M. Hargittai, I. Hargittai - Symmetry through the Eyes of a Chemist (793765), страница 43
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The C6 character table is given in Table 6-8, and, again,contains imaginary characters. These can be handled in the same wayas was done for ammonia, keeping in mind that the solution is right forthe determination of the shape of the SALCs but the derived “quasicharacters” are not real characters.These “quasi-characters” for the two E representations are:21−1−2−11E10110−1−12806 Electronic Structure of Atoms and MoleculesTable 6-8.
The C6 Character TableC6ABE1E2C6C3C2C321111111–1εε∗11− ε∗−ε1–1−1−111−ε− ε∗− ε∗−ε−ε− ε∗11− ε∗−εE220−11ε = exp (2 i/6)C65E1–1ε∗ε−ε− ε∗−1−1z, Rzx2 +y2 , z2(x, y)(Rx , Ry )(xz, yz)(x2 –y2 , xy)−1120−1−1Benzene consists of 30 MOs; only a few of these will beconstructed and shown here. It may be a good exercise for thereader to construct the remaining MOs of benzene by following theprocedure demonstrated here.
The SALCs are sorted according totheir symmetry properties in Table 6-9. Inspection of this table revealsthat the first three group orbitals have common irreducible representations, so they can be mixed with each other. They consist of24 AOs; thus, 24 MOs will be formed. Since each bonding MO hasits antibonding counterpart, there will be 12 bonding and 12 antibonding molecular orbitals.
The former will be the bonding orbitals ofTable 6-9. The Symmetry of the Different Group Orbitals of Benzene⌽2⌽3⌽1⌽4H group orbital C 2s group orbital C 2px , 2pyC 2pz group orbitalgroup orbitalA1gA2gB1gB2gE1gE2gA1uA2uB1uB2uE1uE2u++++++++++++++++++++6.3. Molecules281benzene, since there are six C–C and six C–H bonds. The fourth grouporbital does not belong to any irreducible representation common tothe other three, so it will not be mixed with them.
This representationcorresponds to the orbitals of benzene by itself.Let us now construct the A1g and B1u symmetry orbitals ofbenzene. The totally symmetric representation, A1g , appears threetimes, once in each of ⌽1 , ⌽2 , and ⌽3 . Two A1g representations canbe combined into an MO, and the third one can represent an MO byitself. These three SALCs can be generated by using the projectionoperator pictorially as shown in Figure 6-27. The forms of these grouporbitals are such that ⌽2 (A1g ) can be taken as an MO by itself (C–C bond; cf. also the corresponding orbital, 2A1g , in the contour diagramin Figure 6-28a), and the other two group orbitals can be combinedinto molecular orbitals as shown in Figure 6-29. The contour diagramof the bonding MO is depicted by the 3A1g orbital in Figure 6-28a.The next MO will be of B1u symmetry. This irreducible representation also appears in ⌽1 , ⌽2 , and ⌽3 .
Take this time the corresponding⌽1 and ⌽2 group orbitals and combine them into molecular orbitals:P̂ B1u s1 ≈1 · E · s1 + (−1) · C6 · s1 + 1 · C3 · s1 + (−1) · C2 · s1+ 1 · C32 · s1 + (−1) · C65 · s1 = s1 − s2 + s3 − s4 + s5 − s6 ,or pictorally:The B1u symmetry SALC of ⌽2 , i.e., the group orbital of the six C2s AOs, will have a similar form:The combination of these ⌽1 and ⌽2 SALCs affords the bondingand antibonding combinations shown in Figure 6-30. The contourdiagram corresponding to the bonding MO is the 2B1u orbital inFigure 6-28a.2826 Electronic Structure of Atoms and MoleculesFigure 6-27. Generation of the A1g symmetry group orbitals of benzene.6.3.
Molecules(a)283(b)Figure 6-28. Contour diagrams of some molecular orbitals of benzene. Computerdrawing by Zoltán Varga with Gaussview [29]: (a) orbitals; (b) orbitals.Since there is only one B2u symmetry orbital among the SALCs, theone in ⌽3 , it will be a MO by itself. Let us generate this MO:P̂ B2u p y (C1 ) ≈ 1 · E · p y1 + (−1) · C6 · p y1 + 1 · C3 · p y1+ (−1) · C2 · p y1 + 1 · C32 · p y1 + (−1) · C65 · p y1= p y1 − p y2 + p y3 − p y4 + p y5 − p y6Figure 6-29.
Bonding and antibonding combination of A1g symmetry group orbitalsof benzene.2846 Electronic Structure of Atoms and MoleculesFigure 6-30. Bonding and antibonding combination of B1u symmetry group orbitalsof benzene.This group orbital has the following shape:Compare the above orbital with 1B2u (Figure 6-28a).The orbitals of benzene will be the two doubly degenerate andthe two non-degenerate combinations of the ⌽4 group orbital itself.All of these are shown below.A2u symmetry orbital: this corresponds to the totally symmetricrepresentation in the rotational subgroup C6 ; so, even without usingthe projection operator, its form can be given by:⌽4 (A2u ) = pz1 + pz2 + pz3 + pz4 + pz5 + pz6The corresponding orbital in Figure 6-28b will be the 1A2u orbital.B2g symmetry orbital: using the projection operator, we obtain:P̂ B2g pz (C1 ) ≈ 1 · E · pz1 + (−1) · C6 · pz1 + 1 · C3 · pz1+ (−1) · C2 · pz1 + 1 · C32 · pz1 + (−1) · C65 · pz1= pz 1 − pz 2 + pz 3 − pz 4 + pz 5 − pz 66.3. Molecules285This is the 1B2g orbital of Figure 6-28b.The two E1g symmetry SALCs are constructed in Figure 6-31.Compare them to the contour diagram of the 1E1g orbitals inFigure 6-28b.Finally, the two E2u symmetry orbitals are expressed as follows:1P̂ E2u pz (C1 ) ≈ 2 · E · pz1 + (−1) · C6 · pz1 + (−1) · C3 · pz1+ 2 · C2 · pz1 + (−1) · C32 · pz1 + (−1) · C65 · pz1= 2 pz 1 − pz 2 − pz 3 + 2 pz 4 − pz 5 − pz 6Figure 6-31.
The two E1g symmetry group orbitals formed from the carbon 2pzorbitals in benzene.2866 Electronic Structure of Atoms and MoleculesP̂2E 2upz (C1 ) ≈ 0 · E · pz1 + 1 · C6 · pz1 + (−1) · C3 · pz1+ 0 · C2 · pz1 + 1 · C32 · pz1 + (−1) · C65 · pz1= pz 2 − pz 3 + pz 5 − pz 6 .Their forms are:These SALCs correspond to the contour diagram of the 1E2u orbital(Figure 6-28b). Figure 6-32 shows the relative energies of the benzene orbitals.Figure 6-32. Relative energies of the benzene orbitals.6.3.3.3.
Short Summary of MO ConstructionThe steps of MO construction can now be summarized as follows:1. Identify the symmetry of the molecule.2. List all atomic orbitals that are intended to be used for MOconstruction.3. See whether or not the molecule has a central atom. If it does,then look up in the character table the irreducible representations6.4. Quantum Chemical Calculations4.5.6.7.8.9.287to which its atomic orbitals belong. If there is no central atom inthe molecule, proceed to the next step.Construct group orbitals (SALCs) from the atomic orbitals of likeatoms.Use these orbitals as bases for representations of the point group.Reduce these representations to their irreducible components.Apply the projection operator to the AOs for each of these irreducible representations to obtain the forms of the SALCs.These SALCs will either be MOs by themselves, or they can becombined with other SALCs or central atom orbitals of the samesymmetry.
Each of these combinations will give one bonding andone antibonding MO of the same symmetry.Normalization has been ignored throughout our discussion.However the SALCs must be properly normalized in all calculations [30]. This may be done at the end of the SALC construction,i.e., after step 7 in our list.6.4. Quantum Chemical CalculationsGay-Lussac (1778–1850) wrote in 1809: “We are perhaps not farremoved from the time when we shall be able to submit the bulk ofchemical phenomena to calculaton” [31]. One hundred and ten yearslater, in 1998, John Pople shared the Nobel Prize in Chemistry “forhis development of computational methods in chemistry,” with WalterKohn (in his case, “for his development of the density-functionaltheory”), Figure 6-33.
So even if Gay-Lussac was, perhaps, somewhattoo optimistic, eventually his dreams came true—the development ofcomputational chemistry has been amazing. Quantum chemistry isnot a topic of this book, we are only mentioning it briefly becauseof its inherent relationship to the symmetry concept. For discussionof the topic we refer the reader to a few of the available monographs[32–36].The results of quantum chemical calculations usually yield thewave functions and the energies of a system.
Numerous integralsmust be evaluated even for the simplest molecules. Their number canbe conveniently reduced, however,by applying the theorem accordingto which an energy integral, ⌿i Ĥ⌿j d, is nonzero only if ⌿i and ⌿j2886 Electronic Structure of Atoms and MoleculesFigure 6-33. Walter Kohn (left) and John Pople, the 1998 Nobel laureates in Chemistry.
Photograph by the authors.belong to the same irreducible representation of the molecularpoint group.Most chemical and physical properties of the molecule can becalculated, including the geometry, conformational properties, barrierto internal rotation, relative stabilities of various isomers as well asof different electronic states. Spectroscopic constants, such as dipolemoments, quadrupole moments, and vibrational frequencies can becalculated and thermodynamic quantities determined. In the secondedition of this book, we wrote: “State-of-the-art calculations of molecular geometry involving relatively light atoms are as reliable as theresults of the best experiments”[37].
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