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Co Ordination Chem.

Apr 03, 2018

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    Chapter 12- Coordination Chemistry and Structure(HW: 1,9,10,12,15,34)

    Coordination Number 1: Very few examples.

    Here is an organometallic examplewith Cu(I) and Ag(I).

    Coordination Number 2: Few examples.

    Typical metals that have 2: Cu(I), Ag(I), Au(I), Hg(II)

    But, they can easily go to 4C

    [Cu(NH3)2]+ = [Cu(NH3)4]+

    The metals are all d10, so preference to go Td.

    Coordination Number 3: 3C is not very common in nature.

    Typically, odd symmetry in nature is rare.

    Note that this symmetry was found in a protein,MerR, that is involved in mercury detoxification.

    Week 8-1

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    Coordination Number 4

    Td will be favored if there are highly charged ligands and if theyare large, I- is typical of this.

    Also, if there is no CFSE, Td is preferred (d10, d5 h.s., d0)

    Note, the most important aspect of the chapter is the concept of ISOMERS.

    Td isomers are optically active(chiral), however, very labile andrarely isolated.

    Square Planar Complexes:

    They are less favored sterically than Td

    and can be eliminated if ligands are too large.Pd2+, Pt2+, and Ni2+ with strong, non-bulky ligandsform sq. planar complexes.

    They are enhanced by pi interactions, whichcompensate for the loss of the Oh symmetry.

    cis-(NH3)2M(Cl)2 has a dipole moment even though it is neutral and is moresoluble in water than the trans-complex.

    These complexes are geometrical isomers.

    Week 8-2

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    They are only optical isomers only if they are asymmetrical.

    How can we determine which of thesetwo structures is correct?

    The data show that the product is

    optically activeOnly the sq. planar complex is chiral,the Td is not due to the mirror plane inthe plane of the board.

    Coordination Number 5

    The two common structures of 5C are the following:

    (axial versus equatorial ligands) (apical versus planar ligands)

    Note that 5C complexes are inherently unstable complexes.

    If electrostatic interactions were the only factor then two 5Ccomplexes would disproportionate into one 4C and one 6C.

    However, covalency comes into playand hence 5C are found.

    The energy difference between sq. pyramidal andtrigonal bipyramidal is very small.

    The counter anion can be enough to convertthe two structures as they crystallize.

    Week 8-3

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    Coordination Number 6: Most common geometry

    Octahedral- d3 and d6 are predominately this geometry.

    Geometric isomers of 6C.

    Optical Isomers of 6C.

    Think of the fins of a propeller.

    The two isomers are either delta() or lambda ()

    Week 8-4

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    Coordination Number 7: These are very uncommon.

    C7 is awkward geometry for first row transition elements-too much steric bulk, space problem.

    The bigger the atom the morelikely to have greater then 6Cgeometry.

    Coordination Number 8:

    This coordination is not seen in first transition series.

    It is seen in lanthanides and actinides.

    Coordination Number 9:

    All lanthanides are nana-aquo, [Ln(H2O)9]3+

    Lanthanides are used in MRI imaging agents, withGadolinium(III) having a large magnetic moment, f7

    Week 8-5

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    Linkage Isomerism

    NO2- can bind to the metal in 2 different ways, thru O or N.

    This difference can affect their reactivity.

    Electronic Effects

    Depending on the trans ligand, the atom that binds to the metal can change.

    If Phosphorous is bound then it pulls on the metal d-orbital making Sbonding less likely (S uses d-orbital bonding).

    If ammonia is bound then d-orbital is available for Sulfur binding.

    Can also think of this as a push pull of hard-hard(H3N-Pt orPt-NCS) and soft-soft (Pt-SCN or H3P-Pt)

    Week 8-6

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    Chelate effect:

    We have seen many occasions of ligands being chelates.

    One of the classic examples are the crown ethers.

    These ligands can selectively bind cations due to size.

    15-Crown-5 prefers Li+

    18-Crown-6 prefers Na+

    21-Crown-7 prefers K+

    Oxidation States and their Reduction Potentials:(Ch. 10, HW; 7,8 and pgs 378-381 and Ch. 14, HW; 9,29 and pgs 588-599)

    Does the redox potential of a metaldepend on the d-configuration?

    d2-configuration:Ti2+ is strong reductant

    V3+ is not as strong a reductantFe6+ is a very strong oxidant

    Week 8-7

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    There is no correlation between d-configuration and redox potential.

    The redox potential tells us whether ametal will be an oxidant or a reductant.

    What are some of the rules that help us understand electrochemistry?

    1) The arbitrary standard is that at 1 M H+, and 1 atm of H2 then

    H+ + 1e- = 1/2H2 E = 0

    Therefore, if a metal will reduce protons to hydrogengas there has to be a thermodynamic driving force.

    2) G = -nFEThe E must be positive for the reaction to be spontaneous (-G).

    When comparing two reactions, the difference in Emust be positive for the reaction to be spontaneous.

    For example, will the following reaction proceed?

    Zn 0 = Zn2+ + 2e- E = 0.76 G = -2F(0.76)2H+ + 2e- = H2 E = 0 G = -2F(0)Zn0 + 2H+ = H2 + Zn

    2+ G = -1.52F

    3) The Nerst equation:

    E = E RT/nF(lnQ), Q = Prod/React

    This equation helps you determine if a change in the reactionconditions will change the direction of an electrical reaction.

    For example, with the reduction of water to hydrogen, how will thereduction potential change with acid concentration?

    Week 8-8

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    As the [H+] decreases, the Q gets bigger and E goes more negative (or lessspontaneous because G becomes more positive.

    4) How do we calculate the skip-step potential?

    Remember that we cant add Es but we can add GOne can also do this for reactions in an additive manner.

    MnO4- + 1e- = MnO4

    2- E = +0.90 G = -1F(+0.90)MnO4

    2- + 2e- = MnO2 E = +2.09 G = -2F(+2.09)MnO4

    - + 3e- = MnO2 G = -5.08F

    E = -(-5.08F)/3F = +1.70V

    This is presented to you as a Latimer diagram.

    The main purpose of this chapter is to know how to use Latimer diagramsand to know how to construct one.

    Week 8-9

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    Now we want to be able to answer two questions:

    Will the compound oxidize water or reduce protons to H2?

    HMnO4- + 3H+ +2e- = MnO2 + 2H2O E = +2.09 G = -2F(+2.09)

    H2O = 1/2O2 + 2H+ + 2e- E = -1.23 G = -2F(-1.23)

    HMnO42-

    +H+

    = MnO2 +1/2O2 + H2O G = F(-1.72)E = -(-1.72F)/2F = +1.03V

    The positive E indicates that MnO42- is not stable in acid!

    Will it be more stable at pH 7? Yes it will, less H+

    Note that the E for water oxidation also varies with acid concentration (like

    water reduction).

    Another example of proton reduction to hydrogen.

    Mn(s) = Mn(II) + 2e- E = +1.18 G = -2F(+1.18)2H+ + 2e- = H2 E = 0.00 G = -2F(0.00)Mn(s) + 2H+ = Mn(II) + H2 G = F(-2.36)

    E = -(-2.36F)/2F = +1.18V

    Finally, if you are given a partial Latimer diagram you will be requiredto determine the missing connections of redox potential.

    Week 8-10