Covalent Bonding between the Elements

Question 1

Define cluster compounds.

Answer 1

Species which have 3D shapes and direct element-element bonds.

Typically have a ‘core’ which may be surrounded by a shell of other substituents. If there are no other substituents, then the cluster is said to be ‘naked’.

Lines often only show connectivity, not actual bonds (shows atoms that are close in space).

Structures of clusters are based on the platonic solids.

Question 2

Describe the synthesis of borane clusters.

Answer 2

Traditionally characterised by reactions occurring at high temperature, resulting in low yields with a number of different products.

Question 3

What are the different reactions that borane clusters can undergo?

Answer 3

1. Combustion - forms boron oxides
2. Hydrolysis - forms boron hydroxide compounds
3. Electrophilic substitution - e.g. with Cl₂
4. Base-induced degradation - e.g. with NH₃ or OH^-
5. Deprotonation reactions - e.g. with NaH

Question 4

How do you calculate the total valence electron count (TVEC)?

Answer 4

1. Add up the number of valence electrons on the core atoms
2. Substituents such as H, ^tBu, etc. all count as one electron
3. Add one electron for each negative charge, remove one electron for each positive charge

Question 5

What is the skeletal electron count (SEC) and how is calculated?

Answer 5

The number of electrons which contribute towards the bonding of the cluster core. For clusters based on a deltahedron with each atom bonded to three others:

SEC = TVEC - 2n

(n = number of cluster vertices)

Question 6

What are skeletal electron pairs (SEP) and how are they calculated?

Answer 6

The number of electron pairs contributing towards cluster bonding.

SEP = SEC / 2

Question 7

How can cluster bonding be described?

Answer 7

1. Electron precise - all the bonds in the cluster are 2-electron 2-centre bonds
   
   • TVEC = 5n, SEC = 3n
2. Electron deficient - too few electrons for 2-electron 2-centre bonds
   
   • TVEC < 5n, SEC < 3n
3. Electron rich - too many electrons than needed for 2-electron 2-centre bonds
   
   • TVEC > 5n, SEC > 3n

Question 8

Describe the structure of electron rich / electron deficient clusters.

Answer 8

• Electron rich clusters have open structures
• Electron deficient clusters are much more closed

Question 9

What is Wade’s rule?

Answer 9

For an n vertex cage with n+1 SEP then a closo structure will be adopted (closed structure).

Question 10

Describe the molecular orbital basis of Wade’ rule.

Answer 10

For an n vertex, closo polyhedron, there are n+1 bonding orbitals in the cluster skeleton. This is how we arrive to the SEP = n + 1 rule.

Question 11

Describe the relationships between clusters with similar SEP.

Answer 11

Removing a vertex from a closo structure gives the structure of a nido cluster with the same SEP. Removing a further vertex gives an arachno cluster, also with the same SEP.

Question 12

Describe the different cluster structures in terms of vertices and SEP numbers.

Answer 12

• A closo cluster with n vertices has n + 1 SEP
• A nido cluster with n vertices has n + 2 SEP
• An arachno cluster with n vertices has n + 3 SEP
• A hypho cluster with n vertices has n + 4 SEP

Question 13

Describe how to draw the structure of nido, arachno and hypho clusters.

Answer 13

• They are determined from the shape of the parent closo cluster, which has the same number of SEP and SEP - 1 vertices
• Remove the appropriate number of vertices
  
  • If vertices are different, it is the vertex which is the most connected that is removed
• Additional hydrogen atoms are placed in bridging sites along B-B edges of open faces, or terminal sites (BH -> BH₂) if available

Question 14

Can other groups participate in cluster bonding?

Answer 14

Yes, e.g. C-H can replace B-H in [B₆H₆]^2-. Both the B-H and C-H fragments contribute 3 electrons to give the same octahedral geometry.

In principle, any fragment which has three orbitals available for bonding may form part of the cluster.

Question 15

What is the relationship between many transition metal and main group species?

Answer 15

When the main group species is replaced by a TM, nothing changes except the TVEC which increases by 10 for each TM.

This can be rationalised in two ways: the isolobal analogy and the Wade-Mingos rules.

Question 16

Describe the isolobal analogy.

Answer 16

Replacement of a group by a TM makes no changes to the structure of the cluster so it can be suggested that they are both making the same contributions to the bonding. They must be offering the same types of molecular orbitals/number of electrons.

Two fragments are isolobal if they have the same number of frontier orbitals, with the same symmetry, approximately the same energy and the same number of electrons.

Question 17

How can we predict which fragments are isolobal?

Answer 17

Via homolytic cleavage.

Cleave off groups, e.g. H from C or a ligand from a TM, to obtain the fragments of interest. If they end up with the same number of frontier orbitals (with the same symmetry) and offer the same number of electrons, then they are isolobal.

Question 18

How do the isolobal analogy and Wade’s rules relate?

Answer 18

The isolobal analogy can be applied to Wade’s rules in order to aid in assessing cluster geometry. We can find an easier fragment that is isolobal to the molecule of interest and find the geometry of the simpler fragment.

Question 19

Describe the Wade-Mingos rules for transition metal clusters.

Answer 19

TM’s add a further 10 electrons to the cluster system compared to main group compounds (due to the extra 5x d-orbitals). This changes how the SEC is calculated:

• Remove 2 from the TVEC for each main group vertex
• Remove 12 from the TVEC for each TM vertex

Question 20

Describe capping of TM clusters.

Answer 20

It’s unusual for TM clusters to form large closo-type clusters when n > 6. In these cases, the clusters tend to adopt geometries where additional grops are on the faces of a closo type cluster. This is called face capping.

A capped cluster has the same number of electron pairs for framework bonding as the uncapped cluster.

Question 21

Define the different capped clusters in terms of vertices and SEP.

Answer 21

• A closo cluster with n vertics has n + 1 SEP
• A monocapped cluster with n vertices has n SEP
  
  • based on a closo cluster with n - 1 vertices and one face is capped
• A bicapped cluster with n vertices has n - 1 SEP
  
  • based on a closo cluster with n - 2 vertices and two faces are capped
• A tricapped cluster with n vertices has n - 2 SEP
  
  • based on a closo cluster with n - 3 vertices and three faces are capped

Question 22

What is the general methodology for forming M-M bonds?

Answer 22

1. Determine the oxidation state and d-electron count of the metal - tells us how many electrons are available to form M-M bonds
2. Identify the correct molecular orbital overlap diagram
3. Determine the expected bond order
4. Consider other factors e.g. periodicity

Question 23

Why do you find increased bond dissociation energies as you descend the transition metals?

(3d to 4d to 5d)

Answer 23

• 3d orbitals are core-like so they don’t extend significantly beyond core electrons
  
  • show poor orbital overlap therefore the bonds aren’t strong
• 4d + 5d extend further and have better overlap
  
  • tend to see more M-M bonds in cluster compounds towards the bottom of the periodic table

Question 24

How do metal-metal multiple bonds affect electron counting?

Answer 24

1. M-M bonds don’t affect oxidation state - the electrons are shared evenly between the two metals
2. When counting the TVEC:

M-M -> 1 extra electron

M=M -> 2 extra electrons

M≡M -> 3 extra electrons

Question 25

Describe the bonding in 'bare' M₂ dimers.

Answer 25

* delta (δ) orbitals are formed from face-on interactions between d-orbitals * generally weaker than π-bonds due to poor overlap between precursor orbitals * M-M bonding energy increases down a group * unlike the p-block where overlap gets worse down the group due to becoming more diffuse * bond order in complexes is usually less than 5 because M orbitals are needed for M-L bonds

Question 26

Describe M-M vs. M-L bonding.

Answer 26

There's a competition between M-M and M-L bonding. One orbital can't do both, so if it's involved in M-L bonding then it's factored out of M-M bonding.

Question 27

Describe the M₂X₈ structure.

Answer 27

* quadruple M-M bond * the d_x2-y2 orbital is involved in M-L bonding so it's factored out * d_xy forms the δ-bond - has an orientation dependence and is weak, it can only be formed in an eclipsed conformation * however, sterics can enforce a staggered geometry and there's a low energy to rotation * predicted bond order of 4

Question 28

Describe the M₂X₆ and M₂X₉ structures.

Answer 28

* triple M-M bond * both d_x2-y2 and d_xy are factored out * the d-orbitals, other than d_z2, are hybrids to ensure correct formation of π and π\* orbitals * the d_xy has some d_x2-y2 mixed in, the d_yz has some d_xy, and vice versa * due to the tilting, the π and π\* have some δ and δ\* character * for M₂L₆, the eclipsed conformation gives the best overlap, however most compounds are staggered due to steric reasons

Question 29

Are quintuple M-M bonds possible?

Answer 29

* a Cr₂ complex was synthesised with a quintuple Cr-Cr bond and a very short bond distance * the metal has 9 valence orbitals - the ligands could form bonds with the s- or p-orbitals, not just d-orbitals * the s-orbitals are closer in energy to the d- and ligand-based orbitals and may be available for bonding * the p-orbitals are too high in energy * the ligand formed a bonf with the Cr 4s orbitals, leaving the 5 d-orbitals for Cr-Cr bonding

Question 30

Describe the contribution to σ and π bonds of various elements.

Answer 30

* for the n = 2 row, π-bonds should make a significant contribution for bonding * for the heavier elements, σ-bonding should dominate * until the 1970's, no real examples of π-bonds to heavier elements had been reported

Question 31

What can be used to stabilise double and triple bonds in main group compounds?

Answer 31

Adding bulky substituents. 1. Thermodynamic data shows that, for P, σ-bonding framework is preferred (more stable) 2. However, if R is a very large group, then extra steric repulsions between the R groups disfavours the formation of the σ-framework * primarily due to, in order to form the aggregate structure, the molecules colliding which is inhibited by the bulky groups

Question 32

Describe the synthesis of R₂E=ER₂ compounds.

Answer 32

Question 33

Describe the structures of R₂E=ER₂ compounds.

Answer 33

Apart from carbon, all of these compounds shows ***trans*-bent** structure. The distortion from planarity becomes greater as the periodic table is descended.

Question 34

Describe the synthesis of RE≡ER compounds.

Answer 34

First achieved for Ge, Sn and Pb using the sterically bulk Ar\* and Ar' groups.

Question 35

Describe the structure of Ar\*Pb≡PbAr\*.

Answer 35

Not linear like alkynes, actually has a bent geometry and steric protection of the metals.

Question 36

Describe the synthesis of Si-alkynes.

Answer 36

Power's route is very effective for Ge, Sn and Pb as the ECl₂ compounds are available, however SiCl₂ is not.

Question 37

Describe the general structure of alkynes.

Answer 37

* carbon is the odd element out as it's the only linear alkyne * E-E distance increases as the group is descended * the amount of multiple bond character decreases as the group is descended

Question 38

Describe the bonding in R₂E=ER₂ compounds.

Answer 38

* the σ\* and π orbitals have similar symmetry and are able to mix to form n⁺ and n\* orbitals * the n⁺ orbital (containing the electrons) is lowered in energy by this mixing, hence stabilising the structure in the *C_2h* form * for carbon, the energy gap between the σ\* and π orbitals is very large, therefore no mixing can occur and the molecule stays planar (*D_2h*) * as the group is descended, the difference in energy between the orbitals decreases - more mixing occurs so greater deviations from planarity * for Si-Pb mixing occurs and the structures are *trans*-bent

Question 39

Describe the bonding in RE≡ER compounds.

Answer 39

* mixing between σ\* and π orbitals still occurs, but only one of the two degenerate π orbitals is involved * also, mixing between σ and π\* is possible * the other π and π\* orbitals are unaltered * the unmixed π interactions get weaker down the group, as do the observed changes in bond lengths

Question 40

Describe conductors and insulators in terms of band theory.

Answer 40

In a solid we will have many atoms, giving us many orbitals. The orbitals split into bands as the gap between individual orbitals is very small - valance band and conduction band. In an insulator, the band gap is too large, electrons can't be promoted to the conduction band. In a conductor, the band gap is small (or zero in some cases).

Question 41

Describe intrinsic semi conductors.

Answer 41

A material which formally has a filled valance band and an empty conduction band. The band gap is small so electrons can be thermally promoted. The conductivity of a semi conductor increases with temperature.

Question 42

Describe extrinsic semi conductors.

Answer 42

A material which has been doped to create either a donor band near the conductor band (n-type) or an acceptor band near the valance band (p-type).

Question 43

What are some common intrinsic semi conductors?

Answer 43

* elemental silicon and germanium * III/V materials * contain elements from groups 13 and 15 * II/VI materials * contain elements from groups 12 and 16 * the net number of electrons remains the same, but the band gap is altered

Question 44

What is important in the preparation of semi-conductors?

Answer 44

The purity of the precursor molecules. Even tiny amounts of impurities will alter the properties of the final material, meaning it may no longer be usable.

Question 45

How is electronic-grade silicon synthesised?

Answer 45

Question 46

How is silicon crystallised by the Czochralski process?

Answer 46

The process gives a silicon ingot which is then cut with a diamond saw into wafers. Photolithography is used to etch the silicon to make an integrated circuit.

Question 47

What are alternative routes to III/V materials?

Answer 47

1. Metal-organic chemical vapour decomposition (MOCVD) 2. Replacement group V sources 3. Single souce precursors

Question 48

Describe metal-organic chemical vapour decomposition (MOCVD).

Answer 48

Gasses such as H_2, He or N₂ are used to carry volatile precursors into a chamber which is held at high temperature. The precursors decompose to deposit a film of the desired material. A traditional example is the formation of GaAs.

Question 49

What are the general processes in MOCVD to prepare a range of III/V and II/VI materials.

Answer 49

Question 50

What are some disadvantages to the MOCVD process?

Answer 50

Alkyl metal compounds are pyrophoric and highly toxic, as are compounds such as PH₃ and AsH₃.

Question 51

What properties must the precursors for MOCVD have?

Answer 51

1. Must be volatile so they can travel in the gas phase to the chamber 2. Must be extremely pure, otherwise the materials will be contaminated 3. Must be stable enough to handle and be prepared, but decompose at the correct temperature in the MOCVD reactor

Question 52

Describe replacement group V sources as an alternative route to III/V materials.

Answer 52

* For GaAs, a wide range of arsenic precursors have been investigated, such as AsMe₃, however these give rise to large amounts of carbon contamination * complexes with As-H bonds are better, such as AsEt₂H or As^tBuH₂ * For the corresponding phosphides MP (M = Ga, In), PH₃ is toxic and does not decompose until very high Ts * Like for As, P^tBuH₂ is better as it pyrolyses at a lower temperature

Question 53

Describe single source precursors as an alternative route to III/V materials.

Answer 53

Group 13 (III) alkyl compounds (MMe₃) are Lewis acids while Group 15 (V) compounds (ER₃) are Lewis bases. It should be possible to form adducts, Me₃M 3, that are less air sensitive and easier to prepare/handle to act as precursors. However, the donor/acceptor interaction is weak and easily broken under MOCVD conditions. In order to avoid this problem, a different type of single source precursor has been developed.

Question 54

What different type of single source precursor has been developed?

Answer 54

Precursors with several Ga-E bonds so they are still easy to handle but also have the correct CVD properties.

Question 55

Describe organic vs inorganic polymers and their preparation.

Answer 55

Organic polymers are very well known, despite inorganic materials such as silicon being very abundant. The preparative routes to many organic polymers typically involve polymerising an unsaturated monomer. In the case of main group polymers constructed from heavier elements, preparation of compounds with multiple bonds is difficult and requires sterically demanding groups. This means that different preparative routes are required.

Question 56

Describe polysiloxanes.

Answer 56

Polymers which consist of [-Si(R₂)-O-] repeating groups. The Si-O bond which constitutes the polymer is stronger than a typical C-C bond, making them robust materials. The R group can be altered to alter the polymer properties.

Question 57

Describe polysilanes.

Answer 57

A more ready comparison to organic polymers, [-C(R₂)-] groups, are polysilanes, [-Si(R₂)-]. They can be prepared from Wurtz coupling of dichlorosilanes.

Question 58

Is Wurtz coupling a good method for forming polysilanes?

Answer 58

This is not a great method as it involves the reaction of elemental sodium in toluene at reflux.

Question 59

What is an alternative method to Wurtz coupling for polysilane formation?

Answer 59

Transition metal-catalysed dehydrogenation of primary silanes. This method and the same catalyst can also be applid to other heavier group 14 elements, where materials such as polystannanes may be prepared [-Sn(R)(H)-].

Question 60

Describe the electronic structure of polysilanes.

Answer 60

They are semi conductors and may be doped to become conductors, compared to organic polymers such as polyethylene which are insulators. This is due to σ-conjugation. In polysilanes, the σ-framework is constructed from 3s and 3p atomic orbitals. Because they are diffuse, significant communication between adjacent Si-Si σ-bonds occurs. This is similar to π-conjugation is organic systems.