Test II Flashcards
In the d-block, discuss how atomic size and oxidation state stability changes on going down the group 3d-4d-5d. Illustrate your answer by means of suitable examples.
-Down a group 3d-4d 5d, 4d is always significantly larger in size compared to the 3d
-5d and 4d are typically very similar sizes (sometimes 5d element slightly smaller) due to lanthanide contraction (lanthanide series of elements occurs between the filling of 4d and 5d orbitals – electrons enter into poorly shielding 4f orbitals, resulting in contraction in size of the subsequent 5d elements) plus relativistic effects.
Oxidation states: Stability of elements in higher oxidation states increases on going down a group, e.g. chromium in VI oxidation state (chromate or dichromate) are strong oxidising agents, but Mo and W are stable in the form of MoO42- and WO42- ions (and polyoxoanions).
-For lower oxidation states, it is the 3d element that has a strong preference for these, e.g. Cr, the most stable oxidation state is III (there are a multitude of Cr(III) coordination complexes).
-Mo(III) and W(III) complexes are rare, and lower oxidation state compounds of these metals typically involve metal-metal bonding.
Sketch a plot of the atomisation enthalpy (△Hat) for the (3d, 4d and 5d) transition elements and briefly comment on how this plot can provide information on the metals that are likely to form stable metal-metal bonded compounds.
- The ‘peak’ occurs for the 4d and the 5d elements, with those in the centre of the d block showing the highest atomisation enthalpies (and with 5d higher than 4d).
- Atomisation enthalpy refers to the process M(s) –> M(g) involves breaking of all the metallic bonds in the solid metal.
- Although metal-metal bonded compounds often have oxidation states other than 0 (the OS in metal itself) there is a surprisingly good correlation between the elements with a high atomisation enthalpy and those forming a large number of stable M-M bonded compounds.
- Elements at the start of the d block have many empty orbitals, but too few electrons to form a lot of M-M bonding. (Sc)
- Elements on the RHS of the d block are electron rich but have few empty orbitals (Zn)
- A good compromise is reached for those elements in the middle of the d block which have close to a half filled set of orbitals (Mn, Cr)
Explain why [Re2Cl8]2- contains a quadruple bond, but [Os2Cl8]2- contains a triple bond.
σ > pi > δ
-Re (III) with 4 valence e- –> 8 e- total
-Re complex, the Re(III) centre has a d4 electronic configuration making a total of 8 electrons for the 2Re.
-The σ bonding, degenerate pi bonding and δ bonding molecular orbitals are all filled, resulting in a quadruple bond.
(σ )2 (pi)4 (δ)2
- The osmium complex has 2 additional electrons, because Os is the element after Re in the periodic table, and the charges on the complexes and hence the metal oxidation states are the same.
- These two additional electrons enter the δ anti bonding molecular orbital. There is thus no net delta interaction, leaving an osmium-osmium triple bond.
Briefly explain what would happen to the metal-metal bonding if [Re2Cl8]2- was reduced to [Re2Cl8]4-.
- Reduction is gain of electrons, and in this case the complex has gained 2 electrons.
- The situation is the same as for the Os complex – the new complex will contain a Re-Re triple bond.
- -These two additional electrons enter the δ anti bonding molecular orbital. There is thus no net delta interaction, leaving an Re-Re triple bond.
Outline the synthetic method to:
RhCl(PPh3)3
Wilkinson’s complex
RhCl(PPh3)3= Reaction of hydrated RhCl3 with 3 equivalents of PPh3 in refluxing ethanol.
-The ethanol acts as the reducing agent for rhodium(III) –> rhodium(I), in the process being oxidised to ethanol.
Outline the synthetic method to:
trans-IrCl(CO)(PPh3)2
Vaska’s complex
trans-IrCl(CO)(PPh3)2 Reaction of hydrated IrCl3 with excess PPh3 in refluxing DMF solvent.
DMF = HC(O)NMe2.
-At its boiling point DMF undergoes a small amount of decomposition to form free CO (which is trapped by the Ir) and dimethylamine.
Explain what is meant by the terms “trans effect” and “trans influence” as applied to complexes of the platinum group metals. Clearly state the difference between them.
-Trans effect is a kinetic effect, measured by determining rates of ligand substitution reactions. A high trans effect ligand increases the rate of substitution of the ligand trans to it.
Trans influence= is a thermodynamic, ground-state property. A high trans influence ligand weakens and lengthens the bond trans to it.
Briefly describe one physical method that can be used to provide information on the trans influence of a ligand.
X-ray diffraction – measures bond lengths directly
IR spectroscopy – a longer metal-ligand bond will have a lower IR stretching frequency
NMR spectroscopy – a longer metal-ligand bond will generally have a smaller coupling constant
Discuss the stability of the auride (Au-) anion; include mention of its electronic structure, the reason for its stability, and comparison with an appropriate mercury species
- The auride anion is reasonably stable.
- Gold has a relatively high electronegativity and electron affinity,
- Au- by gain of an electron, giving a 5d10 6s2 electronic config. e.g. Cs is required to do this.
- Caesium auride has a lot of ionic character, i.e. Cs+Au-.
- Electrons in the 6s orbital are significantly stabilised through relativistic effects, which are at a maximum for gold.
- Au- is isoelectronic with elemental Hg; elemental mercury has a relatively low bpt and a moderate vapour pressure, existing as a monoatomic gas with the same 5d10 6s2 electronic configuration.
Two-coordinate gold(I) complexes have a tendency to aggregate in the solid state. Give the name of this interaction, and give examples of species that show this behaviour.
- The aggregation of two coordinate gold(I) species in the solid state is called aurophilicity.
- Such complexes are inevitably linear.
- When the ligands are small, this allows the Au(I) centres to closely approach each other and interact through the aurophilic interaction.
- This is of the same order of magnitude as H bonds. Eg. LAuX (L = neutral ligand e.g. BuNC) X = anion such as halide.
- These types of compounds can form head-to-head or head-to tail dimers, or more complex structures.
Briefly discuss the polymolybdate anion [PMo12O40]3- under the following headings:
Synthesis
Structure
Analysis by mass spectrometry
Application in analytical chemistry
Synthesis
Synthesised by acidification of a mixture of phosphate (PO43-) and molybdate (MoO42-) ions in aqueous solution.
PO43- + MoO42- + H+ –> [PMo12O40]3-
Structure
This polymolybdate ion has the Keggin ion structure, a compact, ball-like structure with a central PO4 group at its core which provides stability.
-4 sets of 3 MoO6 octahedra which share oxygens. These sets then assemble (again by sharing oxygens) around the central PO4 group.
-Each oxygen of the PO4 group is bound to three different Mo centres. There are 3 types of oxygens – terminal, doubly bridging and quadruply bridging.
Analysis by mass spectrometry
- Negative ion ESI mass spectrometry is useful for the analysis of this type of species, because being ionic, the ESI process transfers solution ions to the gas phase readily.
- The parent trianion is observed, as well as the monoprotonated species [HPMo12O40]2-
Application in analytical chemistry
- The polymolybdate anion is used in analytical chemistry in a colorimetric analysis of phosphate e.g. in water samples.
- The phosphate is converted to phosphomolybdate by addition of excess molybdate and acidification.
- Upon reduction [PMo12O40]3- is converted to [PMo12O40]7-.
- This mixed-valence Mo(V)-Mo(VI) species is intensely blue coloured, so the colour is proportional to the amount of phosphate present in the original sample.
High nuclearity metal carbonyl clusters with ≥ 6 metal atoms
- High nuclearity carbonyl clusters (HNCC) especially common for metals such as Os because of strong Os-Os bonds.
- Heating Os3(CO)12 results in formation of mixture of HNCC.
- Metal clusters typically have close packed metal core with CO ligands on outside.
- Many examples known e.g. Os7(CO)21.
- Platinum forms no stable neutral carbonyls but wide range of platinum-CO anions based on the Pt3(CO)6 triangle which stack to form twisted trigonal prismatic clusters with 2- charge, highly reactive, e.g. [Pt6(CO)18]2-.
The use of 4d and 5d metals and their chemical compounds in homogeneous
and heterogeneous catalysis
- Heterogeneous catalysis often preferred industrially due to ease of recovery and recycling of catalyst.
- 4d metal compounds often show the highest activity due to favourable characteristics (a) – high lability (b) roughly equal stabilities of 2 oxidation states 2 units apart which facilitate oxidative addition and reductive elimination.
- 5d elements typically show greater stability in their higher oxidation states, which may result in a less favourable catalytic cycle. Example RhCl(PPh3)3 and RhH(CO)(PPh3)3 used as hydrogenation catalysts, and Pt-Rh alloy supported on an inert inorganic ceramic matrix used in automobile catalytic converters for conversion of NOx to N2 and O2, and complete combustion of partially combusted hydrocarbons.
Molybdate (MoO42-) and tungstate (WO42-) are stable oxyanions, while chromate (CrO42-) is strongly oxidising. Comment on this observation.
Mo (IV) and W(VI) are stable at high oxidation states being 4d and 5d metals.
Cr(VI) is strongly oxidising. 3d metal are less stable at high oxidation states. Cr (III) more stable.
How could [Re2Cl8]2- be synthesised. The reactants should be clearly stated.
[ReO4]- + Zn/HCl or hydrazine –> [Re2Cl8]2-
- Isolated by the addition of a large cation
- Hydrazine is a strong reducing agent.
