Answers to Problems on Symmetry Operations and Point Groups
1) The effect of a symmetry operation on the Cartesian coordinates can be written:
E (x, y, z)  (x , y, z)
C2 z (x, y, z)  (-x , -y, z) (rotation about the z axis by 180o)
…
(a) Give the equivalent expressions for the effect of the operations
C2 x , C2 y , σ (xz) , σ (yz) , σ (xy) , i
C2 x (x, y, z)  (x , -y, -z)
C2 y (x, y, z)  (-x , y, -z)
σ (xz) (x, y, z)  (x , -y, z)
σ (yz) (x, y, z)  (-x , y, z)
σ (xy) (x, y, z)  (x , y, -z)
i (x, y, z)  (-x , -y, -z)
(b) What is the equivalent expression for S1 ( = Sn n) and S2 ( = Sn n/2) (where the S
axis has to be along z and n is an even number)? Why do these operations not appear
in the list of operations of a group?
Sn = σ (xy) x Cn (rotate by Cn z followed by σ (xy) )
therefore:
S1 = σ (xy) x C1 = σ (xy) x E = σ (xy)
S1 = σ (xy)
S2 (x, y, z) = σ (xy) x C2 (x, y, z) = σ (xy) (-x, -y, z) = (-x, -y,- z) = i (x, y, z)
S2 = i
2) Satisfy yourself that all the following molecules belong to the C2v point group,
H2O, pyridine, 1,2 dichloro-benzene, acetone (draw the geometry including the
hydrogen atoms), cis - SCl2 F4. Name two other molecules that have C2v symmetry
that are not simple analogues of these.
H2O, pyridine are both planar and have a C2 primary rotation axis and two mirror
planes perpendicular to the axis. These should be obvious.
Acetone needs to be drawn with the methyl groups with one H atom each in the
molecular plane and symmetrically related to each other. That can be achieved in two
ways – the inplane H’s as close as possible or as far as possible apart. In reality this
makes no difference as the methyl groups are rotating rapidly about the C-C bond.
cis - SCl2 F4 the C2 axis bisects the Cl – S – Cl angle, the rest should be obvious!
3) Match the molecule to the point group.
1,3,5 trichlorobenzene
trans- Pt(NH3)2Cl2 (ignore the hydrogens)
SF5Cl
BFClBr
B(OH)3 (planar and all staggered Hs)
Tennis ball (including the seam)
D3h
D2h
C4v
Cs
C3h
D2d
If you do not see how any one of these works, find someone who thinks they do and
argue with them!
4) On the diagram (or as many copies of it as you need) of the B12H122- ion below,
indicate the position of the symmetry elements of the Ih group.
Ih: E 12C5 12C52 20C3 15C2 i 12S10 12S102 20S6 15σ
12C5 – 6 axes through an atom, the centre of the molecule and the opposite atom and
C5 rotations clockwise and anticlockwise about each axis
12C52 – same and C52 rotations clockwise and anticlockwise about each axis
20C3 – 10 axes through the centre of each triangular face, the centre of the molecule
and the opposite triangular face and C3 clockwise and anticlockwise about each axis
i – obvious
12S10 12S102 20S6 – same axes as the ones above, check that you can see the
staggered pentagons and triangles which accounts for the involvement of C10 and C6
rotations.
15σ – slightly trickier! Take the ‘top’ atom and 5 planes are immediately obvious,
through an adjoining atom and the centre of the molecule. Then work round the 5
atoms in the next ‘level’ identifying additional planes not already found … best of
luck!
5) Identify the Point Groups of the following molecules and ions:
HOCl
Cs
ClO2
C2v
N2O
C∞v
N3
D∞h
SF4
NH4+
PtCl42CH2Cl2
C4v
Td
D4h
C2v
C6F6
D6h
C6HF5
C2v
1,2 -C6H2F4 , C2v 1,3 -C6H2F4 , C2v 1,4 -C6H2F4 , D2h
1,2,3 -C6H3F3 , C2v 1,2,4 -C6H3F3 , C2v 1,3,5 -C6H3F3 , D3h
C6H4F2 same as C6H2F4 ; C6H5F same as C6HF5