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Y2 Inorganic Chemistry > Organometallic Nomenclature > Flashcards

Organometallic Nomenclature Flashcards

1
Q

fill in the blanks

A
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2
Q

Describe the following bond

A
  • σ bond
  • no nodal planes
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3
Q

Describe the following bond

A
  • π bond
  • one nodal plane
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4
Q

Describe the following bond

A
  • δ bond
  • two nodal planes
  • due to the overlap of the 4 lobs of the d-orbitals
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5
Q

What does ηⁿ stand for (eta)?

A
  • Some (unsaturated) ligands can bind through more than one carbon atom (ⁿ)
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6
Q

What does μₙ mean (mu)?

A
  • Some ligands may bridge more than one metal centre (ₙ)
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7
Q

What are the 4 main types of organometallic complexes

A
  • Ionic (charge separated)
  • Electron deficient
  • σ-bonding only
  • π bonding
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8
Q

What is an Ionic (charge seperated) complex?

A

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

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9
Q

What is an Electron deficient complex?

A

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

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10
Q

Describe σ-bonding only complexes

A

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

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11
Q

Describe π-bonding complexes

A
  • 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
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12
Q

How do you work out the formal oxidation state

A

= charges on complex - sum of formal ligand charges

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13
Q

How do you work out the dⁿ electron count

A

dⁿ = metal group number - oxidation state

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14
Q

Which ligands have a -1 formal charge

A
  • Alkyl
  • aryl
  • H
  • η¹-allyl
  • η¹-cyclopentadienyl (Cp)
  • halide
  • NO
  • OR
  • NR₂ (amide)
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15
Q

Which ligands have a 0 formal charge

A
  • CO
  • N₂
  • PR₃
  • NR₃
  • OR₂
  • RNC
  • pyridine
  • CR₂
  • Alkene
  • butadiene
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16
Q

What is Coordinate unsaturation?

A
  • 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
    *
17
Q

What is the Isolobal analogy?

A
  • 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
18
Q

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….

A

Mn(CO)₅ is isolobal to CH₃

19
Q

What is an issue with the isolobal analogy?

A

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

20
Q

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

A

sp₃d₂

21
Q

What is the orbital arrangement for a ML₅ arrangement?

A

sp₃d₂
but one of the hybrids is unused

22
Q

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

A

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

23
Q

What is the orbtial arrangement for the ML₄ arrangment?

A

sp₃d₂
but two of the hybrids are unused

24
Q

What is the core idea of isolobal fragments?

A

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

25
Q

What factor can prevent you switching between two isolobal units?

A

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)

26
Q

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

A

Weak

27
Q

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

A

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

28
Q

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

A

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

29
Q

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

A

Increases
(d & f-block contraction)

30
Q

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?

A
  • 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)
31
Q

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

A
  • Transition metal organometallics have a vacant d-orbital that my decompose by β or reductive elimination (lower energy, hence easier to break)
32
Q

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

A
  • 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

Y2 Inorganic Chemistry (19 decks)

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