Ligand Substitution: A ligand substitution is a reaction in which one ligand
in a complex ion is replaced by another ligand.
Reactions of Aqueous Copper (II) Ions: When copper (II) sulphate is
dissolved in water, pale blue complex ion formed [Cu(H2O)6]
2+
Ligand Substitution- Ammonia + Cu2+: When excess aqueous ammonia (aq) added dropwise.
Four ammonia ligands replace four water ligands. Both complex ions are octahedral.
– In first step of reaction a blue precipitate of Cu(OH)2 formed. Cu(OH)2 then dissolves in
excess ammonia to form the dark blue solution. Ppt reaction. Only first step occurs when a
‘few drops of NH3/ small amount of NH3’- must quote this if referencing this sub step.
– The first step is exactly same as ppt reaction with NaOH because NH3 is acting as a base.
Ligand Substitution- Cl- + Cu2+: When excess conc HCl (provides Cl- ions) added dropwise. Four Cl
ligands replace six water ligands.
– Reaction in equilibrium. So if water added to yellow solution, blue solution forms.
– Turns green and then yellow, as equilibrium moves further right with HCl added.
– Chloride ligands larger than water ligands to fewer Cl fit around central Cu2+ ion so change in
coordination number. Oxidation state of Cu remains +2 in both complex ions. But overall
charge changes as Cl- negative.
– Don’t forget the charge.
Reactions of Aqueous Chromium (III) Ions:
– When chromium potassium sulphate KCr(SO4)2. 12(H2O) dissolved in water, complex ion
[Cr(H2O)6]
3+ formed- a violet solution.
– When chromium (III) sulphate dissolved in water, green solution formed. This forms complex
ion [Cr(H2O)5SO4]
+
. One water ligand of complex ion above replaced by one sulphate ion.
Both this and above contain Cr3+
.
Ligand Substitution- Ammonia + Cr3+: When excess ammonia added dropwise.
– In first step Cr(OH)3 formed, a green grey ppt. Ppt reaction.
– Cr(OH)3 dissolves in excess ammonia to form purple solution [Cr(NH3)6]
3+
.
CoCl4
2- when HCl added.
The complex [Cr(H2O)6]
3+ is violet. However common for Cr(III) to appear green in solution as
contain other mixed complexes of chromium(III) ions. Green is the expected observation when
chromium(III) ions are produced in the oxidation of alcohols and aldehydes using acidified
dichromate.
A complex reacts with silver nitrate. The silver nitrate reacts with the
aqueous halide ions. An excess of silver nitrate reacts with 0.01 mol
of complex to form 2.868g of precipitate. Which complex reacted?
Precipitate must be AgCl because silver nitrate only reacts with the aqueous halide ions- see
formula’s before. The other chloride ions don’t react as bound to metal by coordinate bondionic bond.
Can work out moles of AgCl using 2.868/ Mr = 0.02. So complex (0.01) + AgNO3 -> AgCl (0.02).
Therefore for every 1 complex, 2 free chlorides so must be second one in table.
When trying to work out unknown ligand complexes, better to keep it to what I know rather
than overcomplicating.
When HCl added to , crystals of are formed.
Haemoglobin: Haemoglobin in red blood cells. Has 4 haem groups so can carry four O2. O2
donates lone pair of electrons to central metal ion Fe2+ in haem group to form coordinate bond.
When required O2 substituted and released.
Oxyhaemoglobin: Oxyhaemoglobin forms when haemoglobin bonds to oxygen in the lungs.
Carbon dioxide bonds to haemoglobin and carried to lungs.
Products over reactants. Kstab/ stability constant.
KStab: Water is left out in Kstab because in excess. Don’t forget brackets of concentration and
brackets for complex ion as well- double square brackets. Kstab tells us how stable complex ion
is.
Ligand Substitutions with Haemoglobin: Ligand substitution occurs if oxygen in haemoglobin
replaced by carbon monoxide and forms carboxyhaemoglobin.
CO donates lone pair of electrons to Fe2+ in haem group to form coordinate bond. So carbon
monoxide prevents haemoglobin transporting oxygen, leading to death. Irreversible reaction.
Stability constant/ Kstab value of CO is greater than with complex in O2. Coordinate bond with
CO is stronger than O2.
As well as CO, CN- can substitute in as well.
blue ppt forms.
brown/ yellow solution forms.
Precipitation Reactions: A precipitation reaction occurs when two aqueous solutions containing
ions react to form an insoluble ionic solid- the precipitate. Transition metal ions in aq solution
react with aqueous NaOH and aq ammonia to form ppt.
Must learn equations and colours below.
Ppt Reaction with NaOH: All of these precipitates below are
insoluble in excess NaOH (aq). KOH (aq) can also be used.
Cu2+ is blue solution.
Fe2+ is pale green.
When left to stand, surface
turns brown, as Fe2+ oxidised
to Fe3+
.
Fe3+ is yellow.
Mn2+ is pale pink.
Ppt darkens in air.
Make sure these are balanced.
Cr3+
.
Only one where precipitate is soluble in excess NaOH forming
dark green solution- an octahedral.
Remember charge on OH-
Ppt Reactions with Ammonia: In the reactions above with ammonia + Cu2+/ Cr3+, the first part of
these reactions are precipitation reactions.
– Fe2+, Fe3+ and Mn2+ react with excess aq ammonia the same as they react with aq NaOH
forming those same precipitates. However there is no further reaction so these ppt do not
dissolve/ insoluble.
These are the first steps of ligand sub reactions.
Only have to show ions involved in reaction unless stated otherwise.
Redox Reactions of Titration Elements
All of these are redox reactions.
If a metal in ore, sometimes assume it’s just the element in reactions and equations e.g. Au.
If a ppt, state symbol always (s).
Will be very helpful to learn all below as will be tested on them.
Oxidation of Fe2+ to Fe3+: If solution containing MnO4- reacted with Fe2+, purple MnO4-
+
reduced to colourless Mn2+. Redox titration basis. Fe2+ oxidised to Fe3+
.
Reduction of Fe3+ to Fe2+: If solution containing Fe3+ reacted with I- iodine ions (potassium
iodide), orange brown Fe3+ reduced to pale green Fe2+. I- is oxidised to I2.
Reduction of Cr2O7
2- to Cr3+ to Cr2+: If aqueous acidified Cr2O7
2- (dichromate VI ions) reacted with
zinc ions, orange Cr2O7
2- reduced to green Cr3+. Zn is oxidised to Zn2+
– When excess zinc, green chromium (III) ions are reduced further to blue Cr 2+. Proves that zinc
is a very powerful reducing agent.
Oxidation of Cr3+ to CrO4
2-: Hot alkaline hydrogen peroxide H2O2 is a powerful oxidising agent.
Used to oxidise Cr3+ to chromium (VI) in CrO4
2-
.
Complex is oxidised, charge increases by
1. Electron released.
H2O2 reduced. It can only be reduced to
OH- as reduced to a negative charge.
Reduction of Cu2+ to Cu+
: If aqueous Cu2+
reacted with excess iodine ions, pale blue Cu2+ reduced to Cu+
. I- is oxidised to brown I2.
– Cu+ forms white precipitate of copper (I) iodide. White colour slightly masked through
brown iodide colour, look at bottom for it.
Disproportionation of Cu+: Solid copper oxide with hot dilute sulphuric acid. Brown ppt of
copper formed. Cu+ has been both oxidised and reduced.