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

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D electrons : 3d^x-z ,x= group number z= charge TrendsIonization 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

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Rubrica: : Mono(1 donor atom)dentate

O,N,S and Halides Bidentate,tridentate and polydentate Open chain bidentate and mulitdentate=Chelate=rings=Jackson

Ligands

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Rubrica: : Bidentate

Rubrica: : Tridentate

Geometries

Octaheral square planar tetrahderal Linear

Octahedral 

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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.

Rubrica: : Octahedral complex

Square planar

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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

Rubrica: : Square planar complex

Crystal field theory

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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)

Rubrica: : 5 avaliable oribtals in d metals

The  5 d oribtals

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Rubrica: : 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

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splitting diagram

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So what happens...

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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

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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)

Rubrica: : see worked example

Worked example - Page 23 

Answerif wavelength=493nm the v=1/493=20283.97cm^-1thenhv=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 ligandWeaker 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 

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Rubrica: : 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.

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Rubrica: : 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 momentfancy u=(n(n+2))^1/2 fancy ub