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Home , Quadruple bond

Schedule

• Last week: π-Acceptor and Biology

CO, O2, N2 and NO compp,lexes, introduction to M-M bonds • Lecture 7: M-M bonds δ-bonds and bonding in metal clusters • Lecture 8: Rates of reaction -exchange reactions, labile and inert metal ions • Lecture 9: Redox reactions Inner and outer-sphere reactions

Slide 2/20

Summary of Course – week 6 Summary of Last Lecture

Metal-metal bonding

• be able to predict for M2Lx dimers using d- count and Metal-N2 complexes σ, π and δ diagram •N2 is isoelectronic with CO but M-N2 bonding is much • be able to predict bond order in larger metal-halide clusters using d- weaker electron count shared over edges of cluster • be able to predict bond order in metal carbonyl clusters using 18 e- rule •N2 is non-polar and bond is strong Reaction mechanisms • be able to describe ligand exchange mechanisms NO compl exes - • be able to explain role of metal charge and LFSE in rate of ligand • Can bond as 1 electron donor (NO : bent M-NO) exchange • Can bond as 2 electron donor (NO+ linear M-NO) • be able to describe electron transfer reaction mechanisms • be able to predict relative rate of outer sphere reaction for different metals Resources Today’s lecture • Slides for lectures 7-9 • Metal-Metal bonding in complexes • ShiShriver an dAtkid Atkins “Inorgan ic Chem is try ”Ch” Chap ter 1811212020118.11, 21.20, 20.1-20. 13

Slide 3/20 Slide 4/20 Maximum Bond Order – d- Maximum Bond Order – d-Block

3dσu • the maximum bond order is 5 • but…complexes also contain 3dπg dx2-y2 + dx2-y2 ligands which use some of the d- 2 × 2 × dxy + dxy orbitals reducing number of bonds 3dδu

d 2 2+ d 2 2 2 × x -y x -y d + d 3d 3d d + d xz xz xy xy 2 × 2 × 3dδg dyz + dyz

3dπu d + d 2 × xz xz dyz + dyz dz2 + dz2 y 3dσg 2 dz2 + dz x M M M M z Slide 5/20 Slide 6/20

Complexes Containing d-d Bonds Quadruple Bonds

• In complexes, not all of these molecular orbitals will form as some of the d- • In complexes, not all of these molecular orbitals will form as some of the d- orbitals will be involved in bonding to ligands orbitals will be involved in bonding to ligands

2- 2- • [Re2Cl8] contains two ~square planar ReCl4 units • [Re2Cl8] contains two ~square planar ReCl4 units ¾ there must be a Re-Re bond as the two units are attached ¾ dx2-y2 bonds with the four ligands ¾ the geometry is eclipsed 2- 3+ 4 - y ¾ [Re2Cl8] ≡ 2Re (d ) + 8Cl y

x x

¾ leaving:

dz2 (σ), dxz, dyz (π) and dxy (δ) to form M-M bonds Slide 7/20 Slide 8/20 Quadruple Bonds Quadruple Bonds

•[ReCl ]2- quaduple bond: 3dσ •2 ×Re3+ (d4) Æ 8 e- 2 8 u ¾ (σ)2(π)4(δ)2: a σ-bon d, tw o π-bon ds an d on e δ-bon d 2 4 2 3dπg •(σ) (π) (δ) ¾ the sterically unfavourable eclipsed geometry is due to the δ-bond 2- • quadruple bond: Cl4Re ReCl4 3dδu

dxy + dxy 3d 3d

3dδg

3dπu d + d 2 × xz xz dyz + dyz

3dσ g dz2 + dz2 M M Slide 9/20 Slide 10/20

Quadruple Bonds Larger Clusters – M3

•[ReCl ]2- •ReClexists Re Cl clusters 2 8 3dσ 3 3 9 Cl 2- 3+ 4 - u ¾ [[eRe2Cl8] ≡ 2Re (d ))8C + 8Cl ¾ detailed vi ew of cl uster bon din g m or e inv olv ed… 3+ 4 - Cl2Re ReCl2 ¾ 2 × Re (d ) Æ 8 e 3dπg 2 4 2 3+ 4 - ¾ (σ) (π) (δ) ¾ Re3Cl9 ≡ 3Re (d ) + 9Cl 3+ 4 - Cl Cl ¾ bond order = 4 (quadruple bond) ¾ 3 × Re (d ) Æ 12 e or 6 pairs Re 3dδu ¾ δ-bond: eclipsed geometry ¾ 3 × Re-Re connections in triangle Cl2 ¾ number of pairs / number of edges = 6/3 = 2 ¾ 3 × Re=Re double bonds

4- •[Re2Cl8] 3dδg 4- 2+ 5 - ¾ [Re2Cl8] ≡ 2Re (d ) + 8Cl ¾ 2 × Re2+ (d5) Æ 10 e- 3dπ ¾ (σ)2(π)4(δ)2(δ*)2 u ¾ bond order = 3 () ¾ no δ-bond: staggered geometry 3dσg

Slide 11/20 Slide 12/20 Larger Clusters – M6 Larger Clusters – M6

2- 2+ •[Mo6Cl14] •[Nb6Cl12] ¾ Mo6 octah edr on wi th 8Cl cappin g f aces an d 6 Cl on edges ¾ Nb6 octah edr on wi th 12 Cl on edges

2+ 14+ - ¾ [Nb6Cl12] ≡ [Nb6] + 12Cl 2- 2+ 4 - 5 - ¾ [Mo6Cl14] ≡ 6Mo (d ) + 14Cl ¾ 6 × Nb (d ) Æ 30 e 2+ 4 - 14+ ¾ 6 × Mo (d ) Æ 24 e or 12 pairs ¾ [Nb6] so… ¾ 12 × Mo-Mo connections in octahedron number for M-M bonding is 30 – 14 = 16 e- or 8 pairs ¾ number of pairs / number of edges = 12/12 = 1 ¾ 12 × Nb-Nb connections in octahedron ¾ 12 × Mo-Mo single bonds ¾ number of pairs / number of edges = 8/12 = 2/3 ¾ 12 × Nb-Nb bonds with bond order = ⅔

Slide 13/20 Slide 14/20

Carbonyl Clusters Carbonyl Clusters

• In carbonyl clusters, some of the metal orbitals are used to σ and π-bond • In some cases, the CO ligands bridge two metals – this does not effect to th e CO li gan ds the met hod as CO a lways don ates 2 e- to t he c luster uste • The number and order of the M-M bonds is easily determined by - requiring that all of the metals obey the 18 e rule ¾ Fe2(CO)9 ¾ total number of = 2 × 8(Fe)+98 (Fe) + 9 × 2(CO2 (CO) Æ 34 e- - ¾ Mn2(CO)10 ¾ Fe2 so each Fe has 34/2 = 17 e ¾ total number of electrons = 2 × 7 (Mn) + 10 × 2 (CO) Æ 34 e- ¾ to complete 18 e- configuration, a Fe-Fe is formed - ¾ Mn2 so each Mn has 34/2 = 17 e ¾ to complete 18 e- configuration, a Mn-Mn single bond is formed

Slide 15/20 Slide 16/20 Carbonyl Clusters Summary

• Larger clusters are again possible By now you should be able to

• DtthMOdifLDraw out the MO diagram for L4ML4 complexes ¾ Os3(CO)12 ¾ total number of electrons = 3 × 8 (Os) + 12 × 2 (CO) Æ 48 e- • Complete this diagram by filling in the appropriate - ¾ Os3 so each Os has 48/3 = 16 e number of electrons ¾ to complete 18 e- configuration, each Os makes two Os-Os bonds • Explain the appearance of eclipsed geometries due to δ- ¾ Os triangle results 3 bonding • In larger clusters, use the total number of pairs of metal electrons and number of metal-metal connections to work out the bond order • In carbonyl clusters, use the 18 e- rule to work out how many bonds have to be formed

Next lecture • Ligand-substitution reactions Slide 17/20 Slide 18/20

Practice Practice

• What is the Mo-Mo bond order in the

complex [Mo2(CH3CO2)4]?

• What is the Os -Os bond order in the

cluster Os4(CO)12?

Slide 19/20 Slide 20/20