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The d-block transition elements can be defined as
Metallic elements that have an incomplete d subshell
in at least one of their ions.
Studying the electronic configurations of
the first transition series from scandium
to zinc highlights reveals some important
points regarding the formal definition
shown in the box above
When electrons are filling atomic orbitals, the 4s subshell is filled before the 3d subshell.
When the subshells have no electrons in them the 4s has lower energy than the 3d, even
although the 4s subshell is further from the nucleus.
However, once the electrons are actually in their orbitals, the energy order changes - this
occurs due to the presence of electrons in the 3d subshell, these repel the 4s electrons
even further from the nucleus. Therefore, the 4s electrons are pushed to a higher energy
level, higher than 3d. Consequently, when transition atoms become ions, the electrons from
the 4s subshell are lost before the electrons in the 3d subshell.
In all the chemistry of the transition elements, the 4s subshell behaves as the outermost,
highest energy subshell.
The reversed order of the 3d and 4s subshells 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.
transition metals
The effect of this can be seen in the following examples of atoms and ions
Cobalt Co Co2+
[Ar] 4s2 3d7 [Ar] 3d7
Chromium Cr Cr3+
[Ar] 4s1 3d5 [Ar] 3d3
Iron Fe Fe3+
[Ar] 4s2 3d6 [Ar] 3d5
Note that all three ions shown above have an incomplete d subshell and so fit the definition of
what a transition metal is.
It appears the filling of the d orbitals follows the aufbau principle with the exceptions of
chromium and copper. These exceptions are due to a special stability associated with all
the d orbitals being half filled or completely filled.
Chromium has the electronic configuration [Ar] 4s1 3d5 NOT [Ar] 4s2 3d4
as expected from the aufbau principle.
Copper has the electronic configuration [Ar] 4s1 3d10 NOT [Ar] 4s2 3d9
as expected from the aufbau principle.
The electronic configuration for a scandium atom is
In all its chemistry scandium forms a single ion – the scandium(III) ion – Sc3+
The electronic configuration for the Sc3+ ion is
The electronic configuration for a zinc atom is
In all its chemistry zinc forms a single ion – the zinc(II) ion – Zn2+
The electronic configuration for the Zn2+ ion is
It is clear that scandium and zinc DO NOT comply with the definition of what a transition
metal is because in the only ions these elements form there is NOT an incomplete
d subshell
Scandium and zinc do not have typical transition metal properties {which will be discussed
in the remainder of this section of work} because of this.
Transition metals exhibit the following properties .
The reason behind these properties is largely due
to the electrons in the d subshell of the
transition metals.
An element is said to be in a particular oxidation state when it has a specific oxidation number.
The oxidation number of a species is related to the number of electrons the species has lost or
gained.
The changes in oxidation state during a REDOX reaction also helps us decide if a change is
oxidation or reduction.
Oxidation numbers for a particular atom or ion can be worked out by following the rules set out
below.
The oxidation number of an UNCOMBINED element is ZERO
{the atom or molecule has not lost or gained any electrons
For monatomic ions the oxidation number is equal to the
CHARGE ON THE ION.
E.g. Mg 2+ oxidation number = +2 Cl - oxidation number = -1
In nearly all its compound oxygen has an oxidation number of -2
In nearly all its compound hydrogen has an oxidation number of +1.
Group one metals are always +1 and Group 2 metals are always +2
In polyatomic ions, the sum of all the oxidation numbers for all the
atoms is equal to the overall charge on the ion. For neutral
compounds the sum is equal to zero. (see examples on page 6)
2.
3. What is the oxidation number of manganese in potassium permanganate ?
In this example there is no formula given - so you need to work it out
KMnO4 Overall charge is zero
4.What is the oxidation number and the electronic configuration of vanadium in VO2+ ?
Vanadium atom electron configuration is,
1s2 2s2 2p6 3s2 3p6 3d3 4s2
Four electrons removed
2 from 4s and 2 from 3d
V in VO2+ is 1s2 2s2 2p6 3s2 3p6 3d1
[CrCl6]3- ?
{Overall charge
is ZERO}
1. What is the oxidation number of
nitrogen in nitric acid, HNO3
{Overall charge on
the ion is -3}
You should already know that oxidation is defined as a loss of electrons and reduction is a
gain of electrons (OILRIG). It is easy to interpret the following reactions as oxidation and
reduction if we consider the oxidation numbers of the species involved in the reaction.
Fe2+ Fe3+ + e- Cu2+ + 2e- Cu
This is also easily identified in the balanced redox equation.
2Fe2+ + Cu2+ 2Fe3+ + Cu
Even in less obvious situations oxidation numbers can reveal oxidation or reduction.
For example is the conversion of sulfite to sulfate oxidation or reduction?
SO32- SO4
2-
Work out the oxidation number of sulfur in each ion.
Sulfite Sulfate Ox. No. of S + (3 x Ox. No. of O) = -2 Ox. No. of S + (4 x Ox. No. of O) = -2
Ox. No. of S + (3 x -2) = -2 Ox. No. of S + (4 x -2) = -2
Ox. No. of S = -2+6 Ox. No. of S = -2+8
Ox. No. of S = +4 Ox. No. of S = +6
The change involves an increase in oxidation number from +4 to +6 and so is an oxidation.
Oxidation No
+3 +2
Oxidation No
0 +2
Oxidation as the oxidation
number has INCREASED
Reduction as the oxidation
number has DECREASED
1. Calculate the oxidation of the stated element in each of
the following
a. S in H2SO4 b. Cr in Cr2O72- c. Fe in FeCl42-
d. C in CO32- e. Cu in NaCuCl2 f. Cr in Cr2O3 g. N in NO3
-
h. Fe in K2FeO4 i. I in IO3- j. V in VO3-
2. The equation represents the reaction between permanganate ions and
vanadium (II) ions
a. Use oxidation number to prove that the permanganate ion has been reduced.
b. Which species is the reducing agent?
c. Write the electronic configurations, in terms of s, p and d orbitals of the two vanadium ions
shown in the equation
d. Explain why the V5+ ion is likely to be more stable either the V2+ or the V3+ ions.
e. Why do transition metals like vanadium have multiple oxidation states?
3. The equation shows the reaction of magnesium with hydrochloric acid. The oxidation
numbers of all the reactant and products are also shown.
a. Explain, using the oxidation numbers shown, why this is a redox reaction.
b. (i) Write in the correct oxidation numbers for species a to i
(ii) Is this a redox reaction?
a c
b d
e
f
g
h
i
A very important property of transition metals is their ability to form complexes, often called
coordination compounds.
A complex consists of a central metal ion
surrounded by ligands.
The diagram below shows three
common structures that complexes form.
A covalent bond consists of a shared pair of electrons.
Usually each atom will provide one unpaired electron to
the bond as shown in the diagram opposite.
In the drawing of ammonia, NH3, shown below, each cross
represents a hydrogen electron and each dot represents
a nitrogen electron.
The electronic configuration of a nitrogen atom is 1s2 2s2 2p3
Notice that the nitrogen atom has a lone pair of electrons (the 2s electrons) which are not
involved in bonding to any of the hydrogen atoms. Occasionally these electrons can be used to
from a covalent bond. When this happens a dative bond is formed.
In dative bonding both electrons of the shared pair come from the same atom.
Ligands are electron donors and may be
negative ions or molecules with
non-bonding pairs of electrons (lone pairs).
octahedral
tetrahedral
square planar
When ammonia reacts with an acid, a dative bond is formed between the nitrogen atom in
ammonia and the hydrogen ion of the acid
Once a dative covalent bond has formed it is indistinguishable from any other covalent bond.
This means that, once they are formed, all the covalent bonds in the ammonium ion are
identical.
Ligands are either negative ions or neutral molecules with at least one lone pair of
electrons. Positive ions cannot be classed as ligands. If a ligand uses just one atom to bind
to the central metal it is described as monodentate. If it uses two atoms it is described as
bidentate. EDTA is a hexadentate ligand because it uses six atoms to bind to the central
metal.
Fluoride F-
Chloride Cl-
Cyanide CN-
Water H2O
Ammonia NH3
lone pair on nitrogen
forms this dative
covalent bond.
+
monodentate
negative ions
neutral molecules with
lone pairs of electrons
bidentate
The oxalate ion uses two of its
oxygen atoms and the
ethylenediamine molecule (EN)
uses the two nitrogen atoms to
bind to the metal.
Ethylenediaminetetraacetic acid – EDTA
EDTA is a hexadentate ligand – it uses six atoms (shown in the diagram below) to attach
itself to the central metal atom/ion.
This structure is actually the four
negative ion of EDTA. The acid itself
will have four hydrogen atoms
(one on each of the negative charges
shown on the oxygen atoms).
EDTA is classed as a chelating agent
(from the Greek word chela = claw).
The diagram shows the structure of a cobalt(III) edta complex.
As this ligand binds so strongly to metal ions it is used
to “trap” transition metal ion impurities in many household
products to prevent these ions from hindering the efficacy
of the product. Things like soap, beer and mayonnaise all
have Sodium EDTA added to them.
The total number of bonds from the ligands to the
central transition metal atom/ion is known as the
co-ordination number
EDTA forms an octahedral complex with co-ordination number = 6.
Co-ordination number and oxidation number are often confused.
Try not to mix them up.
co-ordination
number = 4
co-ordination
number = 6
co-ordination
number = 2
co-ordination
number = 4
= 4
1. Ammonia is a monodentate ligand whereas EDTA is a hexadentate ligand. Explain the
difference between these two types of ligand.
2. Dimethylgloxime acts a ligand when it reacts with nickel(II) ions forming the complex
nickel(II) dimethylglyoximate.
a. What structural feature does dimethylglyoxime have which allows it to behave as a
ligand?
b. How many moles of dimethylglyoxime react with one mole of nickel(II) ions?
c. What is the co-ordination number of nickel in the complex?
d. Explain why dimethylgloxime is classed as a bidentate ligand.
3. The complex ion hexaaquacopper(II), [Cu(H2O)6]2+, has an octahedral structure.
a. The oxidation number of the copper ion in this complex is +2. Write the electronic
configuration of this ion in terms of s,p and d orbitals.
b. How many unpaired electron does this copper ion have?
c. Draw a diagram which shows the shape of this complex ion. (Your diagram must show
the Cu2+ ion and all six H2O molecules)
4. The tetrachlorcuprate(II) ion has the structure shown.
a. What is the co-ordination of copper in this complex ion?
b. What name is given to the shape of this complex ion?
c. What is the formula of potassium tetrachlorocuprate(II)?
dimethylglyoxime nickel(II) dimethylglyoximate
At first sight the names and formulae of transition metal complex ions can seem daunting.
Following the rules set out below should help you to name a complex ion given its formula or
write the formula of a complex ion given its name.
Ligand (negative ions) Name in complex ion
bromide, Br- bromo
chloride, Cl- chloro
cyanide, CN- cyano
hydroxide, OH- hydroxo
oxalate, C2O42- oxalato
Ligand (neutral molecule) Name in complex ion
ammonia, NH3 ammine
carbon monoxide, CO carbonyl
water, H2O aqua
Transition metal Name in complex ion
vanadium vanadate
titanium titanate
chromium chromate
manganese manganate
iron ferrate
cobalt cobaltate
nickel nickelate
copper cuprate
zinc zincate
The name of the complex ion consists of TWO PARTS written as one word. This is placed
in square brackets [ -----]. The ligands are named first (if there is more than one ligand
they are placed in alphabetical order) and the central transition metal ion completes the
name.
Ligand part: The ligand name is preceded by a Greek prefix showing the number
of ligands (di-, tri-, etc).
Metal part: The metal name is followed by the
oxidation number in Roman numerals – this is put in
round brackets - (I), (II), (III), (IV), etc
The overall charge on the complex part will be equal
to the sum of the oxidation number and the total
charge on the ligands.
If the complex ion is NEGATIVE the transition metal
name will end in ATE.
When a complex ion combines with oppositely charged ions a coordination compound is formed.
The diagram shown below shows the coordination compound hexaamminecobalt(III) chloride.
Square Brackets [ ] are used to indicate all of the
atomic composition of the coordinate complex:
the central metal atom and the ligands.
If the complex is a positive ion it will appear first
in the formula and the name. If the complex is a
negative ion it will appear last in the formula and the name.
Species outside of the [ ] are not coordinated to the
transition metal but are required to maintain a charge
balance.
Another example -------
1. One of the following nitrogen species is incapable of
acting as a ligand. Identify this species and explain your answer.
NH4+ , NH3 , NH2
-
2. Give the oxidation number and coordination number of the
transition metal in the following complexes.
a. [Co(NH3)4 Cl2]Cl b. K2[Ni(CN)4]
c. Na3[Fe(C2O4)3] d. [Cr(NH3)5SO4]Cl
3. Write the formula for
a. potassium tetrachloroplatinate(II) b. tetraamminediaquairon(II)
c. sodium tetrachlorocobaltate(II) d. tetraaquadichlorochromium(III) chloride
e. pentacarbonyliron(0) f. hexaammineiron(III) nitrate
g. potassium trioxalatoferrate(III) h. hexafluoromanganate(II)
4. Write the electronic configuration, in terms of s,p and d orbitals of the transition
metal in the following complexes
a. [Mn(NH3)6]SO4 b. (NH4)2[Cu(Cl)4]
5. A coordination compound has the formula [X] Cl2. The complex ion is composed of a
central nickel(II) ion, 2 ammonia ligands and 4 water ligands.
a. Write the formula of the complex ion [X]
b. Draw a diagram to illustrate the shape of this complex ion.
c. How many unpaired 3d electrons are there in the nickel(II) ion?
d. What structural feature allows water molecules to act as ligands?
e. The first ionisation energy of nickel is 737 kJ mol-1.
(i) Write an equation which represents the first ionisation of nickel.
(ii) Calculate the wavelength of light associated with this ionisation energy.
The three primary colours of visible light are RED, GREEN and BLUE. When these three
colours mix white light will result. The colour wheels show the colours produced if two
primary colours are mixed. When white light shines on certain chemicals they might absorb
some of the visible light (light of a particular wavelength/frequency) – the colour we see is
the white light minus any absorbed light. This TRANSMITTED light is called the
COMPLEMENTARY colour.
Colour of light absorbed Colours Transmitted Colour observed
Red Blue and green Cyan
Blue Red and green Yellow
Green Blue and red Magenta
If a substance absorbs green and blue light it will appear red – red is the complementary
colour of cyan which is a mixture of red and blue light.
If a substance absorbs only blue light, it will transmit both red light and green light – this
substance will appear yellow. Yellow is the complementary colour of blue.
Colorimetry (see researching chemistry notes – manganese in steel experiment) is an analytical
technique based on complementary colours. Many chemical compounds absorb either
infra – red, visible or ultra – violet radiation. The technique of analysing the amount of light
absorbed is generally known as Absorption Spectroscopy. Colorimetry is very useful in
determining the concentration of a coloured substance in solution. Generally a more
concentrated solution will absorb more light than a dilute solution and be darker in colour.
Many, but not all, transition metal complexes are coloured. Like most transition metal
chemistry the reason for this is largely due to the 3d subshell. The following explanation is
restricted to octahedral complexes in which the transition metal ion has an incomplete 3d
subshell containing at least one 3d electron. This theory is usually referred to as the
“Crystal field theory”
1. In an isolated transition metal atom all five 3d orbitals are degenerate – have equal
energy.
2. As a result of ligands approaching and bonding to the metal, the five 3d orbitals
are no longer degenerate. This is referred to as SPLITTING.
The diagram above shows that two of the 3d
orbitals now have a higher energy value while three
have a lower energy value.
3. This happens because as bonds form due to the
attraction of the electrons on the ligand for the
charge on the metal ion, the electrons belonging
to the ligand repel electrons in the 3d orbitals of
the transition metal.
This electron - electron repulsion is
greater in the two d orbitals which
lie along the axes the ligand
approaches on.
This causes these two orbitals to
have raised energy values.
4. As the d orbitals now have different energy values electrons present in any of the
three lower energy d orbitals can absorb energy and get promoted to one of the two
higher energy d orbitals. This is known as a d to d transition.
5. If the energy absorbed is equal to a wavelength of light in the visible spectrum the
compound will transmit the complementary colour of the light absorbed.
The colour of light absorbed will depend on how much the d orbitals have been split. A
large difference in energy will cause the compound to absorb blue light while a small
difference will absorb red light.
The extent to which the d orbitals split depends largely on three factors;
In this case the d orbitals are split to a small degree.
When compared to case B it will take LESS
energy to promote an electron to the higher
orbital and light of a lower frequency will
be required to do this
In this case the d orbitals are split to a large degree. When compared to case A it
will take MORE energy to promote an electron to the higher orbital and light of a
higher frequency will be required to do this.
The diagram above shows the “POWER” a ligand has to split d orbitals with Iodide (I-)
being the weakest.
1. The colour of [Ti(H2O)6]3+ hexaaquatitanium (III) [Ar] 3d1
The water ligands split the d orbitals.
Visible light shines on the compound.
The green component of visible light is
absorbed causing the electron to move
from a low energy d orbital to a high
energy d orbital – a d to d transition.
The red and blue light is transmitted and
the compound appears magenta.
2. The colours of Ni2+ complex ions
These nickel(II) compounds have different colours because each one has a different
ligand.
Watch the video at http://www.youtube.com/watch?v=hDt2OUnOcug
Don’t forget
1. This theory only works if the transition
metal in the compound has between one and
nine d electrons.
Ions that have no d electrons (d0) or ions
that have a complete d subshell (d10) cannot
have d to d transitions.
2. This explanation of colour IS NOT THE
SAME theory that applies to atomic
emission where the light comes from
electrons returning to lower energy levels
after they have been excited by a flame.
Do not get these two colour theories mixed
up.
hexaamminenickel(II)
deep blue
triethylenediaminenickel(II)
violet
hexaaquanickel(II)
green
tetrachloronickelate(II)
yellow
Catalysis by transition metals
Transition metals and their compounds can acts as catalysts
It is believed that the presence of unpaired d electrons or unfilled d orbitals allows
intermediate complexes to form, providing reaction pathways with lower activation
energies compared to the uncatalysed reaction. The variability of oxidation states of
transition metals is an important factor.
Homogeneous and heterogeneous catalysts should be explained in terms of changing
oxidation states with the formation of intermediate complexes and the
adsorption of reactive molecules onto active sites respectively.
1. Three complex ions of cobalt (III) absorb light at wavelengths at 290 nm, 440 nm and 770
nm. The ions have the formulae;
[Co(CN)6]3- A
[CoF6]3-, B
[Co(NH3)6]3+. C
a. Use the spectrochemical series on page 19 to match each ion with the wavelength of
light it is most likely to absorb.
b. Use your answer to a. to help predict the colour of ion B
c. Write the formula of potassium hexacyanocobaltate(III)
d. Calculate the splitting energy, in kJ mol-1, associated with a wavelength of 290nm.
2. The graph shows the absorption spectrum of the hexaaquacopper(II) ion.
a. Write the formula for this complex ion.
b. Write the electronic configuration in terms of s,p and d orbitals for copper in this
complex ion.
c. Explain why this complex ion absorbs some visible light.
d. Predict the colour of a solution of this complex ion.
e. Suggest what would happen to the maximum absorbed wavelength if the water ligands
were replaced by ammonia.
A list of “learning outcomes” for the topic is shown below. When the topic is
complete you should review each learning outcome.
Your teacher will collect your completed notes, mark them,
and then decide if any revision work is necessary.
State that transition metals have at least one ion with an incomplete 3d subshell.
Scandium and zinc are not transition metals.
When atomic orbitals fill the 4s subshell fills before the 3d subshell.
When transition metals form positive ions electrons are lost from the 4s subshell
before the 3d subshell .
Half filled (d5) and totally filled (d10) 3d subshells have a special stability and as a
result of this chromium atoms and copper atoms both have a 4s1 electronic
configuration.
State that the oxidation number of ion gives an indication of the number of
electrons the ion has lost or gained.
Be able to calculate the oxidation number of an atom/ion in a substance.
State that the oxidation state of an element is zero.
State that in oxidation the oxidation state increases and in reduction the
oxidation state decreases.
State that a ligand is a neutral molecule or negative ion which is able to form a
dative covalent bond with a transition metal.
State that a dative covalent bond is a covalent bond in which the shared pair of
electrons making the bond both originate from the same atom.
Sate that the coordination number in a complex is the number of bonds the
central metal atom/ion makes with the ligand.
Need Help
Understand
Revise
Be able to draw tetrahedral and octahedral diagrams which convey the shapes of
complexes.
Be able to name transition metal complexes and complex ions.
Be able to identify the complementary colours for the visible spectrum.
Be able to explain why octahedral complexes have a variety of colours.
State that changing the ligand in a complex usually changes the colour of the
complex.
Be able to interpret a visible absorption spectrum to deduce the colour of a
compound.
I have discussed the learning outcomes with my teacher.
My work has been marked by my teacher.
Teacher Comments.
Date. __________________________________
Pupil signature. __________________________
Teacher signature. _______________________