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Inorganic and Physical Chemistry: Electromagnetic Radiation and Atomic Spectra
INORGANIC AND PHYSICAL CHEMISTRY
ATOMIC EMISSION SPECTRA
When white light (such as light from an ordinary light bulb) is passed through a prism
a rainbow effect is seen. This is known as a continuous spectrum and includes all the
colours in the visible part of the electromagnetic spectrum (approximately 400–700 nm).
When energy in the form of high voltage electricity is passed through a tube of a gas
such as hydrogen at low pressure coloured light is produced. This light can be analysed
by passing it through a prism or diffraction grating or a spectroscope. This time,
instead of a continuous spectrum being seen, a series of coloured lines is observed.
These lines correspond to certain specific frequencies or wavelengths in the visible
spectrum. This is known as the atomic emission spectrum for that element. No two
elements produce the same atomic emission spectrum. This means that each different
chemical element produces its own unique pattern of lines corresponding to different
frequencies or wavelengths in its emission spectrum. We can see these lines using a hand
held spectroscope if they occur in the visible part of the electromagnetic spectrum but
if they are outwith the visible spectrum then we need instruments such as ultraviolet
spectrometers to help us “see” them.
ONLINE
Head to
www.brightredbooks.net
for more on atomic
emission spectra.
WHAT CAUSES ATOMIC EMISSION SPECTRA?
When energy is transferred to atoms, electrons within these atoms may use this energy
to move to higher energy levels. We say that these electrons have become excited. These
excited electrons can move back to lower energy levels by losing energy. Energy lost from
atoms in this way is released in the form of photons.
Each line in an atomic emission spectrum corresponds to the energy given out when an
excited electron moves to a state of lower energy. This can either be to a lower excited
state or back to the ground state. Atomic emission spectra provide good evidence for
discrete (quantised) energy levels in atoms.
Some atomic emission spectra are shown below:
Hydrogen
DON’T FORGET
Each line in an emission
spectrum corresponds to
the energy given out when
an excited electron moves
to a lower energy level.
Helium
Neon
DON’T FORGET
No two elements produce
exactly the same emission
spectrum.
Sodium
Mercury
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BRP_CfEAdvHigherChemistry_Sample.indd 8-9
SAMPLE PAGES – CFE ADVANCED HIGHER CHEMISTRY
ELECTROMAGNETIC RADIATION AND
ATOMIC SPECTRA
THE ATOMIC EMISSION SPECTRUM OF HYDROGEN
As you can see there are definite distinct lines in the atomic emission spectrum of
hydrogen. In fact these lines are seen in the visible part of the spectrum and there are
also a series of lines “seen” in the infrared and another series in the ultraviolet parts of
the electromagnetic spectrum. So although hydrogen is the simplest element with only
one electron per atom its atomic emission spectrum is quite complicated.
The different series are shown in the energy level diagram below which is not drawn
to scale. These 3 series are named after their discoverers. As you can see the lines
due to excited electrons falling back to the ground state (n = 1) are observed in
the ultraviolet region. This is because the energy involved corresponds to shorter
wavelengths than 400 nm.
Lyman series (ultraviolet)
Name of
series
Paschen series (infra-red)
+
Energy level to which the
excited electron falls
Where the lines
are “seen”
Lyman
n=1
Ultraviolet
Balmer
n=2
Visible
Paschen
n=3
Infrared
1 2 3 4 5 6 7
DON’T FORGET
Balmer series (visible)
These electronic transitions, shown as coloured arrows, correspond to lines with definite values of
frequency and wavelength. The diagram is not to scale and the blue lines going back to the ground state
should be much longer than the other lines because the difference between n = 1 and n = 2 energy
levels is by far the greatest, followed by the difference between n = 2 and n = 3 etc.
Each line in an emission spectrum is due to a transition between definite energy levels.
Using the frequency or wavelength of each line, it is possible to calculate the energy
difference between energy levels.
The structure of the atom as drawn above with electrons orbiting a central positive
nucleus was proposed by the Danish scientist, Neils Bohr. It worked well for hydrogen but
is not a good model for atoms of other elements.
THINGS TO DO AND THINK ABOUT
The energy levels get closer and closer together the further they are from the nucleus. If
you consider the emission spectrum of hydrogen at the high-energy or short-wavelength
end, the lines get closer and closer together until we say that they converge. This is
known as the convergence limit. The convergence limit in the Lyman series for hydrogen
is at 91·2 nm. You can think of this as the excited electron being at its highest energy level
and losing energy as it drops down to the ground state. The highest energy level is as
far as the electron can be from the nucleus without being completely removed from the
hydrogen atom. Consider this happening in reverse, so that the electron moves from the
ground state to just beyond the highest energy level – ionisation would have occurred.
The lines in an emission
spectrum give information
about the difference in
energy levels when an
excited electron moves to
a lower energy level and
energy is released.
VIDEO LINK
You will get more
information on the
atomic emission
spectrum of hydrogen
by watching the clips at
www.brightredbooks.net
ONLINE TEST
Test yourself on this topic
at www.brightredbooks.net
Now consider 1 mole of atoms in the gas state forming 1 mole of ions in the gas state.
The energy required to do this would be the ionisation energy of hydrogen. If you
calculate the energy, in kJ mol–1, corresponding to the wavelength of the convergence
limit in the Lyman series, you should find that it is almost exactly the same as the value
for the ionisation energy of hydrogen given in the SQA Data Booklet. Try this for yourself.
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Inorganic and Physical Chemistry: Reaction Feasibility 1
INORGANIC AND PHYSICAL CHEMISTRY
We know that the enthalpy change (∆H) for the reaction, R → P, is given by:
∆H = HP − HR
DON’T FORGET
The standard state of a
substance is its most stable
form at a pressure of one
atmosphere and usually at a
temperature of 298 K.
where HP and HR are the enthalpies of the products and reactants respectively. This
expression tells us that ∆H could be calculated if we knew the actual enthalpies of all the
reactants and products. However, there is no way we can determine the absolute value of
the enthalpy of a substance. Only values relative to an arbitrary reference point can be
given and for all enthalpy expressions, this reference point is called the standard enthalpy
of formation. The standard enthalpy of formation (∆H°f ) is defined as the enthalpy
change involved when one mole of a substance is formed from its elements in their
standard states. The standard state of a substance is its most stable form at a pressure of
one atmosphere and at a specified temperature which is normally taken as 25°C or 298 K.
It follows from the above definition that the standard enthalpy of formation of an element
in its most stable form is zero.
You will find some standard enthalpies of formation on page 10 of the SQA Data Booklet.
Now let’s consider how standard enthalpies of formation can be used to calculate the
standard enthalpy change for a chemical reaction. First we calculate the total standard
enthalpy of formation of all the products and then we calculate the total standard
enthalpy of formation of all the reactants. The difference between the two totals is the
standard enthalpy change for the reaction. This is expressed as follows:
∆H° = Σ∆H°f (products) – Σ∆H°f (reactants)
where Σ (capital sigma) means ‘the sum of’.
We’ll apply this relationship in the following example:
Calculate the standard enthalpy change for the following reaction:
4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(l)
given that the standard enthalpies of formation of NH3(g), NO2(g) and H2O(l) are −46, +34
and –286 kJ mol–1 respectively.
∆H° =
Σ∆H°f (products) – Σ∆H°f (reactants)
= [4(+34) + 6(-286)] – [4(-46) + 7(0)]
= –1396 kJ mol–1
It is important to remember that the standard enthalpy of formation of each substance
has been multiplied by its corresponding stoichiometric coefficient, e.g. the stoichiometric
coefficient for NH3(g) is 4 since there are 4 mol of NH3(g) in the balanced chemical equation.
ENTROPY AND ENTROPY CHANGE
The entropy of a substance is a measure of the disorder within that system – the larger
the entropy the greater the disorder. Entropy is given the symbol S and the standard
entropy of a substance, S°, is the entropy of one mole of the substance at a pressure of
one atmosphere and normally at a temperature of 298 K. Standard entropy values, S°, for
some selected substances are given on page 17 of the SQA Data Booklet. Notice that the
units of entropy are J K–1 mol–1.
Substances in the solid state tend to have low entropy values. This is not surprising since
the particles in a solid occupy approximately fixed positions. They can vibrate but can’t
move from one place to another. Solids, therefore, are highly ordered. Gases, on the
contd
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BRP_CfEAdvHigherChemistry_Sample.indd 60-61
other hand, have very high entropy values.
Gases contain particles which have complete
freedom of movement and, as a result, are
highly disordered. The entropies of liquids lie
between these two extremes.
The graph on the right shows how the entropy
of a substance varies with temperature.
gas
Entropy/J K–1 mol–1
STANDARD ENTHALPY OF FORMATION
SAMPLE PAGES – CFE ADVANCED HIGHER CHEMISTRY
REACTION FEASIBILITY 1
At 0 K, the particles in a solid are no longer
vibrating and are perfectly ordered. So the
entropy of a substance at 0 K is zero. This is
known as the Third Law of Thermodynamics.
boiling
DON’T FORGET
The Third Law of
Thermodynamics states
that the entropy of a
substance at 0 K is zero.
liquid
solid
melting
Temperature/K
As temperature increases from 0 K, the entropy of the solid substance increases gradually
until its melting point is reached. Here there is a rapid increase in entropy as the
substance changes state from solid to liquid. You can also see that there is an even larger
increase in entropy at its boiling point as the substance changes state from liquid to gas.
VIDEO LINK
Check out the film
clip giving more
information on entropy at
www.brightredbooks.net
So knowing the states of the reactants and products in a chemical reaction should
allow us to predict whether the reaction is accompanied by an increase or a decrease in
entropy. Consider, for example, the reaction 2Na(s) + Cl2(g) → 2NaCl(s). We know that
the entropies of solids are very much smaller than the entropy of a gas and since this
reaction results in a decrease in the number of moles of gaseous molecules (1 to 0), the
entropy will decrease. Similarly, we would predict an entropy increase for the reaction
CaCO3(s) → CaO(s) + CO2(g) since there is an increase in the number of moles of gaseous
molecules (0 to 1). But the entropy change for the reaction CaSiO3(s) → CaO(s) + SiO2(s)
would be difficult to predict since the reactants and products are solids and are likely to have
very similar entropy values. All we can say is that the entropy change is likely to be small.
Changes in entropy have the same sign convention as changes in enthalpy – for increases
in entropy, ∆S° will take a positive sign, i.e. ∆S° > 0, while for decreases in entropy ∆S° will
take a negative sign, i.e. ∆S° < 0.
We have seen that by considering the states of the reactants and products in a chemical
reaction, we can get a qualitative idea of the entropy change but this can be quantified
using the expression:
∆S° = ΣS° (products) − ΣS° reactants)
Consider the following example:
Using information from the SQA Data Booklet and the fact that silver(I) nitrate and
nitrogen dioxide have S° values of 142 and 241 J K–1 mol–1 respectively, calculate the
standard entropy change, ∆S°, for the decomposition of silver(I) nitrate:
2AgNO3(s) → 2Ag(s) + 2NO2(g) + O2(g)
By substituting in the expression ∆S° = ΣS° (products) − ΣS° (reactants) we obtain:
∆S° = [2(43) + 2(241) + (205)] – [2(142)] = +489 J K–1 mol–1.
THINGS TO DO AND THINK ABOUT
1 2ClF3(g) + 2NH3(g) → N2(g) + 6HF(g) + Cl2(g) ∆H° = –1196 kJ mol–1
NH3(g): ∆H°f = −46 kJ mol–1 HF(g): ∆H°f = –271 kJ mol–1
Using the above information, calculate the standard enthalpy of formation of ClF3(g)
DON’T FORGET
When calculating ∆H° and
∆S° values for a reaction
make sure you multiply
the standard enthalpy of
formation and the standard
entropy of each substance
by its corresponding
stoichiometric coefficient.
ONLINE TEST
Head to
www.brightredbooks.net
and take the test
‘Reaction feasibility 1’.
2 a
Predict qualitatively the sign of the entropy change expected for each of the
following reactions:
(i) I2(g) → 2I(g) (ii) Ag+(aq) + Cl−(aq) → AgCl(s) (iii) N2(g) + O2(g) → 2NO(g)
b Using information in the SQA Data Booklet and the fact that the standard
entropy of NO(g) is 211 J K–1 mol–1, calculate the standard entropy change for
reaction (iii) in part (a) above.
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Organic Chemistry and Instrumental Analysis: Molecular Orbitals 3
ORGANIC CHEMISTRY AND INSTRUMENTAL ANALYSIS
MOLECULES AND COLOUR
DON’T FORGET
White light can be thought
of as a combination of three
primary colours – red, green
and blue.
In the inorganic section we looked at transition metal compounds that are coloured due
to their absorption of visible light. The absorption of visible light by these compounds
results in electrons moving from a low energy state to a higher energy state. The colour
of the compound is seen as the complementary colour to the light absorbed. To revise
this topic turn to page 30.
Organic compounds also absorb electromagnetic energy. Most organic compounds are
colourless as the wavelength of light absorbed does not lie in the visible range of the
electromagnetic spectrum. There are some organic compounds that have colour. To
explain this we need to consider the arrangement of electrons in an organic molecule.
MOLECULAR ORBITALS AND ORGANIC COMPOUNDS
When two atomic orbitals
overlap end-on along
the axis of the bond, a σ
molecular orbital forms.
Atomic orbitals that
overlap side-on, form π
molecular orbitals.
As we have already seen, two molecular orbitals form when two atomic orbitals overlap
– a bonding and an anti-bonding molecular orbital. End-on overlap of atomic orbitals
along the axis of the bond result in σ and σ* molecular orbitals forming. Side-on overlap
of atomic orbitals at an angle perpendicular to the axis of the bond result in π and π*
molecular orbitals forming.
π* anti-bonding
Electrons fill the bonding molecular orbitals
first since they have lower energy than the antibonding molecular orbitals. So under normal
conditions the anti-bonding molecular orbitals
will be empty. In a molecule the orbital containing
electrons with the highest energy is known
as the Highest Occupied Molecular Orbital
(HOMO). Similarly, the name given to the lowest
energy molecular orbital that is empty is Lowest
Unoccupied Molecular Orbital (LUMO).
Increasing energy
DON’T FORGET
π* orbital
σ* anti-bonding
σ* orbital
π bonding orbital
σ bonding orbital
This electron transition corresponds to an
electron moving from the HOMO to LUMO
in a simple organic molecule containing
only carbon-to-carbon single bonds.
In organic molecules containing only carbon-to-carbon single bonds and with no nonbonding electrons, the highest occupied molecular orbital is the σ bonding molecular orbital
and the lowest unoccupied orbital is the σ* anti-bonding molecular orbital. The energy
gap between the HOMO and LUMO is large and corresponds to UV light. No visible light is
absorbed and so these compounds are colourless. Even simple molecules with π bonds are
colourless as the energy gap between the π bonding molecular orbital (HOMO) and the σ*
anti-bonding molecular orbital (LUMO) is still too large for visible light to be absorbed.
COLOURED ORGANIC COMPOUNDS
Look at the structures shown of coloured organic compounds.
H
O
H
H
H
H
N
N
H
H
H
H
O
H
contd
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SAMPLE PAGES – CFE ADVANCED HIGHER CHEMISTRY
MOLECULAR ORBITALS 3
HO
O
N
OH
O
You will see that all of these coloured molecules have similar structural features —
relatively large numbers of carbon atoms and multiple double bonds or benzene rings.
These features allow the molecule to have electrons that are delocalised across a number
of carbon atoms. This is known as a conjugated system and for an organic molecule to
be coloured it must contain a large degree of conjugation.
Consider the two molecules below. Molecule A has a system of alternating double and single
bonds and so is a conjugated molecule. Molecule B does not have a conjugated system.
molecule A
When carrying out acid-base titrations you will probably have used phenolphthalein
indicator. It is colourless in acid conditions and pink in alkaline conditions. The structure
of the two forms of phenolphthalein are shown and an equilibrium is established between
the two forms. For more information about pH indicators refer to page 20.
O
C
add alkali
O
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molecular orbitals at
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O–
C
add acid
C
Head to
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for more on bonding.
ONLINE TEST
O–
OH
A system of alternating
single and double bonds
within a molecule results
in conjugation.
ONLINE
THINGS TO DO AND THINK ABOUT
C
A conjugated system
refers to a molecule
which contains delocalised
electrons extending across
a number of carbon atoms.
DON’T FORGET
molecule B
For a molecule to have a conjugated system it must contain a chain of alternating σ and π
bonds or benzene rings that allow electrons to be delocalised across a number of carbon
atoms. For further information about delocalisation in benzene rings see page 52.
HO
DON’T FORGET
O
O
Colourless
Pink
Note: Benzene is normally drawn as
but it can be useful to show the delocalisation
as a system of alternating single and double bonds as shown above.
In both structures electrons are delocalised around the benzene rings. In the structure on
the right hand side delocalisation can occur across the central carbon atom and so covers
the entire structure. The structure on the left has the delocalisation broken – there is no
longer a system of alternating single and double bonds and so there is less conjugation.
The energy difference between HOMO and LUMO is greater and so this molecule does
not absorb light in the visible region.
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