Section 13-16 Determining in Which Molecular Orbital an Atomic Orbital Will Appear
Section 13-16 Determining in Which Molecular Orbital an Atomic Orbital Will AppearFigure 13-14 Valence MOs for staggered ethane. The energy level spacings have been altered for convenience. The hatched areas have a negative sign.
TABLE 13-24 Characters for the D3d Point Group
D3d
E
2C3
3C2
i
2S6
3σd
A1g
1
1
1
1
1
1
A2g
1
1
−1
1
1
−1
Eg
2
−1
0
2
−1
0
A1u
1
1
1
−1
−1
−1
A2u
1
1
−1
−1
−1
1
Eu
2
−1
0
−2
1
0
13-16 Determining in Which Molecular Orbital an AtomicOrbital Will Appear Earlier, we noted that the 2pz AO in ammonia will contribute to a1 MOs because 2pz transforms like z, and z is listed as a basis for the a1 representation. If the basis functions were not listed, we could have reached the same conclusion simply by
Chapter 13 Group TheoryFigure 13-15 taking 2pz and putting it through all the symmetry operations to produce a representation: E2pz = +1 · 2pz σi2pz = +1 · 2pz, i = 1, 2, 3
C±2p
3
z = +1 · 2pz
The representation contains only +1, so 2pz obviously “has” a1 symmetry.
When we come to the 1s AOs on hydrogens in NH3, we cannot use the list of basis functions. It includes only coordinates originating on the principal axis, and the hydrogens are not on that axis. Here we must generate a representation. Let us try to do this by putting one of the hydrogen atoms, H1, through the various symmetry operations (also see Fig. 13-15):
EH1 = +1H1 σ1H1 =+1H1 σ2H1 =
H3
σ3H1 =
H2
C+H
3
1 =
H2
C−H
3
1 =
H3
Since some of these operations interchange H1 with H2 or H3, we are not achieving a one-dimensional representation. We must take all three functions together and work out a three-dimensional representation. Thus,
H
H 1 0 0 1 1
E H2 = H2 , E : 0 1 0 , χ (E) = 3
H3
H3 0 0 1
H
H 1 0 0 1 1
σ1 H2 = H3 , σ1 : 0 0 1 , χ (σ1) = 1
H3
H2 0 1 0
H
H 0 1 0 1 2
C+
H
H
: 0 0 1 ) = 0 3 2 = 3 ,
C+
3
, χ(C+
3
H3
H1 1 0 0