Transition probabilities

The probability of a transition being induced by interaction with electromagnetic radiation is proportional to the square of the modulus of a matrix element of the formh ^k|Qˆ^j| ⁱi, where the state function that describes the initial state transforms as Gⁱ, that describing the final state transforms as G^k, and the operator (which depends on the type of transition being considered) transforms as G^j. The strongest transitions are the E1 transitions, which occur when Qˆ is the electric dipole moment operator,
er. These transitions are therefore often called ‘‘electric dipole transitions.’’ The components of the electric dipole operator trans-form like x, y, and z. Next in importance are the M1 transitions, for which Qˆ is the magnetic dipole operator, which transforms like Rx, Ry, Rz. The weakest transitions are the E2 transitions, which occur when Qˆ is the electric quadrupole operator which, transforms like binary products of x, y, and z.

Example 5.4-1 The absorption spectrum of benzene shows a strong band at 1800 A˚ , two weaker bands at 2000 A˚ and 2600 A˚, and a very weak band at 3500 A˚. As we shall see in Chapter6, the ground state of benzene is ¹A1g, and there are singlet and triplet excited states of B1u, B2u, and E1usymmetry. Given that in D6h, (x, y) form a basis for E1uand z transforms as A2u, find which transitions are allowed.

To find which transitions are allowed, form the DPs between the ground state and the three excited states and check whether these contain the representations for which the dipole moment operator forms a basis:

A1g B1u¼ B1u, A_1g B2u¼ B2u, A_1g E1u¼ E1u:

Only one of these (E1u) contains a representation to which the electric dipole moment operator belongs. Therefore only one of the three possible transitions is symmetry allowed, and for this one the radiation must be polarized in the (x, y) plane (see Table5.2).

The strong band at 1800 A˚ is due to the¹A_1g!¹E_1utransition. The two weaker bands at 2000 A˚ and 2600 A˚ are due to the¹A1g !¹B1uand¹A1g!¹B2utransitions becoming allowed through vibronic coupling. (We shall analyze vibronic coupling later.) The very weak transition at 3500 A˚ is due to¹A1g !³E1u becoming partly allowed through spin–orbit coupling.

Problems

5.1 Find Jones symbols for {R
¹}, R2 D4. Project the function

¼ x²þ y²þ z²þ xy þ yz þ zx

into the G1, G2, and G5subspaces and hence find bases for these IRs. [Hints: Do not refer to published character tables. You will need character sets for the IRs G2and G5

of D4, which were found in Exercise4.6-1.]

5.2 Find the IRs of the point group D4hfor which the following Cartesian tensors form bases:

1, xyz, zðx²
y²Þ, xyðx²
y²Þ, xyzðx²
y²Þ:

[Hint: Use the character table for D4hin AppendixA3, in which the principal axis has been chosen to lie along z.]

5.3 Determine correlation relations between the IRs of (a) Tdand C3v, and (b) Ohand D3d. [Hints: Use character tables from AppendixA3. For (a), choose the C3axis along [1 1 1]

and select the three dihedral planes in Tdthat are vertical planes in C3v. For (b), choose one of the C3axes (for example, that along [1 1 1]) and identify the three C⁰2 axes normal to the C3axis.]

5.4 In the groups C4v, D3h, and D3dwhich E1, M1, and E2 transitions are allowed from a G1

ground state? In each of the three groups, identify the ground state in Mulliken notation. For the E1 transitions, state any polarization restrictions on the radiation.

5.5 Evaluate for the representations i¼ E, T1, and T2of the group O, the DP G^ii, the symmetric DP G^ii, and the antisymmetric DP G^ii. Show that your results satisfy the relation G^ii¼ G^ii G^ii.

Table 5.2. Possible transitions from¹A1gelectronic ground state in benzene.

Symmetry-allowed Symmetry-forbidden Spin-allowed ¹E_1u ¹B_1u,¹B_2u Spin-forbidden ³E_1u ³B_1u,³B_2u

Problems 105

6.1 Hybridization

In descriptions of chemical bonding, one distinguishes between bonds which do not have a nodal plane in the charge density along the bond and those which do have such a nodal plane. The former are called
bonds and they are formed from the overlap of s atomic orbitals on each of the two atoms involved in the bond (ss
bonds) or they are sp or pp
bonds, where here p implies a pzatomic orbital with its lobes directed along the axis of the bond, which is conventionally chosen to be the z axis. The overlap of px or py atomic orbitals on the two atoms gives rise to a p bond with zero charge density in a nodal plane which contains the bond axis. Since it is accumulation of charge density between two atoms that gives rise to the formation of a chemical bond,
or p molecular orbitals are referred to as bonding orbitals if there is no nodal plane normal to the bond axis, but if there is such a nodal plane they are antibonding orbitals. Carbon has the electron configuration 1s²2s²2p², and yet in methane the four CH bonds are equivalent. This tells us that the carbon 2s and 2p orbitals are combined in a linear combination that yields four equivalent bonds. The physical process involved in this ‘‘mixing’’ of s and p orbitals, which we represent as a linear combina-tion, is described as hybridization. A useful application of group theory is that it enables us to determine very easily which atomic orbitals are involved in hybridization. Sometimes there is more than one possibility, but even a rough knowledge of the atomic energy levels is usually all that is required to resolve the issue.

Example 6.1-1 This example describes
bonding in tetrahedral AB₄ molecules. The numb ering of the B atoms is shown in Figur e 6.1 . Den ote by
ra unit vector oriented from A along the bond between A and Br. Withh
1
_2
3
4j as a basis, determine the characters of the representation

. It is not necessary to determine the matrix representatives (MRs)

(T ) from Th
j ¼ h
j
(T ) since we only need the character system 
of the representa-tion

. Every
rthat transforms into itself under a symmetry operator T contributesþ1 to the character of that MR
(T ), while every
rthat transforms into
s, with s6¼ r, makes no contribution to 
(T ). Of course, we only need to determine 
(T ) for one member of each class in the point group. The values of 
(T ) for the point group Tdare given in Table6.1.

This is a reducible representation, and to reduce it we use the prescription c^j¼ g
¹P

T

_jðT Þ^ 
ðT Þ ¼ g
¹P

k

c_k _jðckÞ^ 
ðckÞ: (1)

106

Using the character table of Tdin AppendixA3we find for A1that cðA₁Þ ¼ g
¹½1ð1Þð4Þ þ 8ð1Þð1Þ þ 6ð1Þð2Þ ¼ 1:

We could proceed in a similar fashion for the remaining IRs, A2, E, T1, and T2, but instead we attempt a short-cut by subtracting the character system for A1from that of

:



^A1¼ f3
1 0
1 1g ¼ T₂: (2)

Note that in a character system it is implied that the characters of the classes are given in the same order as in the character table. Also, when a character system is equated to the symbol for a representation, as in eq. (2), it means that it is the character system of that representa-tion. Here then

¼ A₁ T₂: (3)

We know that s forms a basis for A1, and from the character table we see that (x, y, z) and also (xy, yz, zx) form bases for T2. Therefore,
bonds in tetrahedral AB4molecules are formed by sp³and/or sd³hybridization.

In general, an expression for a molecular orbital (MO) would involve linear combinations of s, and px, py, pzand dxy, dyz, dzxatomic orbitals (AOs), but some coefficients might be small or even negligibly small. There are two principles that control the formation of a chemical bond between two atoms: (i) the contributing AOs must be of comparable energy; and (ii) for a bonding MO, the bond should provide maximum overlap of charge density in the region between the atoms. In carbon the 3d orbitals lie about 10 eV above 2p and therefore

X

Y Z

B₃

B₁

B₄ B₂

Figure 6.1. Numbering of the B atoms in a tetrahedral AB₄molecule;
_ris a unit vector pointing from A to atom Br.

Table 6.1. The character system 
for the representation

in the point group Td.

T


4 0 1 0 2

6.1 Hybridization 107

sp³hybridization predominates. But in manganese and chromium the 3d are much closer to the 4s than are the 4p orbitals, and it is likely that sd³hybridization predominates.

Example 6.1-2 This example describes
bonding in the AB₅trigonal bipyramid (e.g. PF5).

As is evident from Figure6.2, the point group is D3h. The character system for

is given in Table6.2. From eq. (1), with the help of the character table for D3hin AppendixA3,

¼ 2A10 A200 E⁰: (4)

Exercise 6.1-1 Verify the reduction of

into the direct sum given in eq. (4).

From the character table for D3hwe find that z forms a basis for A200

while (x, y) form a basis for E⁰. Similarly, 3z²
r², as well as s, form bases for A10and (xy, x²
y²) form a basis for E⁰. The large difference in energy between (nþ 1)s and ns, or between (n þ 1)d_3z²
r²

and nd_3z²
_r², atomic energy levels makes the contribution of two orbitals with different principal quantum numbers to hybrid MOs in AB5very unlikely. We conclude that one s and one d_3z²
r²are involved, together with pz, and (pxpy), and/or (dxyd_x²
y²). In PF5, it is likely that (pxpy) predominate, giving dsp³hybridization, while in molecules in which the central atom has a high atomic number Z, the p and d orbitals will both contribute, giving a mixture of dsp³and d³sp hybridization. For example, in the MoCl5molecule, the molyb-denum 4d AOs are of comparable energy to the 5p orbitals, so that a hybrid scheme dsp³þ d³sp can be expected. It should be remarked that in the abbreviations used for hybridization schemes, specific d orbitals are implied; these may be found very easily by determining the character system for

and using the character table to determine the IRs and their basis functions. The same method may be used to determine the AOs used in

1 2

4

3

5

Figure 6.2. Numbering of the B atoms in the AB5trigonal bipyramid.

Table 6.2. The character system 
for the AB5molecule shown in Figure6.2.

D


5 2 1 3 0 3

p bonding, but we shall not give an example here since p bonding in the ML6octahedral complex will be analyzed later (in Section6.4).