(x, y, z) \rightarrow\left(x^{\prime}, y^{\prime}, z^{\prime}\right) \tag{12.1}
\end{equation}
\)

The operator $\hat{O}{R}$ is defined so that the function $\hat{O}{R} f$ has the same value at $\left(x^{\prime}, y^{\prime}, z^{\prime}\right)$ that the function $f$ has at $(x, y, z)$ :

\(
\begin{equation}
\hat{O}_{R} f\left(x^{\prime}, y^{\prime}, z^{\prime}\right)=f(x, y, z) \tag{12.2}
\end{equation}
\)

For example, let $\hat{R}$ be a counterclockwise $90^{\circ}$ rotation about the $z$ axis: $\hat{R}=\hat{C}{4}(z)$; and let $f$ be a $2 p{x}$ hydrogen orbital: $f=2 p{x}=N x e^{-k\left(x^{2}+y^{2}+z^{2}\right)^{1 / 2}}$. The shape of the $2 p{x}$ orbital is two distorted ellipsoids of revolution about the $x$ axis (Section 6.7). Let these ellipsoids be "centered" about the points $(a, 0,0)$ and $(-a, 0,0)$, where $a>0$ and $2 p{x}>0$ on the right ellipsoid. The operator $\hat{C}{4}(z)$ has the following effect (Fig. 12.9):

\(
\begin{equation}
\hat{C}_{4}(z)(x, y, z) \rightarrow(-y, x, z) \tag{12.3}
\end{equation}
\)

For example, the point originally at $(a, 0,0)$ is moved to $(0, a, 0)$, while the point at $(-a, 0,0)$ is moved to $(0,-a, 0)$. From (12.2), the function $\hat{O}{C{4}(z)} 2 p_{x}$ must have

FIGURE 12.9 The effect of a $\hat{C}_{4}(z)$ rotation is to move the point at $(x, y)$ to $\left(x^{\prime}, y^{\prime}\right)$. Use of trigonometry shows that $x^{\prime}=-y$ and $y^{\prime}=x$.

FIGURE 12.10 Effect of $\hat{O}{C{4}(z)}$ on a $p_{x}$ orbital.

its contours centered about $(0, a, 0)$ and $(0,-a, 0)$, respectively. We conclude that (Fig. 12.10)

\(
\begin{equation}
\hat{O}{C{4}(z)} 2 p{x}=2 p{y} \tag{12.4}
\end{equation}
\)

For the inversion operation, we have

\(
\begin{equation}
\hat{i}(x, y, z) \rightarrow(-x,-y,-z) \tag{12.5}
\end{equation}
\)

and (12.2) reads

\(
\hat{O}_{i} f(-x,-y,-z)=f(x, y, z)
\)

We now rename the variables as follows: $\bar{x}=-x, \bar{y}=-y, \bar{z}=-z$. Hence

\(
\hat{O}_{i} f(\bar{x}, \bar{y}, \bar{z})=f(-\bar{x},-\bar{y},-\bar{z})
\)

The point $(\bar{x}, \bar{y}, \bar{z})$ is a general point in space, and we can drop the bars to get

\(
\hat{O}_{i} f(x, y, z)=f(-x,-y,-z)
\)

We conclude that $\hat{O}{i}$ is the parity operator (Section 7.5): $\hat{O}{i}=\hat{\Pi}$.
The wave function of an $n$-particle system is a function of $4 n$ variables, and we extend the definition (12.2) of $\hat{O}_{R}$ to read

\(
\hat{O}{R} f\left(x{1}^{\prime}, y{1}^{\prime}, z{1}^{\prime}, m{s 1}, \ldots, x{n}^{\prime}, y{n}^{\prime}, z{n}^{\prime}, m{s n}\right)=f\left(x{1}, y{1}, z{1}, m{s 1}, \ldots, x{n}, y{n}, z{n}, m_{s n}\right)
\)

Note that $\hat{O}_{R}$ does not affect the spin coordinates. Thus, in looking at the parity of atomic states in Section 11.5, we looked at the spatial factors in each term of the expansion of the Slater determinant and omitted consideration of the spin factors, since these are unaffected by $\hat{\Pi}$.

When a system is characterized by the symmetry operations $\hat{R}{1}, \hat{R}{2}, \ldots$, then the corresponding operators $\hat{O}{R{1}}, \hat{O}{R{2}}, \ldots$ commute with the Hamiltonian. (For a proof, see Schonland, Sections 7.1-7.3.) For example, if the nuclear framework of a molecule has a center of symmetry, then the parity operator $\hat{\Pi}$ commutes with the Hamiltonian for the electronic motion. We can then choose the electronic states (wave functions) as even or odd, according to the eigenvalue of $\hat{\Pi}$. Of course, not all the symmetry operations may commute among themselves (Fig. 12.7). Hence the wave functions cannot in general be chosen as eigenfunctions of all the symmetry operators $\hat{O}_{R}$. (Further discussion on the relation between symmetry operators and molecular wave functions is given in Section 15.2.)

There is a close connection between symmetry and the constants of the motion (these are properties whose operators commute with the Hamiltonian $\hat{H}$ ). For a system whose Hamiltonian is invariant (that is, doesn't change) under any translation of spatial coordinates, the linear-momentum operator $\hat{p}$ will commute with $\hat{H}$ and $p$ can be assigned a definite value in a stationary state. An example is the free particle. For a system with $\hat{H}$ invariant under any rotation of coordinates, the operators for the angular-momentum components commute with $\hat{H}$, and the total angular momentum and one of its components are specifiable. An example is an atom. A linear molecule has axial symmetry, rather than the spherical symmetry of an atom; here only the axial component of angular momentum can be specified (Chapter 13).

Matrices and Symmetry Operations

The symmetry operation $\hat{R}$ moves the point originally at $x, y, z$ to the new location $x^{\prime}, y^{\prime}, z^{\prime}$, where each of $x^{\prime}, y^{\prime}, z^{\prime}$ is a linear combination of $x, y, z$ (for proof of this see Schonland, pp. 52-53):

\(
\begin{aligned}
& x^{\prime}=r{11} x+r{12} y+r{13} z \
& y^{\prime}=r{21} x+r{22} y+r{23} z \
& z^{\prime}=r{31} x+r{32} y+r{33} z
\end{aligned} \quad \text { or } \quad\left(\begin{array}{l}
x^{\prime} \
y^{\prime} \
z^{\prime}
\end{array}\right)=\left(\begin{array}{lll}
r{11} & r{12} & r{13} \
r{21} & r{22} & r{23} \
r{31} & r{32} & r{33}
\end{array}\right)\left(\begin{array}{l}
x \
y \
z
\end{array}\right)
\)

where $r{11}, r{12}, \ldots, r{33}$ are constants whose values depend on the nature of $\hat{R}$. One says that the symmetry operation $\hat{R}$ is represented by the matrix $\mathbf{R}$ whose elements are $r{11}, r{12}, \ldots, r{33}$. The set of functions $x, y, z$, whose transformations are described by $\mathbf{R}$, is said to be the basis for this representation.

For example, from (12.3) and (12.5), for the $\hat{C}{4}(z)$ operation, we have $x^{\prime}=-y, y^{\prime}=x, z^{\prime}=z$; for $\hat{i}$, we have $x^{\prime}=-x, y^{\prime}=-y, z^{\prime}=-z$. The matrices representing $\hat{C}{4}(z)$ and $\hat{i}$ in the $x, y, z$ basis are

\(
\mathbf{C}_{4}(z)=\left(\begin{array}{rrr}
0 & -1 & 0 \
1 & 0 & 0 \
0 & 0 & 1
\end{array}\right), \quad \mathbf{i}=\left(\begin{array}{rrr}
-1 & 0 & 0 \
0 & -1 & 0 \
0 & 0 & -1
\end{array}\right)
\)

If the product $\hat{R} \hat{S}$ of two symmetry operations is $\hat{T}$, then the matrices representing these operations in the $x, y, z$ basis multiply in the same way; that is, if $\hat{R} \hat{S}=\hat{T}$, then RS $=$ T. (For proof, see Schonland, pp. 56-57.)