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Molecular Symmetry

The symmetry elements of objectsSome objects are ‘more symmetrical’ than others. A sphere is moresymmetrical than a cube because it looks the same after it has beenrotated through any angle about any diameter. A cube looks thesame only if it is rotated through certain angles about specific angles(90°, 180°, or 270° about an axis passing through the centers of anyof its opposite faces or by 120° or 240° about an axis passingthrough any of its opposite corners.

An NH3 molecule is more symmetrical than H2O because NH3 looks thesame after rotations of 120° or 240°, whereas H2O looks the same only aftera rotation of 180°.An action that leaves an object looking the same after it has been carried outis called a symmetry operation. Typical symmetry operations; rotations,reflections, and inversions. Symmetry element for each symmetryoperation – the point, line, or plane with respect to which the symmetryoperation is performed. A rotation (a symmetry operation) is carried outaround an axis (the corresponding symmetry element). We shall see that wecan classify molecules by identifying all their symmetry elements, and

grouping together molecules that possess the same set of symmetryelements.

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Operations and symmetry elementsThe 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 ofsymmetry operation (and five kinds of symmetry element) of this kind.

The identity, E – doing nothing (or rotation by 360°around any axis); thecorresponding symmetry element – the entire object. Because every molecule isindistinguishable from itself if nothing is done to it, every object possesses at leastthe 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 twosymmetry 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 sixC5 axes. A sphere possesses an infinite number of symmetry axes (along anydiameter) of all possible integral values of n. If a molecule possesses severalrotational axes, then the one (or more) with the greatest value of n is called theprincipal axis. The principal axis of a benzene molecule is C6 perpendicular to thehexagonal 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.

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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 thecenter of the molecule, and then moving it out the same distance on theother side: the point (x,y,z) is taken to the point (-x, -y,-z). H2O and NH3 donot have a center of inversion, but a sphere or a cube do. C6H6 or a regularoctahedron do have a center of inversion; a regular tetrahedron (CH4) doesnot.

A n n-fold improperrotation (the operation) about ann-fold improper rotation axis,Sn, (the symmetry element) – twosuccessive transformations: 1) arotation through 360°/n; 2) areflection through a planeperpendicular to the axis of thisrotation; neither of theseoperations alone needs to be a

symmetry operation. CH4 has three S4

axis and the staggered form of ethane aS6 composed of 60° rotation followedby a reflection.

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The symmetry classification of moleculesA molecule belongs to the group 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 thegroup C2v. NH3: the elements are E , C3, and 3σv, so it belongs to the group C 3v. Aheteronuclear diatomic molecule (HCl) belongs to the group C∞v because all rotations aroundthe axis and reflections across the axis are symmetry operations. Other members of the groupC∞v include the linear OCS molecule and a cone.

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Objects that, in addition to the identity and an n-fold principal axis, also have a horizontal mirrorplane 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 presenceof a center of inversion.

A molecule that has an n-fold principal axis andn twofold axes perpendicular to Cn belongs to thegroup Dn. A molecule belongs to Dnh if it alsopossesses a horizontal mirror plane. The planartrigonal BF3 molecule (8) has the elements E, C3,3C2, and σ h (with one C 2 axis along each B-Fbonds) – 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 areend-to-end rotation and end-to-end reflection.

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D∞h is also a group of the liner OCO and HCCH molecules and of a uniform cylinder. Otherexamples 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 staggeredconformation of ethane (13) – D3d.

Molecules which possess one Sn axis belong to the group S n:tetraphenylmethane (14) – S4. Molecules belonging to Sn with n > 4 arerare. 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 theoctahedral groups O and Oh. A few icosahedral (20-faced) moleculesbelonging to icosahedral group, I, are also known: they include some of the boranes(B12H12

2-) and buckminsterfullerene C60 (15).

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

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We can use the flow chart to determine symmetrygroup 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 inthree dimensions), consists of an infinite number ofrotation axes with all possible values of n. A sphereand an atom belong to R3, but no molecule does. Onecan use the R3 group to apply symmetry arguments toatoms, and this is an alternative approach to the theoryof orbital angular momentum.

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A summary of the shapescorresponding to differentpoint groups

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Some immediate consequences of symmetry

PolarityA polar molecule has a permanent dipole moment. If the molecule has

Cn symmetry (n > 1), it cannot possess a charge distribution with a dipolemoment perpendicular to the symmetry axis because the symmetry of themolecule implies that any dipole moment that exists in one directionperpendicular to the axis is cancelled by an opposing dipole. Any dipolemoment in such molecules must be parallel to the rotational axis. Example:H2O has a dipole moment parallel to its twofold axis. The same appliesgenerally 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 takeone end of the molecule into the other. Therefore, such molecules may not have dipolemoment both perpendicular and along the symmetry axis. Only molecules belonging to thegroups Cn, Cnv, and Cs may have a permanent dipole moment. For Cn and Cnv, that dipolemoment must lie along the symmetry axis. Examples: ozone (O3) is angular and belongs tothe group C2v and therefore may be polar (and is), but CO2 is linear and belongs to the groupD∞h and hence is not polar.

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ChiralityA 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 mirrorimage. 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 anenantiometric pair of isomers and rotate the plane of polarization in equal butopposite 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 improperrotation axis. Any molecule containing a center of inversion possesses a S2

axis, because i is equivalent to C2 in conjunction with σh, and that combinationof elements is S2. Hence all molecules with centers of inversion are achiral andoptically inactive. Similarly, because S1 = σ, any moleculewith a mirror plane is achiral. A molecule may be chiral ifit does not have a center of inversion or a mirror plane,which is the case for amino acid alanine (18), but not forglycine (19). However, a molecule may be achiral eventhough 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.

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Character tables. Character tables and symmetry labelsWe saw that molecular orbitals of diatomic and linear polyatomic

molecules are labeled σ, π, etc. These labels refer to the symmetries ofthe orbitals with respect to rotation around the principal axis of themolecule: a σ orbital does not change sign under a rotation through anyangle, a π orbital changes sign when rotated by 180°.The symmetryclassifications σ and π can also be assigned to individual atomic orbitalsin a linear molecule.

Representations and charactersLabels 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 symmetryoperations of the relevant point group of the molecule. A label is assigned to an orbital basedupon the character table of the group – a table that characterizes the different symmetrytypes 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 showsthat the orbital remains the same and the entry –1 show that the orbital changes sign underthe 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 ina complete character table are derived by using the formal techniques of group theory andare called characters, χ.

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The C2v character tableC2v E C2 σv σv’ h = 4A1 1 1 1 1 z z2, y2, x2

A2 1 1 -1 -1 xyB1 1 -1 1 -1 x xzB2 1 -1 -1 1 y yzThe top row: symmetry operations of the group. The left column: irreduciblerepresentations (symmetry species) of the group. All orbitals are subdivided intoirreducible representations according to their behavior with respect to the symmetryoperations of the group. The orbitals, which exhibit the same behavior belong to the sameirreducible 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 ifthe character is –1. Subscripts are used to distinguish irreducible representations if there ismore than one of the same type: A1 is reserved for the fully symmetric representation withcharacter 1 for all operations. When higher dimensional (degenerate) irreduciblerepresentations are permitted, E denotes a two-dimensional (doubly degenerate) irreduciblerepresentation and T a three-dimensional (triply degenerate) irreducible representation; allirreducible 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. Thereare four classes in C2v (four columns in the character table).

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Number of symmetry species = number of classes.h specifies the total number of symmetry operations in the group and iscalled 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 thecorresponding irreducible representation of the group. The columns in a

character are labeled with the symmetry operations of the group. For instance forthe group C3v the columns are headed E, C3, and σv. The numbers multiplying eachoperation are the number of members of each class. In C3v, the two threefoldrotations (clockwise and counter clockwise 120°) belong to the same class: theyare related by reflection.The three reflections also lie in the same class: they are related by the threefoldrotations. On the other hand, the two reflections of the group C2v fall into differentclasses: although they are both reflections, one cannot be transformed into theother by any symmetry operation of the group.

The C3v character tableC3v E 2C3 3σv h = 6A1 1 1 1 z z2, x2 + y2

A2 1 1 -1 xyE 2 -1 0 (x, y) (xy, x2 - y2), (xz, yz)

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The rows under the labels for the operations summarize the symmetry properties of theorbitals. They are labeled with symmetry species. The symmetry species correspond to theirreducible representations of the group, which are the basic types of behavior that orbitalsmay 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 theorbitals 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 inC3v must be labeled e because only E symmetry species have character 2. So, with a glance atthe character table of a molecule, we can state the maximum possible degeneracy of itsorbitals.

Example: BF3 belongs to D3h group. In D3h character table, the maximum number in thecolumn 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 symmetryoperations indicate the behavior of an orbital under the corresponding operation: +1 meansthan 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), thecharacters in a row of the table are the sums of the characters summarizing the behavior ofthe individual orbitals. For example, if one member of a doubly degenerate pair remainsunchanged under a symmetry operations but the other changes sign, the entry is reported asχ = 1 – 1 = 0.

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Consider the O2px orbital in H2O. We need to refer to C2v character table: thelabels 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 alsochanges sign under the reflection σv’, which identifies it as B1. We shall seethat any molecular orbital built from this atomic orbital 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 thesymmetry operations of the molecule is so important that the sym-metryspecies of these orbitals are included in a character table.

The classification of linear combinations of orbitalsThe same technique may be applied to linear combina-tions of orbitals onatoms. Consider the combination

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ψ1 = ψA +ψB +ψC of the three H1s orbitals inNH3 (C3v): this combination remains uncganged under all symmetry operations:χ(E) = 1, χ(C3) = 1, χ(σv) = 1. Comparison with the C3v character table showsthat

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ψ1 is of symmetry species A1 and therefore it contributes to a1 MOs ofNH3.

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Example. Identifying the symmetry species of orbitalsIdentify the symmetry species of the orbital

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ψ = ψA −ψB in a C2v NO2

molecule, where

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ψA is an O2px orbital on one O atom and

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ψB is that on theother O atom. Under C2, ψ changes into itself, implying a character +1.Under reflection σv, both orbitals change sign, so ψ → -ψ, implying acharacter of –1. Under σv’, ψ → -ψ, so the character of for this operation isalso –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 cancontribute to an a2 orbital.

Vanishing integrals and orbital overlapWe need to evaluate the integral

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I = f1 f2∫ dτ where f1, f2 are functions (for

example, AOs centered at different atoms). The value of any integral (and of anoverlap integral) is independent of the orientation of the molecule: I is invariantunder any symmetry operation of the molecule. The volume element dτ isinvariant under any symmetry operation, and so the integral is nonzero only ifthe integrand itself, the product f1f2, is unchanged. If the integrand changed

sign under a symmetry operation, the integral would be the sum of equaland positive contributions, and hence would be zero.

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The only contribution to a nonzero integral comes from functions for which under anysymmetry operation of the molecular point group f1f2 → f1f2, and hence for which thecharacters of the operations are all equal to +1. Therefore, for I not to be zero, the integrandf1f2 must have symmetry species A1 (or its equivalent in the specific molecular group). We usethe following procedure to deduce the symmetry species spanned by the product f1f2 and tosee whether it does indeed span A1:

1) Decide on the symmetry species of the individual functions f1 and f2 by reference to thecharacter 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 writef1: 1 1 1f2: 2 -1 0f1f2: 2 -1 0

Therefore, f1f2 spans E, so the integrand does not span A1 and the integralmust be zero.

Now, let’s take f1 = sN and f2 = s1, where s1 = sA + sB + sC:

f1: 1 1 1f2: 1 1 1 because each of the orbitals involvedf1f2: 1 1 1 span A1

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The characters of the product are those of A1 itself. Therefore, sN and s1 may have anonzero overlap. In general, if f1 and f2 belong to the same symmetry species, their overlapmay be nonzero.

Group theory is specific about when an integral must be zero, but integrals that it allowsto be nonzero may be zero for reasons unrelated to symmetry. For example, if the N-Hdistance in ammonia is very large, the (s1,sN) overlap integral may be zero simply because theorbitals are far apart.

In many cases, the product of functions f1 and f2 spans a sum of irreduciblerepresentations. For example, in C2v we may find the characters 2, 0, 0, -2 when we multiplythe characters of f1 and f2 together. In this case, we note that these characters are the sum ofthe characters for A2 and B1:

E C2 σv σv’A2 1 1 -1 -1B1 1 -1 1 -1A2 + 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.

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The decomposition is not always so obvious as in this case, so a formal procedure existshow 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 thedecomposition.

Let’s find the decomposition of the product f1f2 with characters 8, -2, -6, 4 in C2v. Westart 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 ofcharacters)

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

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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 -4The sum of the numbers in the last line is 8 + (-2) + 6 + (-4) = 8, 8/4 = 2, so A2 occurs twicein 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 -4The sum of the numbers in the last line is 8 + 2 + (-6) + (-4) = 0, so B1 does not occur in thedecomposition.

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 4The sum of the numbers in the last line is 8 + 2 + 6 + 4 = 20, 20/4 = 5, so B1 occurs in thedecomposition five times.

Thus, we find that f1f2 spans A1 + 2A2 + 5B2.

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Orbitals with nonzero overlapThe general rule is: only orbitals of the same symmetry species may have nonzero

overlap, so only orbitals of the same symmetry species form bonding andantibonding combinations. Remember that the selection of atomic orbitals that havemutual nonzero overlap is the central and initial step in the construction of themolecular orbitals by the LCAO procedure. The molecular orbitals formed from aparticular set of atomic orbitals with nonzero overlap are labeled with the lower-case letter corresponding to the symmetry species. For instance, the (sN,s1) orbitalsare called a1 orbitals (or a1* if we wish to emphasize that they are antibonding). Thes2 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 thisconclusion by noting that the charactertable shows that in C3v, the functions x and yjointly belong to the symmetry species E. Therefore, N2px and N2py also belong toE, so they may have nonzero overlap with s2 and s3. Indeed, let’s considerdecomposi-tion of E×E. First, we need to find characters of the product:

C3v E 2C3 3σv

E 2 -1 0E 2 -1 0E×E 4 1 0Now we need to carry out decomposition of the characters 4, 1, 0.

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C3v E 2C3 3σv h = 6E×E 4 1 0A1 1 1 1

4 1 0Summation 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 2and 3 symmetry elements C3 and σv, respectively). Thus, one A1 symmetry species is presentin the decomposition of E×E.C3v E 2C3 3σv h = 6E×E 4 1 0A2 1 1 -1

4 1 0Summation: 4 + 2×1 + 3×0 = 6 6/6 = 1One A2 symmetry species is present in the decomposition of E×E.C3v E 2C3 3σv h = 6E×E 4 1 0E 2 -1 0

8 -1 0Summation: 8 + 2×(-1) + 3×0 = 6 6/6 = 1One E symmetry species is present in the decomposition of E×E and we obtain

E×E = A1 + A2 + E

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Since A1 is present in the decomposition of E×E, we can conclude that two atomic orbitalsbelonging to symmetry species E may have nonzero overlap and therefore to form bondingand antibonding MOs (e).

We can also explore whether any d orbitals on the central atom can take part in bonding:

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dz 2 has A1 symmetry and each of the pairs

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dx2−y2 ,dxy( ) and

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dxz ,dyz( ) has E symmetry.

Therefore, molecular orbitals may be formed by

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s1,dz 2( ) overlap and by overlap of the s2, s3

combinations with the E d orbitals. Whether or not the d orbitals are in fact important, grouptheory cannot answer – the extent of their involvement depends on energy consideration, notsymmetry.

Example. Determining which orbitals can contribute to bonding. The four H1s orbitalsof methane span A1 + T2. With which of the C atom orbitals can they overlap? What bondingpattern would be possible if the C atom had d orbitals available?Td character table:Td E 8C3 3C2 6σd 6S4 h = 24A1 1 1 1 1 1 x2 + y2 + z2

A2 1 1 1 -1 -1E 2 -1 2 0 0 (3z2 – r2, x2 – y2)T1 3 0 -1 -1 1T2 3 0 -1 1 -1 (x, y, z), (xy, xz, yz)

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An s orbital spans A1, so it may have nonzero overlap with the A1 combination of H1sorbitals. 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 remainnonbonding 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 lowestenergy configuration is a1

2t26 with all bonding orbitals occupied.

Vanishing integrals and selection rulesIntegrals of form

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f1 f2 f3dτ∫ are also common in quantum mechanics because they

include matrix elements of operators. For the integral to be nonzero, the product

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f1 f2 f3 mustspan 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 arecompared with the characters of the symmetry species in the group.

Example. Deciding if an integral must be zero. Does the integral

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3dz 2( )∫ x 3dxy( )dτ vanish in

a C2v molecule?

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The C2v character tableC2v E C2 σv σv’ h = 4A1 1 1 1 1 z z2, y2, x2

A2 1 1 -1 -1 xyB1 1 -1 1 -1 x xzB2 1 -1 -1 1 y yz

We find that

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f1 = 3dz 2 spans A1,

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f2 = x spans B1, and

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f3 = 3dxy spans A2 and draw up the

following table:E C2 σv σv’

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f3 = 3dxy 1 1 -1 -1 A2

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f2 = x 1 -1 1 -1 B1

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f1 = 3dz 2 1 1 1 1 A1

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f1 f2 f3 1 -1 -1 1 B2

The

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f1 f2 f3 product span B2. Therefore, the integral is necessarily zero.We’ll see in the Chapters on spectroscopy that the intensity of a spectral line arising

from a molecular transition between some initial state i with wavefunction ψi and a final statef with wave-function ψf depends on the (electric) transition dipole moment, µfi. The z-component of this vector is defined as

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−e ψ f*zψidτ∫ where -e is the charge of the electron.

The transition moment has the form of the integral

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f1 f2 f3dτ∫ ; so, once we know the

symmetry species of the states, we can use group theory to formulate the selection rules forthe transitions.

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Example: investigate whether an electron in an a1 orbital in H2O (C2v symmetry) canmake 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-componentE 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 1f1 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 dipolemoment 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 xdirection.

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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 charactertable.

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

f2 (q) 3 0 -1 -1 1 T2

f1 (py) 3 0 -1 -1 1 T2

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f1 f2 f3 27 0 -1 -1 1

Now we check whether the decomposition of

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f1 f2 f3 contains A1:

Td E 8C3 3C2 6σd 6S4 h = 24

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f1 f2 f3 27 0 -1 -1 1A1 1 1 1 1 1

27 0 -1 -1 1

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