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Brown's Chemistry
Periodic Table Revision Notes
- Part
10. 3d block - Transition Metals
Appendix
8
Complex ion stability, entropy changes and stability constants (Kstab)
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Appendix 8.
Complex ion stability, entropy changes and stability constants (Kstab)
-
Equilibrium expression
for transition metal complex ion formation are readily written out, but
as the case of weak acid ionisation Kc expressions, the
concentration of the solvent water is considered a constant and omitted
from the expression e.g.
-
[Fe(H2O)6]3+(aq) + SCN-(aq) ==> [Fe(H2O)5SCN]2+(aq) + H2O(l)
-
Kstab
= {[Fe(H2O)5SCN]2+(aq)}
/ {[Fe(H2O)6]3+(aq)}
{SCN-(aq)}
= 1.4 x 102 mol-1dm3
-
In this
section {} = concentration, since [] used in complex formulae.
-
The bigger the value of
Kstab the more stable the complex i.e. the equilibrium is more
favoured to the right.
-
Kstab
values vary considerably and are often quoted as lg Kstab
= log10 (Kstab).
-
i.e. for the
above equilibrium: log10 (1.4 x 102) =
lg Kstab = 2.1
-
Complex formation by
polydentate ligands is sometime called chelation or sequestration.
-
What governs the
values of Kstab?
-
The strength of central
metal ion - ligand bond.
-
Substitution with a
bidentate of multidentate ligand tends to give more stable complexes.
-
Trend for
a series/examples of
monodentate ligands to do.
-
Comparing mono and
polydentate ligands
-
Why Kstab values
are much higher for polydentate ligands?
-
Consider three Ni(II) complexes
formed from the hexaaqua ion [Ni(H2O)6]2+:
-
(1)
a monodentate ligand e.g. ammonia
-
[Ni(H2O)6]2+(aq) +
6NH3(aq)
[Ni(NH3)6]2+(aq) + 6H2O(l)
-
Kstab
= {[Ni(NH3)6]2+(aq)}
/ {[Ni(H2O)6]2+(aq)}
{NH3(aq)}
= 4.8 x 107 mol-1 dm3
-
lg
Kstab is 7.7, a typical value for a complex with a
monodentate ligand (compared to the ligand water) and the Ni-NH3
bonds would appear to be stronger than the Ni-OH2
bonds.
-
(2) a bidentate
ligand e.g. 1,2-diamoethane (en = H2N-CH2-CH2-NH2)
-
[Ni(H2O)6]2+(aq) + 3en(aq)
[Ni(en)3]2+(aq) + 6H2O(l)
-
Kstab
= {[Ni(en)3]2+(aq)}
/ {[Ni(H2O)6]2+(aq)}
{en(aq)}3
= 2.0 x 1018 mol-3 dm9
-
lg
Kstab is
18.3, and considerably higher than for the
monodentate ligand like ammonia.
-
(3)
a polydentate ligand e.g. EDTA (a hexadentate ligand)
-
[Ni(H2O)6]2+(aq) +
EDTA4-(aq)
[Ni(EDTA)]2-(aq) + 6H2O(l)
-
Kstab
= {[Ni(EDTA)]2-(aq)}
/ {[Ni(H2O)6]2+(aq)}
{EDTA4-(aq)}
= 1 x 1019 mol-1 dm3
-
lg
Kstab is 19, and even higher than for the bidentate ligand, EDTA is a
hexadentate ligand.
- (4) Theoretically adding EDTA to
any of the other complexes mentioned above would cause the displacement
of the original ligand e.g.
- [Ni(NH3)6]2+(aq) +
EDTA4-(aq)
[Ni(EDTA)]2-(aq) + 6NH3(aq)
- very much to the right since Kstab
(Ni2+/EDTA) >> Kstab (Ni2+/NH3)
- [Ni(en)3]2+(aq) +
EDTA4-(aq)
[Ni(EDTA)]2-(aq) + 3en(aq)
- a bit more to the right than the
left since Kstab (Ni2+/EDTA) just > Kstab
(Ni2+/en).
-
The Kstab is
much higher for polydentate ligands e.g. reactions
(2/3), compared to monodentate ligands e.g. reaction (1),
because of the considerable entropy increase in 'freeing' six small
water molecules in reactions (2/3).
-
In (1) six small molecules
displace six other small molecules (all monodentate ligands), which is likely to involve a much
smaller entropy change, in (2) three larger bidentate ligands displace
six smaller ligand molecules, but in (3) six small molecules are
displaced by one larger hexadentate ligand molecule.
-
In general the more
water molecules of water freed the greater the entropy increase because
there are more ways of distributing the particles and more ways of
distributing the energy of the system.
-
In general, but with
lots of exceptions!, Kstab values tend to be in the order
polydentate > bidentate > monodentate ligands.
Some examples of 3d-block element complex formation and the values of Kstab
NOTE (i) By convention,
the term [ H2O(l)
]n is omitted from equilibrium expressions, because water is the
medium, and the bulk of the solution, therefore its concentration effectively remains
constant. (ii) I'm afraid again, Kstab values to differ
depending on the source!
Scandium
Titanium
Vanadium
Chromium
Manganese
-
The hexa-aqua
manganese(II) ion readily forms complexes with polydentate
ligands.
-
(i) [Mn(H2O)6]2+(aq)
+ 3en(aq) ===> [Mn(en)3]2+(aq)
+ 6H2O(l) (en = H2NCH2CH2NH2)
-
(ii) [Mn(H2O)6]2+(aq)
+ EDTA4-(aq) ===> [Mn(EDTA)]2-(aq)
+ 6H2O(l)
-
The higher Kstab
value for EDTA reflects the greater entropy change. A
simplistic, but not illegitimate argument, shows that in (i) a
net gain of 3 particles, but in (ii) 5 more particles are
formed.
-
-
Iron
-
Complex with
thiocyanate ion
-
Fe(H2O)6]3+(aq) + SCN-(aq) ==> [Fe(H2O)5SCN]2+(aq) + H2O(l)
-
Initial
monosubstituted complex with the fluoride ion ligand
-
[Fe(H2O)6]3+(aq)
+ F-(aq) ==> [Fe(H2O)5F]2+(aq) + H2O(l)
-
-
-
Fe3+ ions
give an anionic complex in concentrated chloride ion solutions
-
[Fe(H2O)6]2+(aq)
+ 4Cl-(aq) ==> [FeCl4]-(aq)
+ 2H2O(l)
-
formation of the
tetrachlroroferrate(III) anion.
-
Kstab = {[FeCl4]-(aq)}
/ {[Fe(H2O)6]2+(aq)} [Cl-(aq)]4
-
Kstab = 8
x 10-1 mol-4 dm12 [lg(Kstab)
= -0.097]
-
Both the hexa-aqua
ions of iron(II) and iron(III) readily complex with EDTA
-
[Fe(H2O)6]2+(aq)
+ EDTA4-(aq) ===> [Fe(EDTA)]2-(aq)
+ 6H2O(l)
-
[Fe(H2O)6]3+(aq)
+ EDTA4-(aq) ===> [Fe(EDTA)]-(aq)
+ 6H2O(l)
-
Note that the more
highly charged Fe3+(aq) ion complexes
more strongly than the Fe2+(aq) ion.
-
Iron(III) ions
complex with the bidentate ligand, the 1,2-diaminoethane molecule (en)
-
[Fe(H2O)6]3+(aq)
+ 3en(aq) ==>
[Fe(en)3]3+(aq)
+ 6H2O(l)
-
Kstab = {[Fe(en)3]3+(aq)}
/ {[Fe(H2O)6]3+(aq)}
[en(aq)]3
-
Kstab
= 3.98 x 109 mol-3 dm9 [lg(Kstab)
= 9.6]
-
-
Cobalt
-
Both the hexa-aqua ions of
cobalt(II) and cobalt(III) readily complex with EDTA
-
[Co(H2O)6]2+(aq)
+ EDTA4-(aq) ===> [Co(EDTA)]2-(aq)
+ 6H2O(l)
-
[Co(H2O)6]3+(aq)
+ EDTA4-(aq) ===> [Co(EDTA)]-(aq)
+ 6H2O(l)
-
Note that the more
highly charged Co3+(aq) ion complexes
more strongly than the Co2+(aq) ion
(see also below with the ammonia complexes).
-
Comparison of the stability
of the hexammine complexes
-
The cobalt(II) ion complexes
with 1,2-diaminoethane, a bidentate ligand
-
-
Nickel
-
[Ni(H2O)6]2+(aq)
+ 6NH3(aq)
[Ni(NH3)6]2+(aq)
+ 6H2O(l)
-
The hexa-aquanickel(II) ion
also forms complexes with other amine ligands
-
e.g. the bidentate
ligand 1,2-diaminoethane (H2N-CH2-CH2-NH2,
often abbreviated to en)
-
[Ni(H2O)6]2+(aq)
+ 3en(aq) ===> [Ni(en)3]2+(aq)
+ 6H2O(l)
-
The complex with EDTA is
also readily formed.
-
[Ni(H2O)6]2+(aq)
+ EDTA4-(aq) ===> [Ni(EDTA)]2-(aq)
+ 6H2O(l)
-
Note that Kstab
for the same ion tend to increase the greater the chelating power of
an individual ligand in terms of the ligand bond formed - mainly due
to the increase in entropy as more particles are formed by the
polydentate ligands
-
e.g. for the same
nicke(II) ion Kstab(EDTA) > Kstab(en)
> Kstab(NH3)
-
Ni2+ forms
the tetrachloronickelate(II) ion, [NiCl4]2-, a
tetrahedral anionic complex
with the chloride ion (Cl-).
-
[Ni(H2O)6]2+(aq) +
4Cl-(aq)
==> [NiCl4]2-(aq) + 6H2O(l)
-
Kstab
= {[NiCl4]2-(aq)}
/ {[Ni(H2O)6]2+(aq)}
[Cl-(aq)]4
-
Kstab
= ? mol4 dm-12 [lg(Kstab)
= ?]
-
Ni2+ forms
the tetracyanonickelate(II) ion, [Ni(CN)4]2-, a
square planar anionic complex
with the cyanide ion (CN-).
-
[Ni(H2O)6]2+(aq) +
4CN-(aq)
==> [NiCN4]2-(aq) + 6H2O(l)
-
Kstab
= {[NiCN4]2-(aq)}
/ {[Ni(H2O)6]2+(aq)}
[CN-(aq)]4
-
Kstab
= 2 x 1031 mol4 dm-12
[lg(Kstab) = 31.3]
-
Its likely that the more
bulky chloride ion (radius Cl > C) 'forces' the formation of the
tetrahedral shape rather than a square planar shaped complex.
-
-
Copper
Zinc
Scandium
* Titanium * Vanadium
* Chromium
* Manganese * Iron * Cobalt
* Nickel
* Copper *
Zinc
* Silver & Platinum
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