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Character tables, the stretching representation, the reduction formula, and reading infrared and Raman activity to predict how many stretching bands a molecule shows.
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By the end of this lesson you will be able to build a molecule's stretching representation, reduce it with a character table, and predict its infrared and Raman stretching bands.
From lesson 21 you can assign a point group and count its order $h$; from lesson 22 you can count a molecule's vibrations and its stretches, one per bond, and you know that a vibration absorbs infrared only if it changes the dipole. This lesson connects the two with the one tool that does most of the work in applied symmetry: the character table.
| Term | What it means |
|---|---|
| Character table | A table listing a point group's classes of operations and its irreducible representations, with a character for each pair. |
| Character, $\chi$ | A number saying how something attached to the molecule behaves under one operation; for a set of bonds, how many stay in place. |
| Irreducible representation | One of the basic ways of behaving under the group's operations, a row of the character table. |
| Reducible representation | A set of characters, such as the bond stretches', that breaks into irreducible pieces. |
| Reduction formula | $n_i = \frac{1}{h}\sum_R g_R\,\chi_\Gamma(R)\,\chi_i(R)$, how many times piece $i$ occurs. |
| Degenerate | Describing an $E$ or $T$ piece, two or three vibrations of equal energy that give one band. |
| Raman active | Showing up in the Raman spectrum, which a vibration does if it transforms like a square or product such as $x^2$ or $xy$. |
| Mulliken labels | $A$ and $B$ for one-dimensional pieces, $E$ for two, $T$ for three. |
A character table has a column for each class of operations and a row for each irreducible representation, each of the basic ways a property of the molecule can behave under those operations. The $C_{2v}$ table, for water, is:
| $C_{2v}$ | $E$ | $C_2$ | $\sigma_v(xz)$ | $\sigma_v'(yz)$ | |
|---|---|---|---|---|---|
| $A_1$ | 1 | 1 | 1 | 1 | $z$; $x^2, y^2, z^2$ |
| $A_2$ | 1 | 1 | -1 | -1 | $xy$ |
| $B_1$ | 1 | -1 | 1 | -1 | $x$; $xz$ |
| $B_2$ | 1 | -1 | -1 | 1 | $y$; $yz$ |
The last column is the key to spectra. A piece listed with $x$, $y$ or $z$ changes the dipole, so a vibration of that symmetry absorbs infrared. A piece listed with a square or a product, $x^2$ or $xy$, changes the polarizability, so a vibration of that symmetry is Raman active.
To use the table, first build a reducible representation from the molecule. For bond stretches this is easy: under each operation, count the bonds that stay where they are. For water, lying in the $xz$ plane, the identity leaves both O-H bonds, the twofold axis swaps them (zero stay), the molecular plane $\sigma_v(xz)$ leaves both, and the other plane swaps them. So $\Gamma = 2, 0, 2, 0$.
Then apply the reduction formula to each row:
$$n_i = \frac{1}{h}\sum_R g_R\,\chi_\Gamma(R)\,\chi_i(R),$$
multiplying, class by class, the number of operations $g_R$, the stretching character and the row's character, adding, and dividing by the order $h$. For $A_1$: $\frac{1}{4}(2 + 0 + 2 + 0) = 1$. For $B_1$: $\frac{1}{4}(2 - 0 + 2 - 0) = 1$. For $A_2$ and $B_2$: zero. So water's stretches are $A_1 + B_1$: a symmetric stretch and an antisymmetric one. Both appear with $z$ or $x$, so both absorb infrared; both also appear with squares, so both are Raman active.
The dimensions of the pieces always add up to the number of bonds, and each piece is one band, however many vibrations it holds: a $T_2$ piece is three vibrations of the same energy and gives a single band.
Another way: picture
Think of the stretching representation as a chord played on a piano and the irreducible representations as the notes. The reduction formula is the ear that picks out which notes are sounding and how often. The character table then says which notes the infrared microphone can hear, and which only the Raman one can.
Another way: steps
Ammonia and boron trifluoride both have three bonds to a central atom, but their spectra differ, and the character tables say why. Ammonia is $C_{3v}$: its stretches give $\Gamma = 3, 0, 1$ across $E$, $2C_3$ and $3\sigma_v$, which reduces to $A_1 + E$. In the $C_{3v}$ table both $A_1$ (with $z$) and $E$ (with $x$ and $y$) are infrared active, so ammonia shows two stretching bands in the infrared.
Boron trifluoride is $D_{3h}$: its stretches reduce to $A_1' + E'$. The $A_1'$ piece has no $x$, $y$ or $z$ beside it in the table, because the flat molecule's symmetric stretch keeps its three dipoles cancelled at every moment; only $E'$ absorbs. So boron trifluoride shows a single infrared stretching band, while both pieces appear in the Raman spectrum. Counting bands is therefore a way to tell a pyramid from a flat triangle without any other information.
In a group with a centre of inversion, every piece is labeled $g$, symmetric under inversion, or $u$, antisymmetric. The coordinates $x$, $y$ and $z$ are all $u$, while their squares and products are all $g$. So in such a molecule, a vibration is either infrared active or Raman active, but never both: the rule of mutual exclusion.
Xenon tetrafluoride, $D_{4h}$, shows it. Its stretches reduce to $A_{1\mathrm{g}} + B_{1\mathrm{g}} + E_u$. The $E_u$ piece absorbs infrared; $A_{1\mathrm{g}}$ and $B_{1\mathrm{g}}$ appear only in the Raman spectrum. Seeing no band shared between the two spectra is strong evidence that a molecule has a centre of inversion, which is how the square-planar structure of xenon tetrafluoride was confirmed soon after the compound was first made.
Bond stretches are the easiest representation to build, but the same method handles all of a molecule's motions. Attach three small arrows to every atom, one along each axis, and ask how many of those arrows each operation leaves pointing the same way. An atom that an operation moves contributes nothing; an atom that stays put contributes a number that depends only on the kind of operation. The result is the representation of all $3N$ motions. Reduce it, then remove the pieces that belong to the three translations, which behave like $x$, $y$ and $z$, and to the three rotations, which the character table lists as $R_x$, $R_y$ and $R_z$. What remains are the vibrations, sorted by symmetry.
For water that gives $2A_1 + B_1$: the two stretches found above plus the bend, which is also $A_1$ because opening and closing the angle keeps every symmetry element. All three are infrared active, and indeed water shows three fundamental bands. The bends of larger molecules are sorted the same way, and subtracting the stretching pieces from the full list is the quickest way to find them.
Chemists rarely do this by hand for large molecules, but the logic never changes, and software that predicts spectra carries out exactly these steps. Knowing them is what lets a chemist see when a program's prediction has gone wrong, for example because the structure it was given had lost a symmetry element through a small distortion.
Three checks catch most slips. First, every $n_i$ must come out as a whole number of zero or more; a fraction or a negative number means a character or a class size has been misread. Second, the dimensions must add up: $A$ and $B$ count one, $E$ two, $T$ three, and the total must equal the character under the identity, which is the number of bonds. Third, the stretching set always contains the totally symmetric piece exactly once, because stretching every bond together keeps the full symmetry.
When counting infrared bands, count pieces, not vibrations: methane's $T_2$ stretch is three vibrations but one band. And remember the class sizes: in $T_d$ the eight $C_3$ operations each contribute, so the $C_3$ column is multiplied by eight in the sum, not by one. Finally, compare the prediction with a molecule of the same shape you have already worked out: every flat triangle of identical bonds gives one infrared stretching band, and every regular tetrahedron gives one too, so a different answer for such a molecule points to an error in the characters or in the class sizes you multiplied them by.
Environmental agencies and industrial plants measure gases in smokestacks and in the open air by their infrared spectra, often from hundreds of meters away with open-path instruments. To pick out one gas among many, the software needs to know exactly which bands each molecule shows, and that list starts with the reduction this lesson teaches.
Sulfur dioxide from burning coal, for instance, is bent like water and $C_{2v}$, so its two S-O stretches are $A_1 + B_1$ and both absorb, near $1{,}150$ and $1{,}360$ cm$^{-1}$. Sulfur trioxide, flat and $D_{3h}$, shows only one S-O stretching band in the infrared. The difference in band counts, not just the positions, helps the instruments tell the two oxides apart in a plume.
Before X-ray crystallography was routine, and still today for compounds that will not crystallize, chemists used the number of infrared and Raman bands to decide between possible structures. A metal carbonyl is the classic case. The number of C-O stretching bands, which appear in an uncrowded part of the spectrum near $2{,}000$ cm$^{-1}$, depends on the arrangement of the carbonyls.
An octahedral $\mathrm{M(CO)_6}$ shows one infrared C-O band; a cis $\mathrm{M(CO)_4L_2}$ shows four; the trans isomer shows only one. Chemists reduce the C-O stretching representation for each candidate structure, compare with the spectrum, and keep the structure whose band count matches. The method is fast, needs only a small sample, and is still used every day in organometallic laboratories.
It is natural to expect four C-H bonds to give four infrared bands. But symmetry does not let the bonds vibrate independently: the four stretches combine into a totally symmetric one and a set of three of equal energy. The first does not change methane's dipole and is invisible in the infrared; the three give one band between them.
The companion error is forgetting the class sizes in the reduction formula. A class such as $8C_3$ contributes eight times its character; leaving out that factor gives fractions that cannot be right. Every $n_i$ must be a whole number.
Build the stretching representation.
$\Gamma = 2, \ 0, \ 2, \ 0$
Bonds left in place by $E$, $C_2$, $\sigma_v(xz)$, $\sigma_v'(yz)$.
Reduce against $A_1$.
$\tfrac{1}{4}(2 + 0 + 2 + 0) = 1$
All characters of $A_1$ are one.
Reduce against $B_1$.
$\tfrac{1}{4}(2 - 0 + 2 - 0) = 1$
Characters $1, -1, 1, -1$.
Reduce against $A_2$ and $B_2$.
$\tfrac{1}{4}(2 + 0 - 2 - 0) = 0$
Neither appears.
Read the spectra.
$A_1 + B_1: \ \text{2 infrared, 2 Raman}$
$A_1$ goes with $z$, $B_1$ with $x$; both also go with squares.
Build the stretching representation.
$\Gamma = 3, \ 0, \ 1$
Under $E$, $2C_3$ and $3\sigma_v$: a plane keeps the one bond it contains.
Reduce against $A_1$.
$\tfrac{1}{6}(1 \times 3 + 2 \times 0 + 3 \times 1) = 1$
Class sizes $1, 2, 3$.
Reduce against $A_2$.
$\tfrac{1}{6}(3 + 0 - 3) = 0$
Characters $1, 1, -1$.
Reduce against $E$.
$\tfrac{1}{6}(3 \times 2 + 0 + 0) = 1$
Characters $2, -1, 0$.
Check the dimensions.
$1 + 2 = 3$
Three bonds, three stretches.
Read the spectra.
$A_1 + E: \ \text{2 infrared bands}$
$A_1$ with $z$, $E$ with $(x, y)$.
Build the stretching representation.
$\Gamma = 3, 0, 1, 3, 0, 1$
Under $E$, $2C_3$, $3C_2$, $\sigma_h$, $2S_3$, $3\sigma_v$.
Reduce against $A_1'$.
$\tfrac{1}{12}(3 + 0 + 3 + 3 + 0 + 3) = 1$
Class sizes times characters of one.
Reduce against $E'$.
$\tfrac{1}{12}(6 + 0 + 0 + 6 + 0 + 0) = 1$
Characters $2, -1, 0, 2, -1, 0$.
Check the dimensions.
$1 + 2 = 3$
No other piece is needed.
Read the infrared spectrum.
$E' \text{ only: 1 band}$
$A_1'$ has no $x$, $y$ or $z$ beside it.
Read the Raman spectrum.
$A_1' \text{ and } E': \text{ 2 bands}$
Both go with squares.
Check the dimensions.
$1 + 3 = 4$
Four bonds.
Find the pieces with $x$, $y$ or $z$.
$T_2$
$A_1$ has none in $T_d$.
Count the bands.
Match each molecule to what its bond stretches reduce to in its point group.
| $A_1 + B_1$ | $A_1 + E$ | $A_1' + E'$ | $A_1 + T_2$ | |
|---|---|---|---|---|
| $\mathrm{SO_2}$ | ||||
| $\mathrm{NH_3}$ | ||||
| $\mathrm{BF_3}$ | ||||
| $\mathrm{SiH_4}$ |
Complete the worked solution: count the totally symmetric stretch of $\mathrm{PCl_3}$, whose stretching characters are $3,\ 0,\ 1$.
Multiply operations, stretching character and symmetric character for each class, and add.
$\text{sum} =$ z
The numerator of the reduction formula.
Add up the operations of the group.
$\text{order} =$ h
The denominator of the reduction formula.
Read the stretches from the identity column.
$\text{stretching vibrations} =$ u
Under the identity every bond stays in place.
The stretching representation of $\mathrm{SO_3}$ in $D_{3h}$ is $\Gamma = 3,\ 0,\ 1,\ 3,\ 0,\ 1$. What does it reduce to?
The stretches of $\mathrm{NH_3}$ reduce to $A_1 + E$ in $C_{3v}$, and those of $\mathrm{[PtCl_4]^{2-}}$ to $A_{1\mathrm{g}} + B_{1\mathrm{g}} + E_u$ in $D_{4h}$. For each, in that order, fill in the number of stretching vibrations, the infrared bands and the Raman bands they give.
| stretching vibrations | infrared bands | Raman bands | |
|---|---|---|---|
| the first molecule | |||
| the second molecule |
For $\mathrm{PCl_3}$ in $C_{3v}$, the stretching representation is $\Gamma = 3,\ 0,\ 1$, and the totally symmetric representation $A_1$ has character $1$ under every class. What is $\sum_R g_R\,\chi_\Gamma(R)\,\chi_{A_1}(R)$, the sum inside the reduction formula?
Answer: sum over the classes
The $\mathrm{B-F}$ stretches of $\mathrm{BF_3}$ ($D_{3h}$) form the representation $\Gamma = 3,\ 0,\ 1,\ 3,\ 0,\ 1$ across the classes of its character table, which reduces to $A_1' + E'$. How many infrared bands do its stretches give?
Answer: infrared stretching bands
An analytical lab identifies gases in an industrial exhaust by their stretching bands, and must know how many to expect in each spectrum. The stretches of $\mathrm{BF_3}$ reduce to $A_1' + E'$, those of $\mathrm{CH_4}$ to $A_1 + T_2$ and those of $\mathrm{XeF_4}$ to $A_{1\mathrm{g}} + B_{1\mathrm{g}} + E_u$. In that order, fill in the infrared and Raman stretching bands for each.
| infrared stretching bands | Raman stretching bands | |
|---|---|---|
| the first compound | ||
| the second compound | ||
| the third compound |
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
The stretches of $\mathrm{NH_3}$ reduce to $A_1 + E$ in $C_{3v}$, and those of $\mathrm{SO_2}$ to $A_1 + B_1$ in $C_{2v}$. For each, in that order, fill in the number of stretching vibrations, the infrared bands and the Raman bands they give.
| stretching vibrations | infrared bands | Raman bands | |
|---|---|---|---|
| the first molecule | |||
| the second molecule |
You can reduce a representation. Explain why boron trifluoride shows one infrared stretching band while ammonia shows two, though both have three bonds.
15. Your turn: methane's C-H stretches give $\Gamma = 4, 1, 0, 0, 2$ in $T_d$ and reduce to $A_1 + T_2$. How many infrared stretching bands?, step 3
$1$
Three vibrations of the same energy, one band.