Point Group And Character Table Practice Problems

Here are some molecular symmetry practice problems. Notice that you actually need to remember how to derive the shapes of some of these molecules- an essential skill that I insist on harping about!

1) Determine the point group for the following molecules:

a) NBr3

Just like NH3’s shape:

A C3 primary axis, vertical mirrors, no h or C2’s, so C3v

b) SeCl6

Do VSEPR, determine that this is a AX6 molecule, so octahedral: Oh

c) CCl4

Do VSEPR, determine that this is a AX4 molecule, so tetrahedral: Td

d) H2C=CH2

Draw the molecule:

You should see a C2 axis, and then find one of the two perpendicular C2’s.

This is a flat molecule, so the plane of the page is a mirror, and it is h (not

v!). Thus, this is D2h

e) H2C=CF2

Draw it:

Now we don’t have C2’s, just one C2 and vertical mirror planes: C2v

f) ozone

Do a VSEPR analysis of ozone (O3), determine the central oxygen is AX2E, so a bent molecule:

A C2 axis bisects the O-O bonds and runs through the central oxygen. O3 is flat, so there is (are!) vertical mirror planes: C2v

g) SCN-

This is a linear molecule with different ends, so it is Cv

h) PCl5

Do a VSEPR analysis: this is AX5, a trigonal bipyramidal molecule:

The equatorial PCl3 plane is a h, we

have C2’s, so D3h

i) Cl3PO.

Do a VSEPR analysis: the central P is AX4, so based on a tetrahedral ideal geometry:

A C3, vertical mirros, so once again C3v

2) Find the point group for the following molecules or ions:

a) FeBr3Cl34- (two isomers!)

Only two isomers: both sets of Three ligands meridional, or both facial. The point groups are C2v and C3v, respectively

b) (ignore H atoms)

Trim away the H’s and look at this: This molecule is D4h

c)

A C4, C2’s, a h, so D4h

d) (ignore Ph groups)

A C2, no C2’s, a v, so C2v

e)

Hard to see unless you look down the right direction: there is a C3 axis,

C2’s, but no mirrors, so D3

f)

The plane of the page is a mirror plane, and there is a C2 perpendicular to

that: C2h

g) a cube:

Once you realize there are multiple C4 axes, it is a special point group: Oh

h)

No perpendicular C2’s, so C3h

i)

As drawn there are no symmetry elements: C1

If you rotate about the C-C bond, you will add a mirror plane between the

two ends, and the point group symmetry is Cs:

j)

As drawn you might miss the mirror plane. Simply rotate so that the unique

Cl atom is in the plane of projection and you’ll see that this is Cs

3) Determine the point group for the following objects:

a) The peace sign: Yet another C2v object

b) The Mercedes-Benz sign:

If the symbol is the same on the backside, there will be a

horizontal mirror, C2’s, and the point group will be D3h. If not, we will

have the C3 and some v’s so the point group would be C3v. When in

doubt, ask about the object in question.

c) An interlaced Jewish star:

The interlacing “destroys” the mirrors and perpendicular C2’s, so all we

have left is the C6 rotational axis: C6.

d) Adamantane (might need a model):

If you look down the 4 points with tertiary carbons, you’ll see they are

actually C3 axes. The only point group that has multiple C3’s (and no

C4’s) is Td

e) The Campanile (Berkeley campus) (ignore the clock faces):

This clock tower has a pointy end (the top) and four sides. If we

ignore the clocks we can see each side has a vertical mirror, so

C4v

Here is a nice website with practice problems, where the molecules are

linked to a 3D graphics program so you can rotate things around to look

them more easily:

http://www.scs.uiuc.edu/~chem315/pointgrouppractice.htm

4) Using character tables to determine the symmetry properties of

molecular bonds, vibrational modes, etc., takes practice. There are

some problems in the textbook to practice on; here’s another one.

Follow the procedure from the lecture notes, step by step.

A new molecule EH3 is prepared. It’s structure (either trigonal planar

or pyramidal) is not known. An infrared spectrum, however shows

two bands for the E-H bond vibrations. From a symmetry analysis of

these two possible geometries, which structure does this molecule

adopt? (HINT: don’t do a full 3N-6 analysis- just consider the

vibrating E-H bonds).

We have two geometries to consider- need to know their point groups. A planar

EH3 is D3h, and a pyramidal EH3 is C3v. Do vibrational analysis in D3h first:

Consider the arrows, representing E-H bond stretches.

Look at D3h character table and consider the action

of each operation.

E: all 3 arrows stay the same

C3: all arrows move, so you get zero for the character

C2: each C2 is along one arrow, so 1 stays the same

h: all three arrows are in the mirror plane, so all stay the same

S3: all arrows still move; the extra reflection doesn’t change this.

v: each vertical mirror contains one arrow, so 1 stays the same.

Thus, vib = 3 0 1 3 0 1

Use vib = 3 0 1 3 0 1, the character table, and the decomposition formula

to decompose vib into its irreducible representations:

n(A1’): 1/12[1(1)(3) + 2(1)(0) + 3(1)(1) + 1(1)(3) + 2(1)(0) + 3(1)(1)] = 1

n(A2’): 1/12[1(1)(3) + 2(1)(0) + 3(-1)(1) + 1(1)(3) + 2(1)(0) + 3(-1)(1)] = 0

n(E’): 1/12[1(2)(3) + 2(-1)(0) + 3(0)(1) + 1(2)(3) + 2(-1)(0) + 3(0)(1)] = 1

n(A1”): 1/12[1(1)(3) + 2(1)(0) + 3(1)(1) + 1(-1)(3) + 2(-1)(0) + 3(-1)(1)] = 0

n(A2”): 1/12[1(1)(3) + 2(1)(0) + 3(-1)(1) + 1(-1)(3) + 2(-1)(0) + 3(1)(1)] = 0

n(E”): 1/12[1(2)(3) + 2(-1)(0) + 3(0)(1) + 1(-2)(3) + 2(1)(0) + 3(0)(1)] = 0

So, vib = 1A1’ + 1E’. Look at the table, see that only E’ is IR-active. Thus, we

conclude that there would be only one E-H stretching IR band.

Now look at pyramidal EH3:

Consider the arrows, representing E-H bond stretches.

Look at C3v character table and consider the action

of each operation.

E: all 3 arrows stay the same

C3: all arrows move, so you get zero for the character

v: each vertical mirror contains one arrow, so 1 stays the same.

Thus, vib = 3 0 1

decompose vib into its irreducible representations:

n(A1): 1/6[1(1)(3) + 2(1)(0) + 3(1)(1)] = 1

n(A2): 1/6[1(1)(3) + 2(1)(0) + 3(-1)(1)] = 0

n(E): 1/6[1(2)(3) + 2(-1)(0) + 3(0)(1)] = 1

So, vib = 1A1 + 1E. Look at the C3v table, see that both are IR-active. Thus, we

conclude that there would be two E-H stretching IR bands. Thus, we decide that

this molecule is pyramidal.