---
title: "Symmetry and Point Groups"
book: "University Chemistry — Year 3"
subject: chemistry
language: en
chapter: 4
exercises: 12
source: https://one-course.com/books/chemistry/4/en/chapter/4-symmetry-and-point-groups
license: CC-BY-NC-SA-4.0
credit: "One Chemistry Book, One Course (one-course.com)"
---

# Chapter 4 — Symmetry and Point Groups

Turn a snowflake by sixty degrees and nothing changes; turn a benzene molecule by the same angle, and its six carbons and six hydrogens fall exactly on each other’s places. The symmetry of a molecule decides several of its properties before any calculation: whether it can have a dipole moment, whether it can exist as two mirror-image forms, which of its vibrations absorb infrared light, which orbitals may mix. To use it, chemists borrowed from mathematics the language of groups. This chapter defines the [symmetry operations](#def-b3-point-groups-operation) of a molecule, collects them into its [point group](#def-b3-point-groups-point-group), gives a procedure to find that group, proves two theorems on polarity and chirality, and introduces the [character tables](#def-b3-point-groups-irrep) on which [Chapter 5](https://one-course.com/books/chemistry/4/en/chapter/5-group-theory-applied#ch-b3-group-theory-applied) builds.

**You already know.**

The Year 1 volume predicted molecular shapes with the VSEPR model, defined the dipole moment of a polar molecule, and defined chirality: a chiral molecule cannot be superimposed on its mirror image. The Year 2 volume used symmetry in words to build the fragment orbitals of $\ce{H2O}$ and $\ce{NH3}$.

![Left: snow crystals photographed by Wilson Bentley around 1902, each with sixfold symmetry. Right: everyday objects with threefold, sixfold and fivefold rotational symmetry.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-point-groups/img-b8cc6167f97d.jpg)

![Left: snow crystals photographed by Wilson Bentley around 1902, each with sixfold symmetry. Right: everyday objects with threefold, sixfold and fivefold rotational symmetry.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-point-groups/img-20996a22b599.jpg)

*Left: snow crystals photographed by Wilson Bentley around 1902, each with sixfold symmetry. Right: everyday objects with threefold, sixfold and fivefold rotational symmetry.*

## 4.1 Symmetry operations and elements

**Definition 4.1 (Symmetry operation, symmetry element).**

A *symmetry operation* is a motion or reflection that leaves a molecule in a configuration indistinguishable from the original (each atom on a position occupied before by an atom of the same kind). A *symmetry element* is the geometric object — a point, a line, a plane — with respect to which the operation is performed.

Every molecule has the identity operation $E$, which does nothing. The others come in four kinds.

**Definition 4.2 (Proper rotation axis, principal axis).**

A *proper rotation axis* $C_n$ is a line about which a rotation by $2\pi/n$ is a [symmetry operation](#def-b3-point-groups-operation), also written $C_n$; its powers $C_n^k$ are rotations by $2\pi k/n$, and $C_n^n = E$. The axis of highest $n$ is the *principal axis*, taken as $z$.

**Definition 4.3 (Mirror planes).**

A *mirror plane* $\sigma$ is a plane through which reflection is a [symmetry operation](#def-b3-point-groups-operation); $\sigma^2 = E$. A mirror plane is a *vertical mirror plane* $\sigma_v$ if it contains the [principal axis](#def-b3-point-groups-rotation), a *horizontal mirror plane* $\sigma_h$ if it is perpendicular to it, and a *dihedral mirror plane* $\sigma_d$ if it contains the [principal axis](#def-b3-point-groups-rotation) and bisects the angle between two $C_2$ axes perpendicular to it.

**Definition 4.4 (Centre of inversion).**

A *centre of inversion* $i$ is a point through which inversion, $(x, y, z) \mapsto (-x, -y, -z)$, is a [symmetry operation](#def-b3-point-groups-operation).

**Definition 4.5 (Improper rotation axis).**

An *improper rotation axis* $S_n$ is a line about which a rotation by $2\pi/n$ followed by reflection in the plane perpendicular to it is a [symmetry operation](#def-b3-point-groups-operation). $S_1 = \sigma$ and $S_2 = i$.

**Example 4.6 (Elements of four molecules).**

$\ce{H2O}$: $E$, a $C_2$ axis bisecting the H–O–H angle, and two vertical planes, the molecular plane and the plane perpendicular to it. $\ce{NH3}$: $E$, a $C_3$ through N (operations $C_3$ and $C_3^2$), and three $\sigma_v$, each containing one N–H bond. $\ce{BF3}$: a $C_3$, three $C_2$ along the B–F bonds, $\sigma_h$ (the molecular plane), three $\sigma_v$, and an $S_3$. $\ce{CH4}$: four $C_3$ along the C–H bonds, three $C_2$ bisecting H–C–H angles, which are also $S_4$ axes, and six $\sigma_d$, each containing two C–H bonds (figure below).

![Symmetry elements. H2O: the C_2 axis (dashed) and two vertical mirror planes. NH3 and BF3 seen down their principal axis (triangle glyph): thick lines are mirror planes seen edge-on; in BF3 the thick circle marks the horizontal mirror plane (the plane of the page) and the lens glyphs the three C_2 axes lying in it.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-point-groups/fig-2eaff266f853.svg)

*[Symmetry elements](#def-b3-point-groups-operation). $\ce{H2O}$: the $C_2$ axis (dashed) and two [vertical mirror planes](#def-b3-point-groups-mirror). $\ce{NH3}$ and $\ce{BF3}$ seen down their [principal axis](#def-b3-point-groups-rotation) (triangle glyph): thick lines are [mirror planes](#def-b3-point-groups-mirror) seen edge-on; in $\ce{BF3}$ the thick circle marks the [horizontal mirror plane](#def-b3-point-groups-mirror) (the plane of the page) and the lens glyphs the three $C_2$ axes lying in it.*

## 4.2 Groups

**Proposition 4.7 (Products of symmetry operations).**

Performing two [symmetry operations](#def-b3-point-groups-operation) of a molecule in succession is again a [symmetry operation](#def-b3-point-groups-operation) of that molecule. Together with $E$ and the inverse of each operation, the set of [symmetry operations](#def-b3-point-groups-operation) is a group: an associative product, an identity, an inverse for each element.

**Proof.** Each operation leaves the molecule indistinguishable from itself, so two in succession do too. The product of operations (composition of maps of space) is associative; $E$ is the identity; an operation undone (a rotation by the opposite angle, a reflection repeated) is also a [symmetry operation](#def-b3-point-groups-operation). ∎

**Proposition 4.8 (A common point).**

All the [symmetry elements](#def-b3-point-groups-operation) of a finite molecule pass through one point.

**Proof.** Every [symmetry operation](#def-b3-point-groups-operation) maps each atom onto an atom of the same mass, so it leaves the centre of mass unchanged. A rotation fixes only the points of its axis, a reflection those of its plane, an improper rotation or an inversion a single point: every element contains the centre of mass. ∎

**Definition 4.9 (Point group, order of a point group).**

The group of all [symmetry operations](#def-b3-point-groups-operation) of a molecule is its *point group*, named because one point stays fixed. The number of its operations is the *order of a point group*, $h$.

**Example 4.10 (The group of water).**

The four operations $E$, $C_2$, $\sigma_v(xz)$, $\sigma_v'(yz)$ of $\ce{H2O}$ form a group of order 4. Its multiplication table (row operation performed after the column one) is

|  | $E$ | $C_2$ | $\sigma_v(xz)$ | $\sigma_v'(yz)$ |
| --- | --- | --- | --- | --- |
| $E$ | $E$ | $C_2$ | $\sigma_v(xz)$ | $\sigma_v'(yz)$ |
| $C_2$ | $C_2$ | $E$ | $\sigma_v'(yz)$ | $\sigma_v(xz)$ |
| $\sigma_v(xz)$ | $\sigma_v(xz)$ | $\sigma_v'(yz)$ | $E$ | $C_2$ |
| $\sigma_v'(yz)$ | $\sigma_v'(yz)$ | $\sigma_v(xz)$ | $C_2$ | $E$ |

For instance a point $(x, y, z)$ reflected in $xz$ goes to $(x, -y, z)$, then rotated by $\pi$ about $z$ to $(-x, y, z)$: the same as one reflection in $yz$.

**Definition 4.11 (Class of symmetry operations, subgroup).**

Two operations $A$ and $B$ belong to the same *class of symmetry operations* if $B = X^{-1}AX$ for some operation $X$ of the group: they are the same kind of operation, seen in a frame moved by $X$. A subset of a group that is a group on its own is a *subgroup*; its order divides the order of the group.

**Example 4.12 (Classes of NHX3\ce{NH3}NHX3​).**

In $C_{3v}$ ($h = 6$), the three reflections form one class (a $C_3$ rotation carries each plane onto another), the two rotations $C_3$ and $C_3^2$ another, and $E$ is a class alone: three classes, written $E$, $2C_3$, $3\sigma_v$. In $C_{2v}$ every operation is its own class: no operation turns one plane into the other.

**Definition 4.13 (Schoenflies symbol).**

A [point group](#def-b3-point-groups-point-group) is named by its *Schoenflies symbol*: $C_n$ (one $C_n$ axis), $C_{nv}$ (adding $n$ vertical planes), $C_{nh}$ (adding $\sigma_h$), $D_n$ (adding $n$ $C_2$ axes perpendicular to $C_n$), $D_{nh}$ and $D_{nd}$ (adding $\sigma_h$, or $n$ dihedral planes, to $D_n$), $S_{2n}$, the cubic groups $T_d$, $O_h$, $I_h$ of the tetrahedron, octahedron and icosahedron, and $C_s$, $C_i$, $C_1$ for molecules with only a [mirror plane](#def-b3-point-groups-mirror), only an inversion centre, or nothing at all. Linear molecules belong to $C_{\infty v}$ or $D_{\infty h}$.

## 4.3 Assigning a point group

**Method 4.14 (Finding the point group of a molecule).**

1. Is the molecule linear? With an inversion centre, $D_{\infty h}$ ( $\ce{CO2}$ , $\ce{N2}$ ); without, $C_{\infty v}$ ( $\ce{HCl}$ , $\ce{HCN}$ ).
2. Does it have several high-order axes ( $n \ge 3$ )? Tetrahedral shape: $T_d$ ; octahedral: $O_h$ ; icosahedral: $I_h$ .
3. Find the [principal axis](#def-b3-point-groups-rotation) $C_n$ (none: $C_s$ if a plane, $C_i$ if a centre, $C_1$ otherwise).
4. Are there $n$ $C_2$ axes perpendicular to $C_n$ ? If yes, the group is $D_{nh}$ (with $\sigma_h$ ), else $D_{nd}$ (with $n$ $\sigma_d$ ), else $D_n$ .
5. If no: $C_{nh}$ (with $\sigma_h$ ), $C_{nv}$ (with $n$ $\sigma_v$ ), $S_{2n}$ (with only an $S_{2n}$ collinear with $C_n$ ), else $C_n$ .

![The search for a point group, in the order of the method.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-point-groups/fig-9f67b7fefbfa.svg)

*The search for a [point group](#def-b3-point-groups-point-group), in the order of the method.*

**Example 4.15 (A gallery).**

$\ce{H2O}$, $\ce{CH2Cl2}$, $\ce{SO2}$: $C_{2v}$. $\ce{NH3}$, $\ce{CHCl3}$: $C_{3v}$. $\ce{BF3}$, $\ce{PCl5}$: $D_{3h}$. $\ce{CH4}$, $\ce{CCl4}$: $T_d$. $\ce{SF6}$: $O_h$. Benzene: $D_{6h}$. Ethene: $D_{2h}$. Ethane: $D_{3d}$ staggered, $D_{3h}$ eclipsed. Allene $\ce{H2C=C=CH2}$: $D_{2d}$, the two $\ce{CH2}$ planes at right angles. Cyclohexane in its chair: $D_{3d}$. trans-$\ce{N2F2}$: $C_{2h}$. $\ce{CHFClBr}$: $C_1$.

![Methane in a cube, its carbon at the centre and its hydrogens on alternate corners. A body diagonal through one hydrogen is a C_3 axis (four of them); the axis through two opposite face centres is a C_2 and an S_4 axis (three of them). The point group is T_d, of order 24.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-point-groups/fig-4685aa8ecf7a.svg)

*Methane in a cube, its carbon at the centre and its hydrogens on alternate corners. A body diagonal through one hydrogen is a $C_3$ axis (four of them); the axis through two opposite face centres is a $C_2$ and an $S_4$ axis (three of them). The [point group](#def-b3-point-groups-point-group) is $T_d$, of order 24.*

## 4.4 Consequences: polarity and chirality

**Theorem 4.16 (Polar molecules).**

A molecule can have a permanent dipole moment only if its [point group](#def-b3-point-groups-point-group) is $C_1$, $C_s$, $C_n$ or $C_{nv}$; the dipole then lies along the [principal axis](#def-b3-point-groups-rotation) (in the plane, for $C_s$).

**Proof.** The dipole moment is a vector property of the molecule, so every [symmetry operation](#def-b3-point-groups-operation) must leave it unchanged. A rotation about an axis leaves only vectors along that axis unchanged; two axes leave no non-zero vector; a $\sigma_h$, an $i$ or an $S_n$ reverses any vector along the axis; a $\sigma_v$ keeps vectors in its plane. Hence a non-zero dipole requires at most one axis, no $\sigma_h$, no $i$, no $S_n$: the groups listed. ∎

**Example 4.17 (Polar or not).**

$\ce{H2O}$, $\ce{NH3}$, $\ce{CHCl3}$ and $\ce{SO2}$ ($C_{2v}$, $C_{3v}$) can be polar, and are. $\ce{BF3}$ ($D_{3h}$), $\ce{CO2}$ ($D_{\infty h}$), $\ce{CH4}$ ($T_d$), $\ce{SF6}$ ($O_h$) and trans-$\ce{N2F2}$ ($C_{2h}$) cannot, whatever their polar bonds.

**Theorem 4.18 (Chiral molecules).**

A molecule is chiral if and only if it has no [improper rotation axis](#def-b3-point-groups-improper) $S_n$ (including $S_1 = \sigma$ and $S_2 = i$).

**Proof.** If the molecule has an $S_n$, then $S_n = \sigma_hC_n$: the mirror image $\sigma_h$ of the molecule equals $C_n^{-1}$ applied to the molecule after $S_n$, that is the molecule rotated — superimposable, hence achiral. Conversely, if the molecule is achiral, its mirror image $\sigma M$ can be brought onto $M$ by a proper motion $R$ (a rotation about the centre of mass): $R\sigma$ is then a [symmetry operation](#def-b3-point-groups-operation). It is improper (it changes handedness, its matrix has determinant $-1$), and every improper operation of a [point group](#def-b3-point-groups-point-group) is an $S_n$ for some $n$ (a rotation followed by a reflection in the perpendicular plane). So the molecule has an $S_n$. ∎

**Example 4.19 (An achiral molecule without a plane).**

Some molecules have neither a [mirror plane](#def-b3-point-groups-mirror) nor a [centre of inversion](#def-b3-point-groups-inversion), yet are achiral because of an $S_4$ axis: the classic case is a spiro compound built from two identically substituted rings at right angles, of [point group](#def-b3-point-groups-point-group) $S_4$. Chirality cannot be judged by looking for a plane alone.

## 4.5 Representations and character tables

Each [symmetry operation](#def-b3-point-groups-operation) moves a point $(x, y, z)$ linearly: it can be written as a $3\times3$ matrix.

**Definition 4.20 (Matrix representation, character).**

A *matrix representation* $\Gamma$ of a [point group](#def-b3-point-groups-point-group) assigns to each operation $R$ a square matrix $D(R)$ such that $D(R_1R_2) = D(R_1)D(R_2)$: products of operations become products of matrices. The *character* $\chi(R)$ of $R$ in $\Gamma$ is the trace of $D(R)$.

**Example 4.21 (The coordinates of C3vC_{3v}C3v​).**

For $C_3$ about $z$, $D = \left(\begin{smallmatrix}\cos120^\circ & -\sin120^\circ &
0\\ \sin120^\circ & \cos120^\circ & 0\\ 0 & 0 & 1\end{smallmatrix}\right)$, of trace $2\cos120^\circ + 1 = 0$; for $\sigma_v(xz)$, $D = \mathrm{diag}(1, -1, 1)$, trace 1; for $E$, trace 3. The [characters](#def-b3-point-groups-representation) of this representation are $(3, 0, 1)$ on the classes $(E, 2C_3, 3\sigma_v)$. Every matrix is block-diagonal, a $2\times2$ block for $(x, y)$ and a $1\times1$ block for $z$: the representation splits into two smaller ones.

**Proposition 4.22 (Characters are class functions).**

Operations of the same class have the same [character](#def-b3-point-groups-representation) in any representation.

**Proof.** If $B = X^{-1}AX$, then $D(B) = D(X)^{-1}D(A)D(X)$, and the trace is unchanged by such a change of basis: $\mathrm{tr}(P^{-1}MP) = \mathrm{tr}(M)$. ∎

**Proposition 4.23 (Reducing by blocks).**

If a basis can be split into subsets that every operation maps into themselves, all matrices are block-diagonal, and the representation is the sum of the smaller representations carried by the subsets; its [characters](#def-b3-point-groups-representation) are the sums of theirs.

**Proof.** An operation that maps each subset into itself has no matrix elements between different subsets; the trace of a block-diagonal matrix is the sum of the traces of its blocks. ∎

**Definition 4.24 (Irreducible representation, character table, Mulliken symbol).**

A representation that cannot be split further by any change of basis is an *irreducible representation*. The *character table* of a [point group](#def-b3-point-groups-point-group) lists the [characters](#def-b3-point-groups-representation) of its irreducible representations, one row each, on its classes, one column each, with the Cartesian functions and rotations that transform like each row. The rows are named by their *Mulliken symbol*: $A$ or $B$ for one dimension (symmetric or antisymmetric under the principal rotation), $E$ for two, $T$ for three; subscripts 1, 2 (symmetric or antisymmetric under a $C_2$ or $\sigma_v$), $g$, $u$ (under inversion), primes (under $\sigma_h$).

**Proposition 4.25 (Size of a character table).**

A [point group](#def-b3-point-groups-point-group) has as many [irreducible representations](#def-b3-point-groups-irrep) as classes, and the squares of their dimensions add up to the order: $\sum_id_i^2 = h$.

**Status.** Proved in [Chapter 5](https://one-course.com/books/chemistry/4/en/chapter/5-group-theory-applied#ch-b3-group-theory-applied) from the orthogonality of [characters](#def-b3-point-groups-representation), itself admitted there. ∎

The tables needed most often in this book are those of water, ammonia and methane:

$$
\begin{array}{l|cccc|l|l}\hline C_{2v} & E & C_2 & \sigma_v(xz) & \sigma_v'(yz) & & \\ \hline 
A_1 & 1 & 1 & 1 & 1 & z & x^2,\ y^2,\ z^2 \\
A_2 & 1 & 1 & -1 & -1 & R_z & xy \\
B_1 & 1 & -1 & 1 & -1 & x,\ R_y & xz \\
B_2 & 1 & -1 & -1 & 1 & y,\ R_x & yz \\
 \hline\end{array}
$$

$$
\begin{array}{l|ccc|l|l}\hline C_{3v} & E & 2C_3 & 3\sigma_v & & \\ \hline 
A_1 & 1 & 1 & 1 & z & x^2 + y^2,\ z^2 \\
A_2 & 1 & 1 & -1 & R_z & \\
E & 2 & -1 & 0 & (x, y),\ (R_x, R_y) & (x^2 - y^2, xy),\ (xz, yz) \\
 \hline\end{array}
$$

$$
\begin{array}{l|ccccc|l|l}\hline T_d & E & 8C_3 & 3C_2 & 6S_4 & 6\sigma_d & & \\ \hline 
A_1 & 1 & 1 & 1 & 1 & 1 & & x^2 + y^2 + z^2 \\
A_2 & 1 & 1 & 1 & -1 & -1 & & \\
E & 2 & -1 & 2 & 0 & 0 & & (2z^2 - x^2 - y^2,\ x^2 - y^2) \\
T_1 & 3 & 0 & -1 & 1 & -1 & (R_x, R_y, R_z) & \\
T_2 & 3 & 0 & -1 & -1 & 1 & (x, y, z) & (xy, xz, yz) \\
 \hline\end{array}
$$

**Method 4.26 (Reading a character table).**

1. The first column of [characters](#def-b3-point-groups-representation) (under $E$ ) is the dimension: the [degeneracy](https://one-course.com/books/chemistry/4/en/chapter/1-quantum-mechanics-for-chemists-model-systems#def-b3-quantum-model-systems-degenerate) of the levels or orbitals labelled by that row.
2. A [character](#def-b3-point-groups-representation) $+1$ means the function is unchanged by the operation, $-1$ that it changes sign; a 0 in a degenerate row means the members of the set are mixed.
3. To find how a function transforms, apply each operation to it and compare with the rows; the right-hand columns give the answer for $x$ , $y$ , $z$ , the rotations, and the quadratic functions (the shapes of $d$ orbitals).

**Example 4.27 (The orbitals of oxygen in water).**

In $C_{2v}$, the $2s$ and $2p_z$ orbitals of oxygen are unchanged by every operation: $a_1$ (lower-case for orbitals). $2p_x$ changes sign under $C_2$ and $\sigma_v'(yz)$: $b_1$. $2p_y$: $b_2$. These are the labels used for the water orbitals in the Year 2 volume; [Chapter 5](https://one-course.com/books/chemistry/4/en/chapter/5-group-theory-applied#ch-b3-group-theory-applied) derives the labels of the hydrogen combinations.

![Stereographic projections. A general point above the equatorial plane (dot) is carried by the operations of the group onto all the points shown; a cross marks a point below the plane. Thick lines and circle: mirror planes. C_2v: 4 points; C_3v: 6; D_3h: 12 (6 above, 6 below, superimposed): the number of points is the order of the group.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-point-groups/fig-ce71b68e649b.svg)

*Stereographic projections. A general point above the equatorial plane (dot) is carried by the operations of the group onto all the points shown; a cross marks a point below the plane. Thick lines and circle: [mirror planes](#def-b3-point-groups-mirror). $C_{2v}$: 4 points; $C_{3v}$: 6; $D_{3h}$: 12 (6 above, 6 below, superimposed): the number of points is the order of the group.*

**History — Schoenflies and the symmetry of crystals.**

Arthur Schoenflies, a mathematician, classified in 1891 the 230 space groups of crystals; the 32 crystallographic [point groups](#def-b3-point-groups-point-group) he named are still written with his symbols. Chemists adopted the notation in the 1930s, when group theory began to be applied to molecular spectra.

**In the lab — Symmetry with a model kit.**

The quickest way to find the elements of an unfamiliar molecule is to build it with a molecular model kit and turn it in the hand, looking for axes down which it looks the same after a fraction of a turn, and for planes that cut it into mirror halves. A computational program will also report the [point group](#def-b3-point-groups-point-group) of an optimised structure, within a tolerance.

## 4.6 Exercises

**Exercise 4.1 ★.**

List the [symmetry elements](#def-b3-point-groups-operation) and give the [point group](#def-b3-point-groups-point-group) of: ethene, $\ce{XeF4}$, $\ce{PCl5}$, 1,3,5-trichlorobenzene.

**Solution of Exercise 4.1.**

Ethene: three perpendicular $C_2$, three planes, $i$: $D_{2h}$. $\ce{XeF4}$ (square planar): $C_4$, four $C_2 \perp C_4$, $\sigma_h$, $2\sigma_v$, $2\sigma_d$, $i$, $S_4$: $D_{4h}$. $\ce{PCl5}$ (trigonal bipyramid): $C_3$, three $C_2$, $\sigma_h$, $3\sigma_v$, $S_3$: $D_{3h}$. 1,3,5-Trichlorobenzene: the same elements as $\ce{BF3}$: $D_{3h}$.

**Exercise 4.2 ★.**

Give the [point groups](#def-b3-point-groups-point-group) of staggered and eclipsed ethane and of the chair of cyclohexane.

**Solution of Exercise 4.2.**

Staggered ethane: $C_3$, three $C_2$ perpendicular, three $\sigma_d$, $i$, $S_6$: $D_{3d}$. Eclipsed: $C_3$, three $C_2$, $\sigma_h$, three $\sigma_v$: $D_{3h}$. Chair cyclohexane: $D_{3d}$.

**Exercise 4.3 ★.**

Which of these can be polar: $\ce{SO3}$, $\ce{SO2}$, $\ce{PCl3}$, $\ce{XeF4}$, $\ce{CH2Cl2}$, cis- and trans-1,2-dichloroethene?

**Solution of Exercise 4.3.**

$\ce{SO3}$ ($D_{3h}$): no. $\ce{SO2}$ ($C_{2v}$): yes. $\ce{PCl3}$ ($C_{3v}$): yes. $\ce{XeF4}$ ($D_{4h}$): no. $\ce{CH2Cl2}$ ($C_{2v}$): yes. cis-1,2-dichloroethene ($C_{2v}$): yes; trans ($C_{2h}$): no.

**Exercise 4.4 ★.**

Write the $3\times3$ matrices of $C_2(z)$, $i$ and $\sigma_h$ acting on $(x, y, z)$, and their [characters](#def-b3-point-groups-representation).

**Solution of Exercise 4.4.**

$C_2(z) = \mathrm{diag}(-1,-1,1)$, $\chi = -1$; $i = \mathrm{diag}(-1,-1,-1)$, $\chi = -3$; $\sigma_h(xy) = \mathrm{diag}(1,1,-1)$, $\chi = 1$.

**Exercise 4.5 ★★.**

Build the multiplication table of $C_{2h}$, with operations $E$, $C_2$, $i$, $\sigma_h$. Is every operation its own class?

**Solution of Exercise 4.5.**

With the diagonal matrices of the previous exercise: $C_2i = \sigma_h$, $C_2\sigma_h =
i$, $i\sigma_h = C_2$, each operation is its own inverse, and every product commutes. The group is commutative, so $X^{-1}AX = A$ for all $X$: each operation is a class alone (four classes, four [irreducible representations](#def-b3-point-groups-irrep) of dimension 1).

**Exercise 4.6 ★★.**

In $C_{3v}$ compute $\sigma_v(1)\,C_3$ and $C_3\,\sigma_v(1)$ by following a point. Are they equal? What does this say about the group?

**Solution of Exercise 4.6.**

Take $\sigma_v(1)$ the $xz$ plane. The point $(1,0)$ goes by $C_3$ to $(-\frac12,
\frac{\sqrt3}2)$, then by $\sigma_v(1)$ to $(-\frac12,-\frac{\sqrt3}2)$; in the other order it goes to $(1,0)$ then $(-\frac12,\frac{\sqrt3}2)$. The two products are reflections in two different planes ($\sigma_v(3)$ and $\sigma_v(2)$): the group is not commutative, which is why its reflections form a class of three.

**Exercise 4.7 ★★.**

Show that biphenyl twisted by an angle between 0 and 90° between its rings belongs to $D_2$, and conclude on its chirality.

**Solution of Exercise 4.7.**

Twisted biphenyl keeps the $C_2$ along the inter-ring bond and two $C_2$ perpendicular to it, bisecting the angles between the ring planes; every plane and the [centre of inversion](#def-b3-point-groups-inversion) of the planar or perpendicular forms are lost. With three perpendicular $C_2$ and no improper element the group is $D_2$: the molecule is chiral (its two twisted forms are enantiomers, separable when bulky ortho substituents prevent rotation).

**Exercise 4.8 ★★.**

Using the $C_{2v}$ table, find the labels of the oxygen $3d$ orbitals in water (take the quadratic functions as their shapes).

**Solution of Exercise 4.8.**

$d_{z^2}$ and $d_{x^2-y^2}$: $a_1$ ($x^2$, $y^2$, $z^2$ are $A_1$); $d_{xy}$: $a_2$; $d_{xz}$: $b_1$; $d_{yz}$: $b_2$.

**Exercise 4.9 ★★.**

Check on the $T_d$ table that $\sum_id_i^2 = h$ and that the rows $A_1$ and $T_2$ are orthogonal when each column is weighted by the size of its class.

**Solution of Exercise 4.9.**

$1 + 1 + 4 + 9 + 9 = 24 = h$. $\sum g\,\chi_{A_1}\chi_{T_2} = 1\cdot3 + 8\cdot0 + 3(-1)
+ 6(-1) + 6(1) = 0$.

**Exercise 4.10 ★★★.**

Show that the product of two reflections in planes making an angle $\theta$ is a rotation by $2\theta$ about their intersection. Deduce that a molecule with two vertical planes at 60° has a $C_3$ axis.

**Solution of Exercise 4.10.**

In the plane perpendicular to the intersection line, a reflection in a line at angle $\alpha$ maps the polar angle $\phi$ to $2\alpha - \phi$. Two reflections, at $\alpha$ then $\alpha + \theta$: $\phi \to 2\alpha - \phi \to 2(\alpha + \theta) - (2\alpha - \phi) =
\phi + 2\theta$, a rotation by $2\theta$ about the line. Two vertical planes at $60^\circ$ thus generate a rotation by $120^\circ$: a $C_3$.

**Exercise 4.11 ★★★.**

Allene $\ce{H2C=C=CH2}$ has the two $\ce{CH2}$ groups in perpendicular planes. Find its elements, show that its [point group](#def-b3-point-groups-point-group) is $D_{2d}$, and explain why 1,3-dichloroallene $\ce{ClHC=C=CHCl}$ is chiral.

**Solution of Exercise 4.11.**

The C=C=C axis is a $C_2$ and an $S_4$ (turn by $90^\circ$, which exchanges the two $\ce{CH2}$ planes, then reflect through the central carbon’s plane); the two $\ce{CH2}$ planes are $\sigma_d$; two $C_2$ perpendicular to the axis bisect them. Order 8: $D_{2d}$, achiral. In $\ce{ClHC=C=CHCl}$ the planes, the $S_4$ and the axial $C_2$ are destroyed by the substitution; only one $C_2$ perpendicular to the axis survives: $C_2$, chiral — an axially chiral allene.

**Exercise 4.12 ★★★.**

Going from $\ce{SF6}$ ($O_h$) to $\ce{SF5Cl}$, then to trans-$\ce{SF4Cl2}$ and cis-$\ce{SF4Cl2}$, give each [point group](#def-b3-point-groups-point-group) and show that each is a [subgroup](#def-b3-point-groups-class) of $O_h$. Which can be polar?

**Solution of Exercise 4.12.**

$\ce{SF6}$: $O_h$ ($h = 48$). $\ce{SF5Cl}$: the $C_4$ through Cl and four $\sigma_v$: $C_{4v}$ ($h = 8$). trans-$\ce{SF4Cl2}$: $D_{4h}$ ($h = 16$). cis-$\ce{SF4Cl2}$: $C_{2v}$ ($h
= 4$). Each keeps a subset of the operations of $O_h$, closed under products, and 8, 16, 4 divide 48. Polar: $\ce{SF5Cl}$ and cis-$\ce{SF4Cl2}$.

## 4.7 Problem: Substituting Methane

**Problem 4.1.**

Weekend problem — the 24 symmetry operations of methane, the point groups of its chlorinated derivatives, polarity and chirality by the theorems, and the counting of isomers

Methane is drawn in a cube of side 2 centred at the origin, with hydrogens on the corners $(1,1,1)$, $(1,-1,-1)$, $(-1,1,-1)$ and $(-1,-1,1)$.

**Part I — The group of methane.**

1. Find the four $C_3$ axes and count the rotations they give.
2. Find the three $C_2$ axes and check that each is also an $S_4$ axis; count the $S_4$ operations.
3. Find the six [mirror planes](#def-b3-point-groups-mirror) and the C–H bonds each contains.
4. Add $E$ : what is the order of the group?
5. Check that there is no [centre of inversion](#def-b3-point-groups-inversion) . (Is $(-1,-1,-1)$ a hydrogen position?)
6. Group the operations into the five classes of the $T_d$ table.

**Part II — Chlorinated methanes.**

7. Give the [point group](#def-b3-point-groups-point-group) of $\ce{CH3Cl}$ , and its order.
8. Same question for $\ce{CH2Cl2}$ .
9. Same question for $\ce{CHCl3}$ and $\ce{CCl4}$ .
10. Same question for $\ce{CH2FCl}$ and $\ce{CHFClBr}$ .
11. Check that each group is a [subgroup](#def-b3-point-groups-class) of $T_d$ , its order dividing 24.
12. Which hydrogens of $\ce{CH3Cl}$ are interchanged by its operations?

**Part III — Polarity and chirality.**

13. Which of the six molecules can be polar? Give the direction of each dipole.
14. Which are chiral?
15. For the chiral one, how many stereoisomers exist?
16. Why is $\ce{CH2FCl}$ achiral although its carbon carries four bonds to three different kinds of atom?
17. Could a methane derivative with four different substituents ever be polar and not chiral?
18. Which representation of $T_d$ do the three $2p$ orbitals of the carbon span?

**Part IV — Counting isomers.**

19. Show that any two hydrogens of methane can be brought onto any other two by an operation of $T_d$ . How many isomers of $\ce{CH2Cl2}$ exist?
20. How many isomers of $\ce{CH2FCl}$ ? Of $\ce{CHFClBr}$ , counting enantiomers?
21. For benzene ( $D_{6h}$ ), how many dichlorobenzenes exist? Give their [point groups](#def-b3-point-groups-point-group) .
22. How many trichlorobenzenes? Give their [point groups](#def-b3-point-groups-point-group) .
23. A planar square “methane” ( $D_{4h}$ ) would have how many isomers of $\ce{CH2Cl2}$ ? What did this argument prove historically?
24. Check that the order of $T_d$ equals the sum of the squared dimensions of its [irreducible representations](#def-b3-point-groups-irrep) , and state the result.

**Solution of Problem 4.1.**

**1.** Along the four body diagonals through C and each H; each gives $C_3$ and $C_3^2$: 8 rotations. **2.** Along $x$, $y$, $z$ through the face centres: $C_2(z)$ maps $(1,1,1)
\to (-1,-1,1)$, a hydrogen. A rotation by $90^\circ$ about $z$ followed by reflection in $xy$ maps $(1,1,1) \to (-1,1,1) \to (-1,1,-1)$, a hydrogen: an $S_4$; with $S_4^3$, 6 operations. **3.** The planes $x = \pm y$, $y = \pm z$, $x = \pm z$; $x = y$ contains $(1,1,1)$ and $(-1,-1,1)$: each plane contains two C–H bonds. **4.** $1 + 8 + 3 + 6 + 6 = 24$. **5.** Inversion maps $(1,1,1)$ to $(-1,-1,-1)$, which is not a hydrogen: no $i$. **6.** $E$; $8C_3$; $3C_2$; $6S_4$; $6\sigma_d$. **7.** $\ce{CH3Cl}$: $C_{3v}$, $h = 6$. **8.** $\ce{CH2Cl2}$: $C_{2v}$, $h = 4$. **9.** $\ce{CHCl3}$: $C_{3v}$; $\ce{CCl4}$: $T_d$, $h = 24$. **10.** $\ce{CH2FCl}$: $C_s$ ($h = 2$); $\ce{CHFClBr}$: $C_1$ ($h = 1$). **11.** Each keeps those operations of $T_d$ that respect the substitution; 6, 4, 6, 24, 2, 1 all divide 24. **12.** The three hydrogens, permuted by $C_3$, $C_3^2$ and the three $\sigma_v$. **13.** All but $\ce{CCl4}$: along the $C_3$ axis for $\ce{CH3Cl}$ and $\ce{CHCl3}$, the $C_2$ axis for $\ce{CH2Cl2}$, in the [mirror plane](#def-b3-point-groups-mirror) for $\ce{CH2FCl}$, in no imposed direction for $\ce{CHFClBr}$. **14.** $\ce{CHFClBr}$ only ($C_1$, no $S_n$). **15.** Two enantiomers. **16.** Its two hydrogens are equivalent: the plane through C, F and Cl bisecting H–C–H is a [mirror plane](#def-b3-point-groups-mirror). **17.** No: four different substituents leave only $E$; with no $S_n$ the molecule is chiral (and may be polar). **18.** $(x, y, z)$: $T_2$. **19.** Two hydrogens are the ends of an edge of the tetrahedron; the 24 operations carry any edge onto any other (the 6 edges form one set): one $\ce{CH2Cl2}$. **20.** $\ce{CH2FCl}$: one. $\ce{CHFClBr}$: two, a pair of enantiomers. **21.** Three: ortho ($C_{2v}$), meta ($C_{2v}$), para ($D_{2h}$). **22.** Three: 1,2,3- ($C_{2v}$), 1,2,4- ($C_s$), 1,3,5- ($D_{3h}$). **23.** Two (Cl atoms adjacent or opposite). Only one $\ce{CH2Cl2}$ is known, which ruled out a planar carbon and supported the tetrahedral carbon proposed in 1874. **24.** $1^2 + 1^2 + 2^2 + 3^2 + 3^2 = 24$: **the order of $T_d$, the [point group](#def-b3-point-groups-point-group) of methane, is $h = 24$**.