Answers to Tutorial Paper - 1

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) K

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_{2g}^{6} e_{g}^{0}. All Co(III) are treated as LOW spin for CHEM1902 (C10K).

The magnetic moment is therefore 0 B.M.

The CN = 6, the OS=3+, the d-orbital occupation is that of a

The magnetic moment is therefore 0 B.M.

(ii) (NH

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 e^{4} t_{2}^{3} and the magnetic moment is 3.87 B.M { sqrt(15) }.

The d-orbital configuration is e

(iii) diamagnetic [NiCl

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.

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)

Cr(III) has a d3 configuration (t_{2g}^{3}).

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.

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

(v) [Mn(H

The aqua group gives rise to HIGH spin complexes so this
octahedral Mn(II) d5 complex is paramagnetic with 5 unpaired electrons (t_{2g}^{3} e_{g}^{2}).

(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.

(ii) does NOT show isomerism

(iii) the known square planar complexes with this formula are

(iv) The

(v) does NOT show isomerism.

(c) Give the IUPAC name for the complex (ii) in part (a).

ammonium tetrafluorocobaltate(II)

Note that the NH_{4}^{+} cation is not coordinated and must be
named before the anion.

Note that the NH

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}(H_{2}O)_{2}]
.3H_{2}O

potassium diaquabis(oxalato)chromate(III) trihydrate

OS= 3, d^{3},
t_{2g}^{3}e_{g}^{0}, CN=6, Shape
=octahedral, 3 unpaired electrons.

b) CrClpotassium diaquabis(oxalato)chromate(III) trihydrate

OS= 3, d

trichloridotripyridinechromium(III)

OS= 3, d

c) K_{4}[Mn(CN)_{6}]

potassium hexacyanidomanganate(II)

OS= 2, d^{5},
t_{2g}^{5}e_{g}^{0}, CN=6, Shape
=octahedral, 1 unpaired electron.

d) [CoCl(NHpotassium hexacyanidomanganate(II)

OS= 2, d

pentaamminechloridocobalt(III) chloride

OS= 3, d

e) Cs[FeCl_{4}]

caesium tetrachloridoferrate(III)

OS= 3, d^{5}, e^{2}t_{2}^{3},
CN=4, Shape =tetrahedral**, 5 unpaired electrons.

f) [NiCl(en)caesium tetrachloridoferrate(III)

OS= 3, d

amminechlorobis(1,2-diaminoethane)nickel(II) chloride

OS= 2, d

g) [Cu(NH_{3})_{4}(H_{2}O)]SO_{4}

tetraammineaquacopper(II) sulfate

OS= 2, d^{9}, 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.
tetraammineaquacopper(II) sulfate

OS= 2, d

** for CHEM1902 (C10K), assume the only square planar complexes are for d

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 d^{5} configuration. In a weak
octahedral crystal field this splits to give
t_{2g}^{3}e_{g}^{2} but in a
strong crystal field it gives
t_{2g}^{5}e_{g}^{0}.

In the first case no pairing of electrons occurs, but in the second 2 pairs of electrons are present leaving only one unpaired electron.

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)

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

b) [Co(en)Two optical isomers can exist

c) [CoCl(NOHint: There are geometric, ionisation and linkage
isomers possible.

d) PtClGeometric (*cis-, trans-*) isomers can
exist.

6) Ans. a) 1.0 x 10^{7}. b) 1.9 x 10^{8}.

a) The total amount of Ni^{2}+ in the
solution is given by:

Total Ni^{2+} = Ni^{2+} + NiL^{2+} +
NiL_{2}^{2+} + NiL_{3}^{2+} +
NiL_{4}^{2+}

If we assume that the only complex formed is NiL_{4}^{2+} then this simplifies to:

Total Ni^{2+} = Ni^{2+} +
NiL_{4}^{2+}

If the solution contains 1.6 x 10^{-4} % of the Ni in the
free form then this means that:

^{2+} is extremely small
then the Total Ni(II) can be approximated to
NiL_{4}^{2+}

The stability constant for the reaction forming the tetraammine is:

_{3} 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

^{7}

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 10^{8}

Total Ni

If we assume that the only complex formed is NiL

Total Ni

If the solution contains 1.6 x 10

Ni ---------- X 100 = 1.6 x 10 ^-4 (equation 5.1) Ni + NiL4But since the amount of free Ni

The stability constant for the reaction forming the tetraammine is:

NiL4 K = --------- (equation 5.2) Ni L^4Given that the NH

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

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

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 10

6)

b)
_{3}]^{2+} is 0.08M and
[en] is 0.40M then given that β3 is 4.07 x 10^{18}

the concentration of free Ni^{2+} can be expressed as:

^{2+} = 3.07 x 10^{-19}

d) If β3 is 4.07 x 10^{18} 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 10^{4}

Return to tutorial paper
[Nien3] K = ---------- (equation 6.1) [Ni] [en]^3If at equilibrium [Ni(en)

the concentration of free Ni

[Nien3] Ni = ---------- (equation 6.2) K [en]^3or Ni

d) If β3 is 4.07 x 10

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 10

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