5.03, Inorganic Chemistry Prof. Daniel G. Nocera Lecture 2 May 11: Ligand Field Theory

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1 5.03, Inorganc Chemstry Prof. Danel G. Nocera Lecture May : Lgand Feld Theory The lgand feld problem s defned by the followng Hamltonan, h p Η = wth E n = KE = where = m m x y z h m Ze r hydrogen atom (e problem) N h m = Z * e r N N = j= j e r j N = ρ (R ) R r dτ R free on or atom lgand feld The Hamltonan contans three types of terms arsng from the electrons nteractng wth the postvely charged nucleus of the atom (defned by e nteractons, r ), e nteractons arsng from e e repulsons (r j ), the nteracton of electrons wth electron cloud of the lgands ( R-r where R s the poston of the lgand),

2 The electronc part of the Hamltonan s spectroscopcally probed n two regmes, dependng on the strength of the lgand feld. In the weak feld, the e energes are greater than the e - energes,.e. Q j >> V r j of whch O s a measure of V Oh. So the strategy here s to frst determne state symmetres followed by applcatons of lgand feld. O h In the strong feld, the e energes (orbtal energes) are frst determned, followed by the perturbaton of e terms begn wth confguratons and see how they are perturbed by Q j. The overall approach to determnng the energy levels of a molecule s: determne free on states determne how atomc states are splt n a weak feld lmt (PE >> O ) determne states arsng from confguratons n a strong feld lmt (PE << O ) correlate between strong and weak felds (Tanabe-Sugano dagrams).

3 The Weak Feld The problem begns by consderng a free on (K h symmetry). Dfferent atomc confguratons gve rse to dfferent terms or energy levels. The many e system s characterzed by a total orbtal and total spn angular momentum, total orbtal angular momentum, L L L = 0,,, 3 = n M m = L, L, L, L L = n total spn angular momentum, S S = n 3,, s S =... M m M s = S, S S, S s = n s A term s symbolzed by L = 0,,, 3, 4 S L J S P D F G J = LS, LS, LS. L S, L S We wll therefore need to ) determne terms of a gven confguraton ) determne the egenfuncton of these terms 3) calculate terms energes 3

4 As an llustratve example, consder the p confguraton, 0 A shorthand notaton for ths confguraton s ( 0), whch s a Slater determnant ( 0) = () (0) () (0) = (, 0) (0,) All the possble confguratons are summarzed n the followng table whch defnes the confguratons n terms of and M S M s 0 (, ) (,0 ) (,0 ) (,0) (,0) 0 (0,0) (, ) (, ) (, ) (, ) (0, ) (0, ) ( 0, ) ( 0, ) (, ) The table summarzng the number of consttuent confguratons n terms of and M S, M S Consderng the defntons of L and S, begn wth hghest wavefuncton and elmnate e wavefunctons from L to L and S to S. The hghest s, M S = 0 =, M S = 0, whch s a D term 4

5 Removng the count of confguratons arsng from a D term (.e., from =,, 0,, and M S =, 0, ), leaves M S 0 0 The hghest s and hghest M S of ths value s also, thus gvng rse to a 3 P term Agan, elmnatng states of the 3 P term, M S 0 0 Ths leaves only one confguraton of = 0 and M S = 0, whch s a S term Thus the terms arsng from a p confguraton are: D, 3 P, and S. May predct the ground state from Hund s rule: ) state wth maxmum M S s lowest n energy ) for states of the same spn multplcty, the state wth largest wll be lowest n energy 3) note, Hund s rule does not address excted state orderng Hund s rule dentfes 3 P as the ground state. Summarzng the energy (term) levels S ths orderng must be determned from spectroscopy p D 3 P 5

6 How about the d on states? d on s straghtforward. There are 0 permutatons of an electron n 5 d orbtals, s = ½ or s = ½ n orbtals of m of to M S 0 0 Ths gves =,, 0,, and M S =, 0,, thus gvng values of L = and S = or a D state. Ths s the only term for the d confguraton d on has many terms that result from the permutaton of e n the orbtal subshells derved from =,m =,, 0,, M s (,) (,0 ) (,0) (,) (,) (,) (,) (,0) (,) (,0) (, ) (,0) 0 (, ) (, ) (, ) (,0) (,0) (, ) (, ) (, ) (, ) ( 0,0) (, ) (, ) (,0) (, ) (, ) where the terms are not shown for the convenence of space. Reducng the terms yelds for a d system: 3 F, 3 P, G, D and S 6

7 The Russell-Saunders terms for all d n free ons: S D P D G D S D S G 3 F 3 P F 4 E, D F G, H G 3 P D F P H P G 4 P F D S 3 D G I 3 P 3 F 3 G 3 H I 4 P, 4 D 4 G D 3 F 4 F 5 D 6 S d,d 9 d,d 8 d 3,d 7 d 4,d 6 d 5 Now need to determne the states that arse from the applcaton of an O h feld on the free on. The degeneracy of the wavefuncton s removed upon applcaton of the lgand feld. The lgand feld causes the terms to splt owng to the change n symmetry from K h to O h. We can use the bass functons of the Oh character table to determne how the fee on states splt, 7

8 0 E 6C 4 3C ( C 4 ) 8C 3 6C l = 0 Γ s A g l = Γ p 3 0 T u l = Γ d 5 E g T g Γ f = A u T u T u Γ g = A g E g T g T g Takng the terms of a p on n a sphercal feld (K h ) to an O h lgand feld leads to the further splttng, The Strong Feld The problem n the strong feld begns wth the electron confguratons derved from the molecular orbtal. As an example, consder the strong feld confguratons for two electrons n an octahedral lgand feld. The two-electron occupancy of the ML 6 (O h ) MO gves rse to three confguratons: (t g ) < (t g ) (e g ) < (e g ) ground state e exctaton e exctaton Must now determne what states from d weak feld correlate to these confguratons. Because the electrons nteract dfferently wth each other, when n dfferent orbtals (dfferent r j ), for a gven electronc confguraton, many states can arse. 8

9 Let s begn wth the e excted state, (t g ) (e g ). The state symmetry takes the symmetry of the ndvdually occuped orbtals, and the electrons can be snglet or trplet pared, (χ χ ) and 3 (χ χ ) t g x e g = T g T g The spn can be S = 0 (snglet) or S = (trplet), thus all potental states that arse from the one electron exctaton of a d on n a O h feld are: T g T g 3 T g 3 T g The (t g ) and (e g ) confguratons are more problematc because the electrons are degenerate. The drect product electrons n n degenerate orbtals of χ and χ symmetry wll gve χ χ orbtal symmetres. (t g ) t g x t g = A g E g T g T g (e g ) e g x e g = A g A g E g But there s a problem, t snglet and trplet spn parng cannot be assgned to each state because some of these states wll volate the Paul excluson prncple. In 5.04 a procedure wll be derved that shows how to elmnate these Paul volatng states. Here the result wll be gven: (t g ) t g x t g = A g E g T g 3 T g (e g ) e g x e g = A g 3 A g E g The states created n the weak feld must be the same as the states obtaned n the strong feld snce we are dealng wth the d problem, just from dfferent perspectves. We see ths s the case. A correlaton dagram relates the weak and strong feld confguratons. The d correlaton dagram s: 9

10 Weak Feld Strong Feld Note that for d (d n ) n O h, the same dagram s obtaned for d 8 (d 0 n ) T d. T d symmetry does not requre the g or u subscrpts. 0