Higham L1-3 Flashcards

1
Q

What does the magnitude of delt o depend on

A

1) metal oxidation state
2) 1st, 2nd, 3rd row metal
3) the nature of the ligand

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2
Q

How does the metal oxidation state affect delta O

A

the higher the oxidation state the larger delta o, as higher oxidation states bring ligands closer and increase electron repulsion

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3
Q

How do the position in the group affect delta o

A

delta O gets bigger descending a group and the big orbitals interact more with the ligand, metals at the top of the group are more likely to be paramagnetic and high spin

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4
Q

what is the problem with CFSE

A

CFSE considers ligands as point negative charge (electrostatic model) but if delta o does change when the ligand changes the model needs to be amended to consider L

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5
Q

What is the spectrochemical series

A

Some ligands are better at causing a larger delta o splitting than others, this leads to a league table of ligands being drawn up (spectrochemical series)

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6
Q

How can you follow transition energies

A

using UV-Vis spectroscopy

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7
Q

Which ligands cause delta o to be small

A

Weak field (high spin) ligands
I-, Br-, F-, OH-

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8
Q

What groups are in the middle of the spectrochemical series

A

H2O and NH3

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9
Q

Which ligands cause delta o to be big

A

strong field (low spin) ligands
PR3, CN-, CO

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10
Q

What sort of character do transition metal complexes have

A

transition metal complexes have considerable covalent character

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11
Q

What does the covalent bonding model involve

A

covalent bonding model involves molecular orbital treatment which must account for splitting of metal d orbitals and incorporate ligand sigma and pi orbitals.
Treat ligands as a group

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12
Q

What does the 4S orbitals give rise to

A

most symmetric combination (ligands group orbitals) interacts with metal 4S orbitals. This gives 1 bonding a1g and 1 anti bonding a1g* MOs

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13
Q

what does the metal 4p orbitals give rise to

A

To generate the interactions with the 4p orbitals we need ligand group orbitals that have t1u symmetry. This gives 3 triply degenerate bonding t1u and 3 anti bonding t1u* MOs

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14
Q

Define gerade

A

The term gerade is given to orbitals that are even or symmetric with respect to an inversion centre so the a1g is gerade while t1u is ungerade

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15
Q

where does the eg and eg* MO come from

A

of the metal 3D orbitals only the dz^2 and the dx^2-y^2 point at the ligands, this gives 2 degenerate bonding eg MOs and 2 degenerate anti bonding eg* MOs

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16
Q

Where does the t2g energy level come from

A

the remaining metal d orbitals t2g do not interact with ligand sigma orbitals because they point in between the ligands and therefore are non-bonding

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17
Q

Describe the MO energy level diagram (sigma bonding only)

A

6 LGOs interact with 6 metal orbitals of corresponding symmetry to give 6 bonding and 6 anti bonding MOs left with 3 non bonding t2g orbitals

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18
Q

Where do the electrons from the lone pair on the ligands fill up

A

the eg, t1u and a1g are perfectly placed to accommodate the 12 electrons from the 6 ligands lone pairs, these will become filled

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19
Q

Describe the eg* orbital

A

eg* orbitals mainly metal based but have some ligand character

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20
Q

Which ligands actually undergo only sigma bonding

A

ligands in the middle of the spectrochemical series
E.g NH3, H2O and NR3 R=alkyl

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21
Q

Describe how M-L pi bonding occurs

A

ligand must have suitable orbitals with pi symmetry, these orbitals do interact with the metal t2g orbitals

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22
Q

What does the nature of the pi interactions depend on

A

depends on the relative energies of the metal t2g and ligand pi orbitals and which orbitals are occupied

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23
Q

How do the Px,Py and Pz pi orbital from the ligand work in concert

A

they work in concert (ligand group orbitals) to form 3 host orbitals of suitable symmetry overlap for the 3x t2g orbitals

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24
Q

What happens if the orbitals on the ligands are full

A

Ligand is acting as a pi base/ pi donor

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25
Q

What happens if the orbitals on the ligands are empty

A

Ligand is acting as a pi acid/ pi acceptor

26
Q

Describe the bonding of pi donor ligands and metals

A

need filled pi orbitals on the ligand and empty metal t2g orbitals, generally the donor atom has two or more lone pairs that are not involved in sigma bonding.

27
Q

Describe the effect of pi donors on sigma only MO diagrams

A

Energy of the filled ligand pi orbital are lower then the metal d orbitals resulting t2g MOs are ligand-based and full. The non bonding t2g becomes metal based and antibonding. Eg and eg* orbitals remain unaffected

28
Q

Give examples of pi donor ligands

A

Cl-, OH-, O2-

29
Q

Which sort if metals have empty t2g orbitals

A

metals which have a high oxidation state

30
Q

Describe the bonding of pi acceptor ligands and metals

A

need empty pi orbitals on the ligands and filled metal t2g orbitals for metal to ligand bonding (metals in low oxidation states)

31
Q

What is metal to ligand pi bonding also called

A

back bonding

32
Q

When do we see back bonding

A

we see back bonding for CO,
alkenes and phosphine
Pi acceptor ligands!

33
Q

Describe the bonding within the carbonyl ligand

A

total of 10 electrons. 6 will form the sigma and 2 pi bonds (the triple bond in CO). The remaining 4 electrons will be present as 2 lone pairs, one on the carbon and one on the oxygen

34
Q

Why do metals bind to the carbon lone pairs and not the oxygen one

A

dative bond from O to C leaves O partially positive. Other bonds almost cancel this out as O pulls electrons towards itself due to it being more electronegative the lone pair on O is more stabilised , therefore the lone pair on carbon is therefore higher in energy and destabilised this is the HOMO. The 2x filled pi orbitals will have more oxygen character as a result of its greater electronegativity, therefore the empty pi* orbitals will have more carbon character so the metal bonds to the carbon via both sigma and pi components

35
Q

Give an example of pi acceptor ligands

A

CO or phosphines

36
Q

What are high oxidation states stabilised by and why

A

High oxidation states are stabilised by pi donor ligands. Electron donation to metal from ligands via sigma and pi for metals which leach electrons in high oxidation states

37
Q

What are low oxidation states stabilised by and why

A

Low oxidation states are stabilised by pi acceptor ligands. High electron density at metal requires mechanism for removal - back bonding via pi system

38
Q

Describe the effect that a pi base/donor has on delta 0

A

Delta 0 decreases

39
Q

Describe the effect that a pi acid/acceptor has on delta 0

A

Delta 0 increases

40
Q

Where does the C-O sigma bond arise from

A

The carbon 2s and one of the 2p will for 2 hybrid SP molecular orbitals. Oxygen 2s and one of the 2p will for 2 hybrid sp molecular orbitals. When the SP hybrid orbital from carbon overlaps with the SP hybrid orbital from the carbon we form the C-O sigma bond

41
Q

How do the pi bonds for in C-O

A

Hybridisation for carbon results in 2SP molecular orbitals and 2 p orbitals remain, this is the same for oxygen. When a p orbital from carbon overlaps with oxygen we form one pi bond but since we have two p orbitals for each element we for 2 pi bonds

42
Q

Describe how all ten electrons in C-O are used

A

4 electrons from carbon, 6 electrons from oxygen.
One sigma and two pi bonds take into account 6 electrons
The second SP hybridised MO for carbon houses a lone pair and the second SP hybridised MO for oxygen houses a lone pair

43
Q

What is found in the 1σ orbital for C-O

A

Lone pair of electrons on the oxygen

44
Q

What is found in the 2σ orbital for C-O

A

Sp hybrid, this is the carbon oxygen single bond, this is populates with two electrons

45
Q

What is found in the 3 σ orbital for C-O

A

Homo of the molecule where the lone pair of the carbon sits

46
Q

What is found in the 1 π orbital for C-O

A

Two degenerate orbitals of π symmetry created by the overlap of the carbon-oxygen py and Pz orbitals, this will have higher oxygen character

47
Q

What is found in the 2 π orbital for C-O

A

The 2 anti bonding orbitals for the pi overlap of py and Pz these as a result have more carbon character

48
Q

How does the carbonyl ligand bond to a metal

A

The 3σ HOMO has the correct symmetry and energy will have good overlap with an empty metal orbital of the same symmetry and we can form a metal-carbon σ bond. If the metal orbital contains electrons it can back donate into the C-O 2π* orbital

49
Q

What would u expect to happen to the length of the bond when you have both sigma and pi bonding to a metal

A

If the bonding from metal to ligand has both a sigma and a pi component this should strengthen the bond, increase the bond order and reduce the bond length. A shorter metal-ligand bond is evidence for pi back bonding

50
Q

What can we use to measure bond lengths

A

X ray crystallography

51
Q

How can measuring bind strengths be evidence for back bonding from metal to C in a carbonyl

A

Back bonding will reduce the C-O bond order to give lower CO stretching frequency than in free CO. The metal carbon bond becomes stronger as the CO bond becomes weaker

52
Q

What do we use to measure bond strengths

A

IR spectroscopy

53
Q

What affects the IR spectra of a metal carbonyl complex

A

1 bonding mode of CO
2 charge on the complex (metal oxidation state:
3 other ligands e.g phosphines

54
Q

Describe the change in the IR spectrum for carbonyls when we go from 1 to 2 to 3 metals attached to the C on the carbonyl

A

When we go from 1 metal to 3 metals we have more pi electron density pumped into the CO antibonding orbital making it weaker. Therefore if we have 3 metals to CO there will be a smaller number observed in its vCO

55
Q

Describe how the charge in the complex/ metal oxidation state affects the IR

A

If the metal is an anionic it will pump more electron density into the CO antibonding orbital making it weaker therefore lower number in vCO

56
Q

Describe the three ways we can prepare metal carbonyls

A

1) direction reaction between M and CO
2) reductive carbonylation
3) photolysis or thermolysis

57
Q

Describe the direct reaction between M and CO

A

Roast the metal in the presence of carbon monoxide, for Ni and Fe

58
Q

Describe reductive carbonylation

A

Reduction of a commonly available metal salt in the presence of CO, the reducing agent may be hydrogen, a metal (Na, Mg, Al) or carbon monoxide itself

59
Q

What is the problem with thermolysis

A

Thermal reactions can lead to a mixture of products

60
Q

What are the benefits of using photochemical reactions to make metal carbonyls over thermal reactions

A

Photochemical reaction conditions can give much cleaner reactions as there js a greater degree of control over substitution