M. Hargittai, I. Hargittai - Symmetry through the Eyes of a Chemist (793765), страница 42
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Another way ofmaking linear combinations is to first combine the 2s and 2pz orbitalsof the same atomand then combine the resulting orbitals into MOs.The result is essentially the same as before.The 2px and 2py orbitals of the two atoms together form a representation that reduces to g and u .
These correspond to two doublydegenerate orbitals, one of which lies in the yz plane2666 Electronic Structure of Atoms and Moleculesand the other in the xz plane. The relative energies of these orbitals areknown from energy calculations. In most cases the order is as follows:1σg < 1σu < 2σg < 2σu < 3σg < 1πu < 1πg < 3σuwhile in some cases 1πu < 3σg .6.3.3.2. Polyatomic MoleculesBefore working out actual examples, let us recall what was said aboutthe symmetry properties of atomic orbitals. If there is a central atomin the molecule, its atomic orbitals belong to some irreducible representation of the molecular point group. For the other atoms of thesemolecules, SALCs are formed from like orbitals.
These new orbitalsare then coupled with the central atom AOs to form MOs.If the molecule does not have a central atom (e.g., C6 H6 ), we beginwith the second step, first forming different group orbitals and thencombining them, if possible, into MOs. Examples will be given forboth cases.a) Water, H2 O. The molecular symmetry is C2v . There are six atomicorbitals available for MO construction: two H 1s, one oxygen 2s andthree oxygen 2p. They can combine to produce six MOs.
As themolecule has a central atom, its AOs will belong to some of the irreducible representations of the C2v point group by themselves. Grouporbitals must be formed from the H 1s orbitals. The symmetry operations applied to them are shown in Figure 6-17. The C2v charactertable was given in Table 6-2. The reducible representation is:C2v2 H(1s)E2C20v (xz)0σν (yz)26.3.
Molecules267Figure 6-17. The C2v symmetry operations applied to the two hydrogen 1s orbitalsof water as basis functions.This representation reduces to A1 + B2 . The projection operator (seeChapter 4) is used to form these SALCs. Since we are interested onlyin symmetry aspects, numerical factors and normalization are omitted.P̂ A1 s1 ≈ 1 · E · s1 + 1 · C2 · s1 + 1 · σ · s1 + 1 · σ · s1= s1 + s2 + s2 + s1 = 2s1 + 2s2 ≈ s1 + s2P̂ B2 s1 ≈ 1 · E · s1 + (−1) · C2 · s1 + (−1) · σ · s1 + 1 · σ · s1= s1 − s2 − s2 + s1 = 2s1 − 2s2 ≈ s1 − s2Thus, the two hydrogen group orbitals (1 and 2 ) will have the forms:ϕ2 = s1 − s2:ϕ1 = s1 + s2:The available AOs are summarized according to their symmetryproperties in Table 6-3. Since only orbitals of the same symmetry canoverlap, two combinations are possible: one has A1 symmetry and the2686 Electronic Structure of Atoms and MoleculesTable 6-3.
The Atomic Orbitals of Water Grouped According to their SymmetryPropertiesO orbitalsH group orbitalsA1A2B1B22s, 2pz12px2py2other has B2 symmetry. The remaining two orbitals of oxygen (onewith A1 and the other with B1 symmetry) will be nonbonding in thewater molecule.If we choose the oxygen 2s orbital for bonding and leave the 2pzorbital nonbonding (from the symmetry point of view the oppositechoice or a mixed orbital would do just as well; actually if the twoarbitals are close in energy, they mix), the MOs of the water moleculecan be constructed as shown in Figure 6-18. These MOs are comparedwith the calculated contour diagrams of the water molecular orbitalsin Figure 6-19.The construction of the molecular orbitals of the watermolecule can also be represented by a qualitative MO diagram(see Figure 6-20).
The relative energies of the orbitals are also indicated in Figure 6-20. What information can be deduced from such adiagram? First, there are two bonding orbitals occupied by four electrons; these correspond to the two O–H bonds of water. There aretwo nonbonding orbitals also occupied; these are the two lone pairs ofoxygen. Finally, there are two antibonding orbitals that are empty, sothere is a net energy gain in the formation of H2 O and the molecule isstable.b) Ammonia, NH3 . This example is given primarily to illustrate theconstruction of degenerate molecular orbitals. The symmetry of themolecule is C3v .
There are seven atomic orbitals available for bonding:three H 1s, one N 2s, and three N 2p AOs; hence, seven MOs must beformed. Since the nitrogen atom is a central atom, the coordinate axescan be chosen so that its AOs lie on all symmetry elements of theC3v point group.
The pertinent character table is given in Table 6-4.The N 2s and 2pz orbitals will have A1 symmetry and the 2px and6.3. Molecules269Figure 6-18. Construction of the molecular orbitals of water.2py orbitals together belong to the E irreducible representation. Grouporbitals must be formed from the three H 1s orbitals. The symmetryelements of the C3v point group applied to these orbitals are shown inFigure 6-21; their representation is given in Table 6-4.This representation can now be reduced by using the reductionformula introduced in Chapter 4:2706 Electronic Structure of Atoms and MoleculesFigure 6-19.
Contour diagrams of the molecular orbitals of water. Computerdrawing by Zoltán Varga with Gaussview [24].Figure 6-20. Qualitative MO diagram for water.6.3. Molecules271Figure 6-21. The C3v symmetry operations applied to the three hydrogen atom 1sorbitals of ammonia as basis function.Table 6-4. The C3v Character Table and the Reducible Representation of theHydrogen Group Orbitals of AmmoniaC3vE2C33vA1A2E11211−11−103 H(1s)301zRz(x, y) (Rx , Ry )x2 + y2 , z2(x2 –y2 , xy) (xz, yz)a A1 = (1/6)(1 · 3 · 1 + 2 · 0 · 1 + 3 · 1 · 1) = 1a A2 = (1/6)(1 · 3 · 1 + 2 · 0 · 1 + 3 · 1 · (−1)) = 0a E = (1/6)(1 · 3 · 2 + 2 · 0 · (−1) + 3 · 1 · 0) = 1Thus, the representation reduces to A1 + E. Next, let us use theprojection operator to generate the form of these SALCs:P̂ A1 s1 ≈ 1 · E · s1 + 1 · C3 · s1 + 1 · C32 · s1 + 1 · σ · s1 + 1 · σ · s1 + 1 · σ · s1= s1 + s2 + s3 + s1 + s2 + s3 = 2(s1 + s2 + s3 ) ≈ s1 + s2 + s3The same procedure is illustrated pictorially in Figure 6-22, afterReference [25].2726 Electronic Structure of Atoms and MoleculesFigure 6-22.
Generation of the A1 symmetry orbital of the 3H group orbitals ofammonia.For the construction of the E symmetry group orbitals, a timesaving simplification will be introduced [26]. First of all, it utilizesthe fact that the rotational subgroup Cn in itself contains all the information needed to construct the SALCs in a molecule that possesses aprincipal axis Cn . The rotational subgroup of C3v is C3 , and its character table is given in Table 6-5. If we perform the three symmetryoperations of the C3 point group and check the generation of the A1symmetry SALC of NH3 (Figure 6-22), we see that the application ofthese three operations suffices to define the form of this orbital.The difficulty in applying the projection operator for this symmetrygroup arises from the fact that the C3 character table contains6.3.
MoleculesC3EA1111E273Table 6-5. The C3 Character TableC3C32 = exp (2 i/3)1εε∗1 ε∗εz, Rzx2 +y2 , z2(x, y) (Rx , Ry )(x2 –y2 , xy) (yz, xz)imaginary characters for the E representation. They can be eliminatedby following the procedure used in Reference [27]. The character corresponds to exp (2i/n), where n is the order of the rotation axis;in our case, 3. Using Euler’s formula, exp(iα) = cos α + i sin α (andthe complex conjugate of will be: ε ∗ = cos α − i sin α), the characters for the E representation will be:11√− 12 + i 3/2√− 12 − i 3/2√ − 12 − i 3/2√− 12 + i 3/2(a)(b)Using two different ways to obtain linear combinations of these characters will make it possible to eliminate the imaginary characters. Oneway may be summing Eqs.
(a) and (b), resulting in2−1−1The other linear combination may be obtained√ by subtraction ofEq. (b) from Eq. (a) and dividing the result by i 3. This linear combination results in the “characters”01−1not satisfying all the relationships of irreducible representations, butit serves our purpose of showing the shape of the SALCs. Our “quasicharacter table” for the C3 point group is now:AE1112 −1 −101 −12746 Electronic Structure of Atoms and MoleculesWhen the projection operator is applied to one of the ls orbitals ofthe hydrogen group orbitals with the two E representations, the two Esymmetry doubly degenerate SALCs result:1P̂ E s1 ≈ 2 · E · s1 + (−1) · C3 · s1 + (−1) · C32 · s1 = 2s1 − s2 − s32P̂ E s1 ≈ 0 · E · s1 + 1 · C3 · s1 + (−1) · C32 · s1 = s2 − s3Figure 6-23 illustrates the same procedure pictorially.Figure 6-23.
Projection of the two E symmetry group orbitals of the three H 1sorbitals in ammonia.The next step is the MO construction. The orbitals used for thispurpose are summarized in Table 6-6. An A1 and a doubly degenerate E symmetry combination is possible here, and there will be anonbonding orbital with A1 symmetry left on nitrogen. Figure 6-24illustrates the building of MOs. Again, they can be compared withTable 6-6. The Atomic Orbitals of Ammonia Sorted According to their SymmetryPropertiesN orbitalsH group orbitalsA1Es, pz(px , py )126.3. MoleculesFigure 6-24. Construction of molecular orbitals for ammonia.2752766 Electronic Structure of Atoms and MoleculesFigure 6-25.
Contour diagrams of the molecular orbitals of ammonia. Computerdrawing by Zoltán Varga with Gaussview [28].the calculated contour diagrams of the ammonia molecular orbitalsin Figure 6-25. The qualitative MO diagram is given in Figure 6-26.The following conclusions can be drawn: (l) There are three bondingorbitals occupied by electrons; these correspond to the three N–Hbonds; (2) there is a nonbonding orbital also occupied by electrons;this corresponds to the lone electron pair; and (3) the three antibonding orbitals are unoccupied, so the MO construction is energetically favorable, and the molecule is stable.c) Benzene, C6 H6 .
The molecular symmetry is D6h . There are 30 AOsthat can be used for MO construction; six H 1s, six C 2s, and 18 C 2porbitals. Since this molecule does not contain a central atom, each AO6.3. Molecules277Figure 6-26. Qualitative MO diagram for ammonia.must be grouped into SALCs in such a way that they can transformaccording to the symmetry operations of the D6h point group. It isstraightforward to combine like orbitals, for example, the hydrogen 1sorbitals, the carbon 2s orbitals, and so on. The following combinationswill be used here:⌽1 (6 H 1s), ⌽2 (6 C 2s), ⌽3 (6 C 2 px , 2 p y ), and ⌽4 (6 C 2 pz ).The next step is to determine how these group orbitals transform in theD6h point group.
The D6h character table is given in Table 6-7. Sincemost of the AOs in the suggested group orbitals are transformed intoanother AO by most of the symmetry operations, the representationswill be quite simple, though still reducible:⌫ ⌽1⌫ ⌽2⌫ ⌽3⌫ ⌽466126000000000202000 −200000000000006060 120 −600002202These representations can be reduced by applying the reductionformula.278E2C62C3C23C2A1gA2gB1gB2gE1gE2gA1uA2uB1uB2uE1uE2u11112211112211–1–11–111–1–11–11111–1–11111–1–111–1–1–2211–1–1–221–11–1001–11–100Table 6-7. The D6h Character Table3C2i2S32S6h1–1–11001–1–1100111122–1–1–1–1–2–211–1–11–1–1–111–111111–1–1–1–1–1–11111–1–1–22–1–1112–23d31–11–100–11–11001–1–1100–111–100x2 +y2 , z2Rz(Rx , Ry )z(x, y)(xz, yz)(x2 –y2 , xy)6 Electronic Structure of Atoms and MoleculesD6h6.3. Molecules279First ⌽1 :a A1g = (1/24)(1 · 6 · 1 + 2 · 0 · 1 + 2 · 0 · 1 + 1 · 0 · 1 + 3 · 2 · 1+ 3 · 0 · 1 + 1 · 0 · 1 + 2 · 0 · 1 + 2 · 0 · 1 + 1 · 6 · 1 + 3 · 0 · 1 + 3 · 2 · 1)= (1/24)(6 + 6 + 6 + 6) = 24/24 = 1a A2g = (1/24)(6 − 6 + 6 − 6) = 0a B1g = (1/24)(6 + 6 − 6 − 6) = 0a B2g = (1/24)(6 − 6 − 6 + 6) = 0a E1g = (1/24)(12 − 12) = 0a E2g = (1/24)(12 + 12) = 1a A1u = (1/24)(6 + 6 − 6 − 6) = 0a A2u = (1/24)(6 − 6 − 6 + 6) = 0a B1u = (1/24)(6 + 6 + 6 + 6) = 1a B2u = (1/24)(6 − 6 + 6 − 6) = 0a E1u = (1/24)(12 + 12) = 1a E2u = (1/24)(12 − 12) = 0Thus, the first representation reduces to the following irreduciblerepresentations:⌫⌽1 = A1g + E 2g + B1u + E 1uWithout giving details of the other three reductions, the results are:⌫⌽2 = A1g + E 2g + B1u + E 1u⌫⌽3 = A1g + A2g + 2E 2g + B1u + B2u + 2E 1u⌫⌽4 = B2g + E 1g + A2u + E 2uSimilarly to the case of ammonia, the rotational subgroup of D6h ,that is C6 , contains enough information to generate the SALCs ofbenzene.
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