Download TRANSITION METAL CHEMISTRY

THE GENERAL FEATURES OF
TRANSITION METAL
CHEMISTRY
General features of transition metals
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This presentation explains what a transition
metal is in terms of its electronic structure,
and then goes on to look at the general
features of transition metal chemistry.
These include
variable oxidation state (oxidation number)
complex ion formation
coloured ions
catalytic activity
What is a transition metal?
• The terms transition metal (or element)
and d block element are sometimes
used as if they mean the same thing.
• They don't - there's a subtle difference
between the two terms.
d block elements
• You will remember that when you are
building the Periodic Table and working
out where to put the electrons, something
odd happens after argon.
• At argon, the 3s and 3p levels are full, but
rather than fill up the 3d levels next, the 4s
level fills instead to give potassium and
then calcium.
• Only after that do the 3d levels fill.
d block elements
The elements in the Periodic Table which correspond
to the d sublevels filling are called d block elements.
The first row of these is shown in the shortened form
of the Periodic Table below.
The electronic structures of the d block elements shown are:
Sc
[Ar] 3d14s2
Ti
[Ar] 3d24s2
V
[Ar] 3d34s2
Cr
[Ar] 3d54s1
Mn
[Ar] 3d54s2
Fe
[Ar] 3d64s2
Co
[Ar] 3d74s2
Ni
[Ar] 3d84s2
Cu
[Ar] 3d104s1
Zn
[Ar] 3d104s2
You will notice that the pattern of filling isn't entirely tidy! It is
broken at both chromium and copper.
This is something that you are just going to have to accept.
There is no simple explanation for it which is usable at this
level. Any simple explanation which is given is faulty!
People sometimes say that a half-filled d level as in
chromium (with one electron in each orbital) is stable, and
so it is - sometimes! But you then have to look at why it is
stable. The obvious explanation is that chromium takes up
this structure because separating the electrons minimises
the repulsions between them - otherwise it would take up
some quite different structure.
But you only have to look at the electronic configuration of
tungsten (W) to see that this apparently simple explanation
doesn't always work. Tungsten has the same number of outer
electrons as chromium, but its outer structure is different 5d46s2. Again the electron repulsions must be minimised otherwise it wouldn't take up this configuration. But in this
case, it isn't true that the half-filled state is the most stable - it
doesn't seem very reasonable, but it's a fact! The real
explanation is going to be much more difficult than it seems
at first sight.
Neither can you use the statement that a full d level (for
example, in the copper case) is stable, unless you can come
up with a proper explanation of why that is. You can't assume
that looking nice and tidy is a good enough reason!
If you can't explain something properly, it is much better just
to accept it than to make up faulty explanations which sound
OK on the surface but don't stand up to scrutiny!
Transition metals
• Not all d block elements count as
transition metals! There are
discrepancies between the various
syllabuses, but the majority use the
definition:
• A transition metal is one which forms
one or more stable ions which have
incompletely filled d orbitals.
SCANDIUM , ZINC and COPPER
On the basis of this definition, scandium and zinc don't count as
transition metals - even though they are members of the d block.
Scandium has the electronic structure [Ar] 3d14s2. When it forms
ions, it always loses the 3 outer electrons and ends up with an
argon structure. The Sc3+ ion has no d electrons and so doesn't
meet the definition.
Zinc has the electronic structure [Ar] 3d104s2. When it forms ions, it
always loses the two 4s electrons to give a 2+ ion with the
electronic structure [Ar] 3d10. The zinc ion has full d orbitals and
doesn't meet the definition either.
By contrast, copper, [Ar] 3d104s1, forms two ions. In the Cu+ ion the
electronic structure is [Ar] 3d10. However, the more common Cu2+
ion has the structure [Ar] 3d9.
Copper is definitely a transition metal because the Cu2+ ion has an
incomplete d orbitals.
Transition metal ions
You have already come across the fact that when the
Periodic Table is being built, the 4s orbital is filled before
the 3d orbitals. This is because before filling orbitals, 4s
orbitals have a lower energy than 3d orbitals.
However, once the electrons are actually in their orbitals,
the energy order changes - and in all the chemistry of
the transition elements, the 4s orbital behaves as the
outermost, highest energy orbital.
Transition metal ions
The reversed order of the 3d and 4s orbitals only
applies to building the atom up in the first place. In all
other respects, you treat the 4s electrons as being the
outer electrons.
This is another of those things that you just have to accept.
The explanation again lies well beyond the level you are
working at. Just remember that once you have the full
electronic structure for one of these atoms, the 4s electrons
are the outermost electrons.
Remember this:
When d-block elements form ions, the 4s electrons are
lost first.
Transition metal ions
To write the electronic structure for Co2+:
Co
[Ar] 3d74s2
Co2+
[Ar] 3d7
The 2+ ion is formed by the loss of the two 4s electrons.
To write the electronic structure for V3+:
V
[Ar] 3d34s2
V3+
[Ar] 3d2
The 4s electrons are lost first followed by one of the 3d
electrons.
Transition metal ions
To write the electronic structure for Cr3+:
Cr
Cr3+
1s22s22p63s23p63d54s1
1s22s22p63s23p63d3
The 4s electron is lost first followed by two of the 3d
electrons.
To write the electronic structure for Zn2+:
Zn
Zn2+
1s22s22p63s23p63d104s2
1s22s22p63s23p63d10
This time there is no need to use any of the 3d electrons.
To write the electronic structure for Fe3+:
Fe
Fe3+
1s22s22p63s23p63d64s2
1s22s22p63s23p63d5
Forming transition metal ions
The rule is quite simple. Take the 4s
electrons off first, and then as many 3d
electrons as necessary to produce the
correct positive charge.
Variable oxidation state (number)
One of the key features of transition metal chemistry is
the wide range of oxidation states (oxidation numbers)
that the metals can show.
It would be wrong, though, to give the impression that
only transition metals can have variable oxidation
states. For example, elements like sulphur or nitrogen
or chlorine have a very wide range of oxidation states in
their compounds - and these obviously aren't transition
metals.
However, this variability is less common in metals apart
from the transition elements. Of the familiar metals from
the main groups of the Periodic Table, only lead and tin
show variable oxidation state to any extent.
Examples of variable oxidation states in the transition
metals
Iron
Iron has two common oxidation states (+2 and +3) in, for
example, Fe2+ and Fe3+. It also has a less common +6
oxidation state in the ferrate(VI) ion, FeO42-.
Manganese
Manganese has a very wide range of oxidation states in its
compounds. For example:
+2 in Mn2+
+3 in Mn2O3
+4 in MnO2
+6 in MnO42+7 in MnO4-
Explaining the variable oxidation
states in the transition metals
We'll look at the formation of simple ions like Fe2+ and Fe3+.
When a metal forms an ionic compound, the formula of the
compound produced depends on the energetics of the
process. On the whole, the compound formed is the one in
which most energy is released. The more energy released,
the more stable the compound.
There are several energy terms to think about, but the key
ones are:


The amount of energy needed to ionise the metal (the sum
of the various ionisation energies)
The amount of energy released when the compound
forms. This will either be lattice enthalpy if you are
thinking about solids, or the hydration enthalpies of the
ions if you are thinking about solutions.
The more highly charged the ion, the more electrons you
have to remove and the more ionisation energy you will have
to provide.
But off-setting this, the more highly charged the ion, the more
energy is released either as lattice enthalpy or the hydration
enthalpy of the metal ion.
Thinking about a typical non-transition
metal (calcium)
Calcium chloride is CaCl2. Why is that?
If you tried to make CaCl, (containing a Ca+ ion), the
overall process is slightly exothermic.
By making a Ca2+ ion instead, you have to supply more
ionisation energy, but you get out lots more lattice energy.
There is much more attraction between chloride ions and
Ca2+ ions than there is if you only have a 1+ ion. The
overall process is very exothermic.
Because the formation of CaCl2 releases much more
energy than making CaCl, then CaCl2 is more stable - and
so forms instead.
Thinking about a typical nontransition metal (calcium)
What about CaCl3? This time you have to remove yet
another electron from calcium.
The first two come from the 4s level. The third one
comes from the 3p. That is much closer to the nucleus
and therefore much more difficult to remove. There is a
large jump in ionisation energy between the second and
third electron removed.
Although there will be a gain in lattice enthalpy, it isn't
anything like enough to compensate for the extra
ionisation energy, and the overall process is very
endothermic.
It definitely isn't energetically sensible to make CaCl3!
Thinking about a typical transition metal (iron)
Here are the changes in the electronic structure of iron
to make the 2+ or the 3+ ion.
Fe
[Ar] 3d64s2
Fe2+
[Ar] 3d6
Fe3+
[Ar] 3d5
The 4s orbital and the 3d orbitals have very similar
energies. There isn't a huge jump in the amount of
energy you need to remove the third electron compared
with the first and second.
The figures for the first three ionisation energies (in kJ
mol-1) for iron compared with those of calcium are:
metal
1st IE
2nd IE
3rd IE
Ca
590
1150
4940
Fe
762
1560
2960
There is an increase in ionisation energy as you take
more electrons off an atom because you have the same
number of protons attracting fewer electrons.
Thinking about a typical transition metal (iron)
However, there is much less increase when you take the
third electron from iron than from calcium.
In the iron case, the extra ionisation energy is
compensated more or less by the extra lattice enthalpy
or hydration enthalpy evolved when the 3+ compound is
made.
The net effect of all this is that the overall enthalpy
change isn't vastly different whether you make, say,
FeCl2 or FeCl3. That means that it isn't too difficult to
convert between the two compounds.
The formation of complex ions
What is a complex ion?
A complex ion has a metal ion at its centre with a
number of other molecules or ions surrounding it.
These can be considered to be attached to the central
ion by co-ordinate (dative covalent) bonds. (In some
cases, the bonding is actually more complicated than
that.)
The molecules or ions surrounding the central metal ion
are called ligands.
Simple ligands include water, ammonia and chloride
ions. What all these have got in common is active lone
pairs of electrons in the outer energy level. These are
used to form co-ordinate bonds with the metal ion.
Some examples of complex ions formed by
transition metals
[Fe(H2O)6]2+
[Co(NH3)6]2+
[Cr(OH)6]3[CuCl4]2Other metals also form complex ions - it isn't
something that only transition metals do. Transition
metals do, however, form a very wide range of complex
ions.
Bonding in simple complex ions
Al(H2O)6 3+
We are going to look in detail at the bonding in the
complex ion formed when water molecules attach
themselves to an aluminium ion to give Al(H2O)63+.
Start by thinking about the structure of a naked
aluminium ion before the water molecules bond to it.
Aluminium has the electronic structure
1s22s22p63s23px1
When it forms an Al3+ ion it loses the 3-level electrons to
leave
1s22s22p6
Bonding in simple complex ions
That means that all the 3-level orbitals are now empty.
The aluminium uses of six of these to accept lone pairs
from six water molecules.
It re-organises (hybridises) the 3s, the three 3p, and two
of the 3d orbitals to produce six new orbitals all with the
same energy.
You might wonder why it chooses to use six orbitals
rather than four or eight or whatever. Six is the
maximum number of water molecules it is possible to fit
around an aluminium ion (and most other metal ions).
By making the maximum number of bonds, it releases
most energy and so becomes most energetically stable.
Only one lone pair is shown on each water molecule.
The other lone pair is pointing away from the
aluminium and so isn't involved in the bonding. The
resulting ion looks like this:
Dotted arrows represent lone pairs
coming from water molecules behind
the plane of the screen or paper.
Wedge shaped arrows represent
bonds from water molecules in front of
the plane of the screen or paper.
Al(H2O)6
3+
Because of the movement of electrons towards the
centre of the ion, the 3+ charge is no longer located
entirely on the aluminium, but is now spread over the
whole of the ion.
Because the aluminium is forming 6 bonds, the coordination number of the aluminium is said to be 6. The
co-ordination number of a complex ion counts the
number of co-ordinate bonds being formed by the metal
ion at its centre.
In a simple case like this, that obviously also counts the
number of ligands - but that isn't necessarily so, as you
will see later. Some ligands can form more than one coordinate bond with the metal ion.
Fe(H2O)6
3+
This example is chosen because it is very similar to
the last one - except that it involves a transition metal.
Iron has the electronic structure
1s22s22p63s23p63d64s2
When it forms an Fe3+ ion it loses the 4s electrons and
one of the 3d electrons to leave
1s22s22p63s23p63d5
Looking at this as electrons-in-boxes, at the bonding
level:
Fe(H2O)6
+3
Looking at this as electrons-in-boxes, at the bonding
level:
Now, be careful! The single electrons in the 3d level are
NOT involved in the bonding in any way. Instead, the
ion uses 6 orbitals from the 4s, 4p and 4d levels to
accept lone pairs from the water molecules.
Before they are used, the orbitals are re-organised
(hybridised) to produce 6 orbitals of equal energy.
Once the co-ordinate bonds have been formed, the ion
looks exactly the same as the equivalent aluminium ion.
Because the iron is forming 6 bonds, the co-ordination
number of the iron is 6.
CuCl4
2-
This is a simple example of the formation of a complex
ion with a negative charge.
Copper has the electronic structure
1s22s22p63s23p63d104s1
When it forms a Cu2+ ion it loses the 4s electron and one
of the 3d electrons to leave
1s22s22p63s23p63d9
To bond the four chloride ions as ligands, the empty 4s
and 4p orbitals are used (in a hybridised form) to accept
a lone pair of electrons from each chloride ion. Because
chloride ions are bigger than water molecules, you can't
fit 6 of them around the central ion - that's why you only
use 4.
Only one of the 4 lone pairs on each chloride ion is
shown. The other three are pointing away from the
copper ion, and aren't involved in the bonding.
CuCl42That gives you the complex ion:
The ion carries 2 negative charges overall. That comes
from a combination of the 2 positive charges on the
copper ion and the 4 negative charges from the 4
chloride ions.
In this case, the co-ordination number of the copper is,
of course, 4.