Transcript Lecture 21
before we did: p2 ML & MS Microstate Table States (S, P, D) Spin multiplicity Terms 3P, 1D, 1S Ground state term 3P For metal complexes we need to consider d1-d10 d2 3F, 3P, 1G, 1D, 1S For 3 or more electrons, this is a long tedious process But luckily this has been tabulated before… Free Ion terms for various d configurations. We seek to understand The “Same as “ relationships What will happen when we put these free ion into a complex, octahedral or tetrahedral. The spectroscopy involved. Explanation of the mirror symmetry behavior of the free ion GS terms. Easy to understand using the hole formalism. dn has same terms as d10-n Recall that we are ignoring “core” electrons. We only work with partially filled subshells. Why? Because the core, complete set of atomic orbitals within a subshell is wholly occupied. All the orbital and spin angular momentum components cancel each other out and add to 0. Hole formalism: Consider d3. That configuration includes many different orbital occupancies. Here is one of those occupancies. It will participate on one or more of the terms. -2 -1 0 1 2 - 1/2 x x x 1/2 Note that Lz is -3 We claim d3 to yield the same terms as d7 Explanation of the mirror symmetry behavior of the free ion GS terms. Aim: d3 has same terms as d10-3 = d7 d7 allows for many different orbital occupancies. Here is one of them. -2 -1 - 1/2 x x 1/2 x x 0 1 2 x x x Note that Lz is -3 We claim this provides the same Lz and Sz as one of the d3 orbital occupancy schemes. Consider the above to be formed from a completed d subshell and thee holes to “create” the missing electrons. The holes look like this -2 - 1/2 1/2 -1 0 1 2 hole hole hole Selection rules (determine intensities) Laporte rule g g forbidden (that is, d-d forbidden) but g u allowed (that is, d-p allowed) Spin rule Transitions between states of different multiplicities forbidden Transitions between states of same multiplicities allowed These rules are relaxed by molecular vibrations, and spin-orbit coupling Group theory analysis of term splitting in octahedral. Notice the correspondence to the symmetry analysis of the atomic orbitals when they are split by an octahedral field. High Spin Ground States Recall that the set of d atomic electrons are split into a t2g set and an eg set by repulsion with the negative octahedral ligands, similarly for tetrahedral environment. So also are the free ion terms split by their environment. dn Free ion GS Oct. complex Tet complex d0 1S t2g0eg0 e0t20 d1 2D t2g1eg0 e1t20 d2 3F t2g2eg0 e2t20 d3 4F t2g3eg0 e2t21 d4 5D t2g3eg1 e2t22 d5 6S t2g3eg2 e2t23 d6 5D t2g4eg2 e3t23 d7 4F t2g5eg2 e4t23 d8 3F t2g6eg2 e4t24 d9 2D t2g6eg3 e4t25 d10 1S t2g6eg4 e4t26 We seek to understand the mirror like GS terms. dn and d10-n have the same GS terms for the free ion.. How to explain this via the ligand splitting that we know about (t2g and eg). High Spin Ground States Hole Formalism: The d10 electronic configurations is a spherical cloud of d electrons. We can move to fewer electrons, d9, d8, etc. by creating holes in the spherical shell. Ex: d9 = d10+ “d-hole” Electronic configuration d1 has the term 2D. Ligand splitting occurs for 2D. for electrons for holes embedded in spherical cloud 2 Eg eg 2 t2g T2g 2 d 2 d D D 2 t2g 2 T2g eg Eg d10-n electron configuration is equivalent to spherical d10 + holen It is the holen which determines the orbital and spin angular momentum and the terms. Think positive electrons. The term can be obtained from either the actual electronic configuration or the hole configuration. . High Spin Ground States Again, d10-n electron configuration is equivalent to spherical d10 + holen It is the holen which determines the orbital and spin angular momentum and the terms. The term can be obtained from either the actual electronic configuration or the hole configuration. But note that ligand splitting of holes will yield inverted splitting relative to electrons. Principle: Electrons are repelled by ligands, but holes attracted to ligands. Ex: d3 has t2g3eg0 electronic configuration (electrons repel ligands). Ex: high spin d7 is either electronic configuration of t2g5 eg2 (ligands destabilize electrons, favoring the t2g . orbitals) or hole3 with t2g1 eg2 hole configuration (ligands stabilize positive holes, favoring the eg) High Spin Ground States dn Free ion GS Oct. complex Tet complex d0 1S t2g0eg0 e0t20 d1 2D t2g1eg0 d2 3F t2g2eg0 e2t20 d3 4F t2g3eg0 e2t21 d4 5D t2g3eg1 e2t22 d5 6S t2g3eg2 e2t23 d6 5D t2g4eg2 e3t23 d7 4F t2g5eg2 e4t23 d8 3F t2g6eg2 e4t24 d9 2D t2g6eg3 d10 1S t2g6eg4 = t2g = eg e1t20 e4t25 e4t26 GS has t2g negative electron, repelling the ligands. Symmetry is T2g for d1 octahedral complex GS has eg positive d-hole, attracted to ligands. Symmetry designation of d9 octahedral complex is Eg High Spin Ground States, d1 and d9 tetrahedral complexes We have obtained the GS sym for octahedral d1 and d9. Now consider tetrahedral. Recall the reversed tetrahedral orbital splitting. For tetrahedral ligand field e electron destabilized less by ligands, t2 more so. Opposite for holes. dn Free ion GS Oct. complex Tet complex d0 1S t2g0eg0 e0t20 d1 2D t2g1eg0 e1t20 d2 3F t2g2eg0 e2t20 d3 4F t2g3eg0 e2t21 d4 5D t2g3eg1 e2t22 d5 6S t2g3eg2 e2t23 d6 5D t2g4eg2 e3t23 d7 4F t2g5eg2 e4t23 d8 3F t2g6eg2 e4t24 d9 2D t2g6eg3 e4t25 d10 1S t2g6eg4 e4t26 For d1 the electron is e GS. =e Symmetry is thus E for the d1 tet complex. = t2 For d9 or d-hole1 the d-hole symmetry is t2. The GS complex symmetry is T 2. High Spin Ground States: d4 and d6 octahedral and tetrahedral The d6 is regarded as a single electron (t2g) on top of the spherical d5 half shell (t2g3eg2) . For tetrahedral we have one electron (e) on top of half shell. But the spherical shell has spin which must spin be allowed for. Using the high multiplicities from the free ion this means 5T2g for octahedral d6 and, similarly, 5dEn for the tetrahedral d6GS . Free ion Oct. complex Tet complex The GS of the octahedral d4 complex will have symmetry of the eg hole, Eg. d0 1S t2g0eg0 e0t20 d1 2D t2g1eg0 e1t20 d2 3F t2g2eg0 e2t20 d3 4F t2g3eg0 e2t21 d4 5D t2g3eg1 d5 6S t2g3eg2 d6 5D t2g4eg2 Spin is taken from the free ion yielding 5Eg. d7 4F t2g5eg2 e4t23 d8 3F t2g6eg2 e4t24 d9 2D t2g6eg3 e4t25 d10 1S t2g6eg4 e4t26 = eg e2t22 = t2 e2t23 = t2g e3t23 =e Reversed splitting in tetrahedral leads to t2 for hole in d4 tetrahedral complexes. Taking spin from the free ion GS yields 5T2. Further hole discussion. A d4 can be considered as spherical d5 with a positive hole embedded in it. In octahedral complex the d-hole is eg symmetry. In tetrahedral the dhole is t2 symmetry. Orgel diagram for d1, d4, d6, d9 in both octahederal and tetrahedral complexes. Here is what we have: d1, d4, d6, d9 all have a d shaped object – either a d electron or d-hole – and thus are designated as a D term for the free ion high spin GS. Ligand splitting causes t2(g) electron to be lower than e(g), reversed in tetrahedral, reversed also for d-holes. Now look at Orgel diagram for these configurations. Look at right part of diagram first Both of these represent an d1 d6 electron on top of a spherical shell. AnToctahederal field splits 2g or T2 Energy d4 d9 D d4 into T2g and Eg (GS) E A tetrahedral splits d4 into T2 (GS) and E d1 into T2 and E (GS) Eg or d9 into T2E(GS) and E d6 into T2 and E (GS) 0 d-hole reversal d9 into T2g and Eg (GS) A tetrahedral splits Both of these represent d1, da6d-hole tetrahedral D (reversal) in a tetrahedral field (a d4, d9 octahedral second reversal). or An octahederal field splits d1 into T2g (GS) and Eg d6 into T2g (GS) and Eg Eg d1, d6 octahedral d4, d9 tetrahedral ligand field strength T2g or T2 Tetrahedral reversal. D High Spin Ground States: d2, d3, d6, and d7 We have taken care of the d0, d1, d4, d5, d6, d9, and d10 configurations. Now have to do d2, d3, d6, and d7 configurations. It turns out that all we have to do is solve d2. d2 and d7 both are electrons on top of a spherical shell yielding a splitting pattern: 1, 2, 3 electrons holes 1 3 d2 d7 = d5 + d2 2 d3=d5-hole2 2 d3 and d8 are both two d-holes in a spherical shell, yielding reversed splitting: 3, 2, 1 3 d8=d10-hole2 1 But it is not so easy. Here is our approach: We know the symmetry of the GS of the free d2 ion. How? We can get the terms for d2 using the methods applied earlier to p2, etc. They are 3F, 1D, 3P, 1G, 1S. We identify the GS as 3F. How? We saw earlier that F in octahedral environment splits to A2g + T1g + T2g; in tetrahedral we would get A2 + T1 + T2. Our problem is the energy ordering. Which is GS? Thus the 3F GS for d2 splits into 3A2g + 3T1g + 3T2g. The 4F GS for d3 splits into 4A2g + 4T 4 1g + T2g. Where did the spin multiplicities come from?? But how do we decide on what becomes the GS after the splitting due to the ligands? We use a correlation diagram. It shows the affect of increasing the ligand field strength from zero (free ion) to very high where energy ordering is determined solely by the occupancy of the t2g and the eg orbitals. d2 F r e e I o n t e r m s We have two electrons in the eg orbitals. It can be shown that these give rise to 1A1g, 1Eg, and 3A2g which have same energy in strong ligand field. Connect the terms of the same symmetry without crossing. Similarly, splitting occurs for these occupancies. Configurations We have included basedthe on 3splitting T1g originating of d from Splitting of free ion electrons. the 3P. We Dominant will need initvery immediately. strong fields. Same Real complexes terms. symmetry as lower energy 3T1g from the 3F. Orgel diagram for d2, d3, d7, d8 ions This curvature will complicate interpretation of spectra. Same symmetry; crossing forbidden First look at Energy And now d2 and d7 in tetrahedral (reversed due to tetrahedral field) T1 or T1g 7 in octahedral (2 elecs Td12Tor T1gT and d1g 1 or on a spherical cloud) P and and T1 or T1g T13 or T1g 8 d and d in tetrahedral (double F reversal: d-holes and tetrahedral) T2 or T2g All states shown are of the same spin. Transitions occur between them but weakly. d3 and d8 in octahedral (reversed due to d-holes). T2 or T2g Note the reversed ordering of the splitting coming from F (T1/T2/A2). The lower T1(g) now aims up and should cross the upper T1(g) but T or T 1 1g does not due to interaction with the upper T1(g). Now have strong curvature to avoid crossing. Note the weak interaction of A A2gT1, the curvature. the 2 ortwo d2, d7 tetrahedral A2 or A2g 0 d2, d7 octahedral d3, d8 octahedral d3, d8 tetrahedral Ligand field strength (Dq) Move to Tanabe-Sugano diagrams. d1 – d3 and d8 – d9 which have only high spin GS are easier. Here is d2. Correlation diagram for d2. Convert to Tanabe-Sugano. Tanabe-Sugano Electronic transitions and spectra d2 Tanabe-Sugano diagram V(H2O)63+, a d2 complex Configurations having only high spin GS d2 d9 d1 d3 Note the two lines curving away from each other. d8 Note the two lines curving away from each other. Configurations having either high or low spin GS The limit between high spin and low spin Determining Do from spectra d1 Exciting electron from t2g to eg d9 Exciting d-hole from eg to t2g Exciting d-hole from eg to t2g One transition allowed of energy Do Exciting electron from t2g to eg Determining Do from spectra mixing Here the mixing is not a problem since the “mixed” state is not involved in the excitation. mixing Lowest energy transition = Do For d2 and d7 (=d5+d2) which involves mixing of the two T1g states, unavoidable problem. Ground state and excited state mixing which we saw earlier. But note that the difference in energies of two excitations is Do. E (T1gA2g) - E (T1gT2g) = Do Make sure you can identify the transitions!! Can use T-S to calculate Ligand Field Splitting. Ex: d2, V(H2O)63+ Observed spectrum u1: 17,800 cm-1 u2: 25,700 cm-1 Technique: Fit the observed energies to the diagram. E/B We must find a value of the splitting parameter, Do/B, which provides two excitations with the ratio of 25,700/17,800 = 1.44 First, clearly u1 should correspond to 3T1 3T2 But note that the u2 could correspond to either 3T1 3A2 or 3T1 3T . 1 The ratio of u2/u1 = 1.44 is obtained at Do / B= 31 DO/B Again, the root, basic problem is that the two T1 s have affected each other via mixing. The energy gap depends to some extent on the mixing! Now can use excitation energies For u1: E/B = 17,800 cm-1 /B = 29 yielding B = 610 cm-1 By using 31 = Do/B = Do/610 obtain Do = 19,000 cm-1 The d5 case All possible transitions forbidden Very weak signals, faint color Jahn-Teller Effect found if there is an asymmetrically occupied e set. octahedral d9 complex b1g x2-y2 z2 a1g x2-y2 z2 b2g xy xy xz xz yz effect of octahedral field yz eg elongation along the x axis Can produce two transitions. This picture is in terms of the orbitals. Now for one derived from the terms. Continue with d9 Eg 2 T2g B2g 2 D A1g 2 Eg B1g Free ion term for d9 effect of octahedral field GS will have d-hole in either of the two eg orbitals. ES puts dhole in either of the three t2g orbitals. effect of elongation along z For example, the GS will have the d-hole in the x2-y2 orbital which is closer to the ligands. Some examples of spectra Charge transfer spectra Metal character LMCT Ligand character Ligand character MLCT Metal character Much more intense bands