Yves Jean - Molecular Orbitals of Transition Metal Complexes (793957), страница 51
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Symmetry-adapted orbitalsWe shall limit ourselves here to the determination of the symmetryadapted A2′′ and E′′ orbitals in the Ŵp⊥ representation. Table 6.25 showsthe action of all the symmetry operations of the D3h point group on thegenerating functions p⊥1 and (p⊥2 − p⊥3 ), as well as the characters ofthe A2′′ and E′′ irreducible representations.The symmetry-adapted orbitals are obtained from formula (6.10),by using the p⊥1 generating function for the φA2′′ orbital and for the firstdegenerate orbital φE′′ (1), and the function (p⊥2 − p⊥3 ) for the seconddegenerate orbital φE′′ (2). We obtain:φA2′′ = 4 × (p⊥1 + p⊥2 + p⊥3 )φE′′ (1) = 2 × (2p⊥1 − p⊥2 − p⊥3 )φE′′ (2) = 6 × (p⊥2 − p⊥3 )so the normalized orbitals are1φA2′′ = √ (p⊥1 + p⊥2 + p⊥3 )31φE′′ (1) = √ (2p⊥1 − p⊥2 − p⊥3 )61φE′′ (2) = √ (p⊥2 − p⊥3 )2(6.28a)(6.28b)(6.28c)Table 6.25.
Action of the symmetry operations of the D3h point group on the p⊥ orbitals (see 6-33) and the characters of the A′′2 and E′′ irreducible representations.−Li bond, and a σv (i) plane contains the M−−Li bond.][See Scheme 6-28 for the definition of the symmetry elements. A C2 (i)-axis is co-linear with the M−D3hEC13C23C2 (1)C2 (2)C2 (3)σhS13S23σv (1)σv (2)σv (3)Rk (p⊥1 )p⊥1p⊥2p⊥3−p⊥1−p⊥3−p⊥2−p⊥1−p⊥2−p⊥3p⊥1p⊥3p⊥2E′′2−1−1000Rk (p⊥2 − p⊥3 )A2′′p⊥2 − p⊥31p⊥1 − p⊥21p⊥3 − p⊥11p⊥2 − p⊥3−1p⊥1 − p⊥2−1p⊥3 − p⊥1−1p⊥3 − p⊥2−1−2p⊥2 − p⊥1−11p⊥1 − p⊥3−11p⊥3 − p⊥21p⊥2 − p⊥11p⊥1 − p⊥31000ExercisesM// (A2⬘)MFigure 6.11. π and π⊥ symmetry-adaptedorbitals in a trigonal-planar ML3 complex.MM// (E⬘)MM⊥(A⬙)2⊥(E⬙)For the A2′ and E′′ symmetry-adapted orbitals (Ŵp ), we obtain (seeExercise 6.12):1φA2′ = √ (p1 + p2 + p3 )31φE′ (1) = √ (2p1 − p2 − p3 )61φE′ (2) = √ (p2 − p3 )2(6.29a)(6.29b)(6.29c)The different symmetry-adapted orbitals are shown in Figure 6.11.−L bonds, the orbitAlthough the σ -type orbitals point along the M−als above are constructed from ligand orbitals that are perpendicular tothese bonds.
This is why they are often written π and π⊥ , even thoughthis notation is not strictly correct according to group theory.ExercisesSymmetry elements and symmetry operations6.1How many reflection planes are there in the following molecules?(1) water (H2 O); (2) ammonia (NH3 ); (3) ethylene (C2 H4 ); (4) (Z)1,2-difluoroethylene; (5) (E)-1,2-difluoroethylene; (6) aluminiumtrichloride (AlCl3 , a trigonal-planar molecule).6.2Which of the following molecules contain an inversion centre?(1) carbon dioxide (CO2 , linear); (2) hydrogen cyanide (HCN, linear);(3) dimethylether (shown in 6-1); (4) benzene (C6 H6 , hexagonal);Elements of group theory and applications(5) (Z)-1,2-difluoroethylene; (6) (E)-1,2-difluoroethylene; (7) aluminium trichloride (AlCl3 , a trigonal-planar molecule); (8) methane(CH4 , tetrahedral); (9) [PtCl4 ]2− (a square-planar complex, Scheme6-10); (10) ethane (C2 H6 ) in the staggered conformation; (11) ethanein the eclipsed conformation.6.31.
Are there any rotation axes in the ammonia molecule (6-7) inaddition to the C3 -axis?2. Locate the three C3 axes and the four C2 axes in the methanemolecule.3. How many C2 axes are there in (i) ethylene; (ii) (Z)-1,2difluoroethylene; (iii) (E)-1,2-difluoroethylene ?4. Which rotation axes are present in AlCl3 ?6.41. Consider a C2 -axis and the co-linear S4 -axis in methane (seeScheme 6-11). Show that S42 = C2 and that S44 = E.2. Carry out the S32 and S34 operations in the PF5 molecule (seeScheme 6-12). Show that these operations are equivalent tooperations associated with the C3 -axis or the σh plane.3. (i) Show that ethane (C2 H6 ) in the eclipsed conformation possessesan improper axis that is co-linear with the C–C bond. What is theorder of this axis? (ii) Repeat the question for ethane in the staggeredconformation.6.5The symmetry elements present in the ammonia molecule (NH3 )are a C3 -axis (6-7) and three planes of symmetry (σv (1), σv (2), andσv (3)) (Exercise 6.1, question 2).1.
Perform the C3 operation followed by σv (1) (written σv (1)C3 );carry out σv (1) followed by C3 (written C3 σv (1)). What conclusion(s) can you draw?2. Perform (C3 σv (1))σv (2) then C3 (σv (1)σv (2)). What conclusion(s)can you draw?3. Which operations are the inverses of E, C3 , C32 , σv (1)?6.6Determine the point-group symmetries of the following molecules:(1) O2 ; (2) HCl; (3) ethylene (C2 H4 ); (4) (Z)-1,2-difluoroethylene;(5) (E)-1,2-difluoroethylene; (6) AlCl3 , a trigonal-planar molecule;(7) tetrachloromethane (CCl4 ); (8) dichloromethane (CH2 Cl2 ); (9)ExercisesNH3 (6-7); (10) [PtCl4 ]2− (6-10); (11) PF5 (6-12); 12) trifluoroethylene (F2 C==CHF, planar).Symmetry-adapted orbitals6.7√Show that in the NH3 molecule, the functions (1/ 3)(1sH1 +1sH2 +√1sH3 ), a basis for the A1 representation, and (1/ 6)(2 × (1sH1 ) −1sH2 − 1sH3 ), one of the basis functions for the E representation, areorthogonal.
The 1sH orbitals are assumed to be normalized, and theoverlap between two 1sH orbitals on different atoms is written S.6.8zy1s H2C1. Reduce the ŴH basis that is constituted by the 1sH orbitals on thehydrogen atoms in the ethylene molecule (the characters of thisrepresentation are given in Table 6.8, and the character table forthe D2h point group is given below).2. Find the linear combinations of these orbitals that are bases forirreducible representations.1s H3C1s H11s H4D2h E C2 (z) C2 (y) C2 (x) iσ (xy) σ (xz) σ (yz)AgB1gB2gB3gAuB1uB2uB3u11−1−1−1−1111111111111−1−111−1−11−11−11−11−11−1−111−1−111111−1−1−1−11−11−1−11−111−1−11−111−1x 2 , y2 , z2xyxzyzzyx6.9Construct the molecular orbitals of NH3 from the symmetryadapted orbitals that are given in Figure 6.3.
To help you constructthe interaction diagram, note that the energies of the orbitals onH3 fragment and of the nitrogen 2pN orbitals are close to −13.5 eV,and that the energy of the 2sN orbital is −26.0 eV.6.1012M34Calculate the coefficients of σ1 , σ2 , σ3 , and σ4 (see below for thenumbering of the atoms) in the symmetry-adapted T2 orbitals givenon the right-hand side of Figure 6.7. You should use the normalization relationships and the fact that these orbitals are mutuallyorthogonal and orthogonal to the symmetry-adapted A1 orbital.Elements of group theory and applications6.11In a trigonal-planar ML3 complex, the Ŵσ representation is reducedthus: Ŵσ = A1′ ⊕ E′ (§ 6.6.3.1, formula (6.18)).13M21.
Characterize the A1′ symmetry-adapted orbital, using σ1 as thegenerating function.2. Characterize the E′ symmetry-adapted orbitals, using σ1 followedby (σ2 − σ3 ) as the generating functions. The projection formula(6.10) and the character table for the D3h point group (Table 6.21)should be helpful.6.12In a trigonal-planar ML3 complex, the Ŵp representation is reducedthus:Ŵp = A2′ ⊕ E′ (§ 6.6.6.1, formula (6.26)).p||1Mp||3p||21. Characterize the A2′ symmetry-adapted orbital, using p1 as thegenerating function.2. Characterize the E′ symmetry-adapted orbitals, using p1 followedby (p2 –p3 ) as the generating functions.The projection formula (6.10) and the character table for the D3hpoint group (Table 6.24) should be helpful.
The complete set ofsymmetry operations is given in Table 6.25.6.13Consider a square-planar ML4 complex in which each ligand possesses a π system made up of two p orbitals that are perpendicularto the M–L bonds. These orbitals are p⊥ and p , depending onwhether they are perpendicular to the plane of the complex or inthat plane. The orientation of these orbitals is shown below.pTp32TMMpTp4p ||2p||31p||4p ||1TTables 6.18 and 6.19, which give the character table for the D4h pointgroup and the full set of the symmetry operations of this group,should be helpful, as should Scheme 6-24 in which the differentsymmetry elements are shown.1. Determine the characters of the Ŵp⊥ representation.2.
Decompose this representation into a sum of irreducible representations of the D4h point group.Exercises3. Characterize the symmetry-adapted orbitals in each of theserepresentations. The generating function p⊥1 should be used forthe one-dimensional representations, and the generating functions p⊥1 and p⊥2 in turn for the two-dimensional representation.4. Sketch the resulting orbitals.5. From an analysis of the interactions between the ligand orbitals,establish the energetic ordering of the symmetry-adapted MO.6. Repeat questions 1–5 for the Ŵp representation.This page intentionally left blankAnswers to the exercises(in skeletal form)Chapter 1: Setting the scene1.1.
The coordination modes are η1 (X-type ligand) and η3 (LX-typeligand).MMη1 -allylη3 -allyl1.2.lxqnodnNt1234567896000d6185000d6165101d6180404d083101d8165101d6184202d6185 00 90 −20 7d8 d018 1810 112707d0181213 14 15 16 17 18 19 20 21 22 23 24400 2 1 0 4 6035 4 4 8 4 00 −2 −3 −1 −1 −2 +1 +1012 3 3 6 5 1d10 d10 d8 d6 d6 d0 d2 d618 16 18 18 18 16 18 184404d0164505d0184202d6180606d0125000d8182202d10181.3. no = 1, d8 , Nt = 16; no = 3, d6 , Nt = 18; no = 3, d6 , Nt = 16;no = 3, d6 , Nt = 18; no = 3, d6 , Nt = 16; no = 3, d6 , Nt = 18;no = 1, d8 , Nt = 16.−− −−1.4. (1) CO, Cl− , Et− , PR 3 , H− , H2 , SiR −3 , SR , CN , I , Me ,−COMe− , F− , O2− , NR3 , C2 H4 , C6 H6 , C5 H5 .Answers to the exercises (in skeletal form)(2)ComplexIonic modelno[Fe(CO)5 ][Ir(CO)(Cl)(PPh3 )2 ][Mn(CO)6 ]+[Ni(CN)5 ]3−[Zn(Cl)4 ]2−[V(Cl)4 ][Cr(CO)3 (η6 -C6 H6 )][Fe(η5 -C5 H5 )2 ][Cu(η5 -C5 H5 )(PR 3 )][Zr(η5 -C5 H5 )2 (CH3 )]+[Ti(PR 3 )2 (Cl)3 (CH3 )][W(PR 3 )2 (CO)3 (η2 -H2 )][Ir(PR 3 )2 (Cl)(H)2 ][Ni(H2 O)6 ]2+[Fe(CO)5 ][Ir+ (CO)(Cl− )(PPh3 )2 ][Mn+ (CO)6 ][Ni2+ (CN− )5 ][Zn2+ (Cl− )4 ][V4+ (Cl− )4 ][Cr(CO)3 (η6 -C6 H6 )][Fe2+ (η5 -C5 H5− )2 ][Cu+ (η5 -C5 H5− )(PR 3 )][Zr4+ (η5 -C5 H5− )2 (CH−3 )][Ti4+ (PR 3 )2 (Cl− )3 (CH−3 )][W(PR 3 )2 (CO)3 (η2 -H2 )][Ir3+ (PR 3 )2 (Cl− )(H− )2 ][Ni2+ (H2 O)6 ]011224021440321.5.1.
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