Higham L4-6 Flashcards

1
Q

Rank the following π acceptor ligands in terms of their π acceptor strengths
CO, CN-, N2, NO+

A

Increasing π acceptor strength
CN- < N2 < CO < NO+
So CN- is the least likely to accept electron donation from the metal

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2
Q

What does isoelectronic mean

A

Have the same number of electrons

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3
Q

Describe the σ bonding/donation for CN-

A

σ donation through lone pair on C therefore a strong σ donor. Negative change of electron rich CN- increases σ donor strength to electropositive metal, high electrostatic attraction. With metals in normal oxidation states it is not necessary to invoke pi back bondinf

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4
Q

Describe the π bonding CN- complexes

A

π accepting tendencies of CN- diminished by the negative charge. Higher energy of π* orbitals of CN- compared to CO - poorer overlap with metal t2g. Therefore a weaker π acceptor than CO. π bondinf does occur with lower oxidation states

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5
Q

Describe the toxicity of cyanide and the impact it has on the body

A

Cyanide is highly toxic it inhibits mitochondrial respiration, can treat cyanide poisoning with hydroxocobalamin which is a form of vitamin B12

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6
Q

Contrast dinitrogen with carbon monoxide

A

N2 is a weaker sigma donor and weaker pi acceptor than CO, N2 is a homonuclear diatomic molecule whereas CO is a Heteronuclear diatomic molecule. As both ends of N2 are the same there is no dipole so there is no distortion of the electron cloud

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7
Q

Describe how dinitrogen can bind

A

N2 can bind end on or be bridged between two metal centres

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8
Q

What are the two ways in which NO can bind

A

NO can bond as either a 3 electron donor which forms a linear M-NO or a 1 electron donor which forms a bent M-NO

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9
Q

When do bent M-NO linkages form

A

Bent M-NO linkages form when back bonding is not an important consideration

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10
Q

What technique can be used to distinguish between bent and linear M-NO

A

Linear and bent NO ligands can be distinguished using IR: linear groups absorb at 1650-1900 bent nitrosyls absorb at 1525-1690

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11
Q

What is a phosphine

A

Phosphines are PR3
R-alkyl or aryl

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12
Q

What is a phosphite

A

Phosphites are P(OR)3
R- alkyl or aryl

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13
Q

Which of the following are more suitable for pi backbonding PR3 or NR3

A

PR3 ligands have suitable empty orbitals for pi back bonding unlike alkyl amines NR3 which have no empty p or d orbitals of low enough energy

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14
Q

Why are phosphines and phosphites a very important group of ligands

A

As we can change their steric and electronic properties

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15
Q

Describe sigma bonding from PR3 to the metal

A

PR3 lone pair sigma bonding from by donating two electrons into empty metal sigma orbital (d orbital)

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16
Q

Describe pi bonding from PR3 to the metal

A

PR3 sigma bond accepts electron density from filled metal pi orbital

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17
Q

Describe how we can we can modify phosphorus ligands

A

We can alter there sterics by making them very big or very small by varying the R groups. We can also alter electronics by varying elements of different electronegativities.

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18
Q

Describe the structures of phosphorus ligands

A

Phosphorus ligand can be monodentate, bidentate or polydentate

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19
Q

What is the benefit of synthesising large phosphines

A

Ligand dissociation is favoured by bulkier groups. The large groups can be used as protecting groups, can stop the metal undergoing decomposition

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20
Q

How can we measure the different steric effects of different phosphines

A

The steric effects is measured in therms of the tolman cone angle

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21
Q

What is the method of determining the tolman come angle

A

Hold the metal and the phosphorus at a distance of 2.28 angstroms, the three R groups are then sat in a cone like shape, you then measure the cone angle. The cone just encloses van der waals surface of all ligand atoms

22
Q

What does an increased cone angle favour

A

Lower coordination numbers
The formation of isomers with minimum sterics interactions
Increases rates and equilibrium constants in dissociative reactions

23
Q

Describe how we can vary the pi acceptor ability for phosphines and phosphites

A

If we contain a complex with both CO and PX3 we can increase the pi acceptor ability by increasing the electronegativity, this causes more back bonding to PX3 therefore less electron density goes into CO pi* antibonding orbital

24
Q

How does ether bond to a transition metal

A

There is a pi bond and a sigma bond between the two carbons in ethene. If the two p orbitals are in the same phase, this pi cloud is then able to sigma bond with an empty metal d orbital accepting electron density. The lumo is formed when these p orbitals are out of phase and overlap with the transition metal and accept electrons

25
Q

What does there need for sigma and pi bonds to form metal to ethene

A

Plane of alkene must be perpendicular to metal alkene bond

26
Q

Explain how the C=C bond weakens when it binds to the transition metal

A

Sigma donation to M and pi back bonding from M causes C=C to lengthen and weaken: more like single bonds. These carbons tends to have more sp3 character/ be more pyramidal in geometry

27
Q

What is η2

A

2 carbons bonded to metal

28
Q

What is η3

A

3 carbons bonded to metal
C3H5 allyl radical or C3H5- anion

29
Q

What is the two ways in which allyl ligand can bond

A

The allyl ligand can bond in a sigma fashion to a which which will also need to donate an electron or pi orbitals can also be involved to give η3 ie bonded through all 3 carbons

30
Q

What is η4

A

4 carbons bonded to the metal
1,3 butadiene

31
Q

What is η5

A

5 carbons bonded to the metal
The cylcopentadienyl anion (cyclopentadienide) ligand cp

32
Q

How do we produce the cyclopentadienyl anion from cyclopentadiene

A

Cyclopentadiene C5H6 is acidic, deprotonation with NaH gives aromatic anion [C5H5]- as Na[C5H5] and H2

33
Q

Describe M-Cp complexes

A

Cp is very stable as it obeys huckles rule and is considered to be aromatic. The resultant complexes when this binds to a metal are very stable and the M-Cp bonds do not easily dissociate and Cl ligand don’t usually get involved in chemical reactions

34
Q

What is the effective atomic number

A

It is the sum of valence electrons on the metal and the electrons donated by the ligands

35
Q

Describe how we use the neutral atom method for electron counting

A

1) count the metal as being in the zero oxidation state: the group its in gives the number of electrons it gives
2) add up the number of electrons that the ligand contributes
3) factor in for the overall charge on the complex

36
Q

Give examples of neutral L type ligands

A

PR3, NH3, H2O, ethene etc

37
Q

give examples of L type ligand

38
Q

What is an L type ligand

A

If can you get them in a bottle or on there own without a metal then they are an L type

39
Q

What is an x type ligand

A

if it is a ligand which is normally bound to something else and not found on its own

40
Q

How many electrons does each Cp contribute

A

5 electrons

41
Q

How many electrons does η4 contribute

42
Q

How many electrons does η3 contribute

43
Q

For the neutral atom method list the ligands which contribute one electron

A

H, Cl, OH, OR, CN, CH3, OH, NO bent

44
Q

For the neutral atom method list the ligands which contribute 2 electrons

A

PR3, NR3, NH3, H2O, CO, alkenes, RCN, imido bent M=NR and M=O

45
Q

For the neutral atom method how many electrons does linear NO donate

46
Q

How many electrons does imido linear M=NR and M=O contribute

47
Q

How many electrons does linear alkoxide donate

48
Q

How many electrons does benzene donate

49
Q

Explain what happens when chlorine is used as a bridging ligand between two metals

A

To form on M-Cl bond the metal has to give up one of its electrons to the chlorine. Because this chlorides is interacting with the second metal, it has a lone pair which is perfectly inline to act as an L electron donor with the second metal