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Topics in inorganic 2 > Mike Whittlesey > Flashcards

Mike Whittlesey Flashcards

1
Q

Explain the ground state electronic configurations of Cr and Cu

A

In other first row metals, 4s fills before 3d, and then the d-orbitals fill sequentially
There is a perturbation at Cr (3d54s1) and Cu (3d104s1), where half-filled s-orbitals are generated
This configuration is most stable due to maximisation of the exchange energy
i.e. the number of parallel spins has been maximised

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2
Q

Exchange energy for 3d44s2

A

6K

also e-e repulsion between the electrons in 4s

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3
Q

Exchange energy for 3d54s1

A

10K

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4
Q

General trend in metallic radius for 1st row metals

A

Metallic radius decreases across the period due to increasing Zeff and the relatively poor shielding ability of d electrons (poor penetrating power - diagram)
Blip in the trend at Mn and Zn (metallic radius increases)
This is because the electrons are in a more spherical arrangement due to the extra electron being placed into an s-obrital (Mn and Zn are after Cr and Cu respectively)

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5
Q

General trend in heat of atomisation for 1st row metals

A

Similar to that for metallic radii

Increases across the period but not a constant trend (blips at Cr/Mn, Cu/Zn)

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6
Q

Heat/enthalpy of atomisation

A

The energy required to remove 1 atom of an element from the bulk of the metal
Measured in kJ/mol
An indicator of M-M bond strength

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7
Q

Why are s-electrons better at shielding than p/d/f?

A

S-electrons can get very close to the nucleus (high penetrating power) so can shield Zeff well
This is not the same for d/f electrons so they are poor shielders

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8
Q

Electronic configurations of 1st row metals

A

3d/4s orbitals are of similar energy so fill sequentially

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9
Q

Electronic configurations of 2nd row metals

A

4d/5s orbitals are similar in energy but not completely the same
They are most similar at the end of row but there is a big difference in their relative energies at the start of the period

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10
Q

Electronic configurations of 3rd row metals

A

There isn’t really a crossover in the energy of 5d/6s orbitals so the pattern in electronic configurations is different

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11
Q

Why is the pattern for the electronic configurations of 1st, 2nd and 3rd row metals different?

A

Because the relative energies of the s- and d-orbitals are different in each row

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12
Q

Trend in metallic radii from 1st to 2nd/3rd row metals

A

Metallic radii of 2nd and 3rd row metals are bigger than those in the 1st row
The electrons are in orbitals of higher n - i.e. the orbitals are larger and further away from the nucleus

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13
Q

Trend in metallic radius from 2nd to 3rd row

A

Not necessarily an increase in metallic radius/size from 2nd to 3rd row
This is due to the lanthanide contraction
The additional row of f-elements between the filling of the 4d and 5d orbitals are poor shielders due to orbital penetration
5s and 5p orbitals penetrate the 4f orbitals, meaning electrons in 4f orbitals are not shielded from the increasing nuclear charge, leading to a decrease in atomic radius across the row of lanthanides

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14
Q

Trend in DeltaHatom from 1st to 2nd to 3rd row metals

A

Enthalpy of atomisation increases from 1st to 2nd to 3rd row metals because the overlap of 5d-5d orbitals is more effective than 4d-4d and 3d-3d overlap
5d orbitals are larger (“greater spatial extent”) so the electron distribution is shifted further out and there is an increased probability of finding electrons further from the nucleus
i.e. bonds increase in strength down the group for d-block elements (in contrast to non-d-block elements, where bonds become weaker down the group)

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15
Q

Metal-metal bonding in d-block elements

A

The trend in enthalpies of atomisation for d-block metals suggests metal-metal bonding is more extensive for the 2nd and 3rd row
Might also anticipate that the highest incidence of metal-metal bonding occurs around the middle of the periods, where enthalpies of atomisation are highest

Heavy d-block elements like M-M bonds

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16
Q

Why is metal-metal bonding extensive in low oxidation state, polynuclear metal complexes of 2nd and 3rd row elements?

A

A low oxidation state means the electrons are readily available in diffuse d-orbitals
i.e. the d-orbitals are more diffuse when the metal is in a low oxidation state so metal-metal bonding is more common (“spatial diffusivity of d-orbitals”)

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17
Q

Metal-metal bonding can be illustrated by comparing the calculated and experimental values for DeltaHf in CrCl2, MoCl2 and WCl2

A

DeltaHf assumes the compound is ionic
DeltaHf(calc) and DeltaHf(exp) are very similar for CrCl2 so this compound can be treated as ionic
There is a discrepancy between the calculated and observed values for MoCl2 and WCl2 so cannot be regarded as ionic solids
(They actually form M-M clusters/aggregates)

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18
Q

Oxidation states

A

In general, higher oxidation states in the 2nd and 3rd row are more stable than those in the 1st row (e.g. WF6 is known but CrF6 is not, WO3 is not readily reduced where CrO3 is a strong oxidising agent)

Low oxidation state organometallic complexes are also increasingly stabilised upon descending a group
e.g. Ru3(CO)12 undergoes substitution of the CO ligands at room temp, whereas Os3(CO)12 only undergoes substitution upon heating, due to better M-L orbital overlap due to the spatial diffusivity of the d-orbitals

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19
Q

Why are higher oxidation states more common on the left hand side of the d-block?

A

Higher oxidation states decrease in stability as you go across the d-block because Zeff increases
(i.e. the nuclear charge is ‘higher’ so wants to retain its electrons)

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20
Q

Group oxidation state

A

Can be reached up to group 8 (e.g. OsO4)

Beyond that, the sum of the ionisation energies is too high to be compensated for by M-L bond formation

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21
Q

Typical coordination numbers for 2nd and 3rd row metals

A

The larger size of the 2nd and 3rd row elements favours CN > 6, and CN = 7, 8, 9 is even more common
High CNs are achieved with small/non-polarisable ligands e.g. OH-, F- or with chelating ligands with low steric requirements

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22
Q

CN = 7,8

A

There are several limiting geometries
And most complexes are fluxional on the NMR timescale, so their geometries must be established in the solid state by X-ray crystallography

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23
Q

Examples of 7-coordinate geometries

A

Pentagonal bipyramid
Capped trigonal prism
Capped octahedron
(draw)

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24
Q

Examples of 8-coordinate geometries

A

Square anti prism
Dodecahedron
Cube (generally unfavourable due to L-L interactions

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25
Q

Electronic structure of d-block elements

A

DeltaOct increases from 1st to 2nd to 3rd row

Meaning 2nd and 3rd row complexes are generally low spin

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26
Q

Evidence for increasing DeltaOct from 1st to 2nd to 3rd row

A

From electronic spectra and magnetic data
Electronic spectra - electron transition energy increases down the group because DeltaOct is larger
Magnetic data [MCl4]2- - when M = Ni, u = 3.89BM (unpaired electrons, tetrahedral geometry), but when M = Pd or Pt, u = 0BM (no unpaired electrons, square planar geometry)

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27
Q

Geometry of low spin complexes

A

Generally square planar

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28
Q

Geometry of high spin complexes

A

Generally tetrahedral

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29
Q

When does the spin-only formula apply?

A

When the magnitude of the magnetic moment is governed only by the spin angular momentum

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30
Q

What is magnetic susceptibility governed by?

A

Total angular momentum

Which = spin angular momentum + orbital angular momentum

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31
Q

Orbital angular momentum in many 1st row octahedral complexes

A

Quenched
Does not contribute to the magnetic susceptibility
Therefore the spin-only formula can be used

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32
Q

Magnetic susceptibility in 2nd and 3rd row octahedral complexes

A

The spin and orbital angular momenta couple in 2nd and 3rd row octahedral complexes (“spin-orbit coupling”)
Therefore the spin-only formula can no longer be used (we need a more complex formula instead)

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33
Q

Why is there a discrepancy between uso and ueff values for 2nd and 3rd row metal complexes?

A

Because spin-orbit coupling is generally large

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34
Q

Effect of temperature on ueff

A

Ueff can be very temperature dependent if the spin-orbit coupling is large (non-Curie behaviour)
This can be illustrated in a Kotani plot (ueff vs -kT/lambda)
(draw)
2nd and 3rd row ions lie on areas of the curve with steep gradients - i.e. their magnetic moments are greatly affected by temperature changes
1st row t2g d4 ions lie on an essentially horizontal part of the curve - i.e. temperature changes have little effect on ueff

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35
Q

What kind of information can be obtained from magnetic measurements?

A

e.g. for [M2Cl9]3-
Cr complex is paramagnetic, Cr-Cr = 3.12 A
W complex is diamagnetic, W-W = 2.42 A
This indicates electronic communication that allows the pairing up of electrons for W but not Cr
i.e. there is a metal-metal bond!
This is consistent with the M-M bond length being less than the sum of the metallic radii for the W complex but not Cr

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36
Q

What dominates the chemistry of the 2nd and 3rd row metals?

A

Metal-metal bonding

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37
Q

2 main classes of compound that display M-M bonding

A
  1. Those with bridging ligands e.g. [W2Cl9]3-
  2. Those with non-bridging ligands e.g. [Mo2Cl8]2-
    (draw structures)
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38
Q

How does the presence of M-M bonding vary within the TM series?

A

M-M bonding requires the presence of at least one d-orbital
The expanded d-orbitals of the elements on the left hand side favour M-M bonding
M-M bonding increases with decreasing oxidation state due to increased spatial overlap of the d-orbitals
Steric crowding around a metal centre can reduce the possibility of the approach of a second metal to form M-M bonds

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39
Q

[Re2Cl8]2- bonding

A

First example of a M-M quadruple bond!
Think of as 2 ReCl4 units brought together along the z-direction
Valence orbitals on Re available for bonding = 6s, 6p and 5d
s, px, py and d(x2-y2) orbitals are used to form the 4 Re-Cl bonds
pz and d(z2) orbitals on each Re mix to form 2 pd(z2) hybrid orbitals - one of these forms a sigma bond between the Re, one points away from the Re-Re bond
dxz and dyz orbitals on each Re overlap side-on to form pi-bonds
dxy on each Re overlap face-on to form a delta bond (providing the 2 ReCl4 units are in an eclipsed conformation)

DRAW

40
Q

How can M-M quadruple bonds be characterised?

A
  1. X-ray crystallography
    (a) Very short M-M bond distances
    (b) Eclipsed ligands - consistent with delta bond formation
    (c) Ligands bend away from M-M bond due to e-e repulsion
  2. UV-Vis spectroscopy - an intense absorption band in the visible region is caused by the excitation of an electron from a sigma2pi4delta2 singlet ground state to give a sigma2pi4deltadelta* singlet excited state
41
Q

Effect of charge on metal centre on length of M-M quadruple bond

A

Decreasing charge (i.e. decreasing oxidation state) means more d electrons, which go into the delta* orbital, therefore lengthening the M-M bond

42
Q

M-M quintuple bonds

A

Reported for 1st row TMs using very bulky ligands (e.g. terphenyl)
Ligand is so bulky that there is only one per metal centre, which therefore frees up the d(x2-y2) orbital to afford an additional M-M bond
sigma2pi4delta4
M-M = 1.84 A

43
Q

Reactions of inorganic systems containing M-M multiple bonds

A

Addition reactions
Elimination reactions
Cluster formation reactions (not polymerisation)

44
Q

What does the reactivity of a compound depend on?

A

Its valence/frontier molecular orbitals (i.e. HOMO/LUMO)

45
Q

Isolobal approach

A

A strategy used in organometallic chemistry to relate the structure/chemistry of organic fragments with inorganic fragments in order to predict the bonding properties of organometallic compounds
Provides a good guide to reactivity, but does not take kinetic/thermodynamic factors into account

46
Q

Isolobal fragments

A

Same number of valence electrons
Same number and shape of frontier orbitals
“Similar” orbital energies

47
Q

.CH3 (methyl radical) is isolobal with…

A

…[ML5].-

.CH3 and [ML5].- have the same number of non-bonding hybrid orbitals occupied by the same number of electrons, therefore they are isolobal

48
Q

:CH2 (carbene) is isolobal with…

A

…[ML4]2-

49
Q

.:CH (carbyne) is isolobal with…

A

…[ML3]3-

50
Q

Reactivity of complexes

A

Metal-metal bonds are weaker than metal-ligand bonds (otherwise complexes would not form)
This means M-M bond orbitals are likely to be the HOMO (because they are the weakest bond)

51
Q

Photochemical M-M cleavage

A

e.g. M2(CO)10 —hv—> 2M(CO)5, where M = Mn, (Tc), Re
Electron is excited from M-M bonding orbital (HOMO) into M-M anti bonding orbital, therefore the M-M bond breaks

NOTE Re-Re bond is stronger than Mn-Mn bond because Re is bigger so there is better overlap of the Re 5d orbitals

52
Q

Element-induced M-M bond cleavage

A

e.g Co2(CO)8 + H2 2HCo(CO)4
Mn2(CO)10 + Br2 2BrMn(CO)5

Reaction is driven by the formation of strong M-L bonds at the expense of weaker M-M and E-E bonds

53
Q

M-M cleavage with a reducing agent

A

e.g. Cp2Fe2(CO)4 —2Na—> [CpFe(CO)2]-
Na adds an electron into the M-M system (i.e. effectively into the M-M bond, therefore the bond breaks)
(same thing can happen with Co2(CO)8 and Mn2(CO)10)

[CpFe(CO)2]- can then react with MeI to give CpFe(CO)2Me and NaI (driving force)

54
Q

Synthesis of M-M multiple bonds via elimination reactions

A

e.g. [CpMo(CO)3]2 —hv—> Cp2Mo2(CO)4 + 2CO(g)

M-M single bond going to M-M triple bond, as orbitals are made available upon loss of CO
Elimination of CO is preferable to breaking the M-M bond

e.g. Os2(CO)9 —hv—> “Os2(CO)8” + CO
M-M single bond going to M-M double bond
“Os2(CO)8” can be ‘trapped’ with ethene
OsCO4 is a 16e fragment so needs 2 more electrons to make it up to 18e

55
Q

Metal clusters

A

All molecules in which 3 or more metal atoms, in addition to being bonded to other non-metal atoms, are bonded to each other
i.e. M-L and M-M bonds are present

56
Q

Early TMs

A

Form pi-donor clusters

57
Q

Late TMs

A

Form pi-acceptor clusters

58
Q

What are the 2 types of cluster?

A

Pi-donor clusters

Pi-acceptor clusters

59
Q

Pi-donor clusters

A

Clusters of early TMs are generally associated with ligands such as O2-, S2-, Cl-, Br-, I- and -OR
Metals are generally in high oxidation states
Based around octahedral geometries or metal triangles
Bonding is often described in terms of 2c-2e bonds

60
Q

Hexanuclear clusters

A

Lower halides of group 5 (e.g. NbF5) are not monomeric
Instead, the metal ions form clusters of atoms based on M6X8 and M6X12 structural cores
The ‘dihalides’ MoCl2/Br2/I2 and WCl2/Br2/I2 also afford M6X12 species (i.e. they are not actually monomeric dihalides)
The cluster core for the group 6 metal halides is based on a central core of [M6X8]4+ with face-capping (u3) halide ligands as well as terminal halides - i.e. the formula can be considered as [M6X8]4+ with X- ligands terminally bonded as well as being shared into large solid-state sheet structures

61
Q

What does MoCl2 exist as?

A

[Mo6Cl14]2-, which can be viewed as [Mo6(u3-Cl)8]4+Cl6]2-

62
Q

Why is the [M6X8]4+ structural motif so common?

A

e.g. for [Mo6(u3-Cl)8]4+ = overall [Mo6]12+ = Mo2+ (d4)
Octahedron of 6 Mo with 24 electrons (6x4 = 24 e = 12 e pairs) to place into 12 edges of the octahedron
i.e. therefore each Mo-Mo bond has a bond order of 1

63
Q

Common types of hexanuclear structures

A

[M6X8]4+
[M6X8]3+
[M6X12]n+

64
Q

[M6X8]3+

A

Hexanuclear cluster

Also based on u3 face-capping ligands

65
Q

[M6X12]n+

A

Hexanuclear cluster
Formed when the ligands edge-bridge (u2)
n is typically 2

66
Q

Hexanuclear clusters…

A

…tend to form materials rather than discrete molecular species, with X being shared into the rest of the surrounding lattice

67
Q

Nb6I11

A

Contains a [Nb6I8]3+ unit
Octahedron of Nb with 8 face-capping I
Extra 3 I balance charge and connect the [Nb6I8]3+ units into a 3D network/lattice

68
Q

Examples of where the [M6X12]n+ unit is found

A

In compounds of type M6X14 and M6X15
Where n = 2 or 3, M = Nb or Ta and X = halide
Octahedron of M with 12 edge-bridging ligands
Extra 2/3 X connect the clusters into a 2D sheet arrangement

69
Q

Consider the case of [Nb6Cl18]4-

A

Treat as [Nb6(u2-Cl)12]2+ cluster
Overall [Nb6]14+, so likely consists of mixed oxidation state metals
e.g. 4 Nb2+ (d3) and 2 Nb3+ (d2)
Octahedron of 6 Nb with 16 electrons (8 e pairs) to place into 12 edges of the octahedron
Therefore each Nb-Nb bond has a bond order of 2/3

70
Q

What are interstitial main group atoms?

A

A main group atom/molecule trapped within the centre of a cluster
e.g. C (carbide)

71
Q

Pi-acceptor clusters

A

i.e. carbonyl clusters
The majority of these are observed for the middle-late TMs (groups 7-10)
Compounds are generally anionic or neutral (cations are rare because they can’t backbond - and CO ligands require backbonding in order to form stable complexes)
Based around triangular building blocks (=deltahedra/polyhedra)

72
Q

Early problem with transition metal carbonyl clusters

A

Inability to predict their structures

It was thought the 18 electron rule would be able to predict structures but this was not the case for Co2(CO)8

73
Q

If we know the structure…

A

…we can predict the extent of M-M bonding

74
Q

Coordination modes of CO ligands

A

Terminal
Bridging
Face-capping (only ever seen in metal clusters, not mono- or binuclear complexes)

DRAW

75
Q

Pi-acceptor ability of bridging vs. terminal CO ligands

A

Bridging carbonyls are better pi-acceptors than terminal CO ligands because there is more effective overlap between the d-orbitals of the 2 metals and the pi* orbital of CO
This increase in backbonding for bridging CO ligands is manifested in the lower IR stretching frequency

76
Q

Effect of charge on the cluster on backbonding

A

The higher the negative charge, the more backbonding there is
Therefore leading to a lower CO stretching frequency

77
Q

Effect of sigma-donors on backbonding

A

Phosphine ligands are better sigma-donors than CO which increases the electron density on the metal
This means there is more backbonding and therefore a lower CO stretching frequency

78
Q

13C NMR of carbonyls

A

Terminal CO ligands display resonances from +180 to +210

Bridging CO ligands are more downfield (higher ppm)

79
Q

Limitations of 13C NMR spectroscopy of carbonyl complexes

A

Samples generally need to be 13C-enriched due to the low natural abundance of 13C
The complexes are often fluxional at room temperature so do not provide much information - need to go to lower temperatures

80
Q

Effective atomic number (EAN) rule

A

An approach for predicting the number of M-M bonds in a metal cluster (NOT predictive of structure)
Assumes that, on average, all metal centres have 18 electrons and therefore all M-M bonds are 2c-2e

81
Q

EAN formula

A

m = (18n-k)/2

Where m = number of M-M bond pairs, k = total number of electrons and n = number of metals

82
Q

Specific valence electron counts have…

A

…characteristic geometries

83
Q

48 valence electrons

A

Triangle cluster cage geometry

84
Q

60 valence electrons

A

Tetrahedron cluster cage geometry

85
Q

62 valence electrons

A

Butterfly OR planar 4-atom raft cluster cage geometry

86
Q

64 valence electrons

A

Square cluster cage geometry

87
Q

72 valence electrons

A

Trigonal bipyramid cluster cage geometry

88
Q

74 valence electrons

A

Square-based pyramid cluster cage geometry

89
Q

86 valence electrons

A

Octahedron cluster cage geometry

90
Q

90 valence electrons

A

Trigonal prism cluster cage geometry

91
Q

Total valence electron count can also be used to…

A

…rationalise 2 electron reductions and oxidations in metal cages

92
Q

Other common ligands in M(CO) cluster complexes

A 
CO
Hydride
Phosphine
Nitric oxide (NO)
Pi-bonded ligands e.g. alkene, alkyne, Cp, arene
93
Q

1H NMR spectroscopy of hydrides in clusters

A

Terminal hydrides 0 to -5 ppm
Edge-bridging hydrides -10 to -15 ppm
Face-capping hydrides -20 to -30 ppm
Interstitial hydrides +20 to -60 ppm (signal is very environment dependent)

94
Q

Summary of metal cluster reactivity

A 
Loss of CO
Breaking M-M bond
Polyhedral rearrangement
Changes to cluster nuclearity
Ligand rearrangements
95
Q

Preparations of hydride clusters using H2 gas

A

Draw

96
Q

Association reactions

A

Association of a phosphine across the ‘double bond’ in H2Os3(CO)10

97
Q

Carbide formation

A

Stabilisation of unusual ligands

Draw

Topics in inorganic 2 (3 decks)

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