Ligand Field Theory Notes

Transcription

1 Ligand Field Theory Notes Read: Hughbanks, Antisymmetry (Handout). Carter, Molecular Symmetry..., Sections Cotton, Chemical Applications..., Chapter 9. Harris & Bertolucci, Symmetry and Spectroscopy..., Chapter 5, pp , pp States and Configurations Each of the electronic states of an atom or a molecule are usually described as being built up from one (or perhaps more than one) configuration. The physical origin of energy differences between atomic states for a single configuration is e - -e - repulsion. Hund s Rules are based on the systematics of differential e - -e - repulsion.

2 Hund s Rules for Atomic Ground States For states of an atom or a molecule built up from one configuration. 1st rule: Ground state has maximum spin multiplicity (largest value of S). 2nd rule: For a given value of S, state with maximum L is most stable. 3rd rule: For a given S and L, the min. J state most stable if shell is < 1 / 2 -filled & max. J most stable is shell is > 1 / 2 -filled. Examples States for a Ti 2+ ion with a [Ar]3d 2 configuration? States for a Cr 3+ ion with a [Ar]3d 3 configuration?

3 Splitting of Atomic States in Chemical Environments How do atomic states split in non-spherical chemical environments? Before addressing this question, we need a general formula for the character for a rotation operation by angle α when operated on a state with angular momentum L: χ(c α ) = sin(l +1 2)α sinα 2 Mathematical Origin of Formula C α e imϕ e im(ϕ α ) = e i( m)α e imϕ C α e e ilϕ i( L 1)ϕ! e 0! e i(1 L)ϕ i( L)ϕ e = e i( L)α e ilϕ e i(1 L)α e! e 0! i( L 1)ϕ e i( L 1)α e i(1 L)ϕ e ilα i( L)ϕ e e i( L)α e i(1 L)α 0! = e 0 0! i( L 1)α e e ilα e e ilϕ i( L 1)ϕ! e 0! e i(1 L)ϕ i( L)ϕ e χ(c α ) = e i( L)α + e i(1 L)α +!+ e 0 +!+ e i( L 1)α + e ilα = z L+ 1 ( 2) z L+ 1 2 z 1 2 z 1 2 [do clever factoring and use geometric series: z n = 1 (1 z) with z = e iα ] 0 sin(l +1 2)α χ(c α ) = sinα 2 A more complete analysis shows that is a result of angular momentum commutation relations. ( z = e iα )

4 Formulas for All Operations Carter, p. 205 χ(c α ) = χ(s α ) = ± χ(e) = 2J + 1 sin[(j + 1 2)α] sin(α 2) χ(i) = ±(2J + 1) sin[(j + 1 2)(α + π )] sin[(α + π ) 2] χ(σ ) = ±sin[(j + 1 2)π ] + signs apply for a gerade atomic state and signs apply for an ungerade atomic state, whether the point group under consideration has inversion symmetry or not. Examples: a 3 P state derived from either p 2 configuration (u u) or a d 2 configuration (g g) give g states. However, a 2 D state derived from a p 3 configuration (u u u) gives a u state, while a 2 D state derived from a d 3 configuration (g g g) gives a g state. The symbol J refers to the angular momentum quantum number of the state under consideration. In the Russell- Saunders scheme, J can be replaced by L when considering a spatial wave function. States from Molecular Configurations; the d 2 case Electronic configurations for which there is no possibility of violating the Pauli principle are easy to handle: For example, the (t 2g ) 1 (e g ) 1 configuration e g t 2g

5 States from Molecular Configurations; the d 2 case Configurations for which Pauli principle violations can occur must be handled carefully: The t 2g 2 or e g 2 configuration e g e g t 2g t 2g Antisymmetric Wavefunctions (Pauli Principle)! Read: Antisymmetry handout; Harris & Bertolucci, Symmetry and Spectroscopy..., Chapter 5, pp ! Wavefunctions describing Fermions (e.g., electrons) must be antisymmetric with respect to interchange of any two particles (electrons) labels. Ψ(1,2, i,, j, N ) = Ψ(1,2, j,,i, N )

6 Singlets and Triplets If two electrons occupy two different orbitals, one can form either a singlet or triplet state. Including spin functions, these have the form: S = 1; 2S + 1 = 3 ( ) 3 Ψ = 1 2 ϕ a (1)ϕ b (2) ϕ b (1)ϕ a (2) S = 0 1 Ψ = 1 2 ϕ a (1)ϕ b (2) + ϕ b (1)ϕ a (2) 1 α 1 α 2 ( ) α β + β α β 1 β 2 ( ) 12 (α 1 β 2 β 1 α 2 ) 0 M S Singlets: Antisymmetric Wavefunctions (Pauli Principle) 1 Ψ = 1!#### Symmetric "#### $! Antisymmetric ##" ## $ ( 2 ϕ a (1)ϕ b (2) +ϕ b (1)ϕ a (2) 1 ) (α 2 1 β 2 β 1 α 2 ) M S 0 Triplets:!##### Antisymmetric "##### $!### Symmetric "### $ 3 Ψ = 1 α 1 α 2 1 ( ϕ a (1)ϕ b (2) ϕ b (1)ϕ a (2)) ( 2 2 α 1 β 2 + β 1 α 2 ) β 1 β

7 Antisymmetric Direct Product If two electrons occupy degenerate orbitals, one can form either a singlet or triplet state. The irreducible representation of the triplet state can be determined by finding the antisymmetric direct product*: χ (R) = 1 ( 2 χ 2 (R) χ(r 2 )) *See Antisymmetry handout for proof. Aside: Some Extra Details The hydrogen atom is a physical problem with much higher symmetry than molecules; the potential energy exerted by the nucleus has spherical symmetry. The role of symmetry is enormous! Nevertheless, that isn t exploited in elementary treatments because the group theory is more demanding. (With great power comes greater difficulty!) We won t go there in this course, but here are a few suggestive slides

8 Aside: SO(3): The Full Rotation Group* SO(3) E! C α! Normalized Basis Functions (Spherical Harmonics) 1 Γ l=0 1! 1! (s orbitals) 2 π sin(3 2)α 1 3 Γ l=1 3!! sinα 2 cosθ, 1 3 sinθ 2 π 2 2π e±iϕ (p orbitals) sin(5 2)α Γ l=2 5!! 1 5 sinα 2 4 π (3cos2 θ 1), 1 15 cosθ sinθ 2 2π e±iϕ, 1 4 " "! "! " sin(l + 1 2)α Γ l 2l + 1!! Y sinα 2 lm (θ,ϕ),m = l, l + 1,!,l 1,l " "! "! " 15 2π sin2 θ e ±2iϕ (d orbitals) The Spherical Harmonics for each l form a basis for (2l +1)-degenerate irreducible representation, Γ l=n. From a group theoretical point of view, that s why the hydrogen-atom s p, d, f,... orbitals are degenerate (within each set). SO(3) is an infinite-order group; in fact, each rotation Cα belongs to an infinite-order class. Consequently, there are an infinite number of irreducible representations, and the order of the I.R. s grows without bound. *Special Orthogonal Group of Order 3 Aside: SO(3) used to get term symbols Group theory can be used to get atomic terms; sin(l +1 2)α χ(c α ) = sinα 2 Example, what are the terms for a d 2 ion (l = 2)? χ 2 (C α ) = sin2 5α 2 ; χ(c 2 sin 2 α ) = sin5α α 2 sinα Since sin2 5α 2 sin3α 2 = 1+ sin 2 α 2 sinα 2 + sin5α 2 sinα 2 + sin7α 2 sinα 2 + sin9α 2 sinα 2, then Γ d Γ d = Γ S Γ P Γ D Γ F Γ G. But sin5α sin3α 2 = 1 sinα sinα 2 + sin5α 2 sinα 2 sin7α 2 sinα 2 + sin9α 2 sinα 2, So + Γ d Γ d = Γ P Γ F and Γ d Γ d = ΓS Γ D Γ G Atomic terms are therefore 3 P, 3 F, 1 S, 1 D, and 1 G

9 Aside, continued: In this case, the group theoretic approach isn t necessarily easier! The trigonometric proofs aren t simple To prove, sin2 (n + 1)α 2 = sinα 2 + sin3α 2 +!+ sin(2n sinα 2 )α = S We'll need the formulae: sin x = eix e ix ; cos x = eix + e ix ; sin 2 x 1 cos x = ; x l = 1 2i l=0 1 x Letting e iα = u and using the formula for sin x, we can write the sum S as S = u1 2 u u3 2 u 3 2 +!+ u2n+1 2 (2n+1 2) u 2i 2i 2i Gathering terms and factoring, S = 1 u 1 2 2n u l u 1 2 2n (u 1 ) l 2i l=0 l=0 Each term can summed using geometric series, u ±1 2 2n u ±l = u ±1 2 u ±l u ±(2n+1) u ±l l=0 l=0 l=0 = (1 u±(2n+1) ) u±1 2 1 u = (1 u±(2n+1) ). ±1 u 1 2 u 1 2 Substituting and rearranging, S = 1 (1 u 2n+1 ) (1 u (2n+1) ) 2i u 1 2 u 1 2 S = 1 2 (u 2n+1 + u (2n+1) ) (1 cos(2n +1)α ) 2 4 (1 2i)(u 1 2 u 1 2 = = sin2 (n + 1)α 2 ) sinα 2 sinα 2 Aside, continued To prove: sin5α sin3α 2 = 1 sinα sinα 2 + sin5α 2 sinα 2 sin7α 2 sinα 2 + sin9α 2 sinα 2 = S sin9α 2 sinα 2 = e 4iα + e 3iα + e 2iα + e iα + 1+ e iα + e 2iα + e 3iα + e 4iα sin7α 2 sinα 2 = e 3iα e 2iα e iα 1 e iα e 2iα e 3iα sin5α 2 sinα 2 = e 2iα + e iα + 1+ e iα + e 2iα sin3α 2 sinα 2 = e iα 1 e iα Summing, S = e 4iα + e 2iα +1+ e 2iα + e 4iα If we use the formula for sin5β 2 sinβ 2 and replace β 2 with α, we obtain the desired result!

10 Slater Determinants! Determinants have the antisymmetry property built-in : φ(i) = ϕ(i)[spinfunction(i)] Ψ(1,2, i,, j, N ) =!########### Different Spin Orbitals "########### in Columns $ (1) φ b (1)! φ p (1)! φ q (1)! φ z (1) (2) φ b (2)! φ p (2)! φ q (2)! φ z (2)!!!!! 1 (i) φ b (i)! φ p (i)! φ q (i)! φ z (i) N!!!!!! ( j) φ b ( j)! φ p ( j)! φ q ( j)! φ z ( j)!!!!! (N ) φ b (N )! φ p (N )! φ q (N )! φ z (N ) Different electrons in rows Slater Determinant - N atom 1s(1) 1s(1) 2s(1) 2s(1) 2 p x (1) 2 p y (1) 2 p z (1) 1s(2) 1s(2) 2s(2) 2s(2) 2 p x (2) 2 p y (2) 2 p z (2) 1s(3) 1s(3) 2s(3) 2s(3) 2 p x (3) 2 p y (3) 2 p z (3) 1 1s(4) 1s(4) 2s(4) 2s(4) 2 p x (4) 2 p y (4) 2 p z (4) 7! 1s(6) 1s(5) 2s(5) 2s(5) 2 p x (5) 2 p y (5) 2 p z (5) 1s(6) 1s(6) 2s(6) 2s(6) 2 p x (6) 2 p y (6) 2 p z (6) 1s(7) 1s(7) 2s(7) 2s(7) 2 p x (7) 2 p y (7) 2 p z (7) which is written as 2s 1s 2 p 1s 1s 2s 2s 2 p x 2 p y 2 p z

11 Slater Determinants-Properties notation: φ(i) = ϕ(i)[α(i) or β(i)] i = electron label Swap any two columns (or rows) and the determinant switches sign: φ A! φ P! φ Q! = φ A! φ Q! φ P! if any two columns (rows) are identical, the determinant vanishes! Columns (rows) can be factored: φ A! φ P + φ Q! φ R! = φ A! φ P! φ R! + φ A! φ Q! φ R! Any constant (including 1) can be factored out: φ A! c φ p! φ R! = c φ A! φ p! φ R! Example: Show that the form of the singlet and triplet wavefunctions I gave in "expanded out" form can be expressed as determinants.: 1 Ψ = 1 ( 2 ϕ (1)ϕ (2) + ϕ (1)ϕ (2) a b b a )( 12 (α β β α ) ) 0 3 Ψ = 1 2 ϕ a (1)ϕ b (2) ϕ b (1)ϕ a (2) Start with four determinants ( ) 1 α 1 α 2 ( ) α β + β α β 1 β 2 M S φ b, φ b, φ b, and φ b expand them out (using the notation and normalizing factors discussed in above), and show that you can recover the "expanded out" forms from the compact determinants. Hint: you will need to take combinations of determinants for both of the MS = 0 cases.

12 Exercise: Answers φ b = 1 2! (1) φ b (1) (2) φ b (2) = 1 2 ϕ a (1)α 1 ϕ b (1)α 1 ϕ a (2)α 2 ϕ b (2)α 2 = 1 ( 2 ϕ (1)α ϕ (2)α ϕ (1)α ϕ (2)α a 1 b 2 b 1 a 2 ) 3 Ψ = 1 ( 2 ϕ (1)ϕ (2) ϕ (1)ϕ (2) a b b a )α 1 α 2 (M S = 1) φ b = 1 2! (1) φ b (1) (2) φ b (2) = 1 2 ϕ a (1)β 1 ϕ b (1)β 1 ϕ a (2)β 2 ϕ b (2)β 2 = 1 ( 2 ϕ (1)β ϕ (2)β ϕ (1)β ϕ (2)β a 1 b 2 b 1 a 2 ) 3 Ψ = 1 ( 2 ϕ (1)ϕ (2) ϕ (1)ϕ (2) a b b a )β 1 β 2 (M S = 1) ( ) = 1 2 φ φ + φ b a b φ b = 1 2! φ b = 1 2! 3 Ψ = 1 2 ϕ a (1)ϕ b (2) ϕ b (1)ϕ a (2) (1) φ b (1) (2) φ b (2) = 1 2 ϕ a (1)β 1 ϕ b (1)α 1 ϕ a (2)β 2 ϕ b (2)α 2 = 1 ( 2 ϕ (1)β ϕ (2)α ϕ (1)α ϕ (2)β a 1 b 2 b 1 a 2 ) = 1 ( 2 ϕ (1)ϕ (2) β α ϕ (1)ϕ (2) α β a b 1 2 b a 1 2 ) (1) φ b (1) (2) φ b (2) = 1 2 ϕ a (1)α 1 ϕ b (1)β 1 ϕ a (2)α 2 ϕ b (2)β 2 = 1 ( 2 ϕ (1)ϕ (2) α β ϕ (1)ϕ (2) β α a b 1 2 b a 1 2 ) ( ) 1 ( 2 α β + β α ) (M S = 0) ( ) = 1 2 φ φ φ b a b ( ) 1 ( 2 α β β α ) (M S = 0) 1 Ψ = 1 2 ϕ a (1)ϕ b (2) + ϕ b (1)ϕ a (2)

13 Summary: Key Features of Slater Determinants Antisymmetry If one swaps any two rows or any two columns, the determinant changes sign. Pauli Principle If any two columns are identical, the determinant must be zero no two electrons can have the same quantum numbers. Adding (or subtracting) one column to (from) another column doesn t change Ψ. Orthonormality Slater determinants are normalized and determinants built from different orthogonal basis orbitals are orthogonal to each other. Symmetry operations on determinants act upon all the spatial parts of the orbitals that make up the determinants. Irreducible Reps and Determinants Closed shells always belong to the totally symmetric representation Example: show that the Slater determinant for a chloride ion belongs to the a 1 rep. in 4mm (C 4v ) symmetry. To determine the state symmetries for open shell molecules, work out orbitals and use properties of determinants. Example: What states arise from the ground configuration of cyclobutadiene (D 4h )?

14 D 4h E 2C 4 C 2 (C 2 4 ) 2C 2 2C 2 i 2S 4 σ h 2σ v 2σ d A 1g x 2 + y 2, z 2 A 2g R z B 1g x 2 y 2 B 2g xy E g (R x, R y ) (xz, yz) A 1u A 2u z B 1u B 2u E u (x, y) D 4 E 2C 4 C 2 (C 4 2 ) 2C 2 2C 2 A x 2 + y 2, z 2 A z, R z B x 2 y 2 B xy E (x, y), (R x, R y ) (xz, yz) Cyclobutadiene Example - Answers M S M S e a e b 1 1 B 1g : 1 2 ( e a e b e a e b ) A 2g e 2 ( a e b + e a e b ) 0 1 A 1g : 1 2 ( e a e a + e b e b ) 0 e a e b 1 1 B 2g : 1 e 2 ( a e a e b e b ) 0

15 A Worked Example: t 2g 2 in O h How many microstates (determinants)? What states arise from this configuration? Find one wavefunction for each state. Use qualitative reasoning to deduce the relative energies of the states (not strictly an exercise in group theory). O E 8C 3 3C 2 (= C 2 4 ) 6C 4 6C 2 A x 2 + y 2 + z 2 A E (2z 2 x 2 y 2, x 2 y 2 ) T (R x, R y, R z ); (x, y, z) T (xy, xz, yz)

16 d2 Correlation Diagram o o 0.0 Ti2+ energies (cm 1) d2 Tanabe-Sugano Diagram Semiquantitative correlation diagram connecting states of free ion and molecule All energies are taken relative to the ground state. In strong-field limit, curves group into families according to slopes, each of which is correlated with a config. States of like symmetry never cross.

17 Three Equivalent Electrons If three electrons occupy orbitals that are at least 3-fold degenerate, one can form either a doublet or quartet states. The characters of the reducible representations of the states are given by: χ S = 3 2 (R) = 1 ( 6 χ 3 (R) 3χ(R)χ(R 2 ) + 2χ(R 3 )) χ S =1 2 (R) = 1 ( 3 χ 3 (R) χ(r 3 )) Harris & Bertolucci, p For derivations using permutation groups, see D. I. Ford, J. Chem. Ed., 49, (1972). Four Equivalent Electrons If four electrons occupy orbitals that are at least 4-fold degenerate, one can form singlet, triplet, and quintet states. The characters of the reducible representations of the states are given by: χ S =0 (R) = 1 ( 12 χ 4 (R) 4χ(R)χ(R 3 ) + 3χ 2 (R 2 )) ( ) χ S =1 (R) = 1 8 χ 4 (R) 2χ 2 (R)χ(R 2 ) + 2χ(R 4 ) χ 2 (R 2 ) χ S =2 (R) = 1 ( 24 χ 4 (R) 6χ 2 (R)χ(R 2 ) + 8χ(R)χ(R 3 ) 6χ(R 4 ) + 3χ 2 (R 2 )) Harris & Bertolucci, p For derivations using permutation groups, see D. I. Ford, J. Chem. Ed., 49, (1972).

18 [Ni(H2O)6]2+ and [Ni(en)3]2+ [Ni(en)3]2+ Ni(H2O)6]2+ ~25,000 ~14,000 The Non-crossing Rule Illustrative Issues: What is the strength of the mixing between the 3F(3T ) and 3P(3T ) 1g 1g atomic states? In the strong-field limit, exactly how much of the two 3F(3T1g) and 3P(3T ) states contribute 1g to the two molecular 3T states? 1g ~9000

19 d8 Tanabe-Sugano Diagram Quantitative correlation diagram connecting states of free ion and molecule In strong-field limit, curves group into families according to slopes, each of which is correlated with a config. States of like symmetry never cross. [Ni(H2O)6]2+ and [Ni(en)3]2+ ~30,000 [Ni(en)3]2+ ~18,500 ~11,000 Ni(H2O)6]2+

20 d8 Tanabe-Sugano Diagram Quantitative correlation diagram connecting states of free ion and molecule In strong-field limit, curves group into families according to slopes, each of which is correlated with a config. States of like symmetry never cross. CrIII d-d spectra All 6-coordinate [Cr(en)3] [Cr(H2O)6] Use energy ratios = λ (nm) = lab_manuals/crtsexptnu/crexptj.html 700