Group 4 Flashcards by Barney Burden

What is the cheapest reducing agent?

Carbon - coal, oil or charcoal

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What does a more electropositive metal result in for reduction?

More electropositive metal = more endothermic reaction

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Why can Ti not be reduced using carbon?

Because Ti is very a electropositive metal, whilst also forming Titanium carbide - which is hard and inert

BAD

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What process is used to obtain Ti?

The Kroll process

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What’re the steps of the Kroll process?

TiO2 is converted to TiCl4 using 2C and 2Cl2

TiCl4 is distilled under vacuum

TiCl4 is then reduced using even more electro+ve metal (Na) to form Ti and 4NaCl. (Mg can also be used.)
- 800 degrees
- batch reactor
- Ar atmosphere

Metal halide is washed out - recycled - giving Ti ‘sponge’
- has to be melted into ingot

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What’re the uses of Ti?

Ti alloys used in aerospace & hi-tech sports equipment - 60% density of steel

Very corrosion resistant

Very high heat resistance - alloys don’t ‘creep’

Twice as ductile with similar tensile strength as steel

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What is yielded when TiCl4 is added to water?

TiO2 and 4HCl

Hydrolysis of TiCl4 is a violent reaction

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What type of ligands does Ti(IV) prefer?

Hard ligands - as is a hard metal ion

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Are Ti(IV) complexes stable under reduction?

Yes

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What does dissolving TiF4 in aqueous HF yield?

Mainly [TiF6]2-

d0 - complex so is colourless

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How do Ti(IV) and V(IV) oxo-complexes differ?

V complexes are monomers - 1005cm-1

Ti complexes are dimers - 880cm-1

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What is the reaction of Ti(IV) halide complexes with neutral monodentate ligands in non-polar solvents?

1 eq gives [TiCl4L]2

2 eq gives [TiCl4L2] - cis and trans variations

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Why does Ti(IV) have no stereochemical preferences?

Because it is a d0 metal ion, and so cannot be controlled through CFT

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What is ‘das’?

It is a very good chelate ligand

Benzene ring provides a rigid backbone

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TiCl4 + 1das gives what in toluene?

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TiCl4 + 2das gives what in toluene?

Whats the difference between a-TiCl3 and ß-TiCl3?

The a form consists of sheets of TiCl6 octahedral.

The ß form consists of single chains of TiCl6, sharing edges

The ß form converts to alpha form at ≥ 300 degrees

What’s the use of ß-TiCl3?

The ß form is a catalyst for alkene polymerisation

What type of polymer is given by polypropylene?

Yields ‘isotactic’ polypropylene - all Me groups oriented same way.

Why must Ti(III) complexes be made in strict absence of oxygen?

So that Ti(III) is not oxidised by oxygen in air.

What does dissolving Ti in dilute HCl give?

[Ti(H2O)6]3+

What does dissolving Ti in conc HCl give?

Both [TiCl(H2O)5]2+ and [TiCl2(H2O)4]+ are formed

What does dissolving Ti in very conc HCl give?

Ti in very concentrated HCl gives Ti(IV) complexes

What are Ti(II) complexes rare?

Because Ti(II) is very strongly reducing

What’re the 2 reasons Ti(II) chemistry has interest?

Organic chemistry - Ti(II) is active in the McMurray coupling reaction Dinitrogen chemistry - Ti(II) reduces N-N triple bond since Ti=NR is isoelectronic with Ti=O.

How are Zr and Hf separated, as they appear in nature together?

By solvent extraction or chromatography

Why is separating Zr and Hf difficult?

Due to their very similar chemical properties

How are Zr and Hf elements obtained?

Elements are obtained by the Kroll process - same as Ti

What’re the uses of Zr?

Used as fuel cladding for nuclear fuels - has low neutron capture cross-section Widely used as alkene polymerisation catalysts

What use has Hf recently found?

As a high dielectric constant insulator layer for the gates of tiny transistors

Is Ti(IV) chemistry similar to Zr(IV) and Hf(IV) chemistry?

Yes

How can stereochemistry be dictated for group 4 M(IV) complexes?

Through sterics, since there is no CFSE in d0 metal ions

What is the difference between TiCl4 / TiBr4 complexes with the Zr/Hf equivalents?

TiCl4 / TiBr4 complexes are tetrahedral and monomeric ZrCl4 / ZrBr4 (and Hf) complexes are polymeric and octahedral

What tendencies do Zr and Hf exhibit with M(II) and M(III) complexes?

Zr and Hf do not form M(II) complexes AT ALL They have lower tendency to form M(III) ALL M(III) complexes are polymeric

How can Zr(III) complexes be diamagnetic?

As they are polymeric, the 2 electrons are shared between the Zr-Zr bond, as though covalent, in the same bonding orbital Hence is diamagnetic (not magnetic)

How are M-M bonds made more stable?

They are made more stable with strong field ligands