Lecture 5 Model Answers to Problems
Self-Study Problems / Exam Preparation
• Draw the MO diagram for Mo2 and show that a sextuple bond order is potentially possible.
o Cr, has the valence configuration [Ar]3d44s2 Mo and W are similar with higher principle
quantum number 5s and 6s respectively, the metals are more stable with half filled shells
[Ar]nd5(n+1)s1, Figure 1
σu+
πu
(n+1)p ->σu+, πu
πu
(n+1)s -> σg+
δu
σ g+
σ g+
nd -> σg+, πg, δg
δg
πu
x
M
dxy
M
z
M
M
dxz
dyz
σ g+
y
dx2-y2
dz2
M
M
Figure 1 M2 showing sextuple bonding
• Use the long method to show that the M2 dimer dxz/dyz combination of AOs has πu symmetry.
o remember the D∞h symmetry elements, Figure 2
C2
M
C2
each C2 has a σv plane
∞σv
∞C 2
C2
i
M
2C∞(ϕ)
M
M
M
σh
M
2S∞(ϕ)
σh plane perpendicular
to the rotation axes along
the bond
C2, C3, C4, ... C∞
each rotation group contains elements from the ones below it
eg C4 contains C2 so some elements are removed
each rotation has 2 unique operations
Figure 2 D∞h symmetry elements
o start building a representation table, Figure 3
D∞h
E
2Cx(ϕ)
∞σv
i
2Sx(ϕ)
∞C2
Figure 3 D∞h empty representation table
Lecture 5 Tutorial Problems
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Lecture 5 Model Answers to Problems
o work out how the degenerate orbitals transform under each symmetry element, Figure 4
dxz
E
dxz
1
dxz
C2(180 º)
dyz
⎡ 1 0 ⎤
⎢
⎥
⎣ 0 1 ⎦
⎡ cos(180º )
Cx(ϕ)
-1 ⎢
⎢⎣
C2(180 º)
0
2
⎤ ⎡
⎤
⎥ = ⎢ −1 0 ⎥
cos(180º ) ⎥⎦ ⎣ 0 −1 ⎦
0
-2
2cos(180º)=-2
∞σv
1
σv
∞σv
dxz
-1
σv
dxz
i
-1
dyz
⎡ 1 0 ⎤
⎢
⎥
⎣ 0 −1 ⎦
dyz
⎡ −1 0 ⎤
⎢
⎥
⎣ 0 −1 ⎦
-2
σh
C2(180 º)
-1
⎡ cos(180º )
0
⎢
0
cos(180º )
⎢⎣
∞C 2
1
C2
dxz
⎤ ⎡
⎤
⎥ = ⎢ −1 0 ⎥
⎥⎦ ⎣ 0 −1 ⎦
-2
2cos(180º)=-2
C2
C2
0
dyz
⎡ 1 0 ⎤
⎢
⎥
⎣ 0 −1 ⎦
∞C 2
0
-1
C2
Figure 4 working out the characters
o and fill in the representation table, Figure 5
D∞h
E
2Cx(ϕ)
2
0
∞σv
i
2Sx(ϕ)
∞C2
2
-2
-2
0
πu
Figure 5 D∞h filled representation table
Lecture 5 Tutorial Problems
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Lecture 5 Model Answers to Problems
•
Clearly show using diagrams that S42 = C2 and S44 = E Thus, showing that there are 2 unique
operations per S4 axis in the Oh point group.
o using one pAO from each of the two sets of symmetry related pAOs for C4 and σh Figure 6
C4
σd
Figure 6 Sets of symmetry related orbitals
o show the effect of two sequential S4 operations, this is the same as 1C2 rotation which is also
the same as 2C4 rotations, Figure 7
C4
C4
σh
C4
σh
S41
C4
σh
S42
C4
σh
structures
are identical
σh
C21 = C42
C21 = C42 = S42
Figure 7 2S4 operations under the Oh point group
o show the effect of four sequential S4 operations, this is the same as E, ie the starting structure,
Figure 8
C4
C4
σh
E
C4
σh
S41
C4
C4
σh
S44
σh
S42
S43
σh
Figure 8 4S4 operations under the Oh point group
o ( S S = S ) are the unique operations (ie a forward and a backward rotation)
1
4
3
4
−1
4
Lecture 5 Tutorial Problems
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Lecture 5 Model Answers to Problems
• Clearly show using diagrams that S61 and S65 = S6−1 are the only unique operations for each S6
axis in the Oh point group.
o there are 8S6 operations in Oh, each C3 axis can also be thought of as having a coincident
C6 and σ h mirror plane perpendicular to this axis. these are not symmetry elements of
Oh because of the staggered arrangement of the three ligands
o all of the S6 operations are shown in Figure 9, only two are unique S61 and S65 = S6−1 .
S62 = C31 , S63 = i , S64 = C32 and S66 = E here I show explicitly that S62 = C31 in Figure 10
S62
S61
S63
S64
6
6
S
S65
Figure 9 The S6 operations
S61
C
S62
structures are
identical
1
3
S62 = C31
Figure 10 2S6 operations under the Oh point group
Lecture 5 Tutorial Problems
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