Organometallic Nomenclature Flashcards
fill in the blanks
Describe the following bond
- σ bond
- no nodal planes
Describe the following bond
- π bond
- one nodal plane
Describe the following bond
- δ bond
- two nodal planes
- due to the overlap of the 4 lobs of the d-orbitals
What does ηⁿ stand for (eta)?
- Some (unsaturated) ligands can bind through more than one carbon atom (ⁿ)
What does μₙ mean (mu)?
- Some ligands may bridge more than one metal centre (ₙ)
What are the 4 main types of organometallic complexes
- Ionic (charge separated)
- Electron deficient
- σ-bonding only
- π bonding
What is an Ionic (charge seperated) complex?
Only with the most electropositive early elements
Highly reactive and unstable
More stable if R can be stabilised by delocalisation or steric bulk
What is an Electron deficient complex?
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
Describe σ-bonding only complexes
Occurs with closed-shell transitional metal and main group centres, often leading to volatile compounds
Describe π-bonding complexes
- 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
How do you work out the formal oxidation state
= charges on complex - sum of formal ligand charges
How do you work out the dⁿ electron count
dⁿ = metal group number - oxidation state
Which ligands have a -1 formal charge
- Alkyl
- aryl
- H
- η¹-allyl
- η¹-cyclopentadienyl (Cp)
- halide
- NO
- OR
- NR₂ (amide)
Which ligands have a 0 formal charge
- CO
- N₂
- PR₃
- NR₃
- OR₂
- RNC
- pyridine
- CR₂
- Alkene
- butadiene
What is Coordinate unsaturation?
- 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
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What is the Isolobal analogy?
- 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
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….
Mn(CO)₅ is isolobal to CH₃
What is an issue with the isolobal analogy?
The isolobal analogy does not say absolutely that a stable compound will result, it only rationalises that bonding may be feasible
What is the orbital arrangement for a ML₆ arrangement?
(metal with 6 ligands)
sp₃d₂
What is the orbital arrangement for a ML₅ arrangement?
sp₃d₂
but one of the hybrids is unused
Going from ML₆, ML₅, ML₄ …. and so on
What happens as a ligand is removed?
The corresponding frontier orbtial pointing towards the vacant site becomes non-bonding
What is the orbtial arrangement for the ML₄ arrangment?
sp₃d₂
but two of the hybrids are unused
What is the core idea of isolobal fragments?
In theory you can switch any isolobal fragments out and it will form the same structure
What factor can prevent you switching between two isolobal units?
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)
Compared woth M-N, M-O, and M-halide bonds, M-C must be considered
Weak
What happens to the M-C enthalpies of transition metal alkyls compared to the main group elements M-C bond?
Equal
(F-block contraction on 2nd and 3rd block in the d-block are same size)
The enthalpy of the M-C bond …….. with increasing atomic number for the main group elements
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
What happens to M-C bond enthalpy for transition metal triad with increasing atomic number
Increases
(d & f-block contraction)
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?
- 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)
Why are the transitional metal organometallics less stable than main group (kinetics)
- Transition metal organometallics have a vacant d-orbital that my decompose by β or reductive elimination (lower energy, hence easier to break)
We don’t want a formed organometallic complex to decompose by β-hydride elimination
So how do we avoid this?
- 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
