Organometallic Nomenclature

Question 1

fill in the blanks

Answer 1

Question 2

Describe the following bond

Answer 2

• σ bond
• no nodal planes

Question 3

Describe the following bond

Answer 3

• π bond
• one nodal plane

Question 4

Describe the following bond

Answer 4

• δ bond
• two nodal planes
• due to the overlap of the 4 lobs of the d-orbitals

Question 5

What does ηⁿ stand for (eta)?

Answer 5

• Some (unsaturated) ligands can bind through more than one carbon atom (ⁿ)

Question 6

What does μₙ mean (mu)?

Answer 6

• Some ligands may bridge more than one metal centre (ₙ)

Question 7

What are the 4 main types of organometallic complexes

Answer 7

• Ionic (charge separated)
• Electron deficient
• σ-bonding only
• π bonding

Question 8

What is an Ionic (charge seperated) complex?

Answer 8

Only with the most electropositive early elements
Highly reactive and unstable
More stable if R can be stabilised by delocalisation or steric bulk

Question 9

What is an Electron deficient complex?

Answer 9

Insufficient electrons to fill valence orbitals and form 2-centre-2-electron bonds between all atoms
Results in multi-centre bonding between R and two or more metal centres

Question 10

Describe σ-bonding only complexes

Answer 10

Occurs with closed-shell transitional metal and main group centres, often leading to volatile compounds

Question 11

Describe π-bonding complexes

Answer 11

• This is the interaction of π and π’ orbitals of organic ligands with metal-based orbitals
• This is especially prevalent with transition metals and zero-valent lanthanides

Question 12

How do you work out the formal oxidation state

Answer 12

= charges on complex - sum of formal ligand charges

Question 13

How do you work out the dⁿ electron count

Answer 13

dⁿ = metal group number - oxidation state

Question 14

Which ligands have a -1 formal charge

Answer 14

• Alkyl
• aryl
• H
• η¹-allyl
• η¹-cyclopentadienyl (Cp)
• halide
• NO
• OR
• NR₂ (amide)

Question 15

Which ligands have a 0 formal charge

Answer 15

• CO
• N₂
• PR₃
• NR₃
• OR₂
• RNC
• pyridine
• CR₂
• Alkene
• butadiene

Question 16

What is Coordinate unsaturation?

Answer 16

• Compounds wwhich have less than 18-electron count are said to be ‘coordinately unsaturated’ and there may be a tendency to add further ligands
• This is because there must be one or more vacant orbital(s) available - hence likely to be reactive and unstable
  *

Question 17

What is the Isolobal analogy?

Answer 17

• This relates two fragments (often an organometallic fragment to an organic fragment)
• The two fragments are said to be isolobal if the:
• Number, Symmetry Properties, Approximate Energy, and Shape (extent in space) of their frontier orbtials, as well as number of electrns occuping them, are similar

Question 18

A consequence of the Isolobal Analogy is that if molecules are contructed from isolobal fragments, they themselves are isolobal
Hence H₃-C-CH₃, (CO)₅Mn-CH₃, and (CO)₅Mn-Mn(CO)₅ are all isolobal to another because….

Answer 18

Mn(CO)₅ is isolobal to CH₃

Question 19

What is an issue with the isolobal analogy?

Answer 19

The isolobal analogy does not say absolutely that a stable compound will result, it only rationalises that bonding may be feasible

Question 20

What is the orbital arrangement for a ML₆ arrangement?
(metal with 6 ligands)

Answer 20

sp₃d₂

Question 21

What is the orbital arrangement for a ML₅ arrangement?

Answer 21

sp₃d₂
but one of the hybrids is unused

Question 22

Going from ML₆, ML₅, ML₄ …. and so on
What happens as a ligand is removed?

Answer 22

The corresponding frontier orbtial pointing towards the vacant site becomes non-bonding

Question 23

What is the orbtial arrangement for the ML₄ arrangment?

Answer 23

sp₃d₂
but two of the hybrids are unused

Question 24

What is the core idea of isolobal fragments?

Answer 24

In theory you can switch any isolobal fragments out and it will form the same structure

Question 25

What factor can prevent you switching between two isolobal units?

Answer 25

Stability: a lack of decomposition of the pure compound in the absence of a reaction partner This can be thermodynamic (stable or unstable) OR kinetic stability (inset or labile)

Question 26

Compared woth M-N, M-O, and M-halide bonds, M-C must be considered

Answer 26

Weak

Question 27

What happens to the M-C enthalpies of transition metal alkyls compared to the main group elements M-C bond?

Answer 27

Equal (F-block contraction on 2nd and 3rd block in the d-block are same size)

Question 28

The enthalpy of the M-C bond ........ with increasing atomic number for the main group elements

Answer 28

Decreases (Down the group the M atom is getting larger, the valance electrons are in more diffuse orbitals and the overlap is not as effective) Hence the bond is weaker

Question 29

What happens to M-C bond enthalpy for transition metal triad with increasing atomic number

Answer 29

Increases (d & f-block contraction)

Question 30

due to the M-C bond enthalpy decreasing with increasing atomic number, you would expect the transition metal alkyl should be more stable than lead alkyl (4th period) However this is not the case, why?

Answer 30

* The reason for the difference in stability is no thermodynamic but **kinetics** * PbEt₄ decomposes by **M-C bond homolysis** (high energy process) as Pb has a full octet (filled 5d shell) * Essentially like hitting with a sledgehammer, forming two radicals (high energy and slow)

Question 31

Why are the transitional metal organometallics less stable than main group (kinetics)

Answer 31

* Transition metal organometallics have a vacant d-orbital that my decompose by β or reductive elimination (lower energy, hence easier to break)

Question 32

We don't want a formed organometallic complex to decompose by β-hydride elimination So how do we avoid this?

Answer 32

* Use ligands which **don't have β-hydride available** (can't eliminate if there is no hydrogen in the first place) * Make the potental **products of decomposition unfavourable** (e.g. ring strain) * Formation of **arynes from arenes** can be prevent by using otho-substituents * **Bulky ligands** * Saturate the Coordination sphere * Obey the **Pauling Electronegativity** Principle (less charged - the less reactive) * Make 18-electron compounds