Inorganic chemistry

Shoneeze Simone Renga
Slide Set by Shoneeze Simone Renga, updated more than 1 year ago
Shoneeze Simone Renga
Created by Shoneeze Simone Renga almost 5 years ago


notes summarizing the propertise of d metals and why we study them.

Resource summary

Slide 1

    Bonding in Transition Metal complex
    Introduction Crystal field theory MO Theory

Slide 2

    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

Slide 3

    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+)

Slide 4

    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

Slide 5

Slide 7

    Octaheral square planar tetrahderal Linear

Slide 8

    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

Slide 9

    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

Slide 10

    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

Slide 11

    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.

Slide 14

    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.

Slide 15

    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

Slide 16

    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

Slide 17

    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

Slide 18

Slide 19

    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

Slide 20

    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
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