Coordination Chemistry - CHEM1902 (C10K)
Answers to Tutorial Paper - 1

From the May 2004 Exam Paper.
1). (a) Give the Oxidation Number, d-orbital occupation, co-ordination number and expected magnetic moment of the central metal ion in the following complexes.
Draw the expected structure.

(i) K3[Co(C2O4)3]
This octahedral Co(III) complex can display optical isomerism.
The CN = 6, the OS=3+, the d-orbital occupation is that of a LOW spin Co(III) complex ie t2g6 eg0. All Co(III) are treated as LOW spin for CHEM1902 (C10K).
The magnetic moment is therefore 0 B.M.
K3[Cr(ox)3] structure

(ii) (NH4)2[CoF4]
This Co(II) complex is tetrahedral (for CHEM1902 (C10K) we have said that square planar complexes will only be seen for d8 configurations and Co(II) is d7.
The d-orbital configuration is e4 t23 and the magnetic moment is 3.87 B.M { sqrt(15) }.

(iii) diamagnetic [NiCl2{P(C6H5)3}2]
The indication that the compound is diagmagnetic means it must be square planar since a Ni(II) d8 configuration in a tetrahedral shape would be paramagnetic.
Examples with this formulation in BOTH tetrahedral and square planar shapes have been found, generally the tetrahedral are green/blue while the square planar are red.
The triphenyl phosphine ligand is a neutral monodentate ligand.

(iv) cis-[CrCl2(bipy)2]Cl
Cr(III) has a d3 configuration (t2g3).
In this octahedral complex we do not need to worry about high/low spin since we always fill from the lower level and there are 3 t2g orbitals and 3 electrons and no electrons left to occupy the eg level.

(v) [Mn(H2O)6]SO4
The aqua group gives rise to HIGH spin complexes so this octahedral Mn(II) d5 complex is paramagnetic with 5 unpaired electrons (t2g3 eg2).

(b) Which of the complexes above can exhibit isomerism? Explain.
(i) can exhibit optical isomerism
(ii) does NOT show isomerism
(iii) the known square planar complexes with this formula are trans- and cis- forms are theoretically possible.
(iv) The cis- form can exhibit optical isomerism and there is a possibility of a trans- form as well.
(v) does NOT show isomerism.

(c) Give the IUPAC name for the complex (ii) in part (a).
ammonium tetrafluorocobaltate(II)
Note that the NH4+ cation is not coordinated and must be named before the anion.


2) Write down the systematic name for each of the following complexes and indicate the oxidation state, electronic configuration, coordination number, stereochemistry and magnetic moment of the central ion.

a) K[Cr(oxal)2(H2O)2] .3H2O
potassium diaquabis(oxalato)chromate(III) trihydrate
OS= 3, d3, t2g3eg0, CN=6, Shape =octahedral, 3 unpaired electrons.

b) CrCl3(pyr)3
trichloridotripyridinechromium(III)
OS= 3, d3, t2g3eg0, CN=6, Shape =octahedral, 3 unpaired electrons.

c) K4[Mn(CN)6]
potassium hexacyanidomanganate(II)
OS= 2, d5, t2g5eg0, CN=6, Shape =octahedral, 1 unpaired electron.

d) [CoCl(NH3)5] Cl2
pentaamminechloridocobalt(III) chloride
OS= 3, d6, t2g6eg0, CN=6, Shape =octahedral, 0 unpaired electrons. *

e) Cs[FeCl4]
caesium tetrachloridoferrate(III)
OS= 3, d5, e2t23, CN=4, Shape =tetrahedral**, 5 unpaired electrons.

f) [NiCl(en)2(NH3)]Cl
amminechlorobis(1,2-diaminoethane)nickel(II) chloride
OS= 2, d8, t2g6eg2, CN=6, Shape =octahedral, 2 unpaired electrons.

g) [Cu(NH3)4(H2O)]SO4
tetraammineaquacopper(II) sulfate
OS= 2, d9, splitting pattern not done for CHEM1902 , CN=5, Shape =square pyramid, 1 unpaired electron. (With 5 d-orbitals that can hold 10 electrons and Cu(II) with 9 electrons then it must mean there is 1 unpaired electron irrespective of the splitting pattern of the energies of the d orbitals)

* for CHEM1902 (C10K), assume all Co(III) complexes are low-spin.
** for CHEM1902 (C10K), assume the only square planar complexes are for d8 configurations

4) The hexaaquamanganese(II) ion contains five unpaired electrons, while the hexacyano- ion contains only one unpaired electron. Explain, using Crystal Field Theory.
Octahedral complexes with between 4 and 7 d electrons can give rise to either high or low spin magnetic properties. Mn(II) has a d5 configuration. In a weak octahedral crystal field this splits to give t2g3eg2 but in a strong crystal field it gives t2g5eg0.
In the first case no pairing of electrons occurs, but in the second 2 pairs of electrons are present leaving only one unpaired electron.

5) Indicate the type of isomerism exhibited by the following complexes and draw structures for these isomers.

a) K [Cr(oxal)2(H2O)2] .3H2O
Both geometric (cis-, trans-) and optical isomers can exist.
b) [Co(en)3]Cl3
Two optical isomers can exist
c) [CoCl(NO2)(NH3)4]Br       Note that there are 10 possible isomers!
Hint: There are geometric, ionisation and linkage isomers possible.
d) PtCl2(NH3)(H2O)
Geometric (cis-, trans-) isomers can exist.
6)  Ans.    a)      1.0 x 107.
            b)      1.9 x 108.
a) The total amount of Ni2+ in the solution is given by:

Total Ni2+ = Ni2+ + NiL2+ + NiL22+ + NiL32+ + NiL42+

If we assume that the only complex formed is NiL42+ then this simplifies to:
Total Ni2+ = Ni2+ + NiL42+

If the solution contains 1.6 x 10-4 % of the Ni in the free form then this means that:
                       Ni
                   ----------   X   100    =  1.6 x 10 ^-4           (equation 5.1)
                    Ni + NiL4
But since the amount of free Ni2+ is extremely small then the Total Ni(II) can be approximated to NiL42+

The stability constant for the reaction forming the tetraammine is:
                            NiL4
                   K  =  ---------             (equation 5.2)         
                          Ni  L^4
Given that the NH3 concentration is 0.5M then equilibrium expression (equation 5.2) can be simplified once it is recognised that the relationship in equation 5.1 is the reciprocal of part of equation 5.2.

In other words the stability constant (K or β4) is

                                 100
                   K  =  ----------------------
                         [1.6 x 10^-4]  [0.5]^4
                or β4 = 1.0 x 107

b) log β6 = log β4 + Log K5 + Log K6

Having just calculated β4 and given the last two stepwise stability constants then the overall stability constant is:

log β6 = 7.0 + 0.85 + 0.42 = 8.27
or β6 = 1.9 x 108


6)
b)
                           [Nien3]
                   K  =  ----------              (equation 6.1)         
                         [Ni] [en]^3
If at equilibrium [Ni(en)3]2+ is 0.08M and [en] is 0.40M then given that β3 is 4.07 x 1018
the concentration of free Ni2+ can be expressed as:
                           [Nien3]
                   Ni  =  ----------             (equation 6.2)         
                           K  [en]^3
or Ni2+ = 3.07 x 10-19

d) If β3 is 4.07 x 1018 then log β3 is 18.61
log β3 = log K1 + log K2 + log K3
Given that log K1 = 7.66 and log K2 = 6.40 then
log K3 = 18.61 - 7.66 - 6.40
or log K3 = 4.55 and K3 = 3.55 x 104
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