Coordination Chemistry - CHEM1902
Answers to Tutorial Paper - 1

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 t_2g⁶ e_g⁰. All Co(III) are treated as LOW spin for CHEM1902 (C10K).
The magnetic moment is therefore 0 B.M.

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) }.

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.

Cr(III) has a d3 configuration (t_2g³).
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 t_2g orbitals and 3 electrons and no electrons left to occupy the eg level.

The aqua group gives rise to HIGH spin complexes so this octahedral Mn(II) d5 complex is paramagnetic with 5 unpaired electrons (t_2g³ e_g²).

(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.

ammonium tetrafluorocobaltate(II)
Note that the NH₄⁺ cation is not coordinated and must be named before the anion.

a) K[Cr(oxal)₂(H₂O)₂] .3H₂O
potassium diaquabis(oxalato)chromate(III) trihydrate
OS= 3, d³, t_2g³e_g⁰, CN=6, Shape =octahedral, 3 unpaired electrons.

c) K₄[Mn(CN)₆]
potassium hexacyanidomanganate(II)
OS= 2, d⁵, t_2g⁵e_g⁰, CN=6, Shape =octahedral, 1 unpaired electron.

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

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 t_2g³e_g² but in a strong crystal field it gives t_2g⁵e_g⁰.
In the first case no pairing of electrons occurs, but in the second 2 pairs of electrons are present leaving only one unpaired electron.

Both geometric (cis-, trans-) and optical isomers can exist.

Two optical isomers can exist

Hint: There are geometric, ionisation and linkage isomers possible.

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

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