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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
Electronic structure of d-block elements
DeltaOct increases from 1st to 2nd to 3rd row | Meaning 2nd and 3rd row complexes are generally low spin
26
Evidence for increasing DeltaOct from 1st to 2nd to 3rd row
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)
27
Geometry of low spin complexes
Generally square planar
28
Geometry of high spin complexes
Generally tetrahedral
29
When does the spin-only formula apply?
When the magnitude of the magnetic moment is governed only by the spin angular momentum
30
What is magnetic susceptibility governed by?
Total angular momentum | Which = spin angular momentum + orbital angular momentum
31
Orbital angular momentum in many 1st row octahedral complexes
Quenched Does not contribute to the magnetic susceptibility Therefore the spin-only formula can be used
32
Magnetic susceptibility in 2nd and 3rd row octahedral complexes
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)
33
Why is there a discrepancy between uso and ueff values for 2nd and 3rd row metal complexes?
Because spin-orbit coupling is generally large
34
Effect of temperature on ueff
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
35
What kind of information can be obtained from magnetic measurements?
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
36
What dominates the chemistry of the 2nd and 3rd row metals?
Metal-metal bonding
37
2 main classes of compound that display M-M bonding
1. Those with bridging ligands e.g. [W2Cl9]3- 2. Those with non-bridging ligands e.g. [Mo2Cl8]2- (draw structures)
38
How does the presence of M-M bonding vary within the TM series?
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
39
[Re2Cl8]2- bonding
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
How can M-M quadruple bonds be characterised?
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
Effect of charge on metal centre on length of M-M quadruple bond
Decreasing charge (i.e. decreasing oxidation state) means more d electrons, which go into the delta* orbital, therefore lengthening the M-M bond
42
M-M quintuple bonds
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
Reactions of inorganic systems containing M-M multiple bonds
Addition reactions Elimination reactions Cluster formation reactions (not polymerisation)
44
What does the reactivity of a compound depend on?
Its valence/frontier molecular orbitals (i.e. HOMO/LUMO)
45
Isolobal approach
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
Isolobal fragments
Same number of valence electrons Same number and shape of frontier orbitals "Similar" orbital energies
47
.CH3 (methyl radical) is isolobal with...
...[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
:CH2 (carbene) is isolobal with...
...[ML4]2-
49
.:CH (carbyne) is isolobal with...
...[ML3]3-
50
Reactivity of complexes
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
Photochemical M-M cleavage
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
Element-induced M-M bond cleavage
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
M-M cleavage with a reducing agent
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
Synthesis of M-M multiple bonds via elimination reactions
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
Metal clusters
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
Early TMs
Form pi-donor clusters
57
Late TMs
Form pi-acceptor clusters
58
What are the 2 types of cluster?
Pi-donor clusters | Pi-acceptor clusters
59
Pi-donor clusters
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
Hexanuclear clusters
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
What does MoCl2 exist as?
[Mo6Cl14]2-, which can be viewed as [Mo6(u3-Cl)8]4+Cl6]2-
62
Why is the [M6X8]4+ structural motif so common?
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
Common types of hexanuclear structures
[M6X8]4+ [M6X8]3+ [M6X12]n+
64
[M6X8]3+
Hexanuclear cluster | Also based on u3 face-capping ligands
65
[M6X12]n+
Hexanuclear cluster Formed when the ligands edge-bridge (u2) n is typically 2
66
Hexanuclear clusters...
...tend to form materials rather than discrete molecular species, with X being shared into the rest of the surrounding lattice
67
Nb6I11
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
Examples of where the [M6X12]n+ unit is found
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
Consider the case of [Nb6Cl18]4-
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
What are interstitial main group atoms?
A main group atom/molecule trapped within the centre of a cluster e.g. C (carbide)
71
Pi-acceptor clusters
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
Early problem with transition metal carbonyl clusters
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
If we know the structure...
...we can predict the extent of M-M bonding
74
Coordination modes of CO ligands
Terminal Bridging Face-capping (only ever seen in metal clusters, not mono- or binuclear complexes) DRAW
75
Pi-acceptor ability of bridging vs. terminal CO ligands
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
Effect of charge on the cluster on backbonding
The higher the negative charge, the more backbonding there is Therefore leading to a lower CO stretching frequency
77
Effect of sigma-donors on backbonding
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
13C NMR of carbonyls
Terminal CO ligands display resonances from +180 to +210 | Bridging CO ligands are more downfield (higher ppm)
79
Limitations of 13C NMR spectroscopy of carbonyl complexes
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
Effective atomic number (EAN) rule
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
EAN formula
m = (18n-k)/2 Where m = number of M-M bond pairs, k = total number of electrons and n = number of metals
82
Specific valence electron counts have...
...characteristic geometries
83
48 valence electrons
Triangle cluster cage geometry
84
60 valence electrons
Tetrahedron cluster cage geometry
85
62 valence electrons
Butterfly OR planar 4-atom raft cluster cage geometry
86
64 valence electrons
Square cluster cage geometry
87
72 valence electrons
Trigonal bipyramid cluster cage geometry
88
74 valence electrons
Square-based pyramid cluster cage geometry
89
86 valence electrons
Octahedron cluster cage geometry
90
90 valence electrons
Trigonal prism cluster cage geometry
91
Total valence electron count can also be used to...
...rationalise 2 electron reductions and oxidations in metal cages
92
Other common ligands in M(CO) cluster complexes
``` CO Hydride Phosphine Nitric oxide (NO) Pi-bonded ligands e.g. alkene, alkyne, Cp, arene ```
93
1H NMR spectroscopy of hydrides in clusters
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
Summary of metal cluster reactivity
``` Loss of CO Breaking M-M bond Polyhedral rearrangement Changes to cluster nuclearity Ligand rearrangements ```
95
Preparations of hydride clusters using H2 gas
Draw
96
Association reactions
Association of a phosphine across the 'double bond' in H2Os3(CO)10
97
Carbide formation
Stabilisation of unusual ligands | Draw

Topics in inorganic 2 (3 decks)

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