Lecture 15 February 15, 2013 Transition metals

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1 Lecture 15 February 15, 2013 Transition metals Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy Course number: Ch120a Hours: 2-3pm Monday, Wednesday, Friday William A. Goddard, III, 316 Beckman Institute, x3093 Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics, California Institute of Technology Teaching Assistants: Ross Fu Fan Liu Ch120a-1

2 Transition metals (4s,3d) Sc---Cu (5s,4d) Y-- Ag (6s,5d) (La or Lu), Ce-Au 2

3 Ground states of neutral atoms Sc (4s)2(3d)1 Sc ++ (3d)1 Ti (4s)2(3d)2 Ti ++ (3d)2 V (4s)2(3d)3 V ++ (3d)3 Cr (4s)1(3d)5 Cr ++ (3d)4 Mn (4s)2(3d)5 Mn ++ (3d)5 Fe (4s)2(3d)6 Fe ++ (3d)6 Co (4s)2(3d)7 Co ++ (3d)7 Ni (4s)2(3d)8 Ni ++ (3d)8 Cu (4s)1(3d)10 Cu ++ (3d)10 3

Dalton (diameter ~ 60A) Four subunits (a1, a2, b1, b2) each with one

4 Hemoglobin Blood has 5 billion erythrocytes/ml Each erythrocyte contains 280 million hemoglobin (Hb) molecules Each Hb has MW=64500 Dalton (diameter ~ 60A) Four subunits (a1, a2, b1, b2) each with one heme subunit Hb Each subunit resembles myoglobin (Mb) which has one heme Mb 4

5 The action is at the heme or Fe-Porphyrin molecule Essentially all action occurs at the heme, which is basically an Fe-Porphyrin molecule The rest of the Mb serves mainly to provide a hydrophobic envirornment at the Fe and to protect the heme 5

6 The heme group The net charge of the Fe-heme is zero. The VB structure shown is one of several, all of which lead to two neutral N and two negative N. Thus we consider that the Fe is Fe 2+ with a d 6 configuration Each N has a doubly occupied sp 2 s orbital pointing at it. 6

7 Energies of the 5 Fe 2+ d orbitals x 2 -y 2 z 2 =2z 2 -x 2 -y 2 yz xz xy 7

8 Exchange stabilizations 8

9 Ferrous Fe II x 2 -y 2 destabilized by heme N lone pairs z 2 destabilized by 5 th ligand imidazole or 6 th ligand CO y x 9

Four coordinate Fe- Heme High spin case, S=2 or q The 5 th axial ligand will destabilize q2 since dz2 is doubly occupied A pi acceptor would stabilize q1

10 Four coordinate Fe- Heme High spin case, S=2 or q The 5 th axial ligand will destabilize q2 since dz2 is doubly occupied A pi acceptor would stabilize q1 wrt q2 Bonding O2 to 5 coordinate will stabilize q3 wrt q1 Future discuss only q1 copyright and denote 2011 William as A. Goddard q III, all rights reserved 10

11 Four coordinate Fe-Heme Intermediate spin, S=1 or t 11

12 Four coordinate Fe-Heme Low spin case, S=0 or s 12

13 Out of plane motion of Fe 4 coordinate 13

14 Add axial base N-N Nonbonded interactions push Fe out of plane is antibonding 14

15 Summary 4 coord and 5 coord states 15

16 Free atom to 4 coord to 5 coord Net effect due to five N ligands is to squish the q, t, and s states by a factor of 3 This makes all three available as possible ground states depending on the 6 th ligand 16

17 Bonding of O 2 with O to form ozone O 2 has available a ps orbital for a s bond to a ps orbital of the O atom And the 3 electron p system for a p bond to a pp orbital of the O atom 17

18 Bond O 2 to Mb Simple VB structures get S=1 or triplet state In fact MbO 2 is singlet Why? 18

19 change in exchange terms when Bond O 2 to Mb O 2 ps O 2 pp Assume perfect VB spin pairing Then get 4 cases Thus average K dd is ( )/4 = K dd 5*4/2 up spin down spin 7 K dd 4*3/2 + 2*1/2 7 K dd 4*3/2 + 2*1/2 6 K dd 3*2/2 + 3*2/2 19

20 Bonding O 2 to Mb Exchange loss on bonding O 2 20

21 Modified exchange energy for q state But expected t binding to be 2*22 = 44 kcal/mol stronger than q What happened? Binding to q would have DH = = + 11 kcal/mol Instead the q state retains the high spin pairing so that there is no exchange loss, but now the coupling of Fe to O 2 does not gain the full VB strength, leading to bond of only 8kcal/mol instead of 33 21

22 Bond CO to Mb H 2 O and N 2 do not bond strongly enough to promote the Fe to an excited state, thus get S=2 22

23 compare bonding of CO and O2 to Mb 23

2 Think of as bond from Tidz2 to H1s Covalent 2 electron CH

24 GVB orbitals for bonds to Ti Ti ds character, 1 elect H 1s character, 1 elect Covalent 2 electron TiH bond in Cl 2 TiH 2 Think of as bond from Tidz2 to H1s Covalent 2 electron CH bond in CH 4 Csp 3 character 1 elect H 1s character, 1 elect 24

25 Bonding at a transition metaal Bonding to a transition metals can be quite covalent. Examples: (Cl 2 )Ti(H 2 ), (Cl 2 )Ti(C 3 H 6 ), Cl 2 Ti=CH 2 Here the two bonds to Cl remove ~ 1 to 2 electrons from the Ti, making is very unwilling to transfer more charge, certainly not to C or H (it would be the same for a Cp (cyclopentadienyl ligand) Thus TiCl 2 group has ~ same electronegativity as H or CH 3 The covalent bond can be thought of as Ti(dz2-4s) hybrid spin paired with H1s A{[(Tids)(H1s)+ (H1s)(Tids)](ab-ba)} 25

26 But TM-H bond can also be s-like Cl 2 TiH + Ti (4s) 2 (3d) 2 The 2 Cl pull off 2 e from Ti, leaving a d 1 configuration Ti-H bond character 1.07 Tid+0.22Tisp+0.71H ClMnH Mn (4s) 2 (3d) 5 The Cl pulls off 1 e from Mn, leaving a d 5 s 1 configuration H bonds to 4s because of exchange stabilization of d 5 Mn-H bond character 0.07 Mnd+0.71Mnsp+1.20H 26

27 Bond angle at a transition metal For two p orbitals expect 90, HH nonbond repulsion increases it What angle do two d orbitals want H-Ti-H plane 76 Metallacycle plane 27

Best bond angle for 2 pure Metal bonds using d orbitals Assume that the first bond has pure d z2 or ds character to a ligand along the z axis Can we make a 2 nd bond, also of pure ds character

28 Best bond angle for 2 pure Metal bonds using d orbitals Assume that the first bond has pure d z2 or ds character to a ligand along the z axis Can we make a 2 nd bond, also of pure ds character (rotationally symmetric about the z axis) to a ligand along some other axis, call it z. For pure p systems, this leads to = 90 For pure d systems, this leads to = 54.7 (or ), this is ½ the tetrahedral angle of (also the magic spinning angle for solid state NMR). 28

29 Best bond angle for 2 pure Metal bonds using d orbitals Problem: two electrons in atomic d orbitals with same spin lead to 5*4/2 = 10 states, which partition into a 3 F state (7) and a 3 P state (3), with 3 F lower. This is because the electron repulsion between say a d xy and d x2-y2 is higher than between sasy d z2 and d xy. Best is ds with dd because the electrons are farthest apart This favors = 90, but the bond to the dd orbital is not as good Thus expect something between 53.7 and 90 Seems that ~76 is often best 29

30 How predict character of Transition metal bonds? Start with ground state atomic configuration Ti (4s) 2 (3d) 2 or Mn (4s) 2 (3d) 5 Consider that bonds to electronegative ligands (eg Cl or Cp) take electrons from 4s easiest to ionize, also better overlap with Cl or Cp, also leads to less reduction in dd exchange (3d) 2 (4s)(3d) 5 Now make bond to less electronegative ligands, H or CH 3 Use 4s if available, otherwise use d orbitals 30

71H ClMnH Mn (4s) 2 (3d) 5 The Cl pulls off 1 e from Mn, leaving a d 5 s 1

31 But TM-H bond can also be s-like Cl 2 TiH + Ti (4s) 2 (3d) 2 The 2 Cl pull off 2 e from Ti, leaving a d 1 configuration Ti-H bond character 1.07 Tid+0.22Tisp+0.71H ClMnH Mn (4s) 2 (3d) 5 The Cl pulls off 1 e from Mn, leaving a d 5 s 1 configuration H bonds to 4s because of exchange stabilization of d 5 Mn-H bond character 0.07 Mnd+0.71Mnsp+1.20H 31

32 Example (Cl) 2 VH 3 + resonance configuration 32

33 Example ClMometallacycle butadiene 33

34 Example [Mn CH] 2+ 34

35 Summary: start with Mn + s 1 d 5 dy2 s bond to H1s dx2-x2 non bonding dyz p bond to CH dxz p bond to CH dxy non bonding 4sp hybrid s bond to CH 35

36 Summary: start with Mn + s 1 d 5 dy2 s bond to H1s dx2-x2 non bonding dyz p bond to CH dxz p bond to CH dxy non bonding 4sp hybrid s bond to CH 36

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