Covalent Bonding Between the Elements

Question 1

Define a main group cluster and give how the cluster is described.

Answer 1

A cluster compound is a species with direct element-element bonds which form a 3D shape. The clusters are commonly described by the number of electrons and number of verticies, n. The bonds between elements are not always 2c-2e bonds, they just represent connectivities.

The cluster have a ‘core’ which is commonly surrounded by substituents (e.g H). If it has no substituents it is called naked.

Question 2

How are borane clusters typically synthesised? What are the product characteristics?

Answer 2

The reactions are typically high temperature with low yield cluster products. They are normally made from borane (B₂H₆), a key intermediate to boron clusters.

Question 3

Give the 5 main reactions Borane clusters undergo.

Answer 3

1. Combustion
2. Hydrolysis
3. Electrophilic substitution
4. Base-induced degradation
5. Deprotonation reactions

Question 4

How do you work out the total valence electron count for a cluster? Use [B₆H₆]^2- as an example.

Answer 4

1. Add up the number of valence electrons on the core atoms (6 borons x 3 valence atoms).
2. Substituents such as H and ^tBu all count as one electron (6 hydrogens x 1).
3. Account for charge (2- so add 2)

TVEC = 26

Question 5

Show how to work out Skeletal Electron Count (SEC) and Skeletal Electron Pairs (SEP) for [B6H6]2- (TVEC = 26).

Define electron precise, deficient and rich. What effect to these have on the shape of the clusters?

Answer 5

SEC = TVEC - 2n where n is the number of vertices in the cluster

SEC = 26 - (2 x 6) = 14

SEP = SEC/2 = 7

• Electron precise is where all the bonds in the cluster are 2c-2e. TVEC = 5n, SEC = 3n
• Electron deficient is where there are too few electrons for 2c-2e bonds. TVEC < 5n, SEC < 3n.
• Electron rich is where there are more electrons that the amount needed for 2c-2e bonds, TVEC > 5n, SEC > 3n.
• Electron rich clusters tend to have open structures, electron deficient clusters tend to be much more closed.

Question 6

Describe the molecular orbital basis of borane clusters, highlighting the BH and B-B interactions of [B₆H₆]^2- leading to its closo strucutre.

How are these generalised to wades rules?

Answer 6

Boron has 3 valence electrons, two of which can bond to hydrogen, one of which is a pi bonding electron. Overall, each BH fragment contributes 2 electrons and 3 orbitals to skeletal bonding.

Three sigma orbitals are formed, one strongly bonding and two antibonding. Six pi bonding orbitals form along with 6 antibonding.

Overall, 7 bonding MOs are formed from the 6 vertex structure, meaning when SEP = n+1 for any borane strucutre, a closo structure is formed.

Question 7

Define closo, nido and arachno shapes. How are they linked by SEP values?

Answer 7

Closo structures are closed shapes (octahedron for 6, trigonal bipyramid for 5, tetrahedron for 4). Nido are ‘nest’ structures, appearing like the closo structures with a vertex missing. Arachno are ‘spider’ structures, appearing like closo structures with two verticies missing.

Structures with the same SEP values will have the same structural basis e.g [B₆H₆]^2- is a closo octahedron, B₅H₉ is a nido octahedron and B₄H₁₀ is an arachno octahedron. All have SEP = 7.

Question 8

Fully describe Wades rules.

Answer 8

A closo cluster with n verticies has n+1 SEP, a nido cluster has n+2 SEP, an arachno cluster has n+3 SEP, an hypho cluster has n+4 SEP.

The structures of the nido, arachno and hypho clusters are determined by the closo parent with the same SEP number (with SEP-1 vertices).

Generally, the most connected vertex is removed (especially from closo to nido).

Additional hydrogen atoms are placed at bridging sites or in terminal sites.

Question 9

Draw the structures of the boron closo clustersfrom B₇ to B₄.

Answer 9

Question 10

Describe how to determine cluster geometry, using B₅H₁₁ as an example.

Answer 10

1. Determine vertices (n), TVEC and SEP. n = 5, TVEC = 26, SEP = 8.
2. Assess if closo, nido, arachno or hypho. n + 3 so arachno.
3. Determine parent closo strucutre. 8 - 1 = 7, so B₇ pentagonal bipyramidal.
4. Remove vertices as appropriate. First remove most connected, then in this case second vertex removed is not most connected.
5. Add remaining hydrogen atoms (terminal H to terminal B, bridging H to open faces).

Question 11

How can boron clusters be characterised?

Answer 11

¹¹B{¹H} NMR spectroscopy will show the number of boron environments to give information about the structure. Additionally, IR and single crystal x-ray diffraction can be used.

Question 12

How can CH groups replace boron vertices in boron clusters? What are the bonding implications?

Answer 12

Introducing carbon groups such as ethyne can replace boron vertices. The CH group is idenitcal to a BH- group (3 electrons given to MOs) meaning C₂B₄H₆ is structurally identical to [B₆H₆]^2-.

This principle can apply to any atom that has three orbitals available for cluster bonding. Group 14 elements (Pb, Sn, Ge) will have 2 lone pairs in place of hydrogen substituents.

Question 13

Define isolobal.

How can the isolobal relationships between transition metals and main group elements be determined?

Answer 13

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.

Transition elements are isolobal with main group elements if they have 10 electons more than the main group equivalent. ^•CH₃ has 7 electrons which makes it isolobal with any metal complex with 17 electrons, such as Mn(CO)₅. Both these species have radicals that occupy an a₁ symmetry orbital.

Question 14

How are Wades rules adapted to become Wade-Mingos rules for transition metal complexes?

Answer 14

As main group elements have 4 orbitals and transition metals have 9, to work out the SEC we do TVEC - 12n (instead of 2n for main group). When both main group and transition metals are present, take 2 for each main group and 12 for each TM.

Question 15

How are transition metal clusters different to main group clusters when increasing n? What implications does this have for increasing numbers of vertices?

Answer 15

Unlike main group clusters, TM clusters form large closo-type strucutres when n > 6 with additional groups ‘capping’ the closo faces.

All the capped clusters have the same number of electron pairs (SEP) as the uncapped structure.

A closo strucutre has n+1 SEP. A monocapped closo has n SEP and is based on the closo structure for n-1 vertices. A bicapped structure has n-1 SEP, an tricapped cluster has n-2 SEP.

Question 16

Describe the Polyhedral Skeletal Electron Pair Theory (PSEPT).

Answer 16

As the number of SEPs increase vs the number of vertices, the structures become more open (closo to nido to arachno). These structures are more electron rich.

Equally, as the value decreases, the structures become more closed (monocapped to bicapped to tricapped). These structures are more electron deficient.

Question 17

What is the trend for metal-metal bond strength and why does it occur?

Answer 17

The bonds are generally stronger down the group. This is because the 3d orbitals are quite core-like and have poor orbital overlap.

Question 18

Generally describe how to determine the bond order of a metal-metal bond. Use V₂ as an example.

Answer 18

1. Work out the oxidation state and d-electron count of the metal. Ox. = 0, d⁵ so 10 electrons in MO.
2. Identify the correct MO overlap diagram. Bonding between d-orbitals, forming 1σ, 2π and 2δ bonding orbitals (along with the same antibonding). (CHECK)
3. Populate the diagram and determine the expected bond order. All bonding orbitals filled, no antibonding - bond order is 5.
4. Consider other factors like periodicity

Question 19

When ligands are present in metal-metal bonding complexes, how do you take into account the overlapping orbitals?

What decides if ligands will be staggered or eclipsed?

Answer 19

Orbitals tend to either be involved in M-M or M-L bonding but not both. Defining the metal-metal bond as the z-axis means d orbitals with more x and y character are likely to be bonding to the ligands.

If δ bonding orbitals are present, the ligands are likely to be eclipsed although the sterics will have an impact.

Question 20

For a complex with a triple metal-metal bond, describe the nature of each of the d-orbitals. How does this affect the MO character?

What are the conformations of these complexes?

Answer 20

d_z² will be purely used in M-M bonding. All the other d-orbitals are hybrids. The d_xz orbital has some d_x²-y² character and the d_yz has some d_xy character and vice versa.

This also translates into the MOs where the π orbitals have some δ character.

These molecules are usually staggered for steric reasons but eclisped gives the best orbital overlap.

Question 21

How can magnatism be used to tell if a triple bond is forming between two metals?

Answer 21

Down group 6, the valence electrons are similar such that if the metals are bonding together, the complex should be diamagnetic. However, especially for Cr (and other 4th row elements) the overlap between d-orbitals is weak and in some conditions, bridging ligands will be favoured over M-M bonds.

If this is the case, the complex will be paramagnetic.

In other cases like Fe₂(CO)₉, a complex with no M-M bonds may still be diamagnetic. The most robust way to test this is by using quantum calculations.

Question 22

Generally, how can unfavourable bonds be stabilised, such as a Cr-Cr quintuple bond and a P-P double bond?

Why are these bonds inherently unfavourable?

Answer 22

Using large substituents prevent reactions at the metals to break the bonds. It is much harder for the reactive centres to come together in these conditions.

As you go down the group, π bonds become much weaker compared to σ bonds. This is because the orbitals are much more diffuse so the overlap will be weaker. Therefore it is rare to see double bonds forming with atoms heavier than O.

Question 23

Draw the preperation of main group double bond species of Sn and Si.

Comment on the stereochemistry of the products. How does this compare to other main group compounds?

Answer 23

In both cases large substituents are needed to stabilise the double bond. In contrast to alkene symmetry, these products are not planar. This deviation from planarity increases down the group (about 45° for Pb).

Question 24

Give the general synthesis of main group triple bond compounds and give which elements this works for.

Describe the product stereochemistry and how the nature of the product changes down the group.

Answer 24

The product has a trans stereochemistry. The amount that the product is linear decreases down the group from C (180°) to Pb (94°). Additionally, the bond length increases and the bond order decreases.

Question 25

Decribe the cause of the break from planarity for main group double and triple bonds down the group with an appropriate diagram.

Answer 25

Unlike carbon, the lower group elements have orbitals close enough in energy to mix. For double bonded elements, the π and σ* obitals mix to form an n⁺ orbital which is lower in energy and in a trans symmetry.

For triple bonds, the change is the same, expect the degenerate π bond stays at the same energy.

At further energy gap decreases, the σ and π* can also occur.

Question 26

Briefly describe the different types of semi-conductors.

Answer 26

Elemental silicon and germanium are both semi conductors. Combinations of elements can also form semi-conductors, III/V and II/VI materials.

Question 27

Describe the synthesis of electronic grade silicon and why it is so difficult.

Answer 27

Any small impurity can make a silicon chip unuseable. The pure Si product is then crystallised and sliced to make Si wafers.

Note: CVD is chemical vapour deposition.

Question 28

Give the general process and reagent requirements to make III/V and II/VI materials.

Answer 28

A carrier gas carries the reagents into the chamber which is held at 600-700 °C. There is little carbon contamination however the reagents are highly toxic and diffcult to handle.

Requirements:

• Compounds must be volatile.
• Reagents must be extremely pure.
• Reagents must be stable enough to handle and prepare but decompose in the CVD.

Question 29

Describe alternative ways to produce III/V materials.

Answer 29

Replacing AsH₃ with AsMe₃ or Et₃ makes it easier to handle however introduces much more carbon contamination. A balance between the two is ideal such as As^tBu₃. Similar changes can be made to PH₃.

The two componants are a lewis acid and base pair meaning they should be possible to prepare as an aduct. However these adducts are weak and tend to fall apart in the CVD conditions. To improve this, compounds with 2 or more M-E bonds can be used. These are more stable and easier to handle.

Question 30

Draw the synthesis of polysiloxanes, polysilanes and polystannanes.

Answer 30

Question 31

Describe how the electronic structure of polysilanes and polystannanes is different to organic polymers and the implications of this.

Answer 31

The orbitals on Si and Sn are much larger than on C. As a result, the atoms in the chain can interact with atoms further along the chain, not just their immediate neighbours. The interaction occurs through the σ-framework, similar to π-conjugation.

This means that both these inorganic polymers are semi-conductors.