Complex metal ions containing simple ligands
What is a complex metal 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 calledligands.
The nature of 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.
All ligands are lone pair donors. In other words, all ligands function as Lewis bases.
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
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When it forms an Al3+ ion it loses the 3-level electrons to leave
1s22s22p6
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:
Note: 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.
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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:
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.
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Because the iron is forming 6 bonds, the co-ordination number of the iron is 6.
CuCl4 2This 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.
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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.
That 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.
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Complex metal ions containing more complicated
ligands
In the examples we've already looked at, each ligand only forms
one bond with the central metal ion to give the complex ion.
Such a ligand is said to be unidentate. That means literally that
it only has one tooth!
It only has one pair of electrons that it can use to bond to the
metal - any other lone pairs are pointing in the wrong direction.
Some ligands, however, have rather more teeth! These are
known generally as multidentate or polydentate ligands, but can
be broken down into a number of different types.
Bidentate ligands
Bidentate ligands have two lone pairs, both of which can bond to
the central metal ion.
The two commonly used examples are 1,2-diaminoethane (old
name: ethylenediamine - often given the abbreviation "en"), and
the ethanedioate ion (old name: oxalate).
In the ethanedioate ion, there are lots more lone pairs than the
two shown, but these are the only ones we are interested in.
You can think of these bidentate ligands rather as if they were a
pair of headphones, carrying lone pairs on each of the "ear
pieces". These will then fit snuggly around a metal ion.
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Ni (NH2CH2CH2NH2)3 2+
You might find this abbreviated to [Ni(en)3] 2+.
Note: It is quite common to wrap square brackets around
complex ions to emphasise that the charge is carried by the
whole ion - as I have done in the abbreviated form, and in the
diagram below.
I have omitted them from the formula in the heading above
because they didn't come out well in the italic font I used. It just
confused things.
The structure of the ion looks like this:
In this case, the "ear pieces" are the nitrogen atoms of the
NH2groups - and the "bit that goes over your head " is the CH2CH2- group. If you were going to draw this in an exam, you
would obviously want to draw it properly - but for learning
purposes, drawing all the atoms makes the diagram look unduly
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complicated!
Notice that the arrangement of the bonds around the central
metal ion is exactly the same as it was with the ions with 6 water
molecules attached. The only difference is that this time each
ligand uses up two of the positions - at right angles to each
other.
Because the nickel is forming 6 co-ordinate bonds, the coordination number of this ion is 6, despite the fact that it is only
joined to 3 ligands. Co-ordination number counts the number of
bonds, not the number of ligands.
Cr (C2O4)3 3This is the complex ion formed by attaching 3 ethanedioate
(oxalate) ions to a chromium(III) ion. The shape is exactly the
same as the previous nickel complex. The only real difference is
the number of charges. The original chromium ion carried 3+
charges, and each ethanedioate ion carried 2-. (3+) + (3 x 2-) =
3-.
The structure of the ion looks like this:
Again, if you drew this in an exam, you would want to show all
the atoms properly. If you need to be able to do this, practice
drawing it so that it looks clear and tidy! Refer back to the
diagram of the ethanedioate ion further up the page to help you.
A quadridentate ligand
A quadridentate ligand has four lone pairs, all of which can bond
to the central metal ion.
An example of this occurs in haemoglobin (American:
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hemoglobin).
The functional part of this is an iron(II) ion surrounded by a
complicated molecule called haem (heme). This is a sort of
hollow ring of carbon and hydrogen atoms, at the centre of
which are 4 nitrogen atoms with lone pairs on them.
Haem is one of a group of similar compounds
called porphyrins. They all have the same sort of ring system,
but with different groups attached to the outside of the ring. You
aren't going to need to know the exact structure of the haem at
this level.
We could simplify the haem with the trapped iron ion as:
Each of the lone pairs on the nitrogen can form a co-ordinate
bond with the iron(II) ion - holding it at the centre of the
complicated ring of atoms.
The iron forms 4 co-ordinate bonds with the haem, but still has
space to form two more - one above and one below the plane of
the ring.
The protein globin attaches to one of these positions using a
lone pair on one of the nitrogens in one of its amino acids. The
interesting bit is the other position.
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Overall, the complex ion has a co-ordination number of 6
because the central metal ion is forming 6 co-ordinate bonds.
The water molecule which is bonded to the bottom position in
the diagram is easily replaced by an oxygen molecule (again via
a lone pair on one of the oxygens in O2) - and this is how oxygen
gets carried around the blood by the haemoglobin.
When the oxygen gets to where it is needed, it breaks away
from the haemoglobin which returns to the lungs to get some
more.
You probably know that carbon monoxide is poisonous because
it reacts with haemoglobin. It bonds to the same site that would
otherwise be used by the oxygen - but it forms a very stable
complex. The carbon monoxide doesn't break away again, and
that makes that haemoglobin molecule useless for any further
oxygen transfer.
A hexadentate ligand
A hexadentate ligand has 6 lone pairs of electrons - all of which
can form co-ordinate bonds with the same metal ion. The best
example is EDTA.
EDTA is used as a negative ion - EDTA4-. The diagram shows
the structure of the ion with the important atoms and lone pairs
picked out.
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Note: The abbreviation EDTA comes from an old name for the
parent acid of this - one where each of the negatively charged
oxygens has a hydrogen atom attached. This used to be
called ethylenediaminetetraacetic acid. Nobody ever calls it
anything other than EDTA!
The EDTA ion entirely wraps up a metal ion using all 6 of the
positions that we have seen before. The co-ordination number is
again 6 because of the 6 co-ordinate bonds being formed by the
central metal ion. The diagram below shows this happening with
a copper(II) ion.
Drawing the product of this clearly while showing all the atoms
defeats me completely! Here is a simplified version. Make sure
that you can see how this relates to the full structure above.
The overall charge, of course, comes from the 2+ on the original
copper(II) ion and the 4- on the EDTA4- ion.
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Some simple shapes for complex ions
6-co-ordinated complex ions
These are complex ions in which the central metal ion is forming
six bonds. In the simple cases we are talking about, that means
that it will be attached to six ligands.
These ions have an octahedral shape. Four of the ligands are
in one plane, with the fifth one above the plane, and the sixth
one below the plane.
The diagram shows four fairly random examples of octahedral
ions.
Note: Remember that the ligands attached to a wedge shaped arrow
are coming out of the screen or paper towards you. Those attached to
a dotted arrow are behind the plane of the screen or paper. The two
ligands attached to the ordinary arrows are above and below the
plane of the rest.
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It doesn't matter what the ligands are. If you have six of them,
this is the shape they will take up. Easy!
4-co-ordinated complex ions
These are far less common, and they can take up one of two
different shapes.
Tetrahedral ions
These are the ones you are most likely to need for A' level
purposes in the UK. There are two very similar ions which crop
up commonly at this level: [CuCl4]2- and [CoCl4]2-.
The copper(II) and cobalt(II) ions have four chloride ions bonded
to them rather than six, because the chloride ions are too big to
fit any more around the central metal ion.
That's not very difficult to remember either!
A square planar complex
Occasionally a 4-co-ordinated complex turns out to be square
planar. There's no easy way of predicting that this is going to
happen. The only one you might possibly come across at this
level is cisplatin which is used as an anti-cancer drug.
Cisplatin is a neutral complex, Pt(NH3)2Cl2. It is neutral because
the 2+ charge of the original platinum(II) ion is exactly cancelled
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by the two negative charges supplied by the chloride ions.
The platinum, the two chlorines, and the two nitrogens are all in
the same plane. We will have more to say about cisplatin
immediately below.
The electron pair repulsion theory
The shape of a molecule or ion is governed by the arrangement
of the electron pairs around the central atom. All you need to do
is to work out how many electron pairs there are at the bonding
level, and then arrange them to produce the minimum amount of
repulsion between them. You have to include both bonding pairs
and lone pairs.
How to work out the number of electron pairs
You can do this by drawing dots-and-crosses pictures, or by
working out the structures of the atoms using electrons-in-boxes
and worrying about promotion, hybridisation and so on. But this
is all very tedious! You can get exactly the same information in a
much quicker and easier way for the examples you will meet if
you are doing one of the UK-based exams for 16 - 18 year olds.
Warning: This method won't work without some
modification for many ions containing metals, and no simple
method gives reliable results where the central atom is a
transition metal. The method will, however, cope with all the
substances that you are likely to meet in this section of the
syllabus. When you deal with transition metal chemistry, you
will be expected to know the shapes of some ions formed by
transition metals, but not to work them out. At that point,
learn the ones your syllabus wants you to know.
It is important to know exactly which molecules and ions
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your syllabus expects you to be able to work out the shapes
for in this part of the syllabus. You should also check past
exam papers. If you are working to a UK-based syllabus for
16 - 18 year olds, and haven't got copies of your syllabus
and past papers follow this link to find out how to get them.
First you need to work out how many electrons there are around
the central atom:



Write down the number of electrons in the outer level of
the central atom. That will be the same as the Periodic
Table group number, except in the case of the noble
gases which form compounds, when it will be 8.
Add one electron for each bond being formed. (This
allows for the electrons coming from the other atoms.)
Allow for any ion charge. For example, if the ion has a 1charge, add one more electron. For a 1+ charge, deduct
an electron.
Now work out how many bonding pairs and lone pairs of
electrons there are:


Divide by 2 to find the total number of electron pairs
around the central atom.
Work out how many of these are bonding pairs, and how
many are lone pairs. You know how many bonding pairs
there are because you know how many other atoms are
joined to the central atom (assuming that only single
bonds are formed).
For example, if you have 4 pairs of electrons but only 3
bonds, there must be 1 lone pair as well as the 3 bonding
pairs.
Finally, you have to use this information to work out the shape:

Arrange these electron pairs in space to minimise
repulsions. How this is done will become clear in the
examples which follow.
Don't panic! This is all much easier to do in practice than it
is to describe in a long list like this one!
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Two electron pairs around the central atom
The only simple case of this is beryllium chloride, BeCl2. The
electronegativity difference between beryllium and chlorine isn't
enough to allow the formation of ions.
Beryllium has 2 outer electrons because it is in group 2. It forms
bonds to two chlorines, each of which adds another electron to
the outer level of the beryllium. There is no ionic charge to worry
about, so there are 4 electrons altogether - 2 pairs.
It is forming 2 bonds so there are no lone pairs. The two bonding
pairs arrange themselves at 180° to each other, because that's
as far apart as they can get. The molecule is described as
beinglinear.
Three electron pairs around the central atom
The simple cases of this would be BF3 or BCl3.
Boron is in group 3, so starts off with 3 electrons. It is forming 3
bonds, adding another 3 electrons. There is no charge, so the
total is 6 electrons - in 3 pairs.
Because it is forming 3 bonds there can be no lone pairs. The 3
pairs arrange themselves as far apart as possible. They all lie in
one plane at 120° to each other. The arrangement is
calledtrigonal planar.
In the diagram, the other electrons on the fluorines have been
left out because they are irrelevant.
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Four electron pairs around the central atom
There are lots of examples of this. The simplest is methane,
CH4.
Note: Elsewhere on the site, you will find the shape of
methane worked out in detail using modern bonding theory.
Here we are doing it the quick and easy way!
If you are interested in the bonding in methane you can find
it in the organic section by following this link, or in a page
oncovalent bonding by following this one.
Carbon is in group 4, and so has 4 outer electrons. It is forming
4 bonds to hydrogens, adding another 4 electrons - 8 altogether,
in 4 pairs. Because it is forming 4 bonds, these must all be
bonding pairs.
Four electron pairs arrange themselves in space in what is
called a tetrahedral arrangement. A tetrahedron is a regular
triangularly-based pyramid. The carbon atom would be at the
centre and the hydrogens at the four corners. All the bond
angles are 109.5°.
Note: It is important that you understand the use of various
sorts of line to show the 3-dimensional arrangement of the
bonds. In diagrams of this sort, an ordinary line represents a
bond in the plane of the screen or paper. A dotted line
shows a bond going away from you into the screen or paper.
A wedge shows a bond coming out towards you.
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Other examples with four electron pairs around the central
atom
Ammonia, NH3
Nitrogen is in group 5 and so has 5 outer electrons. Each of the
3 hydrogens is adding another electron to the nitrogen's outer
level, making a total of 8 electrons in 4 pairs. Because the
nitrogen is only forming 3 bonds, one of the pairs must be a lone
pair. The electron pairs arrange themselves in a tetrahedral
fashion as in methane.
In this case, an additional factor comes into play. Lone pairs are
in orbitals that are shorter and rounder than the orbitals that the
bonding pairs occupy. Because of this, there is more repulsion
between a lone pair and a bonding pair than there is between
two bonding pairs.
That forces the bonding pairs together slightly - reducing the
bond angle from 109.5° to 107°. It's not much, but the examiners
will expect you to know it.
Remember this:
Greatest repulsion
lone pair - lone pair
lone pair - bond pair
Least repulsion
bond pair - bond pair
Be very careful when you describe the shape of ammonia.
Although the electron pair arrangement is tetrahedral, when you
describe the shape, you only take notice of the atoms. Ammonia
ispyramidal - like a pyramid with the three hydrogens at the
base and the nitrogen at the top.
Water, H2O
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Following the same logic as before, you will find that the oxygen
has four pairs of electrons, two of which are lone pairs. These
will again take up a tetrahedral arrangement. This time the bond
angle closes slightly more to 104°, because of the repulsion of
the two lone pairs.
The shape isn't described as tetrahedral, because we only "see"
the oxygen and the hydrogens - not the lone pairs. Water is
described as bent or V-shaped.
The ammonium ion, NH4+
The nitrogen has 5 outer electrons, plus another 4 from the four
hydrogens - making a total of 9.
But take care! This is a positive ion. It has a 1+ charge because
it has lost 1 electron. That leaves a total of 8 electrons in the
outer level of the nitrogen. There are therefore 4 pairs, all of
which are bonding because of the four hydrogens.
The ammonium ion has exactly the same shape as methane,
because it has exactly the same electronic arrangement. NH4+ is
tetrahedral.
Note: To simplify diagrams, bonding electrons won't be
shown from now on. Each line, of course, represents a
bonding pair. It is essential, however, to draw lone pairs.
Methane and the ammonium ion are said to
be isoelectronic. Two species (atoms, molecules or ions) are
isoelectronic if they have exactly the same number and
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arrangement of electrons (including the distinction between
bonding pairs and lone pairs).
The hydroxonium ion, H3O+
Oxygen is in group 6 - so has 6 outer electrons. Add 1 for each
hydrogen, giving 9. Take one off for the +1 ion, leaving 8. This
gives 4 pairs, 3 of which are bond pairs. The hydroxonium ion is
isoelectronic with ammonia, and has an identical shape pyramidal.
Five electron pairs around the central atom
A simple example: phosphorus(V) fluoride, PF5
(The argument for phosphorus(V) chloride, PCl5, would be
identical.)
Phosphorus (in group 5) contributes 5 electrons, and the five
fluorines 5 more, giving 10 electrons in 5 pairs around the
central atom. Since the phosphorus is forming five bonds, there
can't be any lone pairs.
The 5 electron pairs take up a shape described as a trigonal
bipyramid - three of the fluorines are in a plane at 120° to each
other; the other two are at right angles to this plane. The trigonal
bipyramid therefore has two different bond angles - 120° and
90°.
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A tricky example, ClF3
Chlorine is in group 7 and so has 7 outer electrons. The three
fluorines contribute one electron each, making a total of 10 - in 5
pairs. The chlorine is forming three bonds - leaving you with 3
bonding pairs and 2 lone pairs, which will arrange themselves
into a trigonal bipyramid.
But don't jump to conclusions. There are actually three different
ways in which you could arrange 3 bonding pairs and 2 lone
pairs into a trigonal bipyramid. The right arrangement will be the
one with the minimum amount of repulsion - and you can't
decide that without first drawing all the possibilities.
These are the only possible arrangements. Anything else you
might think of is simply one of these rotated in space.
We need to work out which of these arrangements has the
minimum amount of repulsion between the various electron
pairs.
A new rule applies in cases like this:
If you have more than four electron pairs arranged
around the central atom, you can ignore
repulsions at angles of greater than 90°.
One of these structures has a fairly obvious large amount of
repulsion.
In this diagram, two lone pairs are at 90° to each other, whereas
in the other two cases they are at more than 90°, and so their
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repulsions can be ignored. ClF3 certainly won't take up this
shape because of the strong lone pair-lone pair repulsion.
To choose between the other two, you need to count up each
sort of repulsion.
In the next structure, each lone pair is at 90° to 3 bond pairs,
and so each lone pair is responsible for 3 lone pair-bond pair
repulsions.
Because of the two lone pairs there are therefore 6 lone pairbond pair repulsions. And that's all. The bond pairs are at an
angle of 120° to each other, and their repulsions can be ignored.
Now consider the final structure.
Each lone pair is at 90° to 2 bond pairs - the ones above and
below the plane. That makes a total of 4 lone pair-bond pair
repulsions - compared with 6 of these relatively strong
repulsions in the last structure. The other fluorine (the one in the
plane) is 120° away, and feels negligible repulsion from the lone
pairs.
The bond to the fluorine in the plane is at 90° to the bonds
above and below the plane, so there are a total of 2 bond pairbond pair repulsions.
The structure with the minimum amount of repulsion is therefore
this last one, because bond pair-bond pair repulsion is less than
lone pair-bond pair repulsion. ClF3 is described as T-shaped.
Warning! If your syllabus expects you to discuss examples
with more than 4 pairs of electrons around the central atom,
check past exam papers to see if nasty questions like this
one involving ClF3 ever come up. If so, don't leave this
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example until you are sure that you understand it. It is by far
the most complicated one on this page.
Six electron pairs around the central atom
A simple example: SF6
6 electrons in the outer level of the sulphur, plus 1 each from the
six fluorines, makes a total of 12 - in 6 pairs. Because the
sulphur is forming 6 bonds, these are all bond pairs. They
arrange themselves entirely at 90°, in a shape described
as octahedral.
Two slightly more difficult examples
XeF4
Xenon forms a range of compounds, mainly with fluorine or
oxygen, and this is a typical one. Xenon has 8 outer electrons,
plus 1 from each fluorine - making 12 altogether, in 6 pairs.
There will be 4 bonding pairs (because of the four fluorines) and
2 lone pairs.
There are two possible structures, but in one of them the lone
pairs would be at 90°. Instead, they go opposite each other.
XeF4 is described as square planar.
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