Formulating SALCs with Projection Operators

Formulating SALCs with Projection Operators U The mathematical form of a SALC for a particular symmetry species cannot always be deduced by inspection (e.g., e1g and e2u pi-MOs of benzene). U A projection operator is a function that acts on one wave function of the basis set of functions that comprise the SALCs (e.g., one of the six pz orbitals on the carbon atoms in the ring of benzene) to “project out” the SALC function. U A projection operator for each symmetry species must be applied to the reference function to generate all the symmetry-allowed SALCs. U The projection operator for a given symmetry species contains terms for each and every operation of the group (not just each class of operations). Full-Matrix vs. Character Form of the Projection Operator U The full form of the projection operator function for degenerate species, which often directly generates the set of all SALCs belonging to a degenerate symmetry species, requires use of the full operator matrix for each and every operation of the irreducible representation; i.e., the full-matrix form of the irreducible representation. U Because there are no generally available tabulations of the full-matrix forms of the irreducible representations for groups with degenerate species, a simpler form of the projection operator that uses only the characters for each operation is most often used. ; The character form of the projection operator for degenerate species generates only one of the degenerate SALCs, requiring other means to deduce the companion functions. L We will only use the character form of the projection operator function. The Projection Operator in Characters U The projection operator in character form, Pi, acting on a reference function of the basis set, φt, generates the SALC, Si, for the ith allowed symmetry species as d R S i % Pi φt ' i j χi Rjφt h R in which di = dimension of the ith irreducible representation, h = order of the group, χiR = each operation's character in the ith irreducible representation, Rj = the operator for the jth operation of the group. U The term Rjφt gives one of the several basis functions of the set of functions forming the SALCs, in a positive or negative sense. U The summation is taken over all operations of the group, not all classes of operations. U The results Piφt are not the final SALCs; they need “cleaning up”. Requirements for a Wave Function 1. The function must be normalized. N2IΨΨ*dτ = 1, where N is the normalization constant. L We can routinely ignore the factor di /h in the projection operator expression. 2. All wave functions must be orthogonal. IΨiΨjdτ = 0 if i … j Mathematical Simplification When Taking Products of LCAOs U The Piφt products have the general form (aiφi ± ai+1φi+1 ... ± anφn)(bjφj ± bj+1φj+1 ... ±bmφm) U But in general Iφiφjdτ = δij Thus, all φiφj (i … j) terms vanish, and all φiφi or φjφj terms (i = j) are unity. L Ignore the cross terms when normalizing or testing for orthogonality! The σ-SALCs of MX6 (Oh) Example: Generate the SALCs for sigma-bonding of six ligands to a central atom in an octahedral MX6 complex. U From the transformation of six octahedrally arranged vectors pointing toward a central atom, the reducible representation for the six SALCs can be generated and reduced as follows: Γσ = A1g + Eg + T1u c d σ1 a σ 4 b σ6 σ5 a σ3 b σ2 d c Minimizing the Work U Oh has 48 operations, so each projection operator will have 48 terms. U The rotational subgroup O has half as many operations (h = 24) but still preserves the essential symmetry. L Carry out the work in O and correlate the results to Oh. U In O, Γσ = A1 + E + T1, which has obvious correlations to Γσ = A1g + Eg + T1u in Oh. Labeling the Symmetry Elements c d σ1 a σ 4 b σ6 σ5 a σ3 b σ2 d c U 8C3 in O refers to 4C3 and 4C32 whose axes run along the cube diagonals. L Label these by the corners through which they pass: e.g., aa, bb, cc, dd. U 3C2, 3C4, and 3C43 have axes that run through trans-related pairs of ligands. L Label these by the pairs of ligands through which they pass; e.g., 12, 34, 56. U 6C2' have axes that pass through the mid-points of opposite cube edges. L Label these by the two-letter designation of the cube edges through which they pass; e.g., ac, bd, ab, cd, ad, bc. U C3 rotations are clockwise, viewed from the upper cube corner through which the axis passes. U C4 rotations are clockwise, viewed from the lower-numbered ligand. The Effect of the Operations of O on a Reference Function σ1 C3 C3 C3 C3 C3 2 C3 2 C3 2 C3 2 C2 C2 C2 aa bb cc dd aa bb cc dd 12 34 56 Rjσ1 σ1 σ5 σ3 σ6 σ4 σ3 σ6 σ4 σ5 σ1 σ2 σ2 C4 C4 C4 C4 3 C4 3 C4 3 C2 ' C2 ' C2 ' C2 ' C2 ' C2 ' 12 34 56 12 34 56 ac bd ab cd ad bc σ1 σ5 σ4 σ1 σ6 σ3 σ2 σ2 σ3 σ4 σ5 σ6 O E label c d σ1 a σ 4 b σ6 σ5 a σ3 b σ2 d c Projection Operator for the A1 Species in O (A1g in Oh) C3 C3 C3 C3 C3 2 C3 2 C3 2 C3 2 C2 C2 C2 aa bb cc dd aa bb cc dd 12 34 56 Rjσ1 σ1 σ5 σ3 σ6 σ4 σ3 σ6 σ4 σ5 σ1 σ2 σ2 A1 1 1 1 1 1 1 1 1 1 1 1 1 σ5 σ3 σ6 σ4 σ3 σ6 σ4 σ5 σ1 σ2 σ2 C4 C4 C4 C4 3 C4 3 C4 3 C2 ' C2 ' C2 ' C2 ' C2 ' C2 ' 12 34 56 12 34 56 ac bd ab cd ad bc σ1 σ5 σ4 σ1 σ6 σ3 σ2 σ2 σ3 σ4 σ5 σ6 1 1 1 1 1 1 1 1 1 1 1 1 σ1 σ5 σ4 σ1 σ6 σ3 σ2 σ2 σ3 σ4 σ5 σ6 O E label χiRRjσ1 σ1 Summing all the χiRRjσ1 terms gives P(A1)σ1 % 4σ1 + 4σ2 + 4σ3 + 4σ4 + 4σ5 + 4σ6 % σ1 + σ2 + σ3 + σ4 + σ5 + σ6 SALC for A1g Normalizing: N2I(σ1 + σ2 + σ3 + σ4 + σ5 + σ6)2dτ = N 2I(σ12 + σ22 + σ32 + σ42 + σ52 + σ62)dτ = N 2(1 + 1 + 1 + 1 + 1 + 1) = 6N 2 / 1 Y N = 1//6 Therefore, the normalized A1g SALC is Σ1(A1g) = 1//6(σ1 + σ2 + σ3 + σ4 + σ5 + σ6) Projection Operator for the First of Two E SALCs (Eu in Oh) U In O, χR = 0 for 6C4 and 6C2'. Therefore, ignore the last 12 terms. O E label C3 C3 C3 C3 C3 2 C3 2 C3 2 C3 2 C2 C2 C2 aa bb cc dd aa bb cc dd 12 34 56 Rjσ1 σ1 σ5 σ3 σ6 σ4 σ3 σ6 σ4 σ5 σ1 σ2 σ2 E 2 -1 -1 -1 -1 -1 -1 -1 -1 2 2 2 2σ1 -σ5 -σ3 -σ6 -σ4 -σ3 -σ6 -σ4 -σ5 2σ1 2σ2 2σ2 χiRRjσ1 Summing across all χiRRjσ1 gives P(E)σ1 % 4σ1 + 4σ2 – 2σ3 – 2σ4 – 2σ5 – 2σ6 % 2σ1 + 2σ2 – σ3 – σ4 – σ5 – σ6 Normalizing: N 2I(2σ1 + 2σ2 – σ3 – σ4 – σ5 – σ6)2dτ = N 2I(4σ12 + 4σ22 + σ32 + σ42 + σ52 + σ62)dτ = N 2(4 + 4 + 1 + 1 + 1 + 1) = 12N 2 / 1 Y N = 1//12 = 1/(2/3) After normalization: Σ2(E) = 1/(2/3)(2σ1 + 2σ2 – σ3 – σ4 – σ5 – σ6) Test of Orthogonality with Σ1(A1) I[Σ1(A)][Σ2(E)]dτ = I(σ1 + σ2 + σ3 + σ4 + σ5 + σ6)(2σ1 + 2σ2 – σ3 – σ4 – σ5 – σ6)dτ =2+2–1–1–1–1=0 L But Σ2(E) is only the first of two degenerate functions. ; How do we find the partner? Finding the Partner to Σ2(E) - Method I Method I: Apply the E Projection operator to another reference function, here σ3. O E label C3 C3 C3 C3 C3 2 C3 2 C3 2 C3 2 C2 C2 C2 aa bb cc dd aa bb cc dd 12 34 56 Rjσ3 σ3 σ1 σ6 σ2 σ6 σ5 σ1 σ5 σ2 σ4 σ3 σ4 E 2 -1 -1 -1 -1 -1 -1 -1 -1 2 2 2 2σ3 -σ1 -σ6 -σ2 -σ6 -σ5 -σ1 -σ5 -σ2 2σ4 2σ3 2σ4 χiRRjσ3 This gives P(E)σ3 % –2σ1 – 2σ2 + 4σ3 + 4σ4 – 2σ5 – 2σ6 % –σ1 – σ2 + 2σ3 + 2σ4 – σ5 – σ6 Orthogonal to Σ1(A)? I(σ1 + σ2 + σ3 + σ4 + σ5 + σ6)( –σ1 – σ2 + 2σ3 + 2σ4 – σ5 – σ6)dτ = –1 – 1 + 2 + 2 – 1 – 1 = 0 YYes Orthogonal to Σ2(E)? I(2σ1 + 2σ2 – σ3 – σ4 – σ5 – σ6)(–σ1 – σ2 + 2σ3 + 2σ4 – σ5 – σ6)dτ = –2 – 2 – 2 – 2 + 1 + 1 = –6 … 0 YNo! ; What went wrong? Notes on Method I for Finding a Degenerate Partner L Changing the reference function after finding the first member of a degenerate set implicitly changes the axis orientation. The resulting function will be one of the following: • A legitimate partner wave function in either its positive or negative form. [Not this time!] • The negative of the first member of the degenerate set. [Not a useful result, and not what happened this time.] • A linear combination of the first member with its partner(s). If this is the case, try various combinations of the two projected functions to find a function that is orthogonal. [Looks like this must be what we got!] Finding the Partner to Σ2(E) - Method I By trial and error with various combinations we find that this works: P(E)σ1 = 2σ1 + 2σ2 – σ3 – σ4 – σ5 – σ6 2P(E)σ3 = –2σ1 – 2σ2 + 4σ3 + 4σ4 – 2σ5 – 2σ6 3σ3 + 3σ4 – 3σ5 – 3σ6 % σ3 + σ4 – σ5 – σ6 This result is orthogonal to Σ2(E), I(2σ1 + 2σ2 – σ3 – σ4 – σ5 – σ6)(σ3 + σ4 – σ5 – σ6)dτ =0+0–1–1+1+1=0 and also to Σ1(A1), I(σ1 + σ2 + σ3 + σ4 + σ5 + σ6)(σ3 + σ4 – σ5 – σ6)dτ =0+0+1+1–1–1=0 The normalized partner wave function, then, is Σ3(E) = ½(σ3 + σ4 – σ5 – σ6) Finding the Partner to Σ2(E) - Method II Method II: Apply an operation of the group to the first function found. Principle: The effect of any group operation on a wave function of a degenerate set is to transform the function into the positive or negative of itself, a partner, or a linear combination of itself and its partner or partners. L Suppose we perform C3(aa) on our first degenerate function. U Effect of C3(aa) on the functions of the basis set: Before σ1 σ2 σ3 σ4 σ5 σ6 After U σ5 σ6 σ1 σ2 σ3 σ4 Effect of C3(aa) on Σ2(E): (2σ1 + 2σ2 – σ3 – σ4 – σ5 – σ6) 6 (2σ5 + 2σ6 – σ1 – σ2 – σ3 – σ4 ) = (–σ1 – σ2 – σ3 – σ4 + 2σ5 + 2σ6) Is this new function orthogonal to Σ2(E)? I(2σ1 + 2σ2 – σ3 – σ4 – σ5 – σ6)(–σ1 – σ2 – σ3 – σ4 + 2σ5 + 2σ6)dτ = –2 – 2 – 1 – 1 – 2 – 2 = –10 … 0 YNo! Finding the Partner to Σ2(E) - Method II The new function must be a combination of Σ2(E) and the partner function. By trial-and-error: –Σ2: –2{Σ2 × R[C3(aa)]} –2σ1 – 2σ2 + σ3 + σ4 + σ5 + σ6 +2σ1 + 2σ2 + 2σ3 + 2σ4 – 4σ5 – 4σ6 +3σ3 + 3σ4 – 3σ5 – 3σ6 % σ3 + σ4 – σ5 – σ6 This is the same result as found by Method I: Σ3(E) = ½(σ3 + σ4 – σ5 – σ6) Projection Operator for the First of Three T1u SALCs Note: In T1 of the group O, for 8C3 χ = 0; therefore, skip the first eight terms in the operator expression. O E label C3 C3 C3 C3 C3 2 C3 2 C3 2 C3 2 C2 C2 C2 aa bb cc dd aa bb cc dd 12 34 56 Rjσ1 σ1 σ5 σ3 σ6 σ4 σ3 σ6 σ4 σ5 σ1 σ2 σ2 T1 3 0 0 0 0 0 0 0 0 -1 -1 -1 -σ1 -σ2 -σ2 χiRRjσ1 3σ1 C4 C4 C4 C4 3 C4 3 C4 3 C2 ' C2 ' C2 ' C2 ' C2 ' C2 ' 12 34 56 12 34 56 ac bd ab cd ad bc σ1 σ5 σ4 σ1 σ6 σ3 σ2 σ2 σ3 σ4 σ5 σ6 1 1 1 1 1 1 -1 -1 -1 -1 -1 -1 σ1 σ5 σ4 σ1 σ6 σ3 -σ2 -σ2 -σ3 -σ4 -σ5 -σ6 Summing across the χiRRjσ1: P(T1)σ1 % 4σ1 – 4σ2 % σ1 – σ2 We can readily show that this is orthogonal to the previous three functions for A and E, and on normalization we obtain the SALC G4(T1) = 1//2 (σ1 – σ2) Finding the Partner T1u SALCs The partner functions become apparent from considering the form of G4(T1). c d σ1 a b a b σ2 d L c The other two functions must have the same form along the other two orthogonal axes: G5(T1) = 1//2 (σ3 – σ4 ) G6(T1) = 1//2 (σ5 – σ6 ) U Alternately, note the effects of C3(aa) and C32(aa) on (σ1 – σ2): R[C3(aa)] × (σ1 – σ2) = (σ5 – σ6 ) Y G6(T1) R[C32(aa)] × (σ1 – σ2) = (σ3 – σ4 ) Y G5(T1) Summary: The Six σ-SALCs of MX6 (Oh) Σ1(A1g) = 1//6(σ1 + σ2 + σ3 + σ4 + σ5 + σ6) Σ2(Eu) = 1/(2/3)(2σ1 + 2σ2 – σ3 – σ4 – σ5 – σ6) Σ3(Eu) = ½(σ3 + σ4 – σ5 – σ6) G4(T1g) = 1//2 (σ1 – σ2) G5(T1g) = 1//2 (σ3 – σ4 ) G6(T1g) = 1//2 (σ5 – σ6 ) The σ-SALCs of CH4 (Td) z B A y D x C Γ = A1 + T 2 U The A1 SALC is obvious, so only use the projection operator for the T2 SALCs. U The group Td has h = 24, but its rotational subgroup T has h = 12. Therefore, do the work in T, where the T species corresponds to T2 of Td . U For the irreducible representation T, the only non-zero characters are E, C2(x), C2(y), C2(z), so the T operator has only four non-vanishing terms: T E ... C2(x) C2(y) C2(z) RjsA sA ... sC sD sB T 3 ... –1 –1 –1 –sD –sB χiRRjsA 3sA ... –sC The σ-SALCs of CH4 (Td) P(T)sA % 3sA – sC – sD – sB U This function is orthogonal to the A SALC, Φ1 = ½(sA + sB + sC + sD), and could be normalized to give the function Φ(T) = 1/(2/3)(3sA - sC - sD - sB) ; This SALC is supposed to match with one of the 2p AOs on carbon. Φ(T) makes no sense geometrically for such overlap! ( P(T)sA must be a combination of all three functions: P(T z)sA % sA + sB – sC – sD P(T y)sA % sA – sB – sC + sD P(T x)sA % sA – sB + sC – sD ______________________________ P(T)sA % 3sA – sB – sC – sd U After normalization, the three T2 SALCs of MX4 (Td) are: Φ2 = ½{sA + sB – sC – sD} Φ3 = ½{sA – sB – sC + sD} Φ4 = ½{sA – sB + sC – sD}