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The symmetry elements of objects Some objects are ‘more symmetrical’ than others. A sphere is more symmetrical than a because it looks the same after it has been rotated through any angle about any diameter. A cube looks the same only if it is rotated through certain angles about specific angles (90°, 180°, or 270° about an axis passing through the centers of any of its opposite faces or by 120° or 240° about an axis passing through any of its opposite corners. An NH3 is more symmetrical than H2O because NH3 looks the same after rotations of 120° or 240°, whereas H2O looks the same only after a rotation of 180°. An action that leaves an object looking the same after it has been carried out is called a symmetry operation. Typical symmetry operations; rotations, reflections, and inversions. Symmetry element for each symmetry operation – the point, line, or plane with respect to which the symmetry operation is performed. A rotation (a symmetry operation) is carried out around an axis (the corresponding symmetry element). We shall see that we can classify by identifying all their symmetry elements, and grouping together molecules that possess the same set of symmetry elements. 1 Operations and symmetry elements The classification of objects according to symmetry elements that leave at least one common point unchanged gives rise to the point groups. There are five kinds of symmetry operation (and five kinds of symmetry element) of this kind. The identity, E – doing nothing (or rotation by 360°around any axis); the corresponding symmetry element – the entire object. Because every molecule is indistinguishable from itself if nothing is done to it, every object possesses at least the identity element. An n-fold rotation (the operation) about an n-fold axis of symmetry, Cn – a rotation through 360°/n. The operation C1 is equivalent to the identity operator E. H2O has one twofold axis, C2. NH3 has one threefold axis, C3, and there are two symmetry operations associated with it: clockwise and anticlockwise 120° rotations. A pentagon has a C5 axis, a cube has three C4 axes, four C3 axes, and six C5 axes. A sphere possesses an infinite number of symmetry axes (along any diameter) of all possible integral values of n. If a molecule possesses several rotational axes, then the one (or more) with the greatest value of n is called the principal axis. The principal axis of a molecule is C6 to the hexagonal ring. A reflection (the operation) in a mirror plane, σ, (the element) may contain the principal axis or be perpendicular to it. If the plane is parallel to the principal axis, it is called ‘vertical’, σv. H2O has two σv planes, NH3 has three.

2 A vertical mirror plane that bisects the angle between two C2 axis is called a ‘dihedral plane’, σd. When the plane is perpendicular to the principal axis it is called ‘horizontal’, σh. A C6H6 molecule has a C6 principal axis and a horizontal mirror plane. In an inversion (the operation) through a center of symmetry, i (the element), we imagine taking each point in a molecule, moving it to the center of the molecule, and then moving it out the same distance on the other side: the point (x,y,z) is taken to the point (-x, -y,-z). H2O and NH3 do not have a center of inversion, but a sphere or a cube do. C6H6 or a regular do have a center of inversion; a regular (CH4) does not. An n-fold improper rotation (the operation) about an n-fold improper rotation axis, Sn, (the symmetry element) – two successive transformations: 1) a rotation through 360°/n; 2) a reflection through a plane perpendicular to the axis of this rotation; neither of these operations alone needs to be a symmetry operation. CH4 has three S4 axis and the staggered form of a S composed of 60 rotation followed 3 6 ° by a reflection. The symmetry classification of molecules A molecule belongs to the C1 if it has no element other than the identity.

Ci – a molecule has the identity and the inversion alone (structure 3). Cs – a molecule has the identity and a mirror plane alone (4). Cn – a molecule possesses an n-fold axis. An H2O2 molecule (5) has the elements E and C2, so it belongs to the group C2. If, in addition to the identity and a Cn axis, a molecule has n vertical mirror planes σv, then it belongs to the group Cnv. H2O: the elements are E, C2, and 2σv, so it belongs to the group C2v. NH3: the elements are E, C3, and 3σv, so it belongs to the group C 3v. A heteronuclear (HCl) belongs to the group C∞v because all rotations around the axis and reflections across the axis are symmetry operations. Other members of the group

C∞v include the linear OCS molecule and a cone.

4 Objects that, in addition to the identity and an n- fold principal axis, also have a horizontal mirror plane σ h belong to the groups Cnh: trans- CHCl=CHCl (6) – C2h, B(OH)3 (7) – C3h. In C2h the operations C2 and σh jointly imply the presence of a center of inversion. A molecule that has an n-fold principal axis and n twofold axes perpendicular to Cn belongs to the group Dn. A molecule belongs to D nh if it also possesses a horizontal mirror plane. The planar trigonal BF3 molecule (8) has the elements E, C3, 3C2, and σ h (with one C 2 axis along each B-F bonds) – D3h. C6H6: elements are E, C6, 3C2, 3C2’, and σh – D6h. All homonuclear diatomic molecules

(N2) belong to the group D∞h because all rotations around the axis are symmetry operations, as are end-to-end rotation and end-to-end reflection. 5 D∞h is also a group of the liner OCO and HCCH molecules and of a uniform cylinder. Other - examples of Dnh molecules are C2H4 (9) – D2h, PCl5 (10) – D3h, [AuCl4] (11) – D4h. A molecule belongs to the group Dnd if in addition to elements of Dn it possesses n dihedral mirror planes σd. Examples: the twisted, 90° allene (12) – D2d; the staggered conformation of ethane (13) – D3d.

Molecules which possess one S n axis belong to the group S n: tetraphenylmethane (14) – S4. Molecules belonging to Sn with n > 4 are rare. The group S2 is the same as Ci. A number of very important molecules (for example, CH4 and SF6) possess more than one principal axis. Most belong to the cubic groups, and in particular to the tetrahedral groups T, Td and T h or to the octahedral groups O and Oh. A few icosahedral (20-faced) molecules belonging to icosahedral group, I, are also known: they include some of the boranes 2- (B12H12 ) and C60 (15).

6 The groups Td and Oh are the groups of the regular tetrahedron (CH4) and the regular octahedron (SF6), respectively. If the object possesses the rotational symmetry of the tetrahedron or the octahedron, but none of their planes of reflection, then it belongs to the simpler groups T or O. The group Th is based on T but also contains a center of inversion.

7 We can use the flow chart to determine for any molecule. For example, 16 has D 5h symmetry and 17 belongs to the D5d group.

The full rotation group, R3 (the 3 refers to rotation in three dimensions), consists of an infinite number of rotation axes with all possible values of n. A sphere and an atom belong to R3, but no molecule does. One can use the R3 group to apply symmetry arguments to atoms, and this is an alternative approach to the theory of orbital .

8 A summary of the shapes corresponding to different point groups

9 Some immediate consequences of symmetry

Polarity A polar molecule has a permanent moment. If the molecule has Cn symmetry (n > 1), it cannot possess a charge distribution with a dipole moment perpendicular to the symmetry axis because the symmetry of the molecule implies that any dipole moment that exists in one direction perpendicular to the axis is cancelled by an opposing dipole. Any dipole moment in such molecules must be parallel to the rotational axis. Example: H2O has a dipole moment parallel to its twofold axis. The same applies generally to the group Cnv, so molecules belonging to any of the Cnv groups may be polar. In all other groups, such as C3h, D, etc., there are symmetry operations that take one end of the molecule into the other. Therefore, such molecules may not have dipole moment both perpendicular and along the symmetry axis. Only molecules belonging to the groups Cn, Cnv, and Cs may have a permanent dipole moment. For Cn and Cnv, that dipole moment must lie along the symmetry axis. Examples: (O3) is angular and belongs to the group C2v and therefore may be polar (and is), but CO2 is linear and belongs to the group

D∞h and hence is not polar.

10 A chiral molecule (from the Greek word ‘hand’) is a molecule that cannot be superimposed on its mirror image. An achiral molecule can be superimposed on its mirror image. Chiral molecules are optically active in the sense that they rotate the plane of polarized light. A chiral molecule and its mirror image partner constitute an enantiometric pair of and rotate the plane of polarization in equal but opposite directions. A molecule may be chiral, and therefore optically active, only if it does not possess an axis of improper rotation Sn. Molecules belonging to the groups Cnh possess a Sn axis implicitly because they possess both Cn and σh, which are the two components of an improper rotation axis. Any molecule containing a center of inversion possesses a S2 axis, because i is equivalent to C2 in conjunction with σh, and that combination of elements is S2. Hence all molecules with centers of inversion are achiral and optically inactive. Similarly, because S1 = σ, any molecule with a mirror plane is achiral. A molecule may be chiral if it does not have a center of inversion or a mirror plane, which is the case for alanine (18), but not for (19). However, a molecule may be achiral even though it does not have a center of inversion. Example:

the S4 species (20) is achiral andoptically inactive: though it lacks i (S2) it does have and S4 axis. 11 Character tables. Character tables and symmetry labels We saw that molecular orbitals of diatomic and linear polyatomic molecules are labeled σ, π, etc. These labels refer to the of the orbitals with respect to rotation around the principal axis of the molecule: a σ orbital does not change sign under a rotation through any angle, a π orbital changes sign when rotated by 180°.The symmetry classifications σ and π can also be assigned to individual atomic orbitals in a linear molecule. Representations and characters Labels analogous to σ and π (a, a1, e, eg) are used to denote the symmetries of orbitals in polyatomic molecules. These labels indicate the behavior of the orbitals under the symmetry operations of the relevant of the molecule. A label is assigned to an orbital based upon the of the group – a table that characterizes the different symmetry types possible in the point group: C2 (i.e. rotation by 180°) σ +1 (i.e. no change of sign) π -1 (i.e. change of sign) This table is a fragment of the full character table for a linear molecule. The entry +1 shows that the orbital remains the same and the entry –1 show that the orbital changes sign under the operation C2 at the head of the column. To assign the label σ and π to a particular orbital, we compare the orbital’s behavior with the information in the character table. The entries in a complete character table are derived by using the formal techniques of and are called characters, χ. 12 The C2v character table C2v E C2 σv σv’ h = 4 2 2 2 A1 1 1 1 1 z z , y , x A2 1 1 -1 -1 xy B1 1 -1 1 -1 x xz B2 1 -1 -1 1 y yz The top row: symmetry operations of the group. The left column: irreducible representations (symmetry species) of the group. All orbitals are subdivided into irreducible representations according to their behavior with respect to the symmetry operations of the group. The orbitals, which exhibit the same behavior belong to the same irreducible representation. An A or B is used to denote a one-dimensional (non-degenerate) representation; A is used if the character under the principal rotation is +1, and B is used if the character is –1. Subscripts are used to distinguish irreducible representations if there is more than one of the same type: A1 is reserved for the fully symmetric representation with character 1 for all operations. When higher dimensional (degenerate) irreducible representations are permitted, E denotes a two-dimensional (doubly degenerate) irreducible representation and T a three-dimensional (triply degenerate) irreducible representation; all irreducible representations of C2v are one-dimensional. Symmetry operations fall into the same class if they are of the same type (for example, rotations) and can be transformed into one another by a symmetry operation of the group. There are four classes in C2v (four columns in the character table).

13 Number of symmetry species = number of classes. h specifies the total number of symmetry operations in the group and is called the order of the group. The right column shows the types of orbitals (for example, pz or dxy on the central atom) which belong to the corresponding irreducible representation of the group. The columns in a character are labeled with the symmetry operations of the group. For instance for the group C3v the columns are headed E, C3, and σv. The numbers multiplying each operation are the number of members of each class. In C3v, the two threefold rotations (clockwise and counter clockwise 120°) belong to the same class: they are related by reflection. The three reflections also lie in the same class: they are related by the threefold rotations. On the other hand, the two reflections of the group C2v fall into different classes: although they are both reflections, one cannot be transformed into the other by any symmetry operation of the group.

The C3v character table C3v E 2C3 3σv h = 6 2 2 2 A1 1 1 1 z z , x + y A2 1 1 -1 xy E 2 -1 0 (x, y) (xy, x2 - y2), (xz, yz)

14 The rows under the labels for the operations summarize the symmetry properties of the orbitals. They are labeled with symmetry species. The symmetry species correspond to the irreducible representations of the group, which are the basic types of behavior that orbitals may show when subjected to the symmetry operations of the group. By convention, irreducible representations are labeled with upper case Roman letters (A1 and E) but the orbitals to which they applied are labeled with the lower case italic equivalents (a1 and e). The character of the identity operator tells us the degeneracy of the orbitals. In a C3v molecule, any a1 or a2 orbital is non-degenerate. Any doubly degenerate pair of orbitals in C3v must be labeled e because only E symmetry species have character 2. So, with a glance at the character table of a molecule, we can state the maximum possible degeneracy of its orbitals. Example: BF3 belongs to D3h group. In D3h character table, the maximum number in the column headed by the identity E is 2, the maximum orbital degeneracy is 2. Characters and operations The characters in the rows labeled A and B and in the columns headed by symmetry operations indicate the behavior of an orbital under the corresponding operation: +1 means than an orbital is unchanged, -1 indicates that it changes sign. For the rows labeled E or T (which refer to the behavior of sets of doubly and triply degenerate orbitals, respectively), the characters in a row of the table are the sums of the characters summarizing the behavior of the individual orbitals. For example, if one member of a doubly degenerate pair remains unchanged under a symmetry operations but the other changes sign, the entry is reported as χ = 1 – 1 = 0.

15 Consider the O2px orbital in H2O. We need to refer to C2v character table: the labels available for the orbitals are a1, a2, b1, and b2. Under a 180° rotation (C2) the orbital changes sign, so it must be either B1 or B2. The O2px orbital also changes sign under the reflection σv’, which identifies it as B1. We shall see that any built from this will also be a b1 orbital. O2py changes sign under C2 and σv; therefore, it can contribute to b2 orbitals. The behavior of s, p, and d orbitals on a central atom under the symmetry operations of the molecule is so important that the sym-metry species of these orbitals are included in a character table.

The classification of linear combinations of orbitals The same technique may be applied to linear combina-tions of orbitals on

atoms. Consider the combination ψ1 = ψ A + ψB + ψC of the three H1s orbitals in NH3 (C3v): this combination remains uncganged under all symmetry operations: χ(E) = 1, χ(C3) = 1, χ(σv) = 1. Comparison with the C3v character table shows that ψ is of symmetry species A1 and therefore it contributes to a1 MOs of 1 € NH3.

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16 Example. Identifying the symmetry species of orbitals

Identify the symmetry species of the orbital ψ = ψA − ψB in a C2v NO2

molecule, where ψA is an O2px orbital on one O atom and ψB is that on the other O atom. Under C2, ψ changes into itself, implying a character +1. Under reflection σv, both orbitals change sign, so ψ → -ψ, implying a € character of –1. Under σv’, ψ → -ψ, so the character of for this operation is € € also –1: χ(E) = 1 χ(C2) = 1 χ(σv) = 1 χ(σv’) = 1 These characters match the characters of the A2 symmetry species, so ψ can contribute to an a2 orbital.

Vanishing integrals and orbital overlap We need to evaluate the integral where f , f are functions (for I = ∫ f1 f2dτ 1 2 example, AOs centered at different atoms). The value of any integral (and of an overlap integral) is independent of the orientation of the molecule: I is invariant under any symmetry operation of the molecule. The volume element dτ is invariant under any symmetry€ operation, and so the integral is nonzero only if the integrand itself, the product f1f2, is unchanged. If the integrand changed sign under a symmetry operation, the integral would be the sum of equal and positive contributions, and hence would be zero.

17 The only contribution to a nonzero integral comes from functions for which under any symmetry operation of the molecular point group f1f2 → f1f2, and hence for which the characters of the operations are all equal to +1. Therefore, for I not to be zero, the integrand f1f2 must have symmetry species A1 (or its equivalent in the specific molecular group). We use the following procedure to deduce the symmetry species spanned by the product f1f2 and to see whether it does indeed span A1: 1) Decide on the symmetry species of the individual functions f1 and f2 by reference to the character table, and write their characters in two rows in the same order as in the table. 2) Multiply the numbers in each column, writing the result in the same order. 3) Inspect the rows so produced, and see if it can be expressed as a sum of characters from each column of the group. The integral must be zero if this sum does not contain A1. Example: f1 – the sN orbital in NH3, f2 – the linear combination s3 = sB – sC Because sN spans A1 and s3 belongs to E, we write f1: 1 1 1 f2: 2 -1 0 f1f2: 2 -1 0 Therefore, f1f2 spans E, so the integrand does not span A1 and the integral must be zero. Now, let’s take f1 = sN and f2 = s1, where s1 = sA + sB + sC:

f1: 1 1 1 f2: 1 1 1 because each of the orbitals involved f1f2: 1 1 1 span A1 18 The characters of the product are those of A1 itself. Therefore, sN and s1 may have a nonzero overlap. In general, if f1 and f2 belong to the same symmetry species, their overlap may be nonzero. Group theory is specific about when an integral must be zero, but integrals that it allows to be nonzero may be zero for reasons unrelated to symmetry. For example, if the N-H distance in is very large, the (s1,sN) overlap integral may be zero simply because the orbitals are far apart. In many cases, the product of functions f1 and f2 spans a sum of irreducible representations. For example, in C2v we may find the characters 2, 0, 0, -2 when we multiply the characters of f1 and f2 together. In this case, we note that these characters are the sum of the characters for A2 and B1: E C2 σv σv’ A2 1 1 -1 -1 B1 1 -1 1 -1 A2 + B1 2 0 0 -2

To summarize this result, we write the symbolic expression: f1 × f2 = A2 + B1, which is called the decomposition of a direct product.

19 The decomposition is not always so obvious as in this case, so a formal procedure exists how to find it: 1) Write down a table with columns headed by the symmetry operations of the group. 2) In the first row write down the characters of the symmetry species we want to analyze. 3) In the second row, write down the characters of the irreducible representation we are interested in. 4) Multiply the two rows together, add the products together, and divide by the order of the group. The resulting number is the number of times this irreducible representation occurs in the decomposition. Let’s find the decomposition of the product f1f2 with characters 8, -2, -6, 4 in C2v. We start from the irreducible representation A1: E C2 σv σv’ h = 4 (the order of the group) f1f2 8 -2 -6 4 (the characters of the product) A1 1 1 1 1 (the symmetry species) 8 -2 -6 4 (the product of the two sets of characters)

The sum of the numbers in the last line is 8 + (-2) + (-6) + 4 = 4, when that number is divided by the order of the group, we get 1, so A1 occurs once in the decomposition.

20 Now we can repeat the same procedure for the other symmetry species. E C2 σv σv’ h = 4 (the order of the group) f1f2 8 -2 -6 4 (the characters of the product) A2 1 1 -1 -1 (the symmetry species) 8 -2 6 -4 The sum of the numbers in the last line is 8 + (-2) + 6 + (-4) = 8, 8/4 = 2, so A2 occurs twice in the decomposition. E C2 σv σv’ h = 4 (the order of the group) f1f2 8 -2 -6 4 (the characters of the product) B1 1 -1 1 -1 (the symmetry species) 8 2 -6 -4 The sum of the numbers in the last line is 8 + 2 + (-6) + (-4) = 0, so B1 does not occur in the decomposition. E C2 σv σv’ h = 4 (the order of the group) f1f2 8 -2 -6 4 (the characters of the product) B2 1 -1 -1 1 (the symmetry species) 8 2 6 4 The sum of the numbers in the last line is 8 + 2 + 6 + 4 = 20, 20/4 = 5, so B1 occurs in the decomposition five times. Thus, we find that f1f2 spans A1 + 2A2 + 5B2.

21 Orbitals with nonzero overlap The general rule is: only orbitals of the same symmetry species may have nonzero overlap, so only orbitals of the same symmetry species form bonding and antibonding combinations. Remember that the selection of atomic orbitals that have mutual nonzero overlap is the central and initial step in the construction of the molecular orbitals by the LCAO procedure. The molecular orbitals formed from a particular set of atomic orbitals with nonzero overlap are labeled with the lower- case letter corresponding to the symmetry species. For instance, the (sN,s1) orbitals are called a1 orbitals (or a1* if we wish to emphasize that they are antibonding). The s2 and s3 linear combinations have symmetry species E. Intuition suggests that N2px and N2py should suitable to have nonzero overlap with them.We can confirm this conclusion by noting that the charactertable shows that in C3v, the functions x and y jointly belong to the symmetry species E. Therefore, N2px and N2py also belong to E, so they may have nonzero overlap with s2 and s3. Indeed, let’s consider decomposi-tion of E×E. First, we need to find characters of the product: C3v E 2C3 3σv E 2 -1 0 E 2 -1 0 E×E 4 1 0 Now we need to carry out decomposition of the characters 4, 1, 0.

22 C3v E 2C3 3σv h = 6 E×E 4 1 0 A1 1 1 1 4 1 0 Summation of the last row gives 4 + 2×1 + 3×0 = 6 and 6/(h=6) = 1 (The second and third terms are multiplied by 2 and 3, respectively, because C3v group has 2 and 3 symmetry elements C3 and σv, respectively). Thus, one A1 symmetry species is present in the decomposition of E×E. C3v E 2C3 3σv h = 6 E×E 4 1 0 A2 1 1 -1 4 1 0 Summation: 4 + 2×1 + 3×0 = 6 6/6 = 1 One A2 symmetry species is present in the decomposition of E×E. C3v E 2C3 3σv h = 6 E×E 4 1 0 E 2 -1 0 8 -1 0 Summation: 8 + 2×(-1) + 3×0 = 6 6/6 = 1 One E symmetry species is present in the decomposition of E×E and we obtain E×E = A1 + A2 + E

23 Since A1 is present in the decomposition of E×E, we can conclude that two atomic orbitals belonging to symmetry species E may have nonzero overlap and therefore to form bonding and antibonding MOs (e). We can also explore whether any d orbitals on the central atom can take part in bonding:

d 2 has A1 symmetry and each of the pairs d 2 2 ,d and d ,d has E symmetry. z ( x −y xy ) ( xz yz )

Therefore, molecular orbitals may be formed by s ,d 2 overlap and by overlap of the s , s ( 1 z ) 2 3 combinations with the E d orbitals. Whether or not the d orbitals are in fact important, group € theory cannot answer – the extent€ of their involvement€ depends on energy consideration, not symmetry. Example. Determining which orbitals€ can contribute to bonding. The four H1s orbitals of span A1 + T2. With which of the C atom orbitals can they overlap? What bonding pattern would be possible if the C atom had d orbitals available? Td character table: Td E 8C3 3C2 6σd 6S4 h = 24 2 2 2 A1 1 1 1 1 1 x + y + z A2 1 1 1 -1 -1 E 2 -1 2 0 0 (3z2 – r2, x2 – y2) T1 3 0 -1 -1 1 T2 3 0 -1 1 -1 (x, y, z), (xy, xz, yz)

24 An s orbital spans A1, so it may have nonzero overlap with the A1 combination of H1s orbitals. The C2p orbitals span T2, so they may have nonzero overlap with the T2 combination. The dxy, dxz, and dyz orbitals span T2, so they may overlap with the same combi- nation. Neither of the other two d orbitals span A1 or T2 (they span E), so they remain nonbonding orbitals. It follows that in methane there (C2s,H1s)-overlap a1 orbitals and (C2p,H1s)-overlap t2 orbitals. The C3d orbitals might contribute to the latter. The lowest 2 6 energy configuration is a1 t2 with all bonding orbitals occupied.

Vanishing integrals and selection rules Integrals of form are also common in quantum mechanics because they ∫ f1 f2 f3dτ include elements of operators. For the integral to be nonzero, the product f1 f2 f3 must span A1 (or its equivalent) or contain a component that spans A1. To test whether this is so, the characters of all three functions are multiplied together and the resulting characters are compared with€ the characters of the symmetry species in the group. €

Example. Deciding if an integral must be zero. Does the integral 3d 2 x 3d dτ vanish in ∫ ( z ) ( xy ) a C2v molecule?

€

25 The C2v character table C2v E C2 σv σv’ h = 4 2 2 2 A1 1 1 1 1 z z , y , x A2 1 1 -1 -1 xy B1 1 -1 1 -1 x xz B2 1 -1 -1 1 y yz We find that spans A , spans B , and spans A and draw up the f1 = 3dz 2 1 f2 = x 1 f3 = 3dxy 2 following table: E C2 σv σv’ f = 3d 1 1 -1 -1 A 3 xy€ € € 2 f2 = x 1 -1 1 -1 B1

f1 = 3dz 2 1 1 1 1 A1 f f f 1 -1 -1 1 B € 1 2 3 2 The f f f product span B . Therefore, the integral is necessarily zero. € 1 2 3 2 We’ll see in the Chapters on that the intensity of a spectral line arising € from a molecular transition between some initial state i with wavefunction ψi and a final state € f with wave-function ψf depends on the (electric) transition dipole moment, µfi. The z- € * component of this vector is defined as where -e is the charge of the electron. −e ∫ ψ f zψi dτ The transition moment has the form of the integral ; so, once we know the ∫ f1 f2 f3dτ symmetry species of the states, we can use group theory to formulate the selection rules for the transitions. € 26 € Example: investigate whether an electron in an a1 orbital in H2O (C2v symmetry) can make an electric dipole transition to b1. We must examine all three components of the transition dipole moment, and take f2 as x, y, and z. Also, f1 is A1 and f3 is B1. x-component y-component z-component E C2 σv σv’ E C2 σv σv’ E C2 σv σv’ f3 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 B1 f2 1 -1 1 -1 1 -1 -1 1 1 1 1 1 f1 1 1 1 1 1 1 1 1 1 1 1 1 A1 Pr. 1 1 1 1 1 1 -1 -1 1 -1 1 -1

Only the first product (with f2 = x) spans A1, so only the x-component of the transition dipole moment may be nonzero. Therefore the electric dipole transitions between a1 and b1 are allowed. The radiation emitted (or absorbed) is x-polarized and has its electric field vector in the x direction.

27 Example. Deducing a selection rule. Is px→py transition allowed in a tetrahedral environment? We must decide whether the product pyqpx (q = x, y, z) spans A1 by using the Td character table.

Td E 8C3 3C2 6σd 6S4 h = 24 f3 (px) 3 0 -1 -1 1 T2 f2 (q) 3 0 -1 -1 1 T2 f1 (py) 3 0 -1 -1 1 T2 f1 f2 f3 27 0 -1 -1 1

Now we check whether the decomposition of f1 f2 f3 contains A1: € Td E 8C3 3C2 6σd 6S4 h = 24 f f f 27 0 -1 -1 1 1 2 3 € A1 1 1 1 1 1 27 0 -1 -1 1

€ Summation: 27 + 8×0 + 3×(-1) + 6×(-1) + 6×1 = 24 24/24 = 1 A1 occurs once in the decomposition, so the transition is allowed.

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