Download 5.3.1 Transition Elements 2012

Introduction
The first row of transition metals runs from scandium to zinc. All of the elements in this row
have outermost electrons in their 3d sub-shells. The 4s sub-shell is at a lower energy than
the 3d sub-shells and so the 4s sub shell fills before the 3d sub-shell. A d block element is
one which has its highest energy electrons in a d sub-shell. The terms transition metal and
d-block element are not the same.
Definition: A transition metal is a d-block element that forms one or more stable ions
with incompletely filled d-orbitals.
Student Activity
Complete the electronic arrangements in the following table (in pencil). Remember that Cu
and Cr don’t do what you would expect; because half filled and full sub-shells are more
stable than partially filled sub-shells. It is worthwhile, in terms of energy, for Cr and Cu to
promote electrons from the 4s into the 3d to achieve this extra stability.
Element
Scandium
Titanium
Vanadium
Chromium*
Manganese
Iron
Cobalt
Nickel
Copper*
Zinc
No of electrons
21
22
23
24
25
26
27
28
29
30
Electronic Configuration
[Ar]
3d1
2
[Ar] 4s 3d
[Ar] 4s2 3d
[Ar] 4s 3d
[Ar] 4s2 3d
[Ar] 4s2 3d
[Ar] 4s2 3d
[Ar] 4s2 3d
[Ar] 4s 3d
[Ar] 4s2 3d10
4s2
Chromium has one electron in each orbital of the 4s and 3d sub-shells, giving the
configuration [Ar] 4s1 3d5. The 3d is half filled, which is more stable than [Ar] 4s2 3d4.
Copper has a full 3d sub-shell giving the configuration [Ar] 4s1 3d10, which is more stable
than [Ar] 4s2 3d9.
Formation of ions
When a transition metal forms a positive ion, it loses its valence (outer shell) electrons first.
This means that the 4s-electrons are removed before any d-electrons. It’s the other way
round from the filling up.
V+ has an electronic configuration of [Ar] 4s1 3d3
Fe3+ has an electronic configuration of [Ar] 4s0 3d5
1
Student Activity
1. Complete the table below showing the electronic configurations of the ions listed and
the missing ions of elements with given configurations.
Element
Cu
Fe
Mn
Ti
Co
Fe
Cu
Ni
No of electrons
29
26
25
22
27
26
29
28
Ion
Cu2+
Fe2+
Mn4+
Ti3+
Co3+
Configuration
Learn
how to
do!
memthi
s!
[Ar] 4s0 3d5
[Ar] 4s0 3d10
[Ar] 4s0 3d8
2. What do we call ions/atoms with identical electronic configurations? Give an example
………………………………………………………………………….
………………………………………………………………………….
3. Why is zinc a d block element but not a transition element (Hint: it only forms the
Zn2+ ion)?
………………………………………………………………………….
1.Variable oxidation states
Transition metals have more than one stable oxidation state possible and this is seen in
their ions. This is because the electrons in the d sub-shells are close together in energy so
it is almost as easy to remove several outer electrons as it is to remove just one. There is
no easy way to predict the oxidation states which will be displayed by a particular element.
Iron’s common oxidation states are +2 and +3.
Copper’s common oxidation states are + 1 and +2.
Iron and
Copper are
very important
2
2. Formation of coloured compounds and ions
Aqueous solutions of the transition metal ions are coloured.. Remember these ions do not
occur native but as their hexaaqua metal ions eg [Cu(H2O)6]2+. Examples include:
Fe2+ - green
Fe3+ - yellow
Cu2+ - blue
Ni2+ - green
Co2+ - pink
Cr3+ - green
Cr6+ - orange
Mn2+ - pale pink
Mn7+ - purple
3. Catalytic behaviour
Transition metals and their ions are very good catalysts, both homogeneous and
heterogeneous. One reason for this is that they can have different oxidation states and so
can gain or lose electrons easily. Another is that the empty d-orbitals can accept electrons
from other molecules/ions at the transition element surface. Intermediate species can be
made.
a) Reaction: Haber process
Equation:
N2(g) + 3H2(g)
2NH3(g)
b) Reaction : Contact process
Equation:
Catalyst: Iron (s)
2SO2(g) + O2(g)
Catalyst : V2O5(g)
2SO3(g)
Here the V2O5 is believed to react with SO2 to form V2O4 and SO3 and then the V2O4 reacts
with the oxygen to regenerate the V2O5.
c) Reaction : Hydrogenation of alkenes (unsaturated oils can be hydrogenated to make
margarine) Catalyst : Nickel(s)
Equation:
H
H
C
H
+
C
H
H2
H
H
H
C
C
H
H
H
Student activity
Work out the oxidation state of the transition metal species in each of the following
compounds
a) KMnO4
………..
b) Cu2O
……….
c) Fe2O3
……….
d) CuSO4
………..
e) Cr2O72……….
3
Transition metal hydroxides (precipitation reactions)
Transition metal ions react with the hydroxide ion in aqueous
solution to give a precipitate. A precipitate is seen as a solid on the
inside of the test tube. A precipitate is the insoluble product of a
reaction between two aqueous solutions. The colour of the metal hydroxide can be used to
identify the metal. Use Experiment 1 : Precipitation Reactions to help complete the following
table.
You must learn these equations and colours.
Reaction
Co2+(aq) +
2OH-(aq) →
Colour of Precipitate
Co(OH)2(s)
pale blue precipitate
……….(aq) +
………(aq) → Fe(OH)2(s)
foxy red precipitate
Fe3+(aq) +
3OH-(aq) → ……………
Complex ions: structure and bonding
Transition metals form complex ions. A Complex consists of a central metal ion/metal
surrounded by ligands which are bonded co-ordinately.
The co-ordination number is the number of co-ordinate bonds in the complex (not the
number of ligands).
A ligand is a molecule or an ion, which donates a lone pair of electrons to the transition
metal ion by co-ordinate bonding
Example : [Cu(H2O)6]2+
Copper is the
central metal ion
Square brackets go round
the complex ion and the
total charge on the complex
sits outside the square
brackets
Water is the ligand
Dative Bond
Common Ligands include: H2O, Cl-, NH3 and SCN- ,
note that some are neutral ligands and others are charged,
all have a lone pair of electrons. Complex ions can be drawn
in full as shown below when their shape is also displayed.
2+
OH2
H2O
OH2
Fe
H2O
OH2
OH2
Ligand
4
Student Activity 1
State (i) the formula and (ii) the charge of the following complex ions:
(a) one titanium (III) ion and six water molecules;
……………………………………………
(b) one copper (II) ion and four chloride ions;
……………………………………………
(c) one cobalt (II) ion and four chloride ions;
……………………………………………
(d) one copper(II) ion, four ammonia molecules and two water molecules;
……………………………………………
(e) one iron(III) ion, one thiocyanate ion and five water molecules.
……………………………………………
Shapes of complex ions
There are three shapes that are required at this level.
1. Octahedral
2. Tetrahedral
[CuCl4] 22+
OH2
H2O
OH2
Fe
H2O
OH2
OH2
Bond Angles : 90O and 180O
Co-ordination Number = 6
Bond Angle : 109.5O
Co-ordination Number = 4
Bond angles around the oxygen of the water ligand(H-O-H)= ………………
5
3. Square Planar - cis platin (anti-cancer drug) [PtCl2(NH3)2]
Bond Angle(Cl-Pt-N): 90 O; Co-ordination Number = 4
This is a very effective chemotherapy drug which works by binding to the DNA in cancer
cells and causing the shape of the DNA to change such that it can no longer replicate, so
eventually these cells die. However, there are unpleasant side effects for the patient. You
can assume that any 4 co-ordinate complexes are tetrahedral unless they are cis-platin or
[Ni(CN)4]2-, [Ni(NH3)2Cl2] or [AuCl4]- which are square planar.
Student Activity 2
(i)
(ii)
Using 3 dimensional diagrams, draw each of the shapes for the complexes listed
below and predict the bond angles.
For each complex, give the oxidation number of the transition metal and the
coordination number.
a) [Co(NH3)5Cl]2+
b) [Cr(NH3)6]3+
c) [Fe(SCN)(H2O)5]2+
d) [CoCl4]2-
Bidentate or polydentate ligands
Some ligands can donate two (or more electron pairs) and are called bidentate (or
poly/multidentate). Three bidentate ligands will make a six coordinate complex. The
atoms/species which donate the electrons in a ligand are usually: N, O, O -,P, Br-. A neutral
oxygen of a carboxylic acid or an alcohol (bonded to an H) can act as a ligand, but a
carbonyl oxygen will not act as a ligand.
6
Student activity 3
1. State whether each ligand below is monodentate, bidentate or polydentate.
(a)
Glycine (aminoethanoic acid)
(e) Br-
(b)
(f) 1,2-diaminoethane (en)
(c)
(g)
(d) 1,1-cyclobutane dicarboxylate
(h)
7
Student Activity 4
Write the formulae of the complexes formed between:
(a) nickel (II) ions and three 1,2-diaminoethane molecules;
………………………………………………
(b) copper (II) ions and three ethanedioate ions;
………………………………………………
(c) cobalt (III) ions, two chloride ions and two 1,2-diaminoethane molecules.
……………………………………………...
Stereoisomerism
Many complexes can exist as stereoisomers; they have the same formulae but different
spatial arrangements of their atoms. There are two types – cis/trans isomers and optical
isomers.
Cis/Trans isomers
These only happen in square planar or octahedral complexes.
For example, in [Pt(NH3)Cl2] which is square planar, there are two possible arrangements
of the two different ligands:
or
cis
trans
Only the cis isomer has anti-cancer activity. Note that in the cis the bond angles Cl-Pt-Cl
are 90 degrees whereas in the trans it is 180 degrees.
Student Activity 5 In pencil draw the cis and trans isomers of
a) diamminedichloronickel (II);
b) tetraamminedichlorocobalt (III).
The latter is particularly
striking because the cis
crystals are violet
whereas the trans
crystals are green.
8
Optical isomerism
Octahedral complexes which have at least two bidentate ligands also show optical
isomerism. As a result of their 3D shape, there are two possible mirror image isomers,
which are non-super imposable and will rotate the plane of polarised light in opposite
directions. The central ion is not described as chiral; this term is reserved for organic
molecules only. On the next page is a diagram showing the two isomers from a complex
with three ethanediamine (en) ligands around an M+ metal ion
Student Activity 6
In pencil, draw stereoisomers of the following:
(i)
[Cu(H2O)2(NH3)4]2+
(ii)
[CoCl2(en)2] (there are three structures; draw the two optical isomers first, with a
mirror plane between them).
9
Ligand Substitution Reactions
Electron pair donors in one complex can be replaced by other ligands to form more stable
complexes. This is known as ligand substitution. Often the reactions are reversible.
(a) Copper (II) ions and excess ammonia
Observations : Dropwise addition: a pale blue precipitate is observed which dissolves in
excess ammonia to give a royal blue solution.
OH 2
2+
There are two reactions. The first is where ammonia acts a base.
The second is where the ammonia takes part in ligand
substitution.
NH3
HN
Cu
Write an equation to show how ammonia behaves as a base with 3
NH3
H3N
water.
…………………………………………………………………………..
OH 2
1. Cu2+(aq) + 2OH-(aq)→ Cu(OH)2(s)
pale blue ppt
2. [Cu(H2O)6]2+(aq)+ 4NH3(aq) →[Cu(NH3)4(H2O)2]2+(aq) + 4H2O(l)
Royal/deep blue solution
(b) Copper(II) ions and concentrated HCl (aq)
Observations: On addition of the HCl(aq) the blue solution turns yellow.
H+ is spectator ion so does not appear in the equation.
[Cu(H2O)6]2+ (aq) +
Octahedral
4Cl-(aq)
[CuCl4 ]2- (aq) +
Tetrahedral
6H2O(l)
The above reaction is reversible, it is an example of an equilibrium. How could you
change the yellow solution to blue?
(c) Iron (III) and thiocyanate ions (SCN-)- watch your teacher demo this!
Observations: On addition of the SCN- ions to the solution of Iron (III) the yellow solution
turns blood red.
[Fe(H2O)6]3+ (aq) + SCN- (aq)
Yellow Solution
→
[Fe(H2O)5(SCN)]2+(aq) + H2O(l)
Blood Red Solution
(Note: this is not on the specification but is a valid example of ligand substitution)
10
(d) Cobalt (II) ions and concentrated HCl (aq)
When excess chloride ions are added to the hexaaquacobalt (II) ion, there is a colour
change and coordination number change as a substitution reaction occurs.
Write an equation for the reaction here and explain how you would reverse the colour
change using Le Chatelier’s principle.
………………………………………………………………………………………………………
………………………………………………………………………………………………………
………………………………………………………………………………………………………
(e) Demonstrations: Your teacher will demonstrate the following reactions. For each
one you should note down the colour change and write the equation. You do not
need to learn these colour changes but could be asked to write an equation. These
colour changes are valid to use in essay questions.
i)
Copper (II) ions and 1,2 – diaminoethane
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
ii)
Copper (II) ions and Sodium ethanedioate
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
iii)
Copper (II) ions and EDTA
…………………………………………………………………………………………………………
…………………………………………………………………………………………………………
11
Complete the following table for the reactions of copper (II) ions:
Reagent
Formula of
ligand
Water
Formula of
complex
H2O
Colour of
complex
Shape of
complex
Octahedral
[CuCl4 ]2-
Conc HCl
Ammonia
4
NH3
1,2 diaminoethane
Sodium
ethanedioate
EDTA
en
6
[Cu(en)3]2+
[Cu(C2O4)3]4-
EDTA4-
Coordination
number
[CuEDTA]2-
6
Octahedral
Haemoglobin
Ligand substitution is very important in the transport of oxygen by haemoglobin(Hb).
The active bit is the Fe (II) ion which forms four co-ordinate bonds to the four nitrogen
atoms in the haem ligand and to one nitrogen atom in the globin. There is also a weak
bond to a water molecule which can be reversibly substituted by the oxygen.
Hb(aq) + O2
(g)
HbO2
(aq)
This allows haemoglobin to give up oxygen where it is needed in the body.
Sadly, CO can also act as a ligand with the iron (II) ions and can not only bind more
strongly but also irreversibly with the globin which prevents the oxygen from binding.
This can result in death – hence the importance of CO detectors!
12
13
Stability constants
We can write an expression to show how far the equilibrium lies to the left or right of a
reversible reaction. This quantifies equilibria, this is whole topic in itself (more in
Sep/Oct)
[Cu(H2O)6]2+ (aq) +
4Cl-(aq)
[CuCl4 ]2-(aq) +
6H2O(l)
If we divide concentrations products/reactant in an equation we will get a measure of
how far to the right the equilibrium lies. We omit the water as its value is huge in
comparison to the others and it does not change much during the establishment of
equilibrium.
Kstab =
[[CuCl4 ]2- ]
_________________________
[[Cu(H2O)6] 2+] x [Cl- ] 4
You must use two sets of square brackets
for a complex
The value is called the stability constant.
You should be able to deduce expressions for Kstab for any ligand substitution reaction.
Student Activity 7
Write expressions for Kstab for
(i) Hb(aq) + O2 (g)
(ii) [Fe(H2O)6]3+(aq) +SCN-(aq)
HbO2 (aq)
[Fe(H2O)5(SCN)]2+ +H2O
For a complex ion, Kstab is the equilibrium constant for the formation of the complex ion
from its constituent ions in a solvent (usually water). A large value means that the
equilibrium lies to the right hand side and that the new complex is very stable. In order to
avoid awkward powers of 10 we often take logarithms.
Student Activity 8
Compare the stabilities of these three complex ions from the data below and put them in
decreasing order of stability (1 is the most stable 3 is the least).
Ligand
Complex ion
Kstab
Log Kstab
Cl-
[CuCl4 ]2-
4.0 x 105
5.60
edta
[Cu(edta)]2-
6.3 x 1018
18.8
NH3
[Cu(NH3)4(H2O)2]2+ 1.3 x 1013
13.1
14
June 2011: Ques 6b
15