Unless otherwise stated, all images in this file have been reproduced from:

Blackman, Bottle, Schmid, Mocerino and Wille,     Chemistry, 2007 (John Wiley)

     ISBN: 9 78047081 0866

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e CHEM1002 [Part 2]

Dr Michela SimoneLecturer

BSc (I Hons), MSc, D.Phil. (Oxon), MRSC, MRACI

Weeks 8 – 13

Office Hours: Monday 3-5, Friday 4-5Room: 412A (or 416)

Phone: 93512830e-mail: michela.simone@sydney.edu.au

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e

Complexes II

• For octahedral complexes with formulae [MX2Y4], cis and trans geometrical isomers are possible

• For square planar complexes with formulae [MX2Y2], cis and trans geometrical isomers are possible

• For octahedral complexes with bidentate ligands, optical isomerism is also possible

• Metal complex formation can greatly increase solubility

Summary of Last Lecture

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e

Lecture 14• Transition Metals• Electron Configuration• Oxidation States• Colours• Magnetism• Blackman Chapter 13, Sections 13.4 and 13.7

Lecture 15• Metals in Biological Processes• Essential Elements• Toxic Elements• Medicinal Uses• Blackman Chapter 13

Complexes III

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e Transition (or d-block) Metals

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e Electronic Configurations of AtomsIn 4th row, 4s and 3d orbitals are available for electrons• There is one 4s orbital: it can accommodate 2 electrons• There are five 3d orbitals: each can accommodate 2

electrons, giving a total of 10 electrons• Fill 4s then 3d

4s

3d

ener

gy

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e Atomic Configurations

• Fill 4s then 3d• Group number gives number of valence electrons• Electronic configurations:

K: group 1 so [Ar](4s)1(3d)0 Sc: group 3 so [Ar](4s)2(3d)1

Mn: group 7 so [Ar](4s)2(3d)5

Ni: group 10 so [Ar](4s)2(3d)8

Zn: group 12 so [Ar](4s)2(3d)10

[Ar](4s)x(3d)y where x + y = group

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e Electronic Configurations of Cations• Fill 3d only• Group number gives number of valence electrons• Cation has (group number – oxidation number) electrons

Mn7+: group 7 and oxidation number 7 so has: (7 – 7) = 0 electrons: [Ar](3d)0

Mn2+: group 7 and oxidation number 2 so has (7 – 2) = 5 electrons: [Ar](3d)5

Ni2+: group 10 and oxidation number 2 so has (10 – 2) = 8 electrons: [Ar](3d)8

Ni3+: group 10 and oxidation number 3 so has(10 – 3) = 7 electrons: [Ar](3d)7

(4s always empty!)

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e Electronic Configurations of Cations• To minimize repulsion between electrons, they occupy d-

orbitals singly with parallel spins until they have to pair up:

3d

Mn2+: [Ar](3d)5

3d

Ni2+: [Ar](3d)8

• If the metal cation has unpaired electrons, the complex will be attracted to a magnet: paramagnetic

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e Aqueous Oxoanions of Transition Metals

• One of the most characteristic chemical properties of these elements is the occurrence of multiple oxidation numbers, often associated with different colours.

Ion Ox. No. Colour

VO3- +5 yellow

VO2+ +4 green

V3+ +3 blue

V2+ +2 violet

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e Colourful Complexes

• Aqueous solutions of the Co(III) complexes (from left to right): [Co(NH3)5OH2]3+, [Co(NH3)6]3+, trans-[Co(en)2Cl2]+, [Co(en)2O2CO]+ and [Co(NH3)5Cl]2+.

• All contain Co(III): colour influenced by the ligand

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i Absorbed and Observed Colours

Unless the d-orbitals are empty, half full or full, electrons can be excited from one d-orbital to another: absorption of light which we see as colour

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e Summary: Complexes II

Learning Outcomes - you should now be able to:

• Complete the worksheet• Work out the electron configurations of atoms and

cations• Work out the number of unpaired electrons• Answer review problems 13.59-13.62 in Blackman

Next lecture:

• The Biological Periodic Table

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x Practice Examples

1. How many d-electrons and how many unpaired electrons are there in the following complexes?

A.K2[NiCl4]B.[Co(en)3]Cl3C.[CrCl2(OH2)4]+

D.K2[Zn(OH)4]E.[PtCl2(NH3)2]

2. Consider the compound with formula [CoCl2(NH3)4]Br2H2O

(i) Write the formula of the complex ion.

(ii) Write the symbols of the ligand donor atoms.

(iii) What is the d electron configuration of the metal ion in this complex?