Unit 1 - Introduction to Electronic Structure and the
Periodic table.
The work of Rutherford and others in the early part of the 20 th Century resulted in the
model of the present atom.
It is the electrons, which take part in chemical reactions, and so it is necessary to
understand the electronic structure of the atom to explain its chemical properties.
The key to understanding electronic structure and how electrons behave comes from the
study of electromagnetic radiation. This is described as waves of varying length and
these travel at a constant velocity.
1
Electromagnetic Waves - Revision
1.
Wavelength = the distance between wavecrests.
Measured in metres (m) or nanometres (nm).
1nm = 1x10-9m
The symbol for wavelength is the Greek letter  (lambda).



Velocity = Rate of advance of one wavecrest
The symbol for velocity is c = 3x108 ms-1





Frequency = Number of wavelengths passing a point in one
second.
Measured in the reciprocal of time (s-1) or Hertz (Hz).
The symbol for frequency is the Greek letter  (nu) or f.




Wavenumber = Waves in unit length of radiation ( i.e. 1
metre or 1cm)
Often used instead of frequency.
The symbol used is 




Wavenumber =
2
1
Wavelength


The relationship between wavelength, frequency and velocity is the
equation :
Velocity = Wavelength x Frequency
(ms-1)
(m)
(s-1)
OR
C = x 

Remember that as the wavelength increases, the frequency will decrease.
Try these questions for yourself.
1.
A typical microwave oven operates at a frequency of 2.45 x 10 9
Hz. Calculate the wavelength of this radiation. Give your answer
in centimetres.
2.
A beam of light from a sodium street lamp is found to have a
frequency of 5.09 x 1014 Hz. Calculate the wavelength of this
light to the nearest nanometre.
3.
Databook information records the flame colour of potassium as
lilac and with a wavelength of 405nm. Calculate the frequency of
the radiation.
4.
Use the databook to find the wavelength of light emitted by a
sample of copper and then calculate its frequency.
5.
The wavenumber of radiation the C-H bond in an alkane is
2900cm-1. Calculate the wavelength and frequency of this
radiation.
Ans
1) 12.24cm
2) 589nm
3)7.41 x 1014s-1
14
4)9.23 x 10 Hz
5) 8.69 x 1013 s-1
3
Electromagnetic spectrum (cont.)
Notice that gamma rays are electromagnetic rays but alpha and beta radiation
are not. (This is because these are charged particles).
From this it can be seen that visible light (which excites the nerve cells in the
eye) constitutes a very small part of the spectrum between 400-700nm.
When electromagnetic radiation interacts with matter, there is a transfer of
energy to the receiving body. (E.g. sunning ourselves).
Energy can only be transferred in small bundles or packets, which are, called
quanta. These quanta are of a definite size and therefore the transfer of
energy can only occur in definite amounts. Because of this, it is necessary to
regard electromagnetic radiation as not only waves but also as a stream of very
small particles called photons.
Electromagnetic radiation is considered to be a stream of photons with wave
properties, the energy of the radiation being related to the wavelength or
frequency of the radiation by the equation:
For a photon
E = hf
For one mole of photons
E = Lhf or E = Lhc

Where h = Planck’s constant = 6.63x 10-34Js
F = frequency
L = Avogadro constant = 6.02 x 1023
C = velocity = 3x108 ms-1
 = wavelength in metres
4
Electromagnetic spectrum (cont.)
Example
Calculate the energy in kJmol-1 for one mole of photons, which have a wavelength
of 656nm.
E = Lhc



= 6.02 x 1023 x 6.63x 10-34 x 3x108
656 x 10-9m
= 1.82 x 105 Jmol-1
= 182 kJmol-1




Try for yourself
1.
Calculate the energy, in kJmol-1, corresponding to
a) a wavelength of 620nm.
b) A wavenumber of 1000cm-1
2. The bond enthalpy of a Cl-Cl bond is 243 kJmol-1.
Calculate the maximum wavelength of light that would break one mole of
these bonds to form one individual chlorine atom.
Ans
1a) 193.1 kJmol-1
b) 11.974 kJmol-1
5
2) 493nm
Electronic Configuration
Spectroscopy
When a beam of white light is passed through a prism or from a diffraction
grating onto a screen, a continuous spectrum is seen. The same effect can be
seen in a rainbow. (Figure a)
Line Emission Spectra or Atomic Emission Spectra
However, if the light source is supplied by sodium chloride being heated in a
Bunsen burner flame, the spectrum turns out not to be a continuous spectrum
but a series of lines of different wavelengths and thus of different colours.
Spectra that show energy being given out by an atom or ion are called atomic
emission spectra as shown. The pattern of lines in such a spectrum is
characteristic of each element and, like a fingerprint, can be used to identify
the element. (Figure b)
6
Atomic Absorption Spectra
If a beam of continuous radiation like white light is directed through a gaseous
sample of an element, the radiation that emerges has certain wavelengths
missing. This shows up as dark lines on a continuous spectrum and is called an
atomic absorption spectrum. (Figure c)
In both techniques some lines normally occur in the visible region (400-700 nm)
but some applications use the ultraviolet region (200-400 nm). Both emission and
absorption spectroscopy can be used to determine whether a certain species is
present in a sample and how much of it is present, since the intensity of
transmitted or absorbed radiation can be measured.
More on Spectra
A species can only absorb or emit energy in quantised bundles. The energy values
are described as the energy levels and the lowest level is called GROUND
STATE.
If energy is absorbed it can cause an electron in ground state to increase its
energy and move from ground state to an EXCITED STATE.
Excited states
Ground state
ABSORPTION
This ‘excited’ electron can fall back to ground state and in this case this
would show lines in an emission spectra and emit energy. This often
corresponds with the the visible or UV spectrum. Each line in a spectra
corresponds to a specific wavelength or frequency from which the
differences in energy can be calculated.
Excited states
Ground state
EMISSION
7
The Hydrogen Spectrum
Although hydrogen only has 1 electron, the emission spectrum has different
series of lines in different parts of the electromagnetic spectrum.
The differences in energy and hence the part of the electromagnetic
spectrum in which the lines show up depend on the energy level to which the
‘excited’ electron falls back. The full emission spectrum of hydrogen consists
of one series of lines in the UV region, one series in the visible region and
several in the IR region.
Name of Series
Lyman
Balmer
Paschen
Brackett
Pfund
Energy level to which the
excited electron drops to
n=1
n=2
n=3
n=4
n=5
8
Electromagnetic
Spectrum
UV
Visible
IR
IR
IR
The Hydrogen Spectrum (cont)
As the diagram shows, this spectrum would be achieved when the electron,
having been excited into higher energy levels, drop back down to level n=2
and emit radiation in the visible region. This is called the Balmer series.
Electrons dropping to ground state n=1 would emit in the UV region.
This is called the Lyman series.
Also notice that the levels get closer together the further they from the
nucleus.
9
The Hydrogen Spectrum and Ionisation Energy
In the previous diagram, an electron has been shown ‘breaking away’. This
would be the ionisation energy for that e-.
To find the ionisation energy from the spectra details, look for the highest
energy line in the Lyman series where all the lines converge. In hydrogen this
occurs at 91.2nm.
OR look for the line with the lowest wavelength or highest
frequency.
The ionisation energy can then be calculated from this wavelength.

Example
E = Lhc


= 6.02 x 1023 x 6.63x 10-34 x 3x108
91.2 x 10-9m
= 1312.9kJmol-1
Try this question
1.
Calculate the ionisation energy for an atom with the frequency
1.26 x 1015s-1.
10
Summary so far
1.
Matter can only emit energy in small fixed amounts called quanta.
2.
Electrons move around the nucleus in fixed orbits possessing a
certain amount of energy. These orbits (energy levels) are given
numbers
E.g.
Energy Level 1 = Ground State
3.
A photon of light is emitted or absorbed when the electron changes
from one energy level to another.
4.
The energy of the photon is equal to the difference between the 2
energy levels, which is related to the frequency by the equation
E = Lhf.
The scientist Bohr carried out much of the work required to interpret and
explain emission spectra. He developed a model for the structure of the atom
using Planck’s quantum theory, which lead to a new science of quantum mechanics.
This is where electrons as well as waves and particles are considered.
An electron possesses a fixed amount of energy called quanta. The energy
of the electron can be defined as quantum numbers.
11
Quantum Numbers
Emission spectra of elements with more than one electron provide evidence of
sub-levels within each principal energy level above the first. Quantum theory now
defines the allowed energy levels of electrons by four quantum numbers. No two
electrons in an atom can have the same four quantum numbers.
Principal Quantum Number (n)
This is the energy level
e.g. n = 1 is the shell closest to the nucleus.
n = 2 is the second shell. Etc
Second Quantum number (l)
It was observed that other elements, in a spectra analysis, did not just have
single lines but were made up of lots of other lines. This suggests that the shell
is split into sub-shells.
l determines the shape of the sub-shell and is labelled s, p, d, f. This can
have values from zero to (n-1).
Principal Quantum No. (n)
2nd Quantum No.(l)
Sub-shell name
1
0
1s
2
0
1
2s
2p
3
0
1
2
3s
3p
3d
4
0
1
2
3
4s
4p
4d
4f
The sub-shells within a shell have different energies.
Increasing energy
s
p
d
12
f
Shapes of These Orbitals
An orbital is the volume of space where the probability of finding and
electron is more than 90%. Each atomic orbital can hold a maximum of two
electrons
1.
s- Orbitals


2.
Are spherical
As the principal number increases, the distance of the electron from the
nucleus increases and the orbital becomes more diffuse.
p-Orbitals



Are dumb-bell shaped and aligned along an axis
3 types
Are all of equal energy and are described as degenerate.
13
3.
d-Orbitals




4.
5 types
3 lie in the plane of the axes (dxy, dyz, dxz).
The other two lie along the axes (dx2-y2, dz2)
They are degenerate in energy.
f-Orbitals


7 types
Too complex to draw on paper
14
Third Quantum number (m)
This relates to the orientation in space of the orbital. It is dependent on l and
can take the value of:
M = +l………..0……….-l
n
l
m
Type of atomic orbital
1
0
0
1s – 1 type
2
0
1
0
-1, 0, +1
2s – 1 type
2p – 3types
3
0
1
2
0
-1, 0, +1
-2, -1, 0, +1, +2
3s
3p – 3 types
3d – 5 types
Forth Quantum number (s)
There is a spin quantum no. which arises from electrons spinning on their axes.
The spin quantum no gives the direction of the electron spin and it can only occur
in 2 directions – clockwise or anti-clockwise.
In any orbital containing 2 electrons they must be paired, with the spins
opposed.
S= + ½ or
-½
15
Electronic Configuration
Electrons can now be shown more clearly either by the orbital box notation or
the spectroscopic notation by following the rules below.
LEARN THESE RULES
RULE 1
The Aufbau principle (From German 'building up'.)
When electrons are placed into orbitals the energy levels are filled up in order
of increasing energy, e.g. the 1s has filled before 2s.
RULE 2
The Pauli exclusion principle
This states that an orbital cannot contain more than two electrons and they
must have opposite spins, e.g. is NOT allowed.
RULE 3
Hund's rule
This states that when there are degenerate orbitals in a sub shell (as in 2p),
electrons fill each one singly with spins parallel before pairing occurs. Thus
carbon has not paired up the two 2p electrons.
It is necessary to fill the energy levels in order of the lowest first.
1s 2s 2p 3s 3p 4s 3d 4p 5s
Increasing Energy----->
16
Notice that 4s get filled before 3d. Although 4s is further from the nucleus in terms of
space, it is lower than energy and gets filled up first.
Examples
Carbon 1s2 2s2 2p4
Aluminium 1s2 2s2 2p6 3s2 3p1
1s2
Iron
1s2
2s2
2p6
3s2
1s2 2s2 2p6 3s2 3p6 4s2 3d6
2s2
2p6
3s2
OR
3p6
3p1
[Ar] 4s2 3d6
4s2
3d6
Try for yourself
1.
Vanadium
2.
Oxygen
3.
Lithium ion
(Normally electrons are lost or gained from the highest energy level to form the ion)
17
The Periodic Table
The periodic table can be subdivided into 4 blocks (s,p,d and f) corresponding to the
outer electronic configurations of the elements within these blocks.
s-block
elements
p-block
elements
d-block elements
f-block elements
18