Brainscape's Knowledge GenomeTM

Browse over 1 million classes created by top students, professors, publishers, and experts.


See full index

Year 2 Chem > Metal Ligands > Flashcards

Metal Ligands Flashcards

You may prefer our related Brainscape-certified flashcards:
AP® Chemistry flashcards Periodic Table flashcards Chemistry 101 flashcards Organic Chemistry 101 flashcards
1
Q

What is the hapticity (η) of a ligand?

A

The number of contiguous atoms of a ligand that are attached to the metal

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

What is the denticity (K) of a ligand?

A

The number of non-contiguous atoms of a ligand attached to a metal

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

What is the μ of a ligand?

A

The number of metal atoms bridged by a ligand

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

How do you calculate the d-electron count of a metal?

A

Group number - oxidation state

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

How do you calculate TVEC of a complex?

A

TVEC = d-electron count + electrons donated by the ligands + number of metal-metal bonds

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

What are the enthalpic effects of a metal ligand complex forming?

A
  • The more ligands attached to a metal, the stronger the bonds and the greater the thermodynamic stability of the resulting complex, larger metals can accommodate more ligands
  • The number of ligands is limited by ligand-ligand repulsion
  • Large negative and positive charges can’t be easily supported
  • Electronic configuration (CSFE etc.)
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

What are the general bond strengths of metal ligand bonds for M2+ and M3+ metals?

A 
M2+ = 200kJmol^-1
M3+ = 550kJmol^-1
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

What are the entropic effects of a metal ligand complex forming?

A
  • Complexes are more stable when the ligand is a 5 or 6 membered ring rather than multiple monodentate ligands, this is because the complex is less likely to dissociate with a 5/6 membered ring as the entropy increase is so much smaller than if a complex with 6 monodentate ligands is dissociating
  • If the complex requires an ordered solvent cage, the entropy of the system will be lowered, the solvent also stabilises the charge of the complex which makes the system more thermodynamically stable
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

How many orbitals does the valence shell of a transition metal have?

A

9 orbitals, can store up to 18 electrons

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

How do you workout how many non-bonding orbitals a metal ligand complex has?

A

If a complex has n bonding orbitals, then it has 9-n non-bonding orbitals

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

How many electrons does a square planar d8 complex have?

A

16e-

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

What is a σ-donor ligand?

A

A ligand that has a lone pair to donate to the metal

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

What is a σ-donor, π-acceptor ligand?

A

A ligand that has a lone pair to donate to the metal and has an empty π-orbital to accept electrons from a metal, bonding is the same as σ-bonding but with an additional π-interaction between empty ligand orbitals and filled metal orbitals, CO is a common π-acceptor ligand

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

What is a σ-donor, π-donor ligand?

A

A ligand with multiple lone pairs to donate to the metal

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

Which type of metal ligand bonding is stronger?

A

σ-bonding as it has better orbital overlap than π-bonding

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

Which type of orbitals are formed when there is not an orbital of matching symmetry?

A

A non bonding orbital

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
17
Q

Where is ∆o located in an MO diagram for a metal ligand complex?

A

Between e2g and t2g

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
18
Q

How do you determine which orbitals a ligand and a metal have?

A
  • You can find the reducible representation of the ligand/metal orbitals by seeing how many orbitals stay in the same place when you apply each symmetry operation of the point group
  • You can then find the irreducible representations that make up the reducible representation, these are the orbitals that the ligand/metal has
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
19
Q

Why does CO bond to metals as a σ-donor only via the C atom?

A

The 5σ orbital is very slightly non-bonding and has significant carbon character, it gets maximum orbital overlap with the metal when bonding as a σ-donor via the C atom

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
20
Q

How does CO act as a π-acceptor ligand?

A

CO has 2 empty 2π orbitals that can accept electron density from the

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
21
Q

What is the direction of charge transfer in a σ-donor interaction?

A

From the CO molecule towards the metal, the electron density increases on the metal and decreases electron density on the CO ligand

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
22
Q

What is the direction of charge transfer in a π-acceptor interaction?

A

From the metal to the CO molecule, electron density decreases on the metal and increases on the CO ligand. CO ligands can relieve negative charge build up at the metal centre (means they can stabilises metals with low formal oxidation state)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
23
Q

What is the experimental evidence for π-acceptor interactions?

A
  • Pure CO has a v(C-O) of 2143cm^-1, when CO is attached to a metal, v(C-O) = 1984cm^-1
  • v(C-O) when CO is attached to a metal is lower because the electron density is accepted by CO in an anti-bonding orbital which decreases bond strength of the C≡O bond which causes v(C-O) to decrease
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
24
Q

What is the trend in v(CO) of d6 transition metal complexes with 6 CO ligands?

A

As oxidation state decreases the e- density increases leading to greater π-acceptor interactions, this weakens the C≡O bond more therefore v(CO) decreases with oxidation state

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
25
Q

What is the trend in v(CO) of neutral Mo complexes with decreasing numbers of CO ligands attached to them?

A
  • All the complexes are Mo(0) d6, this means that the number of CO ligands attached is the only thing affecting the v(CO)
  • As the number of CO ligands attached decreases, v(CO) decreases as the electron density from the metal is shared between fewer ligands meaning that the CO bonds are weakened more and therefore v(CO) decreases
26
Q

Why does v(CO) decrease going from Ni(CO)(PF3)3 to Ni(CO)(PCl3)3 to Ni(CO)(PMe3)3?

A
  • Electronegativity: F > Cl > Me, this means that PF3 accepts more electron density than PCl3 which accepts more electron density than PMe3
  • The PX3 ligands are competing with CO for electron density so when they can accept more electron density it means CO accepts less and therefore the CO bond is stronger therefore v(CO) is greater for Ni(CO)(PF3)3 than Ni(CO)(PCl3)3 which is greater than Ni(CO)(PMe3)3
27
Q

What is the trend in v(CO) with changing coordination mode?

A
  • Free CO has the highest v(CO) value, Terminal CO which is CO bonded to one metal has the second highest value, µ2-CO has the next highest and µ3-CO has the lowest
  • The order is like this because as the number of metals that CO is bonded to increases, the more electron density it accepts in its anti-bonding orbitals and the weaker the CO bonds become and therefore µ3-CO has the lowest v(CO) value
28
Q

Why are there many more CO metal complexes than N2 metal complexes even though they are isoelectronic?

A

The carbon based orbitals in CO are much larger than the orbitals involved in ligand bonding for N2, this means that CO can form a much larger overlap with the metal than N2 can, this makes it more favourable for metals to form CO complexes

29
Q

For an interaction between M2+ and O2(2-) what is the direction of electron transfer and which orbitals are both full that aren’t full in other oxidation states of oxygen?

A
  • It is a σ-donor interaction so electron density goes from the O2(2-) molecule to the metal
  • 1πgy and 1πgx orbitals are both full in O2(2-)
30
Q

For most complexes, electron density is spread over the entire molecule. How is this different to complexes containing oxygen?

A

Oxygen is very oxidising meaning the electron density lies on the oxygen

31
Q

What is the trend in v(O-O) going from O2 to O2-(K+) to O2(2-)(Na+)2?

A

v(O-O) decreases going from O2 to O2- to O2(2-) because the number of electrons in anti-bonding orbitals increases, this causes the O-O bond length increases which causes v(O-O) to decrease

32
Q

What happens if the σ* (3σu) orbital becomes occupied in O2?

A

When a very reducing complex overlaps with O2 orbitals that have correct symmetry, electron transfer occurs filling the anti-bonding orbitals leading to O=O cleavage

33
Q

Why is η1-O2 bent whereas CO is linear?

A

Because O2 has to accommodate an extra pair of electrons in the 1πg(π*) orbital

34
Q

How many electrons does NO donate when in Linear form?

A

3

35
Q

How many electrons does NO donate in bent form?

A

1

36
Q

How do you determine if NO is bent or linear in a complex?

A
  • Remove NO from complex and calculate electron count and oxidation state of the remaining fragment
  • Add 1 or 3 electrons per NO to increase the electron count to 18 (or as close as possible without exceeding 18), you now know the total electron count and the M-NO geometry
  • Determine the metal oxidation state of the complex including the NO ligands and consider linear NO to be NO+ and bent NO to be NO-, you can determine the d electron count from the new metal oxidation state
  • When checking your answer, you add the new metal d electron count, any other electrons from ligands other than NO and then 2 for each NO ligand whether its bent or linear, your answer should be the same as you calculated originally
  • It is possible if you have multiple NO ligands that some can be bent and some can be linear
37
Q

How do you determine if a dihydrogen is going to form a metal H2 complex or a metal dihydride M(H)2?

A
  • Based on what type of interaction there is between the H2 molecule and the metal
  • If there is a σ-donor interaction then an M(H2) complex is formed
  • If there is a π-acceptor interaction then the H-H bond breaks and an M(H)2 complex is formed
38
Q

What type of ligands are Alkenes?

A

σ-donor, π-acceptor ligands

39
Q

When does an alkene form a η2 bond to a metal?

A

When it forms π-acceptor interactions with the metal as well as σ-donor interactions, the ligand is sp2 hybridised when there are just σ-donor interactions and sp3 hybridised when there both interactions

40
Q

What determines the energy barrier to rotation?

A
  • σ-bonding isn’t affected by rotation because orbital overlap doesn’t change significantly, this means that σ-bonds have a small barrier to rotation that the energy available at room temperature can overcome
  • π-bonding is affected by rotation because the bond has to be broken to rotate, this means that π-bonds have very high barriers to rotation that can’t be overcome at room temperature due to there not being enough thermal energy available. This is why rotation about a double bond doesn’t occur
41
Q

How do PR3 ligands stabilise metals in different oxidation states?

A

They stabilise metals in high oxidation states by strong σ-donor interactions and metals in low oxidation states by π-acceptor interactions

42
Q

How do π-donor interactions occur?

A

A full orbital on the ligand overlaps with an empty d-orbital on the metal

43
Q

When a metal has a multiple bond to a ligand, which type of bonding is involved?

A
  • A σ-donor bond and 1 or 2 π-donor bonds

- Complexes with multiple bonds to N or O have metals in high oxidation states with low d electron counts

44
Q

How many electrons does the O^2- ligand donate when it forms a double bond to the metal?

A

4 electrons, when it forms a triple bond it donates 6

45
Q

What happens when you lower the pH of a system containing an oxo complex?

A

It leads to protonation to form hydroxides and aqua complexes

46
Q

When you have an (η-C5H5)- complex, what do you assume the hapticity is?

A

5 as it is so common for this to be the hapticity, if hapticity is not noted then we assume it is 5

47
Q

What are the possible numbers of electrons donated by NR2- ligands?

A
  • 2e- when bent

- 4e- when linear

48
Q

What are the possible numbers of electrons donated by NR(2-) ligands?

A
  • 4e- when bent

- 6e- when linear

49
Q

What are the possible number of electrons donated by O(2-) ligands?

A
  • 4e- (double bond)

- 6e- (triple bond)

50
Q

What is the effect on the orbitals of having a metal with σ-donor, π-donor ligands attached compared with a metal

A

The t2g d-orbitals are higher in energy in comparison to the σ only case, they’re now closer to the eg level, this results in a reduction of ∆oct

51
Q

What is the effect on the orbitals of having a metal with σ-donor, π-acceptor ligands attached compared with a metal?

A

The t2g d-orbitals are lower in energy in comparison to the σ only case, they’re now further away from the eg level, this results in a increase of ∆oct

52
Q

Which type of ligands cause a transition metal to have small crystal field splitting?

A

Weak field ligands (π-donor ligands), these cause the complex to have a high spin and favour splitting of electrons

53
Q

Which type of ligands cause a transition metal to have large crystal field splitting?

A

Strong field ligands (π-acceptor ligands), these cause the complex to have a low spin and favour pairing of electrons

54
Q

Why is ligand substitution faster for high spin complexes?

A

Because high spin complexes have electrons populating eg orbitals which are anti-bonding

55
Q

Which type of ligands have the strongest trans effect?

A
  • π-acceptor ligands, if there is a ligand trans to a π-acceptor ligand, substitution will occur more quickly at the trans position
56
Q

When CFSE is low does a high or low spin complex form?

A
  • Low spin as the CFSE is greater than the pairing energy making it more energetically favourable to for the electrons to pair up in the lower orbitals before starting to fill the higher energy orbitals
  • Low CFSE forms high spin complexes
57
Q

What determines if a complex is inert or labile?

A
  • If the CFSE of the intermediate is greater than that of the reactant, the complex is labile as there is no activation energy barrier
  • If the CFSE of the intermediate is less than that of reactant, the loss of CFSE becomes the activation energy
58
Q

How do you determine enthalpy and entropy of activation from the Eyring equation?

A

Plot a graph of ln(krate/T) vs 1/T, you can calculate entropy and enthalpy of reaction from this as the gradient = -∆H‡/R and the Y-intercept = ∆S‡/R + lnkB/h

59
Q

Which metal ions are the most common for square planar ligand substitutions?

A

d8 metal ions

60
Q

What is the rate law for square-planar substitution?

A
  • Rate = k1 [ML3X] + K2 [ML3X] [Y]

- Where X is a leaving group and Y is the incoming group (nucleophile)

61
Q

What are the two possible reaction pathways for the substitution of a square planar complex?

A
  • Pseudo first order, where the ML3X complex combines with a large conc. of solvent (often water) to form an intermediate ML3(H2O) complex, the rate equation for this pathway is Rate = k1 [ML3X] as the rate determining step is the formation of the intermediate and not the intermediate being converted to ML3Y. Water doesn’t appear in the rate equation as it is present in a very large conc. so the reaction is pseudo first order
  • Second order, where the ML3X complex reacts in one step with Y to form ML3Y + X hence the rate equation is Rate = [ML3X] [Y]

Year 2 Chem (5 decks)

Brainscape helps you reach your goals faster, through stronger study habits.
© 2024 Bold Learning Solutions. Terms and Conditions