Download Coordination number 4 - INTEC Chemistry Blog

Ligang
H2O
NH3
Ti
h.aqua
No info
OH ppt
OH negative
complex
(i.e. soluble)
Cl-
V
h.
aqua
No info
Cr - amphoteric
h.aqua
Mn
h.aqua
plae pink.
Fe
Co
Fe3+ = yellow
Fe3+ = pale
green
pink
h.aqua
h.
ammine ##
after hydroxide
ppte forms first
No amine
complex
No amine
complex
Cr(OH)3 a green
ppte, then goes
to hexaammine
Mn(OH)2 a sandy
or buff Goes
dark over time
as MnO2 slowly
forms.
Fe(OH)2 = dirty
green
h.aqua
h.
ammine
h.aqua
h.
ammine
Cu
h.aqua
sometimes shown as tetra
aqua as axial H2O’s are
further away than equatorial
H2O’s.
Writing it as hexaaqua is
accepted.
[Cu(NH3)4]2+
Can also show as
[Cu(NH3)4(H2O)2]2+
H2O in axial positions
TiCl4 reacts w.
H2O like SiCl4
does TiO2
formed
Zn - amphoteric
h.aqua
**
[Zn(NH3)4]2+
Zn(OH)2
Fe(OH)3 =
orange/brown
[Cr(OH)6]3- green
solution
[Zn(OH)4]2-
[FeCl4]-
Fe = +III
[FeCl4]2-
[CoCl4]2blue
Tetrahedral
[Cu(Cl)4]2yellow
Tetrahedral
[ZnCl4]2-
[Cu(CN)2]-
[Zn(CN)4]2-
Fe = +II
Tetrahedral
F fluoro
CN (cyano)
[Fe(CN)6]4i.e normal ‘hexa’
[Co(CN)6]4-
Fe(CO)5 unusual
pentavalent
complex Fe in
zero oxidation
state
[Co(Cl)2(NH
+
3) 4]
CO (carbonyl)
Other info
Ni
[Cr(CO)6] Cr in
zero oxidation
state
[Cr(Cl)3(NH3)3]
Cr in +III O.S.
CrO42- yellow
tetrhedra’ –
stable in alkaline
solution
Cr2O72- stable in
acidic solution.
Tetrahedral
about each Cr,
bonded to one
common O atom
[Ni(CN)4]2square
planar
[Ni(CO)4](l)
Used to
extract Ni
from ore.
** George Facer “Make the Grade” says Zn forms [Zn(H2O)4]2+ but G.I.Brown’s “Introduction to Inorganic Chem” (2nd Edition) p391 says it forms the hexaaqua ion.
## George Facer “Make the Grade” says Cr(OH)3 remains on addition of excess NH3. I think that is wrong. I think it forms a hexaammine complex.
Actually the learning requirements for transition metals is quite small.
Test for Fe3+ ions: Add –SCN i.e. thiocyanide ions. The yellow solution goes red. One water ligand is replaced. [Fe(SCN)(H2O)5]+ = blood red
Test for Fe2+ ions: [Fe(H2O)6]2+ (aq) + [Fe(CN)6]3-  [Fe[Fe(CN)6]]- Prussian blue or Fe(III)4[Fe(II)(CN)6]3 This blue was the colour used in
old ‘imprint paper’ but was replaced by carbon imprint paper.
CuCl4]2- = yellow, but often appears green as some blue hexaaqua complex remains yellow and blue look green. Same problem with Vanadium,
Problem with cobalt: (a) = [Co(CN)6]4- , (b) = [Co(NH3)6]2+ , (c) = [Co(H2O)6]3+
[Co(CN)6]3- + e-  [Co(CN)6]4- (a) E = -0.80 V.
(a) will revert back to [Co(CN)6]3- as (a) is reducing and so is oxidized in air
3+
2+

[Co(NH3)6] + e-  [Co(NH3)6] (b) E = +0.10 V
(b) will revert back to [Co(NH3)6]3+ as (b) is reducing (w.r.t. O2 in air) so is oxidized in
air
[Co(H2O)6]3+(c) + e-  [Co(H2O)6]2+ E = +1.82 V –
(c), the 3+ ion (Co also in +III) state will go on to produce [Co(H2O)6]2+ . It is quite a
strong oxidizing agent. (c) will actually react with the surrounding water. O 2 is produced.
A blue/green precipitate of Co(OH)2 will slowly turn into [Co(NH3)6]3+ over time.
Note about Copper:
Copper I oxide has 2 Cu’s in the formula. Cu2O - red
Cupper II oxide has 1 Cu in the formula: CuO - black
Cu+ ions are not stable in solution. If Cu+ is produced in a REDOX reaction, it usually precipitates out. This may take time.
Catalysts:
The catalyst example I gave was with V and Fe but the Fe wasn’t discussed as in the haber process, rather in reaction with S 2O82(peroxydisulphate ions) ions and sulphate ions with Iodide ions (I-) and Iodine (I2).
Deprotonation:
When reacting with dil NH3 , deprotonation occurs as ammonia solutions are basic. NH 3 + H2O  NH4+ + OH-. The presence of the OH- in
ammonia solution deprotonates the H2O ligand. I advise you if showing this reaction, to bring the resulting product to the neutral hydroxide
compound e.g.
[Cu(H2O)6]2+(aq) + dilute ammonia  [Cu(OH)2] or [Cu(OH)2(H2O)4] or Cu(OH)2
Is better than
[Cu(H2O)6]2+ + dilute ammonia  [Cu(OH)(H2O)5]+
The reason why the top line is better is because you are much less likely to spend less time with trying to write the correct number of ligands and
resulting charge. And less likely to make a mistake.
If however the question said something like: aq. Cu 2+ ions have a small amount of dil NH3 added to it and no ppte is seen. Give an equation
describing this reaction, then of course, you MUST write down the second equation! This is rare however, as in real life, it seems the OH- isn’t
shared evenly so some ppte’s out while other Cu2+ remains. The Cu2+ precipitating out iwill be the ones closest to the OH- ions as they are
dropped into the solution. The entropy increase by forming solid probably accounts for some Cu 2+ being greedy for OH- and not sharing it.
Ligand
Ag
Pt
Pd
Al amphoteric
Pb amphoteric
H2O
NH3
hexaaqua
hexaaqua
hexaaqua
hexaaqua
hexaaqua
[Ag(NH3)2]+
[Pt(NH3l)4]2+
OH ppt
AgOH (ppt)
linear
d.n.f
Al(OH)3 ppte
deprotonation
only
Al(OH)3 ppte
Pb(OH)2 ppte
deprotonation
only
Pb(OH)2 ppte
[Al(OH)6]3ppte
[Pb(OH)4]2-
OH negative
complex
Cl-
CN (cyano)
AgCl (ppt)
[Pt(Cl)4]2-
[Pd(Cl)4]2-
[Ag(CN)2]-
Pt(NH3)2(Cl)2
cis-platin an
anti-cancer
drug. Stops
‘unzipping’ of
DNA. Square
planar
Can get [Cu(NH3)2]2+ but not necessary to know this for A-level.
Nice definition of a complex ion: Complex ion: A molecule or ion bonded via a dative covalent bond to a metal ion.
It’s likely you will be asked something about the nature of colour in transition metal complexes. E.g. Account for the fact [Cu(NH3)4]2+ complexes
are a blue colour yet [Co(H2O)6]2+ are pale pink. Questions like this are typically worth about 4-5 marks. You must first describe why they have a
colour to begin with then why that colour is blue (the energy of the blue wavelength doesn’t correspond to d-d transition energy, so is
unabsorbed) and then why the Co complex is pink and then why they differ.
Be prepared for quick Q’s such as: Other than forming coloured complexes, give 2 properties of transition metals.