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 Periodic
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Advanced Level Inorganic Chemistry Periodic Table
Revision Notes stability constants and entropy changes for the formation
of 3d block transition metal complex ions
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Appendix
8
Complex ion stability, entropy changes and stability constants (Kstab)
This page discusses the
relative stability of a wide range of transition metal ion complexes. The
stability is measured in terms of the stability constant Kstab for the
equilibrium expression for the formation of complex ion on interaction
of the hydrated metal ion and another ligand with which it undergoes a
ligand substitution reaction. Other than a variety of the 3d
block transition metals and particular ligands, other factors dealt with
include change in ion charge due to oxidation state, monodentate,
bidentate and polydentate ligands (chelation) are considered for a particular
transition metal ion.
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- 3d block transition metal chemistry Sc Ti V Cr Mn Fe Co Ni Cu Zn |
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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.
-
 Since
[ ] used in depicting complex ion structure, I've used [ ] as
well to denote concentration.
-
[Fe(H2O)6]3+(aq) + SCN–(aq)
 [Fe(H2O)5SCN]2+(aq) + H2O(l)
-
| |
[[Fe(H2O)5SCN]2+(aq)] |
|
Kstab = |
------------------------------------------ |
| |
[[Fe(H2O)6]3+(aq)]
[SCN–(aq)] |
-
On this page I've often
quoted the equilibrium expression and value in one line e.g.
-
Kstab
= [[Fe(H2O)5SCN]2+(aq)]
/ [[Fe(H2O)6]3+(aq)]
[SCN–(aq)]
= 1.4 x 102 mol–1dm3
-
BUT, you do need to be able
to write out the mathematical equilibrium expression as clearly
as possible and note that Kstab is used instead of
the usual Kc when dealing with molar concentration
terms.
-
Note that the water
concentration terms is effectively constant in aqueous solutions
and incorporated into the value of Kstab.
-
In this
section [] = concentration, because [] used in expressing the complex formulae.
-
The Kstab units -
[concentration] / [concentration]2.
-
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 variation of the stability
constant with change in ligand is illustrated with the zinc ion.
-
The data set for zinc compares
five different monodentate ligands.
-
The zinc complex ion data was
one of the more comprehensive set of values (just happened to be a good set of data
I found on the internet).
-
It make no
difference that zinc is NOT a true transition element to illustrate the
point about the relative stability of complex ions formed from a variety of
different ligands.
-
| |
Ligand
substitution reaction to give the new complex ion |
Kstab |
lg Kstab |
| a |
[Zn(H2O)4]2+(aq)
+ 4CN–(aq)
 Zn(CN)4]2–(aq) + 4H2O(l) |
5.0 x 1016 |
16.7 |
| b |
[Zn(H2O)4]2+(aq)
+ 4NH3(aq)
Zn(NH3)4]2+(aq)
+ 4H2O(l) |
3.8 x 109 |
9.58 |
| c |
[Zn(H2O)4]2+(aq)
+ 4Cl–(aq)
 [ZnCl4]2–(aq) + 4H2O(l) |
1.0 |
0.0 |
| d |
[Zn(H2O)4]2+(aq)
+ 4Br–(aq)
[ZnBr4]2–(aq) + 4H2O(l) |
10–1 |
–1.0 |
| e |
[Zn(H2O)4]2+(aq)
+ 4I–(aq)
[ZnI4]2–(aq) + 4H2O(l) |
10–2 |
–2.0 |
-
The stability constant
equilibrium expression for these equilibria
-
|
for a |
[[Zn(CN)4]2-(aq)] |
|
Kstab = |
------------------------------------------ |
| |
[[Zn(H2O)4]2+(aq)]
[CN-(aq)]4 |
- All the Kstab units are [concentration] /
[conentration]5
-
a: Kstab = [[Zn(CN)4]2-(aq)]
/ [[Zn(H2O)4]2+(aq)] [CN-(aq)]4
= 5.0 x 1016 mol–4dm12
-
b: Kstab = [[Zn(NH3)4]2-(aq)]
/ [[Zn(H2O)4]2+(aq)] [NH3(aq)]4
= 3.8 x 109 mol–4dm12
-
c: Kstab = [[ZnCl4]2-(aq)]
/ [[Zn(H2O)4]2+(aq)] [Cl-(aq)]4
= 1.0 mol–4dm12
-
d: Kstab = [[ZnBr4]2-(aq)]
/ [[Zn(H2O)4]2+(aq)] [Br-(aq)]4
= 10–1 mol–4dm12
-
e: Kstab = [[ZnI4]2-(aq)]
/ [[Zn(H2O)4]2+(aq)] [I-(aq)]4
= 10–2 mol–4dm12
-
The very high Kstab value for the tetracyanozincate(II) in reflects the strong of central
metal ion (Zn2+) – ligand (CN) bond.
-
The lower Kstab value for
ammonia indicates on average a weaker dative covalent bond, but still
relatively strong.
-
The ligand bonds are even weaker
for the halide ions possibly due to their larger radius, since there is a
steady decrease in Kstab as the halide ion radius increases (down
the group), making the Zn–X
dative covalent bond longer and weaker (X = halogen).
-
Generally speaking, the stronger the
metal ion-ligand bond, the greater with be Kstab, moving the
equilibrium to the right.
-
Why are Kstab values
are much higher for polydentate ligands? 'The
chelate effect'
-
Here we consider ligand
substitution with a
bidentate of multidentate ligand to give more stable complexes and in doing
so we compare the stability constants for a typical monodentate
ligand, bidentate ligand and a hexadentate ligand.
-
Consider three Ni(II) complexes
formed from the hexaaquanickel(II) 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)]6
= 4.8 x 107 mol–6 dm18
-
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–diaminoethane (en = H2N–CH2–CH2–NH2)
-
old name was
ethylenediamine (abbreviated to 'en')
-
[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) still > 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.
-
These arguments do ignore the
different stoichiometries of the equations.
-
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 (unidentate) 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, and is
incorporated into Kstab, because water is the
medium, and the bulk of the solution, therefore its concentration effectively remains
constant.
(ii) I'm afraid again, Kstab values
do differ
depending on the source!
(iii) log10 = lg and
Kstab = 10Kstab
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)
-
Kstab
= [[Fe(H2O)5F]2+(aq)]
/ [[Fe(H2O)6]3+(aq)]
[F–(aq)]
-
Kstab
= 2.4 x 105 mol–1dm3
-
lg(Kstab) = 5.38
-
The fluoride ion is more
strongly bonded than the thiocyanate ion.
-
–
-
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,
reflecting the effect of the higher charge on the central metal ion
more strongly attracting the lone pairs of same ligand.
-
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 with the EDTA hexadentate ligand than the lower
charged Co2+(aq) ion
(see also below with the ammonia complexes).
-
Comparison of the stability
of the hexammine complexes formed from the monodentate ligand ammonia
-
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 hexaaquanickel(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
nickel(II) ion Kstab(EDTA) > Kstab(en)
> Kstab(NH3)
-
though this argument does ignore
the different stoichiometry of the equations.
-
Ni2+ forms
the tetrachloronickelate(II) ion, [NiCl4]2–, a
tetrahedral anionic complex
with the chloride ion (Cl–).
-
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
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