Yves Jean - Molecular Orbitals of Transition Metal Complexes (793957), страница 48
Текст из файла (страница 48)
We have already show that ŴH can be reduced to the sum oftwo irreducible representations, A1 and B2 (formula (6.6)). We shall nowcontinue by finding the combination of these orbitals that forms a basisfor the A1 representation. We choose one of these orbitals, 1sH1 , forexample, which will act as the generating function φ in equation (6.10).We then construct a table which contains (i) the result of the actionof symmetry operation Rk on the generating function (the term Rk φin equation (6.10)); (ii) the characters of the irreducible representationunder consideration (the term χi (Rk )) and last, the product of these twoterms (Table 6.13).
The sum of these products over all the symmetryoperations (equation (6.10)) gives function φA1 that we seek, that is, thelinear combination of the 1sH1 , and 1sH2 , orbitals that is a basis for theA1 representation.Table 6.13. Quantities necessary for the determinationof the linear combination of the 1sH1 and 1sH2 orbitalsthat has A1 symmetry in the H2 O moleculeC2vEC2zσxzσyzRk (1sH1 )A1Product1sH111 × 1sH11sH211 × 1sH21sH211 × 1sH21sH111 × 1sH1We therefore obtain: φA1 = (1 × 1sH1 ) + (1 × 1sH2 ) + (1 × 1sH2 ) +(1×1sH1 ) = 2×(1sH1 +1sH2 ).
It is clear that we would have obtained thesame result if we had taken the 1sH2 orbital as the generating function,as this would merely have led to an interchange of the subscripts 1 and2 in the second line of Table 6.13.The linear combination that has B2 symmetry is obtained in thesame way, from the data in Table 6.14.
It is: φB2 = (1 × 1sH1 ) − (1 ×1sH2 ) − (1 × 1sH2 ) + (1 × 1sH1 ) = 2 × (1sH1 − 1sH2 ).In summary, the linear combinations of the 1sH1 and 1sH2 orbitalsthat are adapted to the molecular symmetry of H2 O are the sum anddifference of these two. If we include the normalization factor that isTable 6.14.
Quantities necessary for the determination of thelinear combination of the 1sH1 and 1sH2 orbitals that hasB2 symmetry in the H2 O moleculeC2vEC2zσxzσyzRk (1sH1 )B2Product1sH111 × 1sH11sH2−1(−1) × 1sH21sH2−1(−1) × 1sH21sH111 × 1sH1Symmetry-adapted orbitalsgiven by Hückel theory, the expressions are:1φA1 = √ (1sH1 + 1sH2 )21φB2 = √ (1sH1 − 1sH2 )23We have already shown (see Table 6.2)that these linear combinations are stable withrespect to all the symmetry operations of theC2v point group.OOA1B2Figure 6.2.
Symmetry-adapted orbitals forH2 O (C2v point group).(6.11a)(6.11b)This result could have been anticipated, since we have only two equivalent functions to combine.3 These symmetry-adapted orbital are shownin Figure 6.2.CommentIn the standard (non-extended) Hückel method, the overlap between orbitals (S) is neglected in the calculation of the normalization factor. As a result,the sum of the squares of the coefficients is 1. In what follows, when wedescribe symmetry-adapted orbitals as normalized, we shall always adoptthat approximation.6.4.2.2.
The ŴH basis in NH3We now consider a second example, one where the result is more difficultto predict: the three 1sH orbitals in the NH3 molecule. These orbitalsform a basis ŴH = A1 ⊕ E (formula (6.7)). We shall proceed in the sameway, taking the 1sH1 orbital as the generating function. The second lineof Table 6.3 shows how this orbital is transformed by the symmetryoperation of the C3v point group. We may use this result to constructTable 6.15, and hence find the linear combination that has A1 symmetry.By adding the terms in the last line, we obtain: φA1 = 2 × (1sH1 +1sH2 + 1sH3 ).
The (non-normalized) totally symmetric function is therefore simply the sum of the initial functions, the same result as thatobtained for H2 O.For the representation with E symmetry, we must find two independent linear combinations of the 1sH orbitals (a two-dimensionalTable 6.15. Quantities necessary for the determination of the linearcombination of the 1sH1 , 1sH2 , and 1sH3 orbitals that has A1 symmetryin the NH3 moleculeC3vEC31C32σv (1)σv (2)σv (3)Rk (1sH1 ) 1sH11sH31sH21sH11sH31sH2A1111111Product 1 × 1sH1 1 × 1sH3 1 × 1sH2 1 × 1sH1 1 × 1sH3 1 × 1sH2Elements of group theory and applicationsTable 6.16. Quantities necessary for the determination of the first linearcombination of the 1sH1 , 1sH2 , and 1sH3 orbitals with E symmetryin the NH3 moleculeC3vEC31C32σv (1)σv (2)σv (3)Rk (1sH1 ) 1sH11sH31sH21sH11sH31sH2E2−1−1000Product 2 × 1sH1 (−1) × 1sH3 (−1) × 1sH2 0 × 1sH1 0 × 1sH3 0 × 1sH2Table 6.17.
Quantities necessary for the determination of the second linear combination of the 1sH1 , 1sH2 , and 1sH3 orbitalswith E symmetry in the NH3 moleculeC3vEC31C32σv (1)σv (2)σv (3)Rk (1sH2 − 1sH3 ) 1sH2 − 1sH31sH1 − 1sH21sH3 − 1sH11sH3 − 1sH21sH2 − 1sH11sH1 − 1sH3E2−1−1000Product2 × (1sH2 − 1sH3 ) (−1) × (1sH1 − 1sH2 ) (−1) × (1sH3 − 1sH1 ) 0 × (1sH3 − 1sH2 ) 0 × (1sH2 − 1sH1 ) 0 × (1sH1 − 1sH3 )4The second function can also be foundby another method, from the requirementthat it be orthogonal to thesymmetry-adapted orbitals already found(φA1 and φE (1)) (see Exercise 6.7).representation).
If we use the 1sH1 orbital once again as the generating function, and the characters of the E representation (Table 6.16), weobtain φE (1) = 2 × (1sH1 ) − 1sH2 − 1sH3 .In this way, we have found one of the two linear combinationsthat we need. If we now change the generating function to 1sH2 or1sH3 , we shall clearly obtain the same type of function, but where thesubscripts 1, 2, and 3 have been permuted. This function is acceptable,but it is not orthogonal to the previous one, and in general we preferto use linear combinations of the orbitals that are mutually orthogonal.4In this example, the second linear combination can be obtained if thecombination (1sH2 − 1sH3 ) is used as the generating function.
Action ofthe symmetry operations on this function gives the results in Table 6.17.Taking the characters of the E representation into account, weobtain the function: φE (2) = 2 × (1sH2 − 1sH3 ) + (−1) × (1sH1 −1sH2 )+(−1)×(1sH3 −1sH1 ) = 3×(1sH2 −1sH3 ), that is, the generatingfunction itself (without considering normalization).In conclusion, the normalized linear combinations of the 1sH1 , 1sH2 ,and 1sH3 orbitals that are adapted to the molecular symmetry of NH3are:1φA1 = √ (1sH1 + 1sH2 + 1sH3 )31φE (1) = √ (2 × (1sH1 ) − 1sH2 − 1sH3 )61φE (2) = √ (1sH2 − 1sH3 )2(6.12a)(6.12b)(6.12c)Construction of MO: H2 O as an exampleNA1H3H2H1A1H3H2H12s2pzNH3H2H1E (1) = exE2pxNH1Figure 6.3.
Symmetry-adapted orbitals forNH3 (C3v point group) and orbitals of thesame symmetry on the central atom.E (2) = eyH3H2H12pyThese orbitals are shown in Figure 6.3, together with the orbitals onthe central atom whose symmetry is given in the character table for theC3v point group (Table 6.6).The set of orbitals that we have found for the E representations,φE (1) and φE (2), is not unique.
Any pair of independent linear combinations of these two orbitals also constitutes a basis for this representation,and the same is true for the 2px and 2py orbitals on the central atom.The φE (1) and φE (2) orbitals shown here are, however, the set that isused most frequently. The first is transformed in the same way as 2px byall the symmetry operations of the C3v point group and the second like2py .
They are therefore often written ex and ey , respectively.6.5. Construction of MO: H2 O as an exampleThe problem of allowing AO, or SALCO of these orbitals, to interact toform molecular orbitals (MO), is simplified considerably if the symmetryproperties of the system are taken into account. The use of symmetryallows us to identify rapidly those interactions which are exactly zero,and therefore to consider only those which really do contribute to theformation of the MO.6.5.1. Symmetry and overlapTwo orbitals φ1 and φ2 interact if their overlap is non-zero (Chapter 1, §1.3).
This overlap is equal to the integral over all space of the product ofthe functions φ1 and φ2 :S12 =spaceφ1∗ φ2 dτ(6.13)It can be shown that this integral is non-zero if the product functionis a basis for the totally symmetric representation of the group (or of aElements of group theory and applicationsreducible representation that contains the totally symmetric representation).
It is therefore necessary for the two orbitals to have the same symmetry(see § 6.3.4) for the integral to be non-zero. If their symmetries are different, the integral is rigorously equal to zero, and their overlap is saidto be zero by symmetry.CommentThese rules (that are not proved here) can be illustrated by two simpleexamples. Consider the function with a single variable y = x 2 . The integral of this symmetric (or even) function between −a and +a is non-zero333([x 3 /3]+a−a = (a /3) − (−a /3) = 2a /3). However, the integral of theantisymmetric (or odd) function y = x 3 , which changes sign when x is44replaced by −x, is equal to zero ([x 4 /4]+a−a = (a /4) − (a /4) = 0).6.5.2. Molecular orbitals for H2 OTo construct the MO of the H2 O molecule, we shall allow the atomicorbitals of the central oxygen atom to interact with the symmetryadapted orbitals on the hydrogen atoms (the fragment method).
Wehave already established the symmetry properties of the various orbitals(C2v point group), and they are repeated in Figure 6.4. It is easy to checkvisually that the orbitals with the same symmetry on two different fragments have a non-zero overlap, whereas those with different symmetrieshave zero overlap (‘by symmetry’).The interaction diagram is shown in Figure 6.5, where the orbitalson the fragments and in the full molecule are labelled by symmetry(a1 , b2 , . . .).There is an interaction between the two orbitals with B2 symmetry that leads to the formation of a bonding MO, written 1b2 , andan antibonding MO, written 2b2 . As there are three orbitals with A1OA1H1H21H1H2B2pyAOB2Figure 6.4. Symmetry-adapted orbitals forH2 O (C2v point group) and orbitals with thesame symmetry on the central atom.H1H22pz2sB1H2H12px2Symmetry-adapted orbitals in several MLn complexes3a12b2b2b11b1b2a1a12a11b2a1Figure 6.5.
Construction of the MOs forH2 O from AO on oxygen and thesymmetry-adapted orbitals on thehydrogen atoms.1a1symmetry, three molecular orbitals are formed: 1a1 (bonding), 2a1 (nonbonding), and 3a1 (antibonding). The atomic orbital with B1 symmetry(2px ) cannot, by symmetry, interact with any other, and so it staysunchanged in shape and in energy (1b1 , nonbonding).6.6. Symmetry-adapted orbitals in severalMLn complexesThe aim of this section is to construct the symmetry-adapted orbitalsfor the principal ligand fields, making use of the reduction (6.5) andprojection (6.10) formulae. In what follows, with the exception of §6.6.6, we shall only consider a single orbital on each ligand, the onethat is used to create the σ bond with the metal (Chapter 1, § 1.5.1).The orbital on ligand Li will be written σi .
![](https://s.studizba.com/z.php?f=/templates/si/i/noavatar.png&w=100&h=100&t=1"){kind=avatar}
