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344 10MOLECULAR SYMMETRY
Similarly
n(A2) = 1
6 (1×1×3 + 2×1×0 + 3×(−1)×1) = 0
n(E) = 1
6 (1×2×3 + 2×(−1)×0 + 3×1×1) = 1
�e three H1s orbitals sB, sC, and sD therefore span A1 +E; adding this to
the result from above that sA spans A1 gives 2A1 + E.
As explained in Section 10C.2(a) on page 409, only orbitals of the same
symmetry species may have a nonzero overlap. �e C3v character table
also shows that the dz2 orbital spans A1 in C3v, dx y and dx2−y2 jointly
span E, and dxz and dyz also jointly span E. It follows that dz2 can form
molecular orbitals with the A1 combinations of hydrogen orbitals while
the other d orbitals can all form molecular orbitals with the E combi-
nations of hydrogen orbitals. So yes , more d orbitals can be involved in
formingmolecular orbitals than in the case of tetrahedralmethanewhere,
as shown in the previous exercise, only dx y , dyz and dxz are able to form
molecular orbitals.
(b) �e irreducible representations spanned by the H1s orbitals in the C2v
distorted molecule, shown in the right-hand side of Fig. 10.18, are found
in the same way as above. It is convenient to consider sA and sB separately
from sC and sD.�is is because sA and sB are mixed with each other but
not with sC or sD by the operations of the group. Similarly sC and sD are
mixed with each other but not with sA or sB. For sA and sB the matrix
representatives are
D(E) = ( 1 0
0 1 ) D(C2) = ( 0 1
1 0 )
D(σv) = ( 1 0
0 1 ) D(σ ′v) = ( 0 1
1 0 )
for which the characters are
χ(E) = 2 χ(C2) = 0 χ(σv) = 2 χ(σ ′v) = 0
As before, this result is obtained with much less e�ort by simply counting
the number of orbitals which are not moved by each operation. Using
the decomposition formula, or simply by inspection of the C2v character
table, this representation decomposes to A1 +B1. Analysing sC and sD in
the same way gives the representatives as
D(E) = ( 1 0
0 1 ) D(C2) = ( 0 1
1 0 )
D(σv) = ( 0 1
1 0 ) D(σ ′v) = ( 1 0
0 1 )
�e characters are
χ(E) = 2 χ(C2) = 0 χ(σv) = 0 χ(σ ′v) = 2