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Inorganic Chemistry (AH) – Slides
Lesson 1
ELECTROMAGNETIC SPECTRUM
The relationship between these quantities is given by c=fλ where c represents the speed of light, f
represents the frequency of the wave (s-1) and λ represents the wavelength (m). When radiation in
the visible part of the electromagnetic spectrum is being described, it is common to specify
wavelength in terms of nanometres (nm).
One nanometre is 10-9
Velocity of electromagnetic radiation in a vacuum is 3 x108 ms-1
velocity
c
=
frequency
=
f
x
wavelength
x

and so
c
f
=

The energy associated with a single photon is given by E=hf where ‘h’ represents Planck’s
constant. For Chemists, it is more convenient to express the energy associated with a mole
of photons which is given by E=Lhf giving the energy in J mol-1 where L represents
Avogadro’s constant.
The relationship between energy and frequency is
E = hf
where J is Planck’s constant and has a value of 6.63 x 10-34 Js. In chemistry energy values
are normally expressed in KJmol-1, so to convert from J to kJ, must divide by 1000.
To obtain moll-1 in units of KJmol-1, we change relationship to
E = Lhf
where L is Avogadro’s Constant = 6.02 x 1023
Therefore E = Lf gives the energy value in J, E = Lhf gives the energy in Jmol-1 and
E=Lhf/1000 gives the energy value in KJmol-1
It is often useful to relate energy, in KJmol-1 to wavelength, as f = c/
𝐸=
𝐿ℎ𝑐
1000. 𝜆
E = Lhc
1000
Note – Put that wavelength into the equation in metres!
The electromagnetic spectrum is the full range of electromagnetic radiation which travels
through space with a constant velocity of 3 x 108 ms-1. This radiation can be described in
terms of wave motion.
SPECTRA (CONTINOUS)
When a beam of pure white light is passed through a prism a continuous spectrum is seen
(all the colours of the rainbow).
Above shows all the colours in the visible part of the spectrum.
SPECTRA (EMISSION)
Atomic emission spectra are produced when atoms of elements, usually in their gaseous
state, are excited by heat or electrical discharge so that they emit radiation. The radiation
emitted is passed through a prism and the spectrum obtained is a series of sharp coloured
lines on a black background.
When energy is transferred to atoms, electrons within the atoms may be promoted to higher
energy levels. To allow the electrons to return to their original levels, energy must be lost
from the atom. This energy is released in the form of a photon.
An atom can be considered as emitting a photon of light energy when an electron moves
from a higher energy level to a lower energy level. Each element produces a unique pattern
of frequencies of radiation in its emission spectrum.
Emission Spectra have coloured lines on a dark background
Visible part of spectrum from 400 to 700nm wavelength
ATOMIC EMISSION SPECTRUM
Examination of the atomic emission spectrum of hydrogen shows that this consists of a
number of lines (a line spectrum) of very precise frequency, corresponding to precise
amounts of energy.
The lines in the visible part of the spectrum (the Balmer series) are produced by electrons
falling back down to the second energy level from higher levels within the atom. The lines
in the ultraviolet part of the spectrum (the Lyman series) are produced by electrons falling
back down to the first energy level (ground state) and the lines in the infra red part of the
spectrum (the Paschen series) are produced by electrons falling back down to the third &
fourth energy levels.
HYDROGEN SPECTRUM
The hydrogen spectrum shown below consists of the three series of lines, the Lyman
series in the Ultraviolet (higher energy) region of the spectrum, the Balmer series in the
visible region and the Paschen series in the Infra Red (lower energy) region.
POINTS TO NOTE
1. The spectrum appears as lines at very precise frequencies which indicates that the
energy levels are themselves fixed – the electrons have fixed energies – the energy of
the electrons are said to be quantised.
2. The fact that the lines in all the series (UV, visible & IR) all converge towards the violet
(high energy) end of the spectrum shows that the energy levels come closer together
until they converge. i.e. energy levels further from the nucleus get closer together.
3. By determining the wavelength at which the lines converge in the ultraviolet series of
lines in a hydrogen spectrum we can determine the amount of energy given out when
an electron falls from the outermost energy level to the first energy level (ground state).
This is, therefore, also the amount of energy that has to be supplied to remove an
electron from an isolated, gaseous atom – the (1st) ionisation energy of hydrogen.
Q1) Calculate the energy, in kJ mol –1 ,
corresponding to a wavelength of 620 nm
Note! Put the wavelength in as metres please, so you do have to multiple the
nanometre wavelength by 10-9
Answer
 = 620 nm = 620  10–9 m
Lhc 6.02 10 23  6.63 10 34  3.00 108
E

λ
620 10  9
= 193125 J mol
–1
= 193.1 kJ mol
–1
(correct)
Q2) Calculate the ionisation energy for hydrogen if the
wavelength of the line at the convergence limit is 91.2 nm
(in chemistry, the convergence limit is the short
wavelength limit of spectral lines)
Note! Put the wavelength in as metres please, so you do have to multiple the
nanometre wavelength by 10-9
Answer
Lhc 6.02 10 23  6.63 10 34  3.00 108 10 3
E

λ
91.2 10  9
= 1312.9 kJ mol –1
Q3) The bond enthalpy of a Cl—Cl bond is 243 kJ
–1
mol . Calculate the maximum wavelength of light
that would break one mole of these bonds to form
individual chlorine atoms
Answer
E
Lhc

and so,

Lhc
E
34
6.02 10  6.63 10  3.00 10
λ
3
243 10
23
= 4.93  10–7 m
= 493 nm
8
Lesson 2
Electrons within atoms are said to be quantised i.e. can only possess fixed amounts of
energy called quanta – as a result, the electrons can be defined in terms of quantum
numbers.
Within the atom, the electrons behave as waves – different shapes and sizes of these
waves are possible around the nucleus. These are known as orbitals.
The subshell has the lowest energy, then p, then d and so on. Each type of subshell
contains one or more orbitals.
An orbital can hold a maximum of 2 electrons
ELECTRONS and QUANTUM NUMBERS
The energy possessed by an electron in an atom can be defined in terms of 4 quantum
numbers. These are used to describe the movement and trajectories of each electron in an
atom. Each electron in an atom has a unique set of all four quantum numbers i.e. as
according to the Pauli Exclusion Principle – no two electrons can share the combination of
all four quantum numbers.
Quantum numbers are important as they can be used to determine the electronic
configuration of an atom and the probable location of an atom’s electrons.
Remember that quantum numbers designate specific shells, subshells, orbitals and spins
of electrons i.e. the characteristics of an electron in an atom
The principle quantum number n describes the energy of the electron and the most
probable distance of the electron from the nucleus i.e. it refers to the size of the orbital and
the energy level of an electron it is placed in.
The orbital angular momentum quantum number (l) described the shape of the orbital.
The magnetic quantum number (m) describes the energy levels in the subshell and the
spin quantum number (s) describes the spin on the electron i.e. which can be either up or
down.
Principal Quantum Number (n)
This quantum number (n) designates the principle electron shell. Because this describes
the most probable distance from the nucleus, the larger then value of n, then the further
the electrons are from the nucleus, the larger the size of the orbital and the larger the size
of the atom is.
n=1 designates the first principal shell (the innermost shell) which is also the Ground State
i.e. the lowest energy state. This explains why n cannot be zero as there exists no atoms
with zero or a negative amount of energy levels/principle shells.
Remember that electrons can emit light as they jump to lower principle shells.
The principle quantum number, symbol n, can be any positive integral number i.e.
1,2,3,4…..n. The principle quantum number tells us which shell or energy level the
electron is in.
Closer examination of emission spectra under high resolution shows that the lines are
often not single lines but doublets, triplets etc. This suggests that the electron shells are
further sub-divided into sub-shells. These are defined by a further quantum number,
symbol l.
Angular Momentum Quantum Number (l)
The Angular momentum quantum number determines the shape of the orbital and
therefore the angular distribution. Each value of l indicates a specific s,p, d, f subshell
(each unique in shape.)
The values of l are related to those of the principal quantum number n. For any given
value of n, l may take the value of 0, 1, 2…..(n-1). To avoid confusion the values of l
corresponding to 0, 1, 2 and 3 are given the letters s, p, d and f.
s, p, d and f come from the old spectroscopic terms of sharp, principal,
diffuse and fundamental
Magnetic Quantum Number (m)
The magnetic quantum number m determines the number of orbitals and their
orientation within a subshell. Consequently, its value depends on the angular
momentum quantum number l. Given a certain value for l, ml is an interval ranging
from –l to +l, so it can be zero.
Electron Spin Quantum Number (s)
Unlike n, l and m, the electron spin quantum number does not depend on
another quantum number.
It designates the direction of electron spin and may have a spin of +1/2,
represented by an upward pointing arrow or -1/2 represented by a
downwards pointing arrow.
This means when S is positive, the electron has an upward spin which can
be referred to as “spin up”. When it is negative, the electron has a
downward spin, so it is “spin down.”
The significance of the electron spin quantum number is to determine the
atom’s ability to generate a magnetic field or not.
(s-orbitals)
1s
2s
3s
The shape of the orbital is governed by the value of the 2 nd quantum number, l. All s
orbitals are spherical in shape – the diameter increasing as the value of n increases.
Outside the boundary represented by the surface of the sphere, the probability of finding
an electron is low (although it can never be zero)!
Note
The number of values of the orbital angular momentum quantum number l can also be
used to identify the number of subshells in a principal electron shell.
i.e. When n =1, l = 0 (takes on one value and there can only be one subshell)
When n = 2, l = 0,1 (takes on two values thus there are two possible
subshells)
When n = 3, l = 0,1,2 (takes on three values and there are three possible
subshells)
Note
From above, it can be seen that the value of n is equal to the
number of subshells in a principal electronic structure
i.e. Principal shell with n = 1, has one subshell
Principal shell with n = 2, has two subshells
Principal shell with n = 3, has three subshells
To identify which type of subshells n has, these subshells have been assigned letter
names. The value of l determines the name of the subshell.
Name of subshell
Value of l
s subshell
0
p subshell
1
d subshell
2
f subshell
3
Note
The number of orbitals in a subshell is equal to the number of
values of the magnetic quantum number ml takes on.
Helpful equation to determine the number of orbitals in a subshell
is 2l + 1
s orbitals
p orbitals
d orbitals
f orbitals
l
0
1
2
3
ml
0
-1,0,+1
-2,-1,0,+1,+2
-3,-2,-1,0,+1,+2,+3
1
3
5
7
The last line shows the number of orbitals in a designated subshell
(p-orbitals)
Each of the p orbitals, unlike the s orbitals, are not spherical in shape but are dumb-bell
shaped and they lie along the x, y and z axes as shown below.
For the p orbitals (l = 1) there are three possible values for m, namely –1, 0, +1. This
gives rise to three p-orbitals which have equal energy in an isolated atom. These orbitals
of equal energy are said to be degenerate. The three p-orbitals are arranged in space
along the three mutually perpendicular axes x, y and z. The value of m governs the
orientation in space, it has no effect on the energy of the orbital.
(d orbitals)
With the d-orbitals (l = 2) there are 5 possible values of m (-2, -1, 0, +1, +2) and so there
are 5 d-orbitals. Like the p-orbitals, the d-orbitals are all of equal energy (they are
degenerate) in an isolated atom.
Note – dz2 has the orbital on the z-axis and dx2-y2 on the x and y axes, the three
others have their orbitals between the respective axes.
SUMMARY OF 4 QUANTUM NUMBERS
Quantum
Name
Symbol
Meaning
Value
Number
1st
Principal
n
Energy level
1, 2, 3…..
2nd
Angular momentum
l
Shape (s, p, d, f)
0, 1, 2…..(n-1)
3rd
Magnetic
Orientation in
- l…….0……+ l
m
space
4th
Spin
s
Electron spin
-½,+½
PAULI EXCLUSION PRINCIPLE
The Pauli exclusion principle states that no two electrons in the same atom can have the
same set of all 4 quantum numbers. This is important in two ways:
1.
The total number of electrons which may occupy any one orbital is two.
2.
If there are two electrons in an orbital then the spins must be opposed.
Spin Quantum Numbers (s) have a value of +1/2 and -1/2
Degenerate orbitals have the same energy
Remember that the 4S orbital always has a lower energy than the 3d orbital, so fill up first
and will be shown by the Aufbau Principle.
SHOWING ELECTRON ARRANGEMENTS
There are a number of ways in which we can express the various orbitals in atoms. Using
the hydrogen atom we can see that the single electron will occupy the lowest energy level.
This can be represented as H = 1s1.
Another way is by using the notation where an orbital is represented by a box and each
electron by an arrow. We can represent the electron configuration of helium as He
s=+½
s=-½
1s2
The significance of one arrow pointing up and the other down is to show the opposed
spins of the electrons.
AUFBAU PRINCIPLE
Spectroscopic data gives the following arrangement of the energies of the orbitals.
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p6 7s2 5f14 6d10 7p6
The filling order can be remedied by using the diagram shown below.
Practice Drawing this Aufbau Triangle now
HUND’S RULE OF MAXIMUM MULTIPLICITY
When the situation is reached that there is more than one degenerate orbital available for
the electrons, it is necessary to apply Hund’s rule of maximum multiplicity which states
that “when two electrons occupy degenerate orbitals they do so in such a way as to
maximise the number of parallel spins” – i.e. electrons do not pair together until they have
to.
Thus for a nitrogen atom containing a total of 7 electrons, the electron configuration can be
written in two ways;
N = 1s2 2s2 2p3
or
N
Note – in reality the last electron box notation above shows an empty orbital, but if it
was to be empty, then it would not exist.
The periodic table of the elements
Element
H
He
Li
Be
B
C
N
O
F
Ne
Na
Mg
Al
Electron arrangement
1s1
1s2
1s22s1
1s22s2
1s22s22p1
1s22s22p2
1s22s22p3
1s22s22p4
1s22s22p5
1s22s22p6
1s22s22p63s1
1s22s22p63s2
1s22s22p63s23p1
Element
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Si
1s22s22p63s23p2
Ge
P
1s22s22p63s23p3
As
S
1s22s22p63s23p4
Se
Cl
1s22s22p63s23p5
Br
Ar
1s22s22p63s23p6
Kr
Electron arrangement
1s22s22p63s23p64s1
1s22s22p63s23p64s2
1s22s22p63s23p64s23d1
1s22s22p63s23p64s23d2
1s22s22p63s23p64s23d3
1s22s22p63s23p64s13d5
1s22s22p63s23p64s23d5
1s22s22p63s23p64s23d6
1s22s22p63s23p64s23d7
1s22s22p63s23p64s23d8
1s22s22p63s23p64s13d10
1s22s22p63s23p64s23d10
1s22s22p63s23p64s23d10
4p1
1s22s22p63s23p64s23d10
4p2
1s22s22p63s23p64s23d10
4p3
1s22s22p63s23p64s23d10
4p4
1s22s22p63s23p64s23d10
4p5
1s22s22p63s23p64s23d10
4p6
Present day periodic table
Trends in the first ionisation values
Bring in trends across period here for last questions and stress the decrease in ionisation
energy in the graph above.
Be and B
N and O
Mg and Al
P and S for the decreases above
Mg(g)
Mg+(g) + e-
H = (+) 744 kJ.mol-1
Mg+(g)
Mg2+(g) + e-
H = (+) 1460 kJ.mol-1
Mg2+ (g)
Mg3+(g) + e-
H = (+) 7750 kJ.mol-1
MORE ON ION FORMATION
The overlapping of energy levels leads to the 4s orbital filling before the 3d (an electron in
a 4s orbital has a lower energy than an electron in a 3d orbital), when the atom ionises it is
the 4s electrons that are lost first.
Note – there is a special stability associated with a filled subshell or a half filled subshell –
for example the p subshell when it contains 3 or 6 electrons. Likewise the d-subshell is
most stable when it contains 5 or 10 electrons.
So following on from this, the more stable the electronic configuration, the more difficult it
is to remove the outer electron and therefore the ionisation energy is higher.
Questions
Q1. The number of orbitals and the number of electrons in an energy level or sublevel is
limited.
a) Give the number of orbitals that make up:
i) the s sublevel
ii) the d sublevel.
b) Give the number of electrons that are needed to completely fill:
i) the p sublevel
ii) the first energy level
iii) the third energy level.
c) Give the sublevels in:
i) the first energy level
ii) the fourth energy level.
Answers
Q1. a) i) 1
ii) 5
b) i) 6
ii) 2
iii) 18
c) i) 1s
ii) 4s 4p 4d 4
Q2. The electron configuration of an atom of element Y in the ground state can be
represented as:
a) Identify element Y
b) The electron configuration of an atom or ion may also be expressed in
another form,
e.g. 1s2 2s2 2p1 for boron.
Give the electron configuration for Y in this form.
Q2. Answers
a) Manganese
b) 1s2 2s2 2p6 3s2 3p6 4s2 3d5 or [Ar] 4s2 3d5
Q3. The first 20 elements show many periodic properties, e.g. the variation in first
ionisation energy (IE).
Using the above, your knowledge of higher and your data booklet to observe the positions
of elements in periods and groups:
a) Predict, from the graph, the first IE of rubidium (note Rb is below K in Group 1)
b) Explain why the noble gases have the highest values of IE in each period.
c) i) Explain why the Group 1 metals have the lowest value of IE.
ii) Explain why the values of IE decrease Li to Na to K.
d) Explain the general increase in value of IE from Li to Ne.
e) i) Explain the drop in value of IE from Be to B.
ii) Explain the drop in value of IE from N to O.
Answers
a) 380 kJ mol (±20 kJ mol )
b) There is a huge energy requirement to break the noble gases in a stable octet
or It is very difficult to remove an electron from a full energy level.
c) i) The Group 1 metal has the largest radius in that period and has the
smallest nuclear charge in that period.
Both facts lead to a lesser attraction for the outermost electron.
ii) Each new energy level means a larger radius (less attraction for the
outer most electron) and provides a greater shielding effect
(again reduced attraction by the nucleus).
d) Two factors apply: the steady increase in nuclear charge and the slight
decrease in atomic radius from Li to Ne makes the attraction of the
nucleus for outer electrons greater.
e) i) Be 1s2 2s2
B 1s2 2s2 2p1, i.e. B has started a new subshell so
its outermost electron is relatively easier
to remove than that of Be, where a complete
subshell has to be broken into.
ii) Half-full shells are relatively stable so N (with a half-filled p subshell)
has a higher IE than O, which has one electron more.
–1
–1
Lessons 3
VSEPR stands for Valence Shell Electron Pair Repulsion and these electron pair
repulsions are responsible for the shapes of molecules and polyatomic ions such as NH 4+
Step 1 – Calculate the number of outer electrons (or valence) on the central atom of
molecule or ion –achieved by taking the number of electrons on the centre atom and
adding one electron for each of the other atoms attached.
Note – if we are dealing with an ion with a 1+ charge, then we subtract an electron from
the total to account for this charge. For an ion with a -1 charge, we add an electron to the
total. For example:
NH4+ number of outer electrons on N atom = 5 (from N) + 4 (1 from each H) 1 = 8 AIH4- number of outer electrons on Al atom = 3 (from AI) + 4 (1 from
each H) + 1 = 8. Dividing the total number of electrons by two gives us the
number of electron pairs surrounding the central atom. This means that the
shape of the ammonium ion is tetrahedral just like methane. So BH3 will have
three electron pairs surrounding the central atom, whereas NH4+ and AIH4will each have four.
As electron pairs are negatively charged, they will repel each other and will
be arranged in such a way as to minimise their repulsion and maximise their
separation. Some of the arrangements of electron pairs around the central
atom are outlined in the table.
Note
Non-bonding pair/non-bonding pair repulsion is greater than non-bonding pair/bonding pair
repulsion, which, in turn, is greater than bonding pair/bonding pair repulsion.
SHAPES OF MOLECULES AND IONS
So far we have considered the shapes adopted by electron pairs surrounding
the central atom or ion and have taken no account of the two different types
of electron pairs.
To predict the shape of a molecule, you first calculate the number of
electron pairs and their arrangement. However to obtain the actual
“molecular shape”, you must take into account whether these electron
pairs are bonding or non-bonding pairs.
H
H
..
180
..
109.5
:
Cl
:
Be
..
C
Cl
..
Linear
H
..
..
..
..
H
Bent
..
H
..
..
N
120
..
B
Cl
O
H
Tetrahedral
Cl
..
104.5
..
H
Cl
106.7
Trigonal Planar
..
..
H
F
..
H
..
Pyramidal
Linear
Note - Ammonia has one lone pair of electrons
..
H
In PH3, for example, (not in the diagram) there are four electron pairs, but three of them
are bonded pairs and one is a non-bonded pair. The four electron pairs adopt a tetrahedral
shape but the three bonded pairs adopt a pyramidal shape. So the PH 3 molecule is
described as pyramidal, not tetrahedral.
ELECTRONIC CONFIGURATION
The d-block transition metals are metals with an incomplete d subshell in at least one of
their ions. The first row is scandium to zinc, the second is yttrium to cadmium and the third
row would be platinum to gold.
Across the first transition series (Sc to Zn) the 3d orbitals (subshell) are being filled by the
Aufbau Principle, the 4s orbital having already been filled (according to the Aufbau
Principle). The electronic configurations of the elements of the first transition series are
shown below table.
The table shows the electronic configuration in spectroscopic and orbital box notation for
the elements from scandium to zinc. [Ar] represents the electronic configuration of argon,
which is Is2 2S2 2p6 3s2 3p6. It is okay to use this shorthand here instead of writing out the
full electron shells up to 3p. However, in the exam you should write out the spectroscopic
notation for each element in full.
Element
Atomic Number
Electron Arrangement
Scandium
21
1s22s22p63s23p64s23d1
Titanium
22
1s22s22p63s23p64s23d2
Vanadium
23
1s22s22p63s23p64s23d3
Chromium
24
1s22s22p63s23p64s13d5
Manganese
25
1s22s22p63s23p64s23d5
Iron
26
1s22s22p63s23p64s23d6
Cobalt
27
1s22s22p63s23p64s23d7
Nickel
28
1s22s22p63s23p64s23d8
Copper
29
1s22s22p63s23p64s13d10
Zinc
30
1s22s22p63s23p64s23d10
There is a special stability associated with half-filled or completely filled d orbitals. Bear this in mind
when looking at the orbital box notation and you can understand why chromium is [Ar] 3d54s1 and
copper is [Ar] 3d104S1, rather than the [Ar] 3d4 4s2 and [Ar]3d9 4s2 as you might have expected.
ELECTRONIC CONFIGURATIONS OF THE 1ST SERIES TRANSITION METALS
Spectroscopic Orbital box notation Element notation
3d
4s
Sc
[Ar] 4s2 3d1


Ti
[Ar] 4s2 3d2


V
[Ar] 4s2 3d3



Cr
[Ar] 4s1 3d5






Mn
[Ar] 4s2 3d5






Fe
[Ar] 4s2 3d6






Co
[Ar] 4s2 3d7






Ni
[Ar] 4s2 3d8






Cu
[Ar] 4s1 3d10






Zn
[Ar] 4s2 3d10








Note - However, when any transition metal atom forms an ion, the electrons that are lost first are
those in the outer subshell, the 4s electrons. Therefore the electronic configuration of the Co2+ ion
is [Ar] 3d7.
TYPICAL PROPERTIES OF TRANSITION METALS
The typical properties of transition metals are;
1. They are metallic so conduct both heat and electricity
2. They show marked catalytic ability both as the metal and in compounds
3. They exhibit variable valency (exception : zinc – always has a valency of 2)
4. They form many complex salts
5. They form coloured ions (exception – zinc)
(VARIABLE OXIDATION STATE)
The oxidation state is similar to the valency that an element has when it is part of a compound. For
example, in iron (lI) chloride we might say that the iron has a valency of 2. However, it is actually
more accurate to say that iron is in oxidation state (II) or has oxidation number +2.
There are certain rules to be followed when assigning an oxidation number to an element:

the oxidation number of an uncombined element is 0

for ions containing single atoms such as Na+ or 02-, the oxidation number is the same
as the charge on the ion - in the examples given, these would be + 1 and - 2

in most of its compounds, oxygen has oxidation number -2

in most of its compounds, hydrogen has oxidation number + 1

fluorine has oxidation number -1 in all its compounds

the sum of all the oxidation numbers of all the atoms in a molecule or neutral compound must
add up to zero

the sum of all the oxidation numbers of all the atoms in a polyatomic ion must add up to the
charge on the ion
To work out the oxidation number for sulphur in sulphuric acid;
In sulphuric acid the overall oxidation number is zero, the 4 oxygen atoms give a total
oxidation number of –8, the hydrogen atoms give a total oxidation number of +2, so the
sulphur atom must have an oxidation number of +6 to leave the overall charge of H 2SO4
neutral.
If we want to find the oxidation number of manganese in Mn04 -, the sum of the oxidation numbers
of the one manganese atom and the four oxygen atoms must add up to -1 as this is the charge on
the ion. Each oxygen atom has oxidation number -2 and so sum of the oxidation numbers of the
four oxygen atoms must be -8. Therefore the oxidation number of manganese must be 7, as 7 - 8 =
-1. We can say the manganese has oxidation number + 7 or is in oxidation state (VII).
Note – Iron (lII) is usually more stable than iron(II); iron(l)
Sometimes transition metal compounds have different colours depending on the oxidation state of
the metal. For example, iron (ll) compounds are often a pale green colour, which slowly changes to
the familiar yellow-orange colour of iron (lII) compounds as oxidation occurs.
OXIDATION can be defined as an increase in oxidation number, while REDUCTION can
be defined as a decrease in oxidation number.
In general, compounds containing metals in high oxidation states tend to be
good oxidising agents (electron acceptors )as the ions are easily reduced to
lower oxidation states.
Likewise, compounds containing metals in low oxidation states tend to be
reducing agents (electron donors).
Spectroscopic Orbital box notation element notation
3d
4s
Fe2+
[Ar] 3d6





Fe3+
[Ar] 3d5





Changing from one oxidation state to another is an important aspect of transition metal
chemistry, often characterised by a distinct colour change, as shown in the table below.
Ion
Oxidation state of transition metal
Colour
VO3-
+5
Yellow
VO2+
+4
Blue
V3+
+3
Green
V2+
+2
Violet
The common oxidation states for the elements in the first transition series are shown
below;
7+
3+
Sc
6+
6+
5+
5+
5+
4+
4+
4+
4+
4+
4+
4+
3+
3+
3+
3+
3+
3+
3+
3+
2+
2+
2+
2+
2+
2+
2+
2+
1+
1+
1+
1+
1+
1+
1+
V
Cr
Mn
Fe
Co
Ni
Cu
Ti
6+
5+
2+
Zn
Oxidation states greater than +4 are observed but never in simple ions. In high oxidation
states, the central ion (e.g. vanadium) is usually bonded to an electronegative element like
oxygen.
Formula of ion
Oxidation state
V2+
+2
V3+
+3
VO2+
+4
VO3-
+5
Questions
Q1 a) Sketch the shapes of NH3 and BCl3 molecules, showing clearly all the
bond angles and their values.
Answer
(1)
(1)
b) Since both nitrogen and boron have three bonding electrons, why do
NH3 and BCI3 not have the same molecular shape?
Answer
a) Nitrogen has an extra pair of electrons (a lone pair).
These exert a strong repulsive force downwards
on the bonding pairs hence they are pushed down
‘below’ the tetrahedral angle, creating a pyramid:
b) Boron has only three outermost electrons, so BCI3 has only three pairs of
bonding electrons. They spread themselves symmetrically (or as far from each
other as possible), i.e. pointing to the corners of an equilateral triangle.
Q2) Chlorine and fluorine react to produce a compound of formula
CIF3 . This molecule contains three chlorine-fluorine single bonds.
Each fluorine atom contributes one electron to the bonding.
a) How many electron pairs (both bonding and non-bonding) surround the
central chlorine atom in the molecule?
b) What would be the three-dimensional arrangement of electron pairs
(both bonding and non-bonding) around the chlorine atom?
c) The fluorine atoms may occupy different positions in this shape, giving
rise to three possible shapes for the molecule. Draw two of these,
showing the angles between the bonds.
Tip – think about the position of the lone pair of electrons and bond
angles
Answer
a) There are five pairs of electrons around the central chlorine atom (since an
atom of Cl has seven outermost electrons and each F atom contributes
one electron to the total ==> 7 + 3 = 10 electrons ==> 5 pairs).
Answer
b) Five pairs lead to a trigonal bipyramid
c) Answer
(Any two of these three shapes are acceptable)
Lessons 4
(TRANSITION METAL COMPLEXES)
An important feature of the transition metals is their ability to form complex ions and
molecules, often called coordination compounds. A complex consists of a central metal
ion surrounded by ligands. A ligand is a molecule or negative ion with at least one lone
(non-bonding) pair of electrons available for bonding with the metal ion. The electron
orbitals of the ligand have an effect on the electron distribution in the central ion and this in
turn results in significant effects in physical properties of the complexes e.g. colour.
Typical ligands involved in the formation of complexes are negative ions such as;
 CN F-, Cl-, Br-, I NO2 OH-
and molecules such as;
 H2O
 NH3
EDTA (ethylenediaminetetraacetic acid) contains the EDTA4- ion which is a hexadentate
ligand which forms very stable complexes with metal ions;
EDTA4-
-.. ..
O..
O
C
Nickel EDTA complex
.....O.
2-
O
C
H2C .
. CH2
.
.
H2C N
N CH2
... -...
C C
C
C
.O
..
.O..
H2 H2
O
O
O
N
Ni
N
O
O
O
HAEMOGLOBIN
Iron forms the red complex, haemoglobin, responsible for the transport of oxygen in the
blood.
C
C
C
HC
C
C
C
C
H
C
C
C
C
.N.
..
H.
N
.
C
N
C
C
C
..
NH
C
H
CH
C
C
The porphyrin ring (a tetradentate ligand) attaches itself to a central iron(III) ion via its 4
nitrogen atoms. In haemoglobin the nitrogen atoms in the porphyrin ring occupy four
ligand sites. The other two sites are occupied by the protein, globin, and a molecule of
oxygen.
SHAPES OF COMPLEXES
The number of bonds from the central metal ion to the ligand(s) is known as the
coordination number of the central ion. The same term was used in a similar way in
describing the arrangement of ions in a crystal lattice earlier in this topic. The coordination
number will determine the shape of the complex ion.
With coordination number of 6 a complex will have an octahedral shape;
With a coordination number of 4 the complex could have a tetrahedral shape;
or a square planar shape,
see table below;
Coordination number
Shape
4
square planar
F
F
Xe
F
4
H
tetrahedral
H
6
F
C H
H
Cl
octahedral
Cl
Cl
S
Cl
Cl
Cl
NAMING COMPLEXES
chloride, Cl-
chloro
oxide, O2-
oxo
cyanide, CN-
cyano
oxalate, C2O42-
oxalato
ammonia, NH3
ammine
water, H2O
aqua
carbon monoxide, CO
carbonyl
Vanadium
Vanadate
Chromium
Manganese
Chromate
Iron
Cobalt
Manganate
Ferrate
Cobaltate
Nickel
Copper
Nickelate
Tin
Lead
Stannate
Cuprate
Plumbate
For example, naming K3[Fe(CN)6]
1. Since there are three potassium ions (each 1+), overall charge on the complex ion
must be 32. There are 6 cyanide ions surrounding the central metal ion , each with a charge of
1- so;
(Oxidation number of Fe)
+
(6 x -1) = -3
Oxidation number of Fe
= -3 + 6
Oxidation number of Fe
= +3
3. Six cyanide ions gives ‘hexacyano-‘
4. Iron is the central metal ion. Since the complex ion is a negative ion (anion), this
gives ‘ferrate(III)’
The positive ion name precedes the negative ion name, giving the name;
(COLOUR OF TRANSITION METAL IONS AND COMPLEX IONS)
red
yellow
green
white
cyan
magenta
blue
The reason that compounds of transition metals absorb white light is due to the loss of
degeneracy of the d orbitals in these compounds. In the free ion, e.g. Ti 3+
(1s22s22p63s23p63d1), the 5 d orbitals (dxy, dyz, dxz, dz2, dx2-y2) are degenerate;
However in a complex ion such as [Ti(H2O)6]3+ the metal ion is no longer isolated but
surrounded by 6 water ligands. The complex has an octahedral shape and the water
molecules can be considered to be approaching the central Ti3+ ion along the x-, y- and zaxes.
The d orbitals are split differently in complexes of different shapes. The amount of
splitting of the d orbitals is given by the spectrochemical series;
small orbital splitting
I- <
Br-
<
Cl-
<
increasing
large orbital splitting
F-
<
OH-
<
H2O
<
NH3
<
CN-
<
CO

The colour of the complex can be found by subtracting the colour of the light absorbed
from white light and adding together the remaining colours of transmitted light.
UV and VISIBLE ABSORPTION SPECTROSCOPY
A schematic diagram of a uv/vis spectrometer is shown in the diagram;
Samples are in solution and placed in a cell (a small cuvette). Another identical cell
containing the pure solvent used for the solution is also placed in the machine. Radiation
across the whole range is scanned continuously through both sample and pure solvent.
The spectrometer compares the two beams, the difference is the light absorbed by the
compound in the sample. This data is produced as a chart of wavelength against
absorbance.
E.g. for Ti3+(aq) visible spectrum
In the case of visible light being absorbed, the colour transmitted is white light minus the
absorbed light and the complimentary colour is observed. In this example, from the
position of the main peak, it can be seen that mainly green light is absorbed, hence the
purple colour of the solution (red and blue are transmitted).
Watch the Royal Society of Chemistry video on ‘UV/vis Spectrophotometry’
http://my.rsc.org/video/56
A colorimeter is a simpler device which can also be used in analysis. In this, a light source
used with suitable filters provides light with a narrow band of wavelengths. This is passed
through the sample in a cell as before and the absorbance measured.
The absorbance is directly proportional to the concentration of the absorbing species. By
choosing a range of known different concentrations a calibration curve be drawn and the
concentrations of unknown solutions can then be calculated from the calibration curve.
CATALYSTS
Transition metals and their compounds are important catalysts in biological and industrial
chemical reactions. Elements such as iron, copper, manganese, cobalt, nickel and
chromium are essential for the effective catalytic activity of certain enzymes. Some
examples of transition metals and their compounds used as catalysts in industrial
reactions are shown in the table;
Process
Catalyst
Haber
Iron
Contact
Vanadium(V) oxide
Ostwald
Platinum
Catalytic converters in car exhausts
Platinum, palladium and rhodium
Preparation of methanol
Copper
Preparation of margarine
Nickel
Polymerisation of alkenes
Titanium compounds
It is thought that the presence of unpaired d electrons or unfilled d orbitals allows
intermediate complexes to form, providing reaction pathways of lower energy compared to
the uncatalysed reaction. Many transition metals act as catalysts because of their ability to
exist in a variety of different oxidation states, the transition metal reverts to its original
oxidation state once the reaction is complete.
A good example is the homogenous catalysis by cobalt (II) chloride of the reaction
between hydrogen peroxide and potassium sodium tartrate (Rochelle salt). In the reaction
the cobalt changes oxidation state from +2 (pink) to +3 (green) and then back to +2 (pink);
Co2+
pink
→
Co3+
green
→
Co2+
pink
Questions
Q1)Account for the green colour of an aqueous solution of V3+ ions.
(Make reference to ligands, electrons and the visible spectrum in your
answer)
Answer
The H2O ligands split the degenerate d orbitals
Energy from the red end of the visible spectrum is absorbed
as electrons are promoted across the small energy gap, ΔE,
now existing in the d orbitals. Hence the green colour is
seen
Q2) The ability of a ligand to split the d orbitals when forming a complex ion is
given in the spectrochemical series. Three ligands from this series and their
relative ability to split the d orbitals are:
NH3 > H2O > Cl
-
A study of part of the absorption spectrum for the complex ion,
hexaaquanickel(II) shows a broad absorption band that peaks at about 410
nm.
a) Explain the origin of the absorption band at this wavelength.
b) State towards which end of the visible spectrum the wavelength of the
absorption band would move if the water ligands were replaced with
chloride ions. Give an explanation for your answer.
Answer
2a) Five degenerate orbitals are split into two orbitals of
higher energy and three of lower energy. Electrons can
be promoted across this gap by absorbing energy from
the visible spectrum.
The peak around 410 nm represents the wavelengths absorbed
(and equals the energy value of the d–d split).
b) Since the chloro complex leads to a lower value of d–d splitting
less energy is needed to make the jump, so absorption moves to the lower or
red end of the spectrum.