1 a MnO4–
+7
b crO42-
+6
c cr2O72-
+6
2
Acid-base. Il is not redox as the chromium is in the +6 state before and after. One of the oxygen atoms accepts two protons (H+ ions).
1
2 a
b Bromide ions are bigger than water molecules, so fewer can fit around the copper ion. The shape is tetrahedral.
3
At each step two entities produce three, therefore the entropy change of each step is likely to be positive.
4
No, it remains at +2.
b
[Co(H2O)6]2+ is octahedral and [CoCI4]2- is tetrahedral.
c
The ligands have changed as has the co-ordination number.
1 They are smaller and more highly charged and therefore more strongly polarising. They can thus weaken one of the 0-H bonds in one of the water molecules that surround them, so releasing a H+ ion.
2 Aluminium is not a transition metal. It has no pan-filled d-orbitals. Most transition metal compounds arc coloured because of the electrons moving between pan-filled d-orbitals that absorb light.
3 Both NH3 and H2O are of similar size and are neutral ligands.
4 a
i
ii
b a i
The orange brown solution changes to a reddish brown solid.
ii?
The pale blue solution turns to dark blue.
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(a)
N in Cu(NO3)2 oxidation state: +5
N in NO2 oxidation state: +4
Oxidation product: oxygen
(b)
[Cu(H2O)6]2+
Octahedral
(c)
Cu(H2O)4 (OH)2 OR Cu(OH)2
[Cu(H2O)6]2+ + 2NH3 → Cu(H2O)4 (OH)2 + 2NH4+
(d)
[Cu(NH3)4(H2O)2]2+ deep blue Cu(H2O)4(OH)2 + 4NH3 → [Cu(NH3)4(H2O)2]2+ + 2H2O + 2OH– |
(e)
[CuCl4]2-
Yellow-green
Tetrahedral
(f) (i)
1s2 2s2 2p6 3s2 3p6 3d10
(ii) A reducing agent
2 (a)
Forms blue or pink precipitate.
Co(H2O)4 (OH)2
Precipitate dissolves in excess ammonia.
Forms yellow or pale brown ‘straw’ coloured solution.
[Co(NH3)6]2+
Darkens on standing in air.
[Co(NH3)6]3+ formed.
Due to oxidation by O2 in air.
(b)
Fe3+ has a larger charge and smaller size than Fe2+.
The Fe3+ polarises a ligand water molecule to a greater extent.
The solution of Fe3+ contains more H+ ions.green precipitate with Fe2+
FeCO3
brown or red/brown precipitate with Fe3+
[Fe(H2O)3(OH)3]
Effervescence as carbon dioxide is evolved from the Fe3+ reaction.
3
(a)
Reaction 1
Ammonia (NH3) (solution) / NaOH
(b)
Reaction 2
Ammonia (conc/xs)
[Cu(H2O)4(OH)2] + 4NH3 → [Cu(H2O)2(NH3)4]2+ + 2H2O + 2OH–
(c)
Reaction 3
Na2CO3 / any identified soluble carbonate /
NaHCO3
(d)
Reaction 4
HCl (conc/xs) / NaCl
[Cu(H2O)6]2+ + 4Cl– → [CuCl4]2- + 6H2O
4
(a)
W is CuCl42-
Yellow-green/yellow/green
[Cu(H2O)6]2+ + 4Cl– → CuCl42- + 6H2O
(b)
X is Cu(H2O)4(OH)2
Blue precipitate/solid
[Cu(H2O)6]2+ + 2NH3 → Cu(H2O)4(OH)2 + 2NH4+
(c)
Y is [Cu(NH3)4(H2O)2]2+
Deep/dark/royal blue solution
Cu(H2O)4(OH)2 + 4NH3 → [Cu(NH3)4(H2O)2]2+ + 2H2O + 2OH–
(d)
Z is CuCO3
Green solid/precipitate
[Cu(H2O)6]2+ + CO32- → CuCO3 + 6H2O
(e) (i)
Cu2+(aq) + Fe(s) → Cu(s) + Fe2+(aq)
Blue
Green
(ii)
Any two correct points about copper extraction from two of these three categories:
Any relevant mention of lower energy consumption
Any relevant mention of benefits of less mining (of copper ore)
Less release of CO2 (or CO) into the atmosphere
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