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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 345
which decomposes to A1 + B2.
Combining these results, the four H1s orbitals therefore span 2A1 + B1 +
B2. Inspection of the C2v character table indicates that z2, and hence
dz2 , spans A1, so this orbital can form molecular orbitals with the A1
combinations of hydrogen orbitals. Because both x2 and y2 span A1, x2−
y2 and hence dx2−y2 also spans A1 and can therefore also formmolecular
orbitals with the A1 combination of hydrogen orbitals. �e orbitals dzx
and dyz span B1 and B2 respectively, so can form molecular orbitals with
the B1 and B2 combinations of hydrogen orbitals. However, dx y spans A2
and therefore remains non-bonding because the hydrogen orbitals do not
span A2.
�erefore yes , more d orbitals can formmolecular orbitals than for tetra-
hedral methane, but not as many for the C3v distorted molecule where all
�ve d orbitals could be involved in forming molecular orbitals.
P10C.4 �e character table for the Ih point group is available in the Online resource
centre.
(a) As explained in Section 10C.3 on page 411, a photon-induced transition
froma statewith symmetry Γ(i) to onewith symmetry Γ(f) is only allowed
if the direct product Γ(f) × Γ(q) × Γ(i) contains the totally symmetric
irreducible representation, A1g. Because the ground state is A1g the direct
product becomes Γ(f) × Γ(q) ×A1g.�is is simply Γ(f) × Γ(q) because the
direct product of the totally symmetric irreducible representation with
any other representation is the latter representation itself.
If the product Γ(f)×Γ(q) is to beA1g, then Γ(f)must equal Γ(q) because the
direct product of two irreducible representations only contains the totally
symmetric irreducible representation if the two irreducible representa-
tions are identical. However, inspection of the Ih character table shows
that Γ(q) = T1u, because x, y and z together span the T1u irreducible rep-
resentation. Because neither of the lowest-lying excited states are of T1u
symmetry, photon-induced transitions to these states are not allowed .
(b) If the centre of inversion is removed then the distorted molecule has I
symmetry. For this point group the character table in the Resource section
shows that Γ(q) spans T1. By the same reasoning as above it follows that
transitions to an excited state with symmetry T1 are allowed but transi-
tions to a state with symmetry G are not .
P10C.6 As explained in Section 10C.3 on page 411, a photon-induced transition from
a state with symmetry Γ(i) to one with symmetry Γ(f) is only allowed if the
direct product Γ(f) × Γ(q) × Γ(i) contains the totally symmetric irreducible
representation, A1g. For a transition from an A1g ground state to an Eu excited
state the direct product becomes Eu×Γ(q)×A1g.�is is simply Eu×Γ(q) because
the direct product of the totally symmetric irreducible representation A1g with
any other representation is the latter representation itself.