Reactions of inorganic compounds in aq Flashcards
Lewis theory of acids
Electron pair acceptors in the formation of co-ordinate bonds
Lewis theory of bases
Electron pair donors in the formation of co-ordinate bonds
Why is the Lewis definition of acids wider than the Bronsted-Lowry one?
Bronsted-Lowry applies only to exchange of protons so compounds have to contain hydrogens
Aqua ions
6 water molecules act as ligands bonding to the metal ion in an octahedral arrangement
Why is Fe3+ (aq) acidic at all and why the difference with Fe2+(aq)?
The Fe3+ ion is both smaller and more highly charged than Fe2+, making it more strongly polarising. So in the [Fe(H2O)6]3+ ion the iron strongly attracts electrons from the oxygen atoms of the water ligands, thus weakening the O-H bonds in of the water molecules. This complex ion will then readily release an H+ making the solution acidic. Fe2+ is less polarising and hence less O-H bonds break in solution.
Hydrolysis reaction
O-H bonds are broken and a new species are formed.
How to distinguish between Fe2+ and Fe3+ in aq
[Fe(H2O)6]2+ = pale green
[Fe(H2O)6]3+ = pale brown
Add dilute alkali, which precipitates the hydroxides whose colours are more obviously different.
Reactions with the base CO32-
In general, carbonates of transition metal ions in oxidation state +2 exist, while those of ions in the +3 state don’t
2 Amphoteric hydroxides
Al(H2O)3(OH)3 + Cr(H2O)3(OH)3
CrO42-
Yellow
Cr2O72-
Orange
How can water molecules acting as ligands be replaced by other ligands?
- Other ligands form stronger co-ordinate bonds (better Lewis bases)
- Present in higher conc. and an equilibrium is displaced
[Co(H2O)4(OH)2]
Blue precpitate
What happens if more conc. NH3 is added to [Co(H2O)4(OH)2]?
Both OH- and all 4 water ligands are replaced by ammonia because :
- Ammonia is a better ligand than water
- High conc. of ammonia displaces equilibria, displacing water and OH-
Observations if more conc. NH3 is added to [Co(H2O)4(OH)2]
The blue precipitate dissolves to form a pale yellow solution which is oxidised by oxygen in air to a brown mixture with Co(III)
[Cu(H2O)6]2+ + 4NH3
[Cu(NH3)4(H2O)2]2+ pale blue to very deep blue
Shape of [Cu(NH3)4(H2O)2]2+
Octahedral, 4 NH3 square planar arrangement with H2O above and below. Cu-O bonds are longer and weaker than Cu-N
[Cu(H2O)6]2+ + 4Cl-
[CuCl4] 2-, Pale blue replaced by yellow [CuCl4]2- -> co-ordinate number 6-> 4
Chelation
Formation of complexes with multidentate ligands
Why do multidentate ligands form more stable complexes than monodentate ligands?
Entropy change
[Fe(H2O)6]2+
Pale green
[Co(H2O)6]2+
Pink
[Cu(H2O)6]2+
Pale blue
[Al(H2O)6]3+
Colourless
[V(H2O)6]3+
Green
[Cr(H2O)6]3+
Pale green
[Fe(H2O)6]3+
Yellow
Deprotonation
The loss of a proton by a water ligand to form a hydroxo ligand
Fe(OH)2
Green
Co(OH)2
Blue
Cu(OH)2
Pale Blue
Al(OH)3
White
Cr(OH)3
Green/grey
Fe(OH)3
Brown
Al(OH)63-
Colourless
Cr(OH)63-
Green
FeCO3
Green ppt 2+ / Brown ppt 3+
CoCO3
Pink ppt
CuCO3
Green-Blue ppt
[Co(OH)2(H2O)4]
Blue ppt
[Co(NH3)6]2+
Straw
[Co(NH3)6]3+
Brown
[Cu(NH3)4(H2O)2]2+
Dark Blue
CoCl42-
Blue
CuCl42-
Green/yellow
The chelate effect
The entropy of the system thus increases, and multidentate complexes are therefore more stable than complexes involving monodentate ligands.
VO2+
Yellow
CrCO3
Pale green ppt
CoCO3
Pink ppt
AlCO3
White ppt
3 changes to a transition-metal complex which result in a change in colour
- Oxidation states
- Ligands
- Co-ordination number
Outline a plan for experiments which would enable you to determine the conc. of a transition metal ion in a solution of unknown conc.
- Add appropriate ligand to intensify colour
- Make up solutions of known conc.
- Measure absorption
- Plot graph of results
- Measure absorption of unknown + compare
Linear complex formed when an excess of conc. HCl is added to silver chloride
[AgCl2]2-
Reagents for VO2+ -> [V(H2O)6]2+
HCl, Zn
Reagent for {Ag(NH3)2]+ -> Ag
Aldehyde
Reagent for [Cr(H2O)6]3+ -> CrO42-
H2O2, NaOH
Describe an experiment to show that vanadium has several oxidation states
- Xs Zn, HCl
- Absence of air
- Colours seen
- V(IV), V(III), V(II) seen
2 equations to show that chromium (III) hydroxide is amphoteric
- Cr(OH)3 + 3H+ -> [Cr(H2O)6]3+
2. Cr(OH)3 + 3OH- -> [Cr(OH)6]3-