Inorganic chemistry

Bonding in Transition Metal complex

• Introduction
• Crystal field theory
• MO Theory

Paper 1 Break down

• Six topics with the average lecture time indicated in brackets.
• Wave Mechanics and Spectroscopy(12) - Bourne
• Inorganic bonding(10) -Egan
• Introduction to liquids(10)-Venter
• Crystal structure analysis(9)-Oliver
• Inorganic Reaction Mechanism(8)-Jackson
• Inorganic complex equilibra (8)-Jackson

Introduction

• 
• D electrons : 3d^x-z ,x= group number z= charge
• Trends
• Ionization energy-increases from left to right , the changes are small due atomic size decrease from left to right.=D oribtals are furtheraway from the nuclues,(less effective nuclear pull) and thus are more effectively shielded than s and p oribtals.
• Ionic radii
• Innic size depends on charge of ion and magnetic propertise.
• M(2+) is >>M(3+)

Introduction-Acid-bases

• Lewis theory
• Ligand-Base-A pair of electron donor
• Metal-Acid-A pair f elctron acceptor
• Lewis bases-Bronsted-Lowery bases
• Not all lewis bases are sig. Bronsted-Lowery
• Exception;CO,CN even though they are organic moclues which have relatively high pka they form coordinative complexes
• Metal-ligand=Lewis Acid-Lewis base
• Bronsted-Lowery theory
• Ligand-A proton acceptor
• Metal-A proton donor
• Exceptions Cl,Br,I and CO not sig.bronsted lowey bases but for complexes.
• Metal-ligand=Lewis Acid-Lewis base

LIGANDS

• 
• Caption: Mono(1 donor atom)dentate
• O,N,S and Halides
• Bidentate,tridentate and polydentate
• Open chain bidentate and mulitdentate=Chelate=rings=Jackson

Ligands

• 
• 
• Caption: Bidentate
• Caption: Tridentate

Geometries

• Octaheral
• square planar
• tetrahderal
• Linear

Octahedral

• 
• The metal centre want a maxium charge around it .
• The d block has a number of 6 ligands9max) due to steric factors .
• Has a centre of inversion
• Can have a pesudoctahedral form.
• Caption: Octahedral complex

Square planar

• 
• Large ligands e.g halides and small metals e.g Ni(11) and Zn(11).
• Tetragonal and square planar =e;ectronic reasons
• Tetragonal = d4 (HS) and d7(LS) ,d9(both)
• Square planar=d8
• Caption: Square planar complex

Crystal field theory

• 
• A Electrostatic model
• uses ligands to create a electric field around the metal centre
• Ligands are treated as negative point charges which are attracted to postive charged metal.
• The ligand electrons and metal electrons repel one another.
• There is no metal-ligand covalent interaction
• What do we use this theory for?
• Colour-which is due the presence of d--d transtions in electronic spectra of complexes.
• Magnetic propertise of complexes
• High spin/Low spin---Metals with the same d count can give complexes with different number of unpaireed electrons.
• since CFT-ignores covalent interaction,one must take into account in advanced ligand filed theory (Jackson)
• Caption: 5 avaliable oribtals in d metals

The 5 d oribtals

• 
• Caption: The 5 types of oribtals you can get
• Since the distribution of ligands around the metal is not spherically symmetrical ,there are electrons which are more stable than others.
• Those which point along the dx2y2 and dz2 are less stable than those in the dxy,dxz,dyz.
• Thus if it were spherical the energies of 5 oribtals would be risen (destablized by the same amount) however this is not the case.

Re-enforce the idea

splitting diagram

So what happens...

• 
• Replusion of electrons from 6 ligands spilt the d complex into:
• 1.doubly degenerate (eg)
• 2.Triple degenerate (t2g)
• Octahedreal and square planar complexes are centrosymmetric while tetragonal and teterahedral are not.
• NB!!!!
• 1.-dz2 and dx2-dy2 have subscript g =gerade=even
• if they have n centre of iversion=ungendree=uneven
• The energy seperation between the energy levels is delta oct
• An electron in one of this oribtals is stablized by 2/5 delta oct called ligand field stabalization energy(LFSE)
• CFSE- a measure of the net energy of occupation of the d oribtals realitive to the mean energy
• Formula: CFSE=-(0.4X+0.6Y)delta oct
• The reason we see colour is because of the delat oct value which carries the right energy to oxlate electrons from t2g-eg levels .
• The E of the visible peak max corresponds to delta oct.

Colour

• 
• If the lower t2g levels are filled and the upper eg levels are vacant or partially vacant .
• you have a energy jump from the lower level to the higher level =d-d transitions.
• colour dependency is on the the change of energy-change of wavelength.
• 1.REQUIREMENT
• must have a spin state change from (+0.5-0.5)
• Caption: see worked example

Worked example - Page 23

• Answer
• if wavelength=493nm the v=1/493=20283.97cm^-1
• then
• hv=delta oct
• =20284cm^-1 convert to kj.mol^-1
• =243.4kj.mol^-1

Variations in the Crystal splitting Field

• Nature of the ligand
• Small degree of splitting- small delta value-weak ligand.
• Large degree of splitting-large delta value-strong ligand .
• SPECTROCHEMICAL SERIES
• I
• Weak ligands-LHS and vice versa
• For a ligand colour depends on the oxidation state of metal.
• For a given metal colour depends on the ligand
• Weaker field-smaller delta-longer wavelength and vice versa
• Oxidation state of metal
• Delta values increase with oxidation state of the metal ion for a given metal and given ligand.
• For a given metal and given ligand delta oct increases down the group .
• Remember high oxidation state-Smaller ion , Shorter M-L bond.
• Mn(2)
• NB Delta Oct /tet=determined by strength of crystal field.
• THE AMOUNT OF CRYSTAL SPILITING IS DETERMINED BY STRENGTH OF THE LIGAND

The idea of HS Vs LS

• 
• Caption: The energy of the electron is >> than the splitting of the energy levels.I.e it is easier to promote electrons to upper eg levels than it is to pair them in t2g level.
• 
• Caption: VICE VERSA

This means Hs/Ls is determine by CFS

• if delta oct(spilting of d oribtals)
• General rules:
• Forumla for HS: CFSE=nt2g*-0.4delta oct +(ng*0.6)deltaoct
• LS:CFSE:nt2g*-.04delta oct +neg*0.6delta oct
• LS-STRONG FIELD LIGAND ,HS-weak field ligand(Given metal,Given ligand) and it is metal dependant

How do we show we have HS/LS?

• vIa magnetic measurements using magnetic susceotibility .
• Achieved through spin-only magnetic moment
• fancy u=(n(n+2))^1/2 fancy ub