Inorganic Chemistry (AH) – Slides (RG) – 1 st sub- topic Part 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 x10 8 ms -1 velocity = frequency x wavelength c = f 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
Transcript
Inorganic Chemistry (AH) – Slides (RG) – 1 st sub-topic
Part 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 = frequency x wavelength
c = f x
and soc
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= Lhc1000. 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
= 193125 J mol–1 = 193.1 kJ mol –1 (correct)
9
83423
10620103.00106.63106.02
λLhc
E
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
= 1312.9 kJ mol–1
Q3) The bond enthalpy of a Cl—Cl bond is 243 kJ
9
383423
1091.210103.00106.63106.02
λLhc
E
mol–1. Calculate the maximum wavelength of light that would break one mole of these bonds to form individual chlorine atoms
Answer
and so,
= 4.93 10–7 m
= 493 nm
Part 2
LhcE
ELhc
3
83423
10243103.00106.63106.02
λ
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.
Note
Heisenberg Uncertainty Principle
This essentially states that it is impossible to know the precise location of an electron in a certain moment of time, because the electron is behaving as a particle and a wave at the same time i.e. wave/particle duality.
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 subshellsTo 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 ls 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.
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 Electron arrangement Element Electron arrangementH 1s1 K 1s22s22p63s23p64s1
He 1s2 Ca 1s22s22p63s23p64s2
Li 1s22s1 Sc 1s22s22p63s23p64s23d1
Be 1s22s2 Ti 1s22s22p63s23p64s23d2
B 1s22s22p1 V 1s22s22p63s23p64s23d3
C 1s22s22p2 Cr 1s22s22p63s23p64s13d5
N 1s22s22p3 Mn 1s22s22p63s23p64s23d5
O 1s22s22p4 Fe 1s22s22p63s23p64s23d6
F 1s22s22p5 Co 1s22s22p63s23p64s23d7
Ne 1s22s22p6 Ni 1s22s22p63s23p64s23d8
Na 1s22s22p63s1 Cu 1s22s22p63s23p64s13d10
Mg 1s22s22p63s2 Zn 1s22s22p63s23p64s23d10
Al 1s22s22p63s23p1 Ga 1s22s22p63s23p64s23d104p1
Si 1s22s22p63s23p2 Ge 1s22s22p63s23p64s23d104p2
P 1s22s22p63s23p3 As 1s22s22p63s23p64s23d104p3
S 1s22s22p63s23p4 Se 1s22s22p63s23p64s23d104p4
Cl 1s22s22p63s23p5 Br 1s22s22p63s23p64s23d104p5
Ar 1s22s22p63s23p6 Kr 1s22s22p63s23p64s23d104p6
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 BN and OMg and AlP 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.
Tip – Use the Aufbau Triangle
Answers
Q1. a) i) 1 ii) 5 b) i) 6 ii) 2 iii) 18 c) i) 1s ii) 4s 4p 4d4f
Q2. The electron configuration of an atom of element Y in the ground state can be represented as:
a) Identify element Yb) 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
Q3a) 380 kJ mol–1 (±20 kJ mol–1 ) b) There is a huge energy requirement to break the noble gases in a stable octetor 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 thesmallest 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 theouter 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 slightdecrease in atomic radius from Li to Ne makes the attraction of thenucleus for outer electrons greater.e) i) Be 1s2 2s2
B 1s2 2s2 2p1, i.e. B has started a new subshell soits outermost electron is relatively easierto remove than that of Be, where a completesubshell 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.
Part 3VSEPR 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 AIH4- will
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 into account the two different types of electron pairs.
RECAP
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.
Note - Ammonia has one lone pair of electrons
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
180
Be ::Cl Cl
Linear
C
. .
. .. .. .
H
H
H
H
109.5
Tetrahedral
N
. .
. .. .. .HH
H
106.7Pyramidal
F
. .
. .. .. .
H
LinearTrigonal Planar
B
. .
. .. .
Cl
Cl Cl
120
Bent
O
. .
. .
H
H
104.5. .
. .
pyramidal shape. So the PH3 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 1s22s22p63s23p64s 1 3d 5
Manganese 25 1s22s22p63s23p64s 2 3d 5
Iron 26 1s22s22p63s23p64s23d6
Cobalt 27 1s22s22p63s23p64s23d7
Nickel 28 1s22s22p63s23p64s23d8
Copper 29 1s22s22p63s23p64s 1 3d 10
Zinc 30 1s22s22p63s23p64s 2 3d 10
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 1 ST 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.
Questions
Q1 a) Sketch the shapes of NH3 and BCl3 molecules, showing clearly all the
bond angles and their values.
b) Since both nitrogen and boron have three bonding electrons, why do NH3 and BCI3 not have the same molecular shape?
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) 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
Answers1a)
1b)
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
2) a) There are five pairs of electrons around the central chlorine atom (since anatom of Cl has seven outermost electrons and each F atom contributesone electron to the total ==> 7 + 3 = 10 electrons ==> 5 pairs).
b)
(Any two of these three shapes are acceptable)
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 ColourVO3
- +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+6+ 6+ 6+
5+ 5+ 5+ 5+
4+ 4+ 4+ 4+ 4+ 4+ 4+
3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+ 3+
2+ 2+ 2+ 2+ 2+ 2+ 2+ 2+ 2+1+ 1+ 1+ 1+ 1+ 1+ 1+
Sc Ti V Cr Mn Fe Co Ni Cu 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
Part 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.
Ligands are electron donors, donating their non- bonding electrons into unfilled metal
orbitals and forming what are known as dative covalent bonds i.e. where the two electrons
are donated by one of the atoms, rather than one electron from each atom.
The lone pair or non-bonding pair of electrons are shown as a pair of dots on the atoms.
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- (nitrite ion)
OH-
and molecules such as;
* H2O (two non-bonding pairs of electrons from oxygen)
NH3 (one non-bonding pair of electrons from nitrogen)
NOTE
1. Ligands that donate one pair of electrons to the central metal ion are said to be
monodentate (Cl-, H2O)
2. Bidentate ligands – donate two pairs of electrons to the central metal ion (oxalate
ion (C2O42-) and 1,2-diaminoethane (C2H4 (NH2)2)
3. EDTA is a hexadentate ligand which has six pairs of non-bonding electrons that
bind to a central ion in a 1:1 ratio, a nickel (II) ion being involved. Note that the
EDTA ion has a 4- charge.
The total number of bonds from the ligand to the central metal ion is known as the
coordination number. In the Ni 2+ -EDTA complex, the coordination number is six.
EDTA (ethylenediaminetetraacetic acid) contains the EDTA4- ion which is a hexadentate ligand which forms very stable complexes with metal ions;
EDTA4- Nickel EDTA complex
CH2
CH2
N N CH2
CH2
CH2
CH2
CCOO O
O
C CO O OO
..
......
.. ...... --
.. ...... ....- -
NiN
OO
NO
O
2-
HAEMOGLOBIN
Iron forms the red complex, haemoglobin, responsible for the transport of oxygen in the
blood.
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.
CC
C
N
CH
CC
NH C
C
CCH
C NHC
CC
CH
C
N CCH
CC
....
....
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
4 tetrahedral
6 octahedral
XeFF
F F
C
H
H HH
SClCl
Cl Cl
Cl
Cl
NAMING COMPLEXES
Complexes and complex ions are named and written according to IUPAC rules - the International Union of Pure and Applied Chemistry is the world authority on chemical nomenclature.
When writing the formula of a complex, the symbol of the metal is written first, then the negatively charged ligands, followed by the neutral ligands. Finally, the formula of the complex ion is enclosed within square brackets as in, for example, [Fe(OH)2(H2O)6) +. This ion has an overall charge of + 1 as this is the sum of the Fe 3+ ion and the two OH- ions.
Rules for naming complexes are:
the ligands are named first in alphabetical order, followed by the name of the metal and its oxidation state
if the ligand is a negative ion ending in -ide , then in the complex the ligand name changes to end in '0' - examples include chloride, which becomes chloro, and cyanide, which becomes cyano
If the ligand is ammonia (NH 3), it is named ammine in the complex
water as a ligand is called aqua
if the complex is a negative ion overall, the name of the complex ion ends in -ate , e.g. cobalt ate for a negative ion containing cobalt, cuprate is used if the complex contains copper and ferrate if it contains iron, from their Latin names (not copperate or ironate)
Examples of names of complexes
the complex ion [Cu(H2O)4]2+ is the tetraaquacopper(II) ion
[Co(NH3)6]2+ is hexaamminecobalt(II)
[Fe(CN)6]4- is hexacyanoferrate(II) as it is a negative ion and the iron is in oxidation state (II)
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 Chromate
Manganese Manganate
Iron Ferrate
Cobalt Cobaltate
Nickel Nickelate
Copper Cuprate
Tin Stannate
Lead Plumbate
For example, naming K3[Fe(CN)6]
1.There are three potassium ions (each 1+), overall charge on the complex ion must be
3-
2. There are 6 cyanide ions surrounding the central metal ion, each with a charge of 1-
so;
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 greenyellow
whitemagenta cyan
blue
In an isolated transition metal atom or ion (one that is not complexed with any ligands), the five different d orbitals in the 3d subshell are degenerate.
Now consider the formation of an octahedral complex such as [Ni(H2O)]62+. Think of six water
ligands approaching the Ni2+ ion along the x-, y- and z-axes. The electrons in the d orbitals of the nickel ion that lie along the axes will be repelled by the electrons of the approaching ligands.
As a result, these d orbitals now have a higher energy than the d orbitals that lie between the axes . Therefore the d orbitals are no longer degenerate in the complex ion, as the d-orbitals differ very slightly in energy.
Note – electrons can jump from one d-orbital to another if they absorb energy. For most transition metal complexes, the frequency of light absorbed in those energy transitions is in the visible region of the spectrum. And the ion appears coloured.
The d orbitals that lie on the axes are dx 2 -y 2 (a double dumbbell lying on both the x- and y-axes) and dz 2 , which lies on the z-axis. The lower energy orbitals are the d xy, dxz and dyz orbitals (double dumbbells that lie between the axes).
We call this 'splitting' of the d orbitals. The splitting is different in octahedral complexes compared with tetrahedral and other shapes of complexes.
The energy difference between the different subsets of d orbitals depends on the position of the ligand in the spectrochemical series. This is a series that puts in order the ability of different Iigands to split the d orbitals. Those ligands that cause a large difference in energy in the d orbitals are said to be 'strong field' ligands, in contrast with 'weak field' ligands where the energy difference is small.
A short form of the spectrochemical series is CN- > NH3 > H2O > OH- > F- > Cl- > Br- > I-, in which the cyanide ion causes the greatest energy difference.
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. Ti3+
(1s22s22p63s23p63d1), the 5 d orbitals (dxy, dyz, dxz, dz2, dx
2-y
2) 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 z-
axes.
The d orbitals are split differently in complexes of different shapes. The amount of splitting
of the d orbitals depends on the ligand. The ability of ligands to cause the splitting of the d
orbitals is given by the spectrochemical series;
small orbital splitting large orbital splittingI- < Br- < Cl- < F- < OH- < H2O < NH3 < CN- < CO
increasing
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.
COLOUR IN TRANSITION METAL COMPLEXES
Many transition metal compounds are coloured. For example, solutions of copper (II) compounds are usually blue and solutions of nickel(II) compounds are usually green. To explain how these colours arise, we must examine simple colour theory.
white light can be thought of as a combination of three primary colours - red, green and blue if red light is absorbed, green and blue light are transmitted and we see this as a blue/green
colour, or cyan if green light is absorbed, a combination of red and blue light is transmitted and we I see this
as purple or magenta if blue light is absorbed, red and green light are transmitted, which we see as yellow
Why do transition metal compounds absorb light?
Note – it is the light absorbed that excites the electrons.
Think back to the split d orbitals. Electrons in the lower energy d orbitals can absorb energy and move to the higher energy d orbitals. If the energy absorbed in these so-called d-d transitions is in the visible part of the electromagnetic spectrum, the colour of the transition metal compound will be the complementary colour of the absorbed colour. So the colour we see will be white light minus the colour absorbed.
Note – The colour of transition metal complexes is due to the presence of incomplete d-subshells.
The electrons in d-subshells undergo d-d transitions and thus impart colour to the compound. Thus
for the compound to be coloured, an essential requirement is an incompletely filled d-subshell.
Ultraviolet and visible spectroscopy
Atomic Absorption SpectroscopyIn atomic absorption spectroscopy, the electromagnetic radiation is directed through a gaseous sample of the substance.
Radiation corresponding to certain wavelengths is absorbed as electrons are promoted to higher energy levels.
The wavelength of the absorbed radiation is measured and used to identify each element, as the element has a characteristic absorption spectrum in terms of absorption spectral lines.
A transition between two electronic energy levels should give rise to a line in the absorption spectrum.
The amount of species present can also be determined by measurement of the amount of light absorbed. The measured absorbance is proportional to the concentration of the element in the sample.
The effects of d-d transitions can be studied using spectroscopy. If the absorbed energy is
in the visible part of the electromagnetic spectrum, giving a coloured compound, visible spectroscopy is used. If the absorbed energy is in the ultraviolet part of the electromagnetic spectrum, the compound will be colourless and ultraviolet spectroscopy is used.
When the ligands surrounding the transition metal ion are strong field ligands such as CN-, d-d transitions are more likely to occur in the ultraviolet region. The wavelength range of the ultraviolet light is approximately 200-400nm.
Complexes containing weak field ligands such as H2O are more likely to absorb visible light, making them coloured. The wavelength range for the visible region is approximately 400-700 nm.
-B
A colorimeter is a device used in colorimetry. In scientific fields the word generally refers to the device that measures the absorbance of particular wavelengths of light by a specific solution. This device is most commonly used to determine the concentration of a known solute in a given solution by the application of the Beer-Lambert law, which states that the concentration of a solute is proportional to the absorbance.
An ultraviolet spectrometer is a bit more complicated than a colorimeter. Different wavelengths of ultraviolet light from 200 to 400 nm are passed through the sample and the amount of ultraviolet light absorbed at different wavelengths is recorded. The results are plotted automatically as an ultraviolet spectrum. As with a colorimeter, the absorbance is directly proportional to the concentration of the absorbing species.
CATALYSTS
Transition metals and their compounds are used as catalysts. Catalysts you may already know are: iron in the Haber process (industrial production of ammonia); platinum in the Ostwald process (industrial production of nitric acid); and platinum, rhodium and palladium in catalytic converters.
These are all examples of heterogeneous catalysts as they are in a different physical state to the reactants in the reactions being catalysed.
The transition metal atoms or ions on the surface of the active sites of the catalyst form weak bonds with the reactant molecules. It is thought that the presence of unpaired d electrons or unfilled d orbitals allows intermediate complexes to form. The effect of this is to weaken the covalent bonds inside the reactant molecules and, as these reactant molecules are now held in a favourable position, they are more susceptible to attack by molecules of the other reactant. The overall effect is that an alternative reaction pathway with a lower activation energy is provided and so the rate of the reaction is increased.
Another reason that transition metals are good catalysts may be because they have variable oxidation states. Again, this allows the transition metal to provide an alternative reaction pathway with a lower activation energy, so speeding up the reaction.
Note
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 CatalystHaber 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
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+ → Co3+ → Co2+
pink green 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)
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 orbital 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.
AnswersQ1) The H2O ligands split the degenerate d orbitals
Energy from the red end of the visible spectrum is absorbedas electrons are promoted across the small energy gap, ΔE,now existing in the d orbitals. Hence the green colour isgreen.
2a) Five degenerate orbitals are split into two orbitals ofhigher energy and three of lower energy. Electrons canbe promoted across this gap by absorbing energy fromthe 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.
