Topic 12 ATOMIC STRUCTURE: Electronic configuration
12. 1. Ionization energy
If an electron obtains enough energy it jumps beyond the highest energy level and becomes a free
electron: the atom itself becomes a positive ion.
This electron transition is represented as follows: n = 1 to n = . Notation n =  refers to the “world”
outside the atom – beyond the control of the nucleus; any electron outside an atom has no potential
energy (energy = 0).
With hydrogen the ionization energy can be calculated from the frequency at the point in the emission
spectrum where the lines in the Lyman series converge. (see extension in topic 2 notes)
(http://www.physchem.co.za/Atomic/Hydrogen%20Spectrum.htm: shows energy levels)
There are different types of ionization energies:
 first ionization energy (IE1):
 the minimum energy needed to remove one (first/highest energy level)/most loosely held electron
electron from 1 mole of gaseous (or free/by itself) atoms to make a gaseous cations;
X (g)
 X+ (g)
+ e-
 the more strongly the electron is attracted to the nucleus, the greater the amount of energy
needed
 ionization energy is usually measured in kilojoules per mole Kj mol-1 of atoms;
 ionization energies are positive values (endothermic process) as energy is needed to remove an
electron.
 successive ionization energy:
 energy needed to remove second, third,... electrons from 1 mole of gaseous ions, e.g.
second ionization energy (IE2)
X+ (g)
third ionization energy (IE3):
X2+ (g)


X2+ (g)
X3+ (g)
+ e+ e-
 the lower the energy level an electron is on, the closer it is to the nucleus, the more strongly it is
attracted, the more energy is needed;
 in the same atom, successive ionization energies increase as a result of a combination of the
following factors:
 the electron-electron repulsion decreases (this repulsion puts electrons onto higher energy
levels) and this increases the attraction from the nucleus, but also because
 the first electron is removed from an atom whilst the second and successive electrons are
removed from ions with an increasing positive charge !!!!
 nucleus has greater effect on smaller number of electrons and they are held more tightly
These ionization energies provide evidence on how an atom is structured and on the number of
electrons on the energy levels.
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12. 2 Successive ionization energies: evidence for the main energy levels and the
number of electrons that can occupy an energy level.
12.1.2 Explain how successive ionization energy data is related to the electron configuration of an atom.
All the successive ionization energies for the same element can be shown on a graph of successive
ionization energies against electrons removed.
In the space below sketch the graph for sodium.
.
Notice that the scale for ionization energies is a logarithm scale so that the large differences between
the ionization energies can be shown onto the same graph.
Observations:
Explanation:
1 electron has a much smaller ionization
energy than the rest which means it needs a lot
less energy for the electrons to escape to n =
.
8 electrons require a lot more energy than the 2
electrons at the higher energy levels
This electron must be on a much higher energy level
than the rest; most of the time it is the furthest away
from the nucleus and held less strongly.
2 electrons need an even higher amount of
ionization energy
there are 2 electrons on an even lower energy level;
most of the time they are closer to the nucleus than
the other electrons and are held very tightly.
there are 8 electrons in the same lower energy level
closer to the nucleus.
Write equations for the first 4 ionization energies of magnesium.
Sketch graphs showing successive ionization energies for Mg, Cl, F, Al.
Study the table (from http://www.shodor.org/chemviz/ionization/students/background.htm ) below
showing the first seven ionization energies for the first 12. For each element write down what the
ionization energies say about the electron arrangement.
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Ionization Energies in kJ/mol
1
2
3
4
5
6
7
8
H
1312
He
2372
5250
Li
520
7297
11810
Be
899
1757
14845
21000
B
800
2426
3659
25020
32820
C
1086
2352
4619
6221
37820
47260
N
1402
2855
4576
7473
9442
53250
64340
O
1314
3388
5296
7467
10987
13320
71320
84070
F
1680
3375
6045
8408
11020
15160
17860
92010
Ne
2080
3963
6130
9361
12180
15240
Na
496
4563
6913
9541
13350
16600
20113
25666
Mg
737
1450
7731
10545
13627
17995
21700
25662
12.3 First ionization energies: evidence for main energy levels and sub-levels
12.1.1 Explain how evidence from first ionization energies across periods accounts for the existence of main energy levels
and sublevels in atoms.
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Observation
Evidence for:
There is a repeated pattern (=periodic trend)
in the peaks/highest values for noble gases
and lowest values (alkali metals).
Overall increase across a period.
Decrease in ionization energy between He
and Li (Ne and Na; Ar and K)
There is no continuous increase between Li
and Ne ( this period of 8 elements is made
up of smaller groups of 2, 3 and 3)
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Existence for main energy levels:
 same pattern is repeated as al 8 first electrons in
period 3 are in a higher main energy level than
those in period 2
 Noble gases have full main energy levels
Existence of main energy level:
Electrons are added to same main energy level but
nuclear charge increases.
Existence of main energy level:
The large drop from He to Li can be explained in that
the 3rd Li electron (the electron in the highest energy
level) is at a higher main energy level than the 2nd
electron (the electron in the highest energy level) in
He; any large decrease in ionization energy means
that the electron is in a higher energy level/less
strongly attracted/further away from the nucleus.
Existence of sub-level:
If all the valence electrons of the elements in the
second period were in the same main energy level
(n=2), a normal steady rise in first ionisation energy
would be expected. This is because the electrons
would be attracted more strongly by the greater
nuclear charge and decrease in atomic radius.
However, instead there are 2 drops in ionisation
energy between Li and Ne; not all the most outer
electrons of the 8 elements in that period 2 are the
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same as they are not on the same energy level
A decrease from Be to B; this decrease
occurs after the second element in the
period
Existence of sub-level:
 the drop from Be to B indicates that the 5th B
electron is at a higher energy level than the 4th Be
electron both of which should be on the 2nd main
energy level; this has been explained by the fact
that the 5th B electron is on a higher sub-level within
the 2nd main energy level;
 the lowest level at this main energy level is full after
2 electrons
There are 6 elements between Be and Ne
(Ne including) after which there is again a
great decrease
A decrease from N to O (also P to S)
The second level in this main energy level can hold 6
electrons
Existence of sub-level:
The electron arrangement ensures that the most outer
electron in oxygen is at a higher energy level than the
most outer electron in nitrogen. This is the case
because the electron removed in O is a paired electron
(two electrons in same orbital) and this causes greater
repulsion and less attraction.
Further discussion: section between Ar and Kr (see extra photocopies):
 Ar - Ca: 2 points = filling up of a sub-level with room for two electrons (4s)
 Sc - Zn: 10 points = filling up of sub-level with room for ten electrons (3d)
 Ga - Kr: 6 points = filling up of a sub-level with room for 6 electrons (4p)
Websites:
 http://www.tutorvista.com/content/chemistry/chemistry-iii/periodicity/ionization-energy.php
 http://www.chemguide.co.uk/atoms/properties/moreies.html
 http://www.vias.org/genchem/img/ionization_energy.png (shows trends without transition block
elements)
12. 4 Relative energies of sub-levels within a single main energy level:
12.1.3 State the relative energies of s, p, d and f orbitals in a single energy level.
relative energies: s  p  d  f
This makes sense as an s electron tends to be on average more closely to the nucleus than electrons
on any of the other sub-levels. Remember that the more closely to the nucleus the more strongly an
electron is attracted, the lower its energy!
12. 5. Orbitals
12.1.4 State the maximum number of orbitals in a given energy level.
12.1.5 Draw the shape of an s orbital and the shapes of the px, py and pz orbitals.
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Bohr’s model was unable to explain the emission spectra of more complex elements so a new way of
thinking was needed.
This new thinking was based on the Heisenberg Uncertainty Principle which states that it is impossible
to pinpoint accurately both the position and the momentum of a small particle like an electron.
In simpler language, this means we can never be certain of where exactly an electron is in an atom.
To deal with this new idea, the concept of atomic orbitals was introduced.
An atomic orbital refers to a volume of space around the nucleus in which there is a high probability
(90%) (not an absolute certainty) of finding an electron of a given energy.
An orbital therefore is an area of high electron density; it is an electron charge cloud which can be
occupied by a maximum of two electrons. Atomic orbitals can have different shapes; the shape
depends on the electron distribution.
These orbitals are found at the sub-levels.
Each sub-level has its own electron distribution both in terms of shape of the orbital and the number of
orbitals.
Orbitals in the same sub-level are equivalent in energy and are called degenerate orbitals.
The orbital in the s-subshell has a spherical symmetrical shape: probability of finding electron is the
same in all directions. The 3 orbitals in the p-subshell have an hourglass shape; the three orbitals are
labeled px, py and pz.
Just like the main energy levels which have been given numbers; each orbital is also given a number
depending on the main energy level it is in and its shape g 1s, 2p, 3d, 4f and so on. The larger the
principal quantum number, the larger the orbital.
main
energy level
sub-level (+ in
brackets number of
electrons it can hold)
maximum number
of orbitals in sublevel
n=1
s (2)
1
spherical
2
n=2
s (2)
p (6)
1
3
spherical
hourglass
8
n=3
s (2)
p (6)
d (10)
1
3
5
spherical
hourglass
lobes
n=4
s (2)
p (6)
d (10)
f (14)
1
3
5
7
spherical
hourglass
lobes
lobes
shape of orbital
total number of
electrons in the
main energy level
18
32
Websites showing orbitals and their shapes:



http://en.wikipedia.org/wiki/Atomic_orbital
http://www.chemistry.wustl.edu/~edudev/Orbitals/movie.html
www.shef.ac.uk/chemistry/orbitron/AOs/ (best one)
12. 6. Filling up of orbitals: electronic configuration
12.1.6 Apply the Aufbau principle, Hund’s rule and the Pauli exclusion principle to write electron configurations for
atoms and ions up to Z = 54.
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Electrons spin on their own axis which can be clockwise or anticlockwise, relative to the orbital of the
electron. Whenever a charge spins it produces a magnetic field; as a result electrons have their own
magnetic fields and they acts as magnets. The direction of the magnetic field of an electron depends
on the direction of its spin. Depending on their spin, two electrons either attract (both have opposite
spin) or repel (have same spin) each other.
Any repulsion between its electrons always increases the potential energy of an atom.
The greater the potential energy, the more unstable an atom becomes so electrons in atoms always
adopt the lowest possible energy configuration which is referred to as the ground state.
To determine the electron arrangement which achieves the lowest possible energy we need to
apply 3 principles which are described below:
1. The Aufbau principle:
When filling sub-levels with electrons, we start with the lowest level, 1s. Each sub-level is filled before
starting to fill the sub-level of the next highest energy.
2. Pauli Exclusion principle:
Remember that two electrons can occupy the same atomic orbital but they can only do so if they have
opposite or paired spins as this reduces the repulsion between them and therefore the total amount of
potential energy in the atom.
The Pauli Exclusion Principle states that an atomic orbital can only hold two electrons if they have an
opposite spin (no repulsion caused by opposite magnetic spin of each electron).
Evidence for Pauli Principle includes the study of the magnetic effect of atoms; for instance consider
the helium atom which could have two possible electronic configurations:
  in this case the magnetic effect of each electron is cancelled out; the atom is
diamagnetic and should only be slightly repelled by a magnet; or
  in this case the magnetic fields of each electron reinforce each other making the atom
paramagnetic and which would cause it to be attracted by a magnet (strength of
paramagnetism indicates number of unpaired electrons);
The experimental evidence shows that a magnet has very little effect on the helium atom suggesting
that is diamagnetic and that electrons cancel out magnetic fields.
3. Hund’s rule
When filling up a sub-level with 3 or more atomic orbitals, the favoured arrangement is where electrons
occupy different atomic orbitals and have the same spin (or “electrons prefer to have parallel spins
which is only possible - as a result of Pauli’s exclusion principle - if they go into a different orbital of the
same sub-level” ) (or “in a sub-level with more than one orbital the electrons are spread out over these
degenerate sub-levels with their spins in the same direction”).
Exercise
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Draw ‘electron-in-the boxes’ and show the electron configuration for the following species: B, F, Al, K,
K+, Cl-, O2-, Al3+, H-, S2-, Mg2+.
The electron structure of atoms: Aufbau, Hund, Pauli
You need to be able to write an electron configuration for the first 56 elements.
1.
Electrons always fill from the lowest energy level upwards so that their total energy is kept at a
minimum. Called the Aufbau principle.
2.
When one energy level is full the electrons start filling up the next highest energy level.
3.
Electrons fill up the orbital in pairs. Each electron in the pair spin in opposite directions to one
another (represented by
). In the p and d orbitals the electrons fill up singularly first and then
as pairs. Called Hund’s rule.
4.
It is impossible for two electrons with the same spin to be in the same orbital and an orbital can
only hold a maximum of two electrons. Called the Pauli Exclusion principle.
5.
The electron structure is written in order of increasing energy level
number of electrons
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 etc
energy
level
orbital
A short hand version can be used for more than 20 electrons by using an argon, Ar core (18
electrons.
e.g. for titanium with 22 electrons the electron configuration is 1s2 2s2 2p6 3s2 3p6 4s2 3d2 or
Ar 4s2 3d2
6.
The 4s orbital is lower in energy than the 3d orbital and so the 4s orbital will be filled before the
3d.
7.
One exception to the Aufbau principle is chromium, Cr (atomic number = 24)
You would expect its electron configuration to be Ar 4s2 3d4 (4s __ 3d __ __ __ __ __)
but its electrons configuration is actually Ar 4s1 3d5 (4s __ 3d __ __ __ __ __)
This is because the energy of the 4s and 3d orbital are very similar and they almost act as one
orbital. ½ filled orbitals or completely full orbitals are more stable than a mixture of partially
filled orbitals so the most stable configuration is Ar 4s1 3d5.
Another exception is copper, Cu (atomic number = 29)
You would expect its electron configuration to be Ar 4s2 3d9 (4s __ 3d __ __ __ __ __)
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but its electrons configuration is actually Ar 4s1 3d10 (4s __ 3d __ __ __ __ __)
Here a full 3d orbital and a ½ filled 4s orbital is more stable than a partially filled 3d orbital.
IB past paper questions on both topic 2 and 12
1. (N05) Which equation represents the third ionization energy of an element M?
2. (N06) What are valence electrons?
A. Electrons in the energy level closest to the nucleus
B. Electrons in the highest main energy level
C. The number of electrons required to complete the highest main energy level
D. The total number of electrons in the atom
3. (N06) What is the total number of electrons in p orbitals in an atom of iodine?
A. 5
B. 7
C. 17
D. 23
4. (M06) Which statement is correct about electron orbitals and energy levels?
A. Yttrium, Y, (Z = 39) is the first element in the periodic table with an electron in a f sub-level.
B. The maximum number of electrons in one d orbital is 10.
C. The maximum number of electrons in the 4th main energy level is 18.
D. In a main energy level, the sub-level with the highest energy is labeled f.
5. (M06) Which is correct about the element tin (Sn) (Z = 50)?
6. (N01) In which of the following ground-state electron configurations are unpaired electrons
present?
I. 1s2 2s2 2p2
II. 1s2 2s2 2p3
III. 1s2 2s2 2p4
A. II only
B. I and II only
C. II and III only
D. I, II and III
7. (N01) Which atom or ion has the electron configuration 1s2 2s2 2p6 3s2 3p6 3d7 ?
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A. Co
C. Co2+
B. Mn
D. Fe3+
8. (N02) A transition metal ion X3+ has the electronic configuration [Ar] 3d4 . What is the atomic
number of element X?
A. 22
B. 24
C. 25
D. 27
9. (N02) Which of the following electronic configurations gives rise to the largest increase between
the second and third ionization energies?
10. (M99) Using the Aufbau Principle, deduce which element below has the greatest number of
unpaired electrons in its ground state.
A. Z = 13
B. Z= 14
C. Z = 15
D. Z = 16
11. (M99) Values for the first ionization energies (IE) for five successive elements in the Periodic
Table are given below. Based on these values, which statement is correct?
Element
IE (kJ mol-1)
A.
B.
C.
D.
K
1060
L
1000
M
1260
N
1520
O
418
P
?
the outermost electron in element K is in a higher energy level than that in element L
element M is probably a metal
element N is probably a non-metal
element P has a lower first ionization than element O
12. (specpap) An atom of which element in its ground state will have the largest number of unpaired
electrons?
A. Si
B. P
C. S
D. Cl
13. (M00) The first 4 ionization energies (kJ mol-1 ) for a particular element are 550, 1064, 4210 and
5500 respectively. This element should be placed in the same group as
A. Li
B. Be
C. B
D. C
14. (N98) Which ionization energy requires the greatest amount of energy?
A. Na (g)  Na+ (g) + eC. Mg (g)  Mg+ (g) + e-
B. Na+ (g)  Na2+ (g) + eD. Mg+ (g)  Mg2+ (g) + e-
15. (M98) All of the following effect the value of the ionization energy of an atom except the
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A. mass of the atom
B. size of the atom
B. charge on the nucleus
D. main energy level from which electron is removed.
16. (N00) A solid element, X, contains unpaired electrons in its atoms and forms an ionic chloride,
XCl2. Which electron configuration is possible for element X?
A. [Ne] 3s2
B. [Ar] 3d2 4s2
C. [He] 2s2 2p2
D. [Ne] 3s2 3p4
PAPER 2
1. (N05) A sample of germanium is analyzed in a mass spectrometer. The first and last
processes in mass spectrometry are vaporization and detection.
(a) (i) State the names of the other three processes in the order in which they occur in a
mass spectrometer.
[2]
(ii) For each of the processes named in (a) (i), outline how the process occurs.
[3]
(b) The sample of germanium is found to have the following composition:
(i) Define the term relative atomic mass.
[2]
(ii) Calculate the relative atomic mass of this sample of germanium, giving your answer to
two decimal places.
[2]
(c) Use the Aufbau principle to write the electron configuration of an atom of germanium.
[1]
(d) The successive ionization energies of germanium are shown in the following table:
(i) Identify the sub-level from which the electron is removed when the first ionization
energy of germanium is measured.
[1]
(ii) Write an equation, including state symbols, for the process occurring when measuring
the second ionization energy of germanium.
[1]
(iii) Explain why the difference between the 4th and 5th ionization energies is much greater than
the difference between any two other successive values.
[2]
2. (N06) Some vaporized magnesium is introduced into a mass spectrometer. One of the ions that
reaches the detector is 25Mg+
(a) Identify the number of protons, neutrons and electrons in the 25Mg+ion.
(b) State how this ion is accelerated in the mass spectrometer.
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[1]
[1]
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(c) The 25Mg 2+ion is also detected in this mass spectrometer by changing the magnetic field.
Deduce and explain, by reference to the m/z values of these two ions of magnesium,
which of the ions 25Mg+ and 25Mg 2+ is detected using a stronger magnetic field.
[2]
3. (N01)
(a) The isotopes of sulfur occur naturally in the following percentages:
(i) Calculate the relative atomic mass of sulfur to three significant figures.
[2]
(ii) Determine the number of neutrons in the atom of the least abundant sulfur isotope.
[1]
(b) (i) Describe how the first four ionization energies of aluminium vary. (You may wish to
sketch a graph to illustrate your answer.)
[2]
(ii) State the electronic configurations of aluminium, boron and magnesium. Explain how
the first ionization energy of aluminium compares with the first ionization energies of
boron and magnesium.
[5]
(c) When hydrogen gas is placed in an electric discharge tube, an emission spectrum is obtained.
Sketch the spectrum, labeling its high energy end. Explain why such a spectrum is obtained. [3]
4. (N98) The successive ionization energies for boron are given below in kJ mol-1.
1st
799
2nd
2420
3rd
3660
4th
25000
5th
32800
(a) Explain the reason why there is a large increase between the third and fourth values.
(b) Explain the reason why the increase between the first and second ionization energies is
more than the increase between the second and third ionization energies.
(c) State, with reasons, how the value for the second ionization energy of carbon would
compare with that of the second ionization energy of boron.
[6]
5. (M99) Use the modern theory of the atom to answer each of the following.
(a) List the d, f, p and s orbitals in order of increasing relative energy.
[2]
(b) Give the number of each type of orbital, d, f, p and s ate ach energy level.
[2]
(c) Describe the changes which occur when hydrogen produces a line spectrum.
[2]
(d) Explain why the electron configuration of the nitrogen atom is written as N:
1s22s22p12p12p1 rather than 1s22s22p22p12p0. Write the electron configuration of titanium.
[3]
(e) (i) name the instrument used to determine the atomic masses of the two naturally
occurring isotopes of gallium. Briefly describe each step involved in the operation of the
instrument.
[3]
(ii) A certain sample of gallium contains 60% Ga-69 and 40% Ga-71. Give the nuclear
structures of these isotopes and calculate the relative atomic mass of gallium in this
sample.
(f) Explain the difference in the two values of ionization energy for each of the following
pairs:
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i.
The 1st ionization energy of beryllium is 900 kJ mol-1 whereas the 2n ionization energy of
beryllium is 1757 kJ mol-1.
ii.
The 1st ionization energy of aluminium is 577 kJ mol-1 whereas the 1st ionization energy
of magnesium is 736 kJ mol-1.
iii.
The 1st ionization energy of aluminium is 577 kj mol-1 whereas the 1st ionization energy
of boron is 799 kJ mol-1.
Mark scheme
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Paper 1
1
2
3
4
5
6
B
B
D
B
C
D
7
8
9
10
11
12
C
D
A
C
C
B
13
14
15
16
17
18
B
B
A
B
19
20
21
22
23
24
25
26
27
28
29
30
Paper 2.
1. (N05)
2. (N06)
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3. (N01)
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5.
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