IB Chemistry
ATOMIC THEORY
Atomic Structure
Atomic Structure
Atoms are very small ~ 10-10 meters
All atoms are made up of three sub-atomic
particles: protons, neutrons and electrons
The
protons and neutrons form a small
positively charged nucleus
The electrons are in energy levels outside the
nucleus
Atomic Structure
The actual values of the masses and charges of the
sub-atomic particles are shown below:
A meaningful
way to consider the masses of the subatomic particles is to use relative masses
Atomic Structure -
Definitions
Atomic number (Z) is the number of protons
in the nucleus of an atom. The number of
protons equals the number of electrons in a
neutral atom
 N.B. No. of protons always equals the no. of
electrons in any neutral atom of an element.
Mass number (A) is the sum of the number of
protons and the number of neutrons in the
nucleus of an atom.
So
how can you work out the number of neutrons in an atom?
No. of neutrons = Mass number – atomic number
Atomic Structure -
Example
So
how can you work out the number of neutrons in
an atom?
Example
No. of neutrons = Mass number – atomic number
No. of neutron = Mass No. – Atomic No.
= 23 – 11
= 12
Atomic Structure -
Questions
1. What are the three sub atomic particles that make
2.
3.
4.
5.
up the atom?
Draw a representation of the atom and labelling the
sub-atomic particles.
Draw a table to show the relative masses and
charges of the sub-atomic particles.
State the atomic number, mass number and
number of neutrons of: a) carbon, b) oxygen and
c) selenium.
Which neutral element contains 11 electrons and
12 neutrons?
Atomic Structure -
Questions
5. Copy and complete the following table:
Summary Slide
All atomic masses are relative to the
mass of carbon-12.
Eg one hydrogen atom weighs 1/12 the
mass of a carbon-12 atom.
Isotopes
Isotopes are atoms of the same element with the
same atomic number, but different mass numbers,
i.e. they have different numbers of neutrons.
Each atom of chlorine
contains the following:
35 Cl
17
17 protons
17 electrons
18 neutrons
37 Cl
17
17 protons
17 electrons
20 neutrons
The isotopes of chlorine are often referred to as
chlorine-35 and chlorine-37
Isotopes
Isotopes of an element have the same chemical
properties because they have the same number of
electrons. When a chemical reaction takes place, it
is the electrons that are involved in the reactions.
However isotopes of an element have the slightly
different physical properties because they have
different numbers of neutrons, hence different
masses.
The isotopes of an element with fewer neutrons
will have:


Lower masses
Lower densities
•
•
faster rate of diffusion
lower melting and boiling points
Isotopes -
Questions
1. Explain what isotopes using hydrogen as an
2.
example.
One isotope of the element chlorine, contains 20
neutrons. Which other element also contains 20
neutrons?
3. State the number of protons, electrons and
neutrons in:
a) one atom of carbon-12
b) one atom of carbon-14
c) one atom of uranium-235
d) one atom of uranium-238
Mass Spectrometer
The mass spectrometer is an instrument used:
 To measure the relative masses of isotopes
 To find the relative abundance of the isotopes in a
sample of an element
When charged particles
pass through a magnetic
field, the particles are
deflected by the magnetic
field, and the amount of
deflection depends upon
the mass/charge ratio of
the charged particle.
Mass Spectrometer –
5 Stages
Once the sample of an element has been
placed in the mass spectrometer, it undergoes
five stages.
Vaporisation – the sample has to be in
gaseous form. If the sample is a solid or
liquid, a heater is used to vaporise some of
the sample.
X (s)  X (g)
or X (l)  X (g)
Mass Spectrometer –
5 Stages
Ionization – sample is bombarded by a
stream of high-energy electrons from
an electron gun, which ‘knock’ an
electron from an atom. This produces a
positive ion:
X (g)  X + (g) + e-
– an electric field is used to accelerate
the positive ions towards the magnetic field. The
accelerated ions are focused and passed through a
slit: this produces a narrow beam of ions.
Acceleration
Mass Spectrometer –
5 Stages
Deflection –
The accelerated ions are deflected into
the magnetic field. The amount of
deflection is greater when:
• the mass of the positive ion is less
• the charge on the positive ion is greater
• the velocity of the positive ion is less
• the strength of the magnetic field is
greater
Mass Spectrometer
If all the ions are travelling at the same
velocity and carry the same charge, the
amount of deflection in a given magnetic field
depends upon the mass of the ion.
For a given magnetic field, only ions with a
particular relative mass (m) to charge (z)
ration – the m/z value – are deflected
sufficiently to reach the detector.
Mass Spectrometer
Detection – ions that reach the detector
cause electrons to be released in an ioncurrent detector
The number of electrons released, hence the
current produced is proportional to the
number of ions striking the detector.
The detector is linked to an amplifier and
then to a recorder: this converts the current
into a peak which is shown in the mass
spectrum.
Atomic Structure –
Mass Spectrometer
Name the five stages which the sample
undergoes in the mass spectrometer
and make brief notes of what you
remember under each stage.
Complete Exercise 4, 5 and 6 in the
handbook. Any incomplete work to be
completed and handed in for next
session.
Atomic Structure – Mass
Spectrometer
 Isotopes
of boron
m/z value
Relative
abundance %
11
10
18.7
81.3
Ar of boron = (11 x 18.7) + (10 x 81.3)
(18.7 + 81.3)
= 205.7 + 813
100
= 1018.7 = 10.2
100
Mass Spectrometer –
Questions
A mass spec chart for a sample of neon
shows that it contains:



90.9%
0.17%
8.93%
20Ne
21Ne
22Ne
Calculate the relative atomic mass of neon
You must show all your working!
Mass Spectrometer –



90.9%
0.17%
8.93%
Questions
20Ne
21Ne
22Ne
(90.9 x 20) + (0.17 x 21) + (8.93 x 22)
100
Ar= 20.18
Mass Spectrometer –
Questions
Calculate the
relative atomic
mass of lead
You must show all
your working!
52.3
23.6
22.6
1.5
204 206 207 208 m/e
Mass Spectrometer –




Questions
1.5% 204Pb
23.6% 206Pb
22.6% 207Pb
52.3% 208Pb
(1.5 x 204) + (23.6 x 206) + (22.6 x 207)+(52.3 x
208)
100
306 + 4861.6 + 4678.2 + 10878.4
20724.2
100
100
Ar= 207.24
Energy Levels
Electrons go in shells or energy levels.
The energy levels are called principle
energy levels, 1 to 4.
The energy levels contain sub-levels.
Principle
energy level
1
2
Number of
sub-levels
1
2
3
4
3
4
These sub-levels
are assigned the
letters, s, p, d, f
Energy Levels
Each type of sub-level can hold a
different maximum number of electron.
Sub-level
s
p
d
f
Maximum
number of
electrons
2
6
10
14
Energy Levels
The energy of the sub-levels increases
from s to p to d to f. The electrons fill
up the lower energy sub-levels first.
Looking at this table can you
work out in what order the
electrons fill the sub-levels?
Energy Levels
Let’s take a look at the Periodic Table to
see how this fits in.
Electronic Structure
So how do you write it?
1s2
Energy level
Sub-level
Example
For magnesium:
1s2, 2s2, 2p6, 3s2
Number of
electrons
Electronic Structure
The electronic structure follows a pattern – the order
of filling the sub-levels is 1s, 2s, 2p, 3s, 3p…
After this there is a break in the pattern, as that the
4s fills before 3d.
Taking a look at the table below can you work out
why this is?
• This is because the 4s
sub-level is of
lower energy than the
3d sub-level.
Electronic Structure
The order in this the energy levels are
filled is called the Aufbau Principle.
Example
(Sodium – 2, 8, 1)
Electronic Structure
There are two exceptions to the Aufbau
principle.
The electronic structures of chromium and
copper do not follow the pattern – they are
anomalous.
Chromium – 1s2, 2s2, 2p6, 3s2, 3p6, 3d5, 4s1
Copper – 1s2, 2s2, 2p6, 3s2. 3p6, 3d10, 4s1
Write the electronic configuration for the following elements:
a) hydrogen c) oxygen
e) copper
b) carbon
d) aluminium f) fluorine
Electronic Structure
– of ions
When an atom loses or gains electrons
to form an ion, the electronic structure
changes:

Positive ions: formed by the loss of e1s2 2s2 2p6 3s1  1s2 2s2 2p6
Na atom

Na+ ion
Negative ions: formed by the gain of e1s2 2s2 2p4  1s2 2s2 2p5
O atom
O- ion
Electronic Structure – of transition metals
With the transition metals it is the 4s
electrons that are lost first when they
form ions:

Titanium (Ti) - loss of 2 e-
1s2 2s2 2p6 3s2 3p6 3d2 4s2  1s2 2s2 2p6 3s2 3p6 3d2
Ti atom

Ti2+ ion
Chromium (Cr) - loss of 3 e-
1s2 2s2 2p6 3s2 3p6 3d5 4s1  1s2 2s2 2p6 3s2 3p6 3d3
Cr atom
Cr3+ ion
Electronic Structure
- Questions
Give the full electronic structure of the
following positve ions:
a) Mg2+
b) Ca2+ c) Al3+
Give the full electronic structure of the
negative ions:
a) Clb) Brc) P3-
Electronic Structure
- Questions
Copy and complete the following table:
Atomic
no.
Mass
no.
No. of
No. of
No. of
protons neutrons electrons
Mg
12
Al3+
27
S2Sc3+
Ni2+
21
1s2 2s2 2p6 3s2
10
16
16
45
30
Electronic
structure
26
Orbitals
The energy sub levels are made up of
orbitals, each which can hold a maximum of 2
electrons.
Different sub-levels have different number of
orbitals:
Sub-level
No. of
orbitals
Max. no. of
electrons
s
1
2
p
3
6
d
5
10
f
7
14
Orbitals
The orbitals in different sub-levels have
different shapes:
• s orbitals
1s
• p orbitals
2s
Orbitals
Within a sub-level, the electrons occupy
orbitals as unpaired electrons rather than
paired electrons. (This is known as Hund’s
Rule).
We use boxes to represent orbitals:
2p


1s



2s

Electronic structure of
carbon, 1s2, 2s2, 2p2
Orbitals
The arrows represent the electrons in the
orbitals.
The direction of arrows indiactes the spin of
the electron.
Paired electrons will have opposite spin, as
this reduces the mutual repulsion between
the paired electrons.
2p


1s


2s


Electronic structure of
carbon, 1s2, 2s2, 2p2
Orbitals
Using boxes to represent orbitals, give the full
electronic structure of the following atoms:
a) lithium
d) nitrogen
b) fluorine
e) oxygen
2p
2s
1s
c) potassium
Orbitals
Using boxes to represent orbitals, give the full
electronic structure of the following atoms:
a) lithium
d) nitrogen
b) fluorine
e) oxygen
Electronic structure of
lithium: 1s2, 2s1
2p
2s


1s

c) potassium
Orbitals
Using boxes to represent orbitals, give the full
electronic structure of the following atoms:
b) fluorine
e) oxygen
2p 


1s


2s

Electronic structure of
fluorine: 1s2, 2s2, 2p5

c) potassium

a) lithium
d) nitrogen

Orbitals
Using boxes to represent orbitals, give the full
electronic structure of the following atoms:
a) lithium
d) nitrogen
b) fluorine
e) oxygen
4s 
Electronic structure of potassium:
1s2, 2s2, 2p6, 3s2, 3p6, 4s1
3p   
3s 
2p   
2s 
1s 
c) potassium
Orbitals
Using boxes to represent orbitals, give the full
electronic structure of the following atoms:
a) lithium
d) nitrogen
b) fluorine
e) oxygen
Electronic structure of
nitrogen: 1s2, 2s2, 2p3
2p


1s


2s


c) potassium

Orbitals
Using boxes to represent orbitals, give the full
electronic structure of the following atoms:
a) lithium
d) nitrogen
b) fluorine
e) oxygen
Electronic structure of
oxygen: 1s2, 2s2, 2p4
2p   


1s


2s
c) potassium

Ionization Energy
Ionization of an atom involves the loss of an electron
to form a positive ion.
The first ionization energy is defined as the
energy required to remove one mole of electrons
from one mole of atoms of a gaseous element.
The first ionization energy of an atom can be
represented by the following general equation:
X(g)  X+ + e- ΔH > 0
Since all ionizations requires energy, they are
endothermic processes and have a positive enthalpy
change (ΔH) value.
Ionization Energy
The value of the first ionization energy
depends upon two main factors:
The size of the nuclear charge
The energy of the electron that has
been removed (this depends upon its distance from the nucleus)
Ionization Energy
As the size of the nuclear charge increases the force
of the attraction between the negatively charged
electrons and the positively charged nucleus
increases.
Small
nuclear
charge

Small force
of attraction

Smaller
ionization
energy

+
+

Large
nuclear
charge

Large force
of attraction

Greater
ionization
energy
Ionization energy
As the energy of the electron increases, the electron
is farther away from the nucleus. As a result the
force of attraction between the nucleus and the
electron decreases.
Electrons closer to
positive nucleus

Large force of
attraction

Greater
ionization
energy
+
+
Electrons further
away from
positive nucleus

Small force of
attraction

Smaller
ionization
energy
Ionization energy -
Questions
Write an equation to represent the first
ionization of:
a) aluminium
b) lithium
c) sodium
Trends across a Period
Going across a period, the size of the 1st
ionisation energy shows a general increase.
This is because the electron comes from the
same energy level, but the size of the nuclear
charge increases.
+
+
+
Going across a Period
+
Trends across a Period
(2 exceptions)
The first ionisation of Al is less than that of Mg,
despite the increase in the nuclear charge.
The reason for this is that the outer electron removed
from Al is in a higher sub-level: the electron removed
from Al is a 3p electron, whereas that removed from
Mg is a 3s.
Electronic structure
Ionisation energy/kJ mol-1
Na
1s2, 2s2, 2p6, 3s1
494
Mg
1s2, 2s2, 2p6, 3s2
736
Al
1s2, 2s2, 2p6, 3s2, 3p1
577
Si
1s2, 2s2, 2p6, 3s2, 3p2
786
P
1s2, 2s2, 2p6, 3s2, 3p3
1060
S
1s2, 2s2, 2p6, 3s2, 3p4
1000
Cl
1s2, 2s2, 2p6, 3s2, 3p5
1260
Ar
1s2, 2s2, 2p6, 3s2, 3p6
1520
Trends across a Period
(2 exceptions)
The first ionisation energy of S is less than that of P,
despite the increase in the nuclear charge.
In both cases the electron removed is from the 3p sublevel. However the 3p electron removed from S is a
paired electron, whereas the 3p electron removed from P
is an unpaired electron.
When the electrons are paired the extra mutual
repulsion results in less energy being required to
remove an electron, hence a reduction in the ionisation
energy.
Sulphur
Phosphorus
3p  
3s 

3p  
3s 

Trends across a Period -
Questions
There is a break in this general trend going across a
Period.
Look at the table below and point out where the break in
the the trend is and try to give an explanation.
Electronic structure
Ionisation energy/kJ
mol-1
Na
1s2, 2s2, 2p6, 3s1
494
Mg
1s2, 2s2, 2p6, 3s2
736
Al
1s2, 2s2, 2p6, 3s2, 3p1
577
Si
1s2, 2s2, 2p6, 3s2, 3p2
786
P
1s2, 2s2, 2p6, 3s2, 3p3
1060
S
1s2, 2s2, 2p6, 3s2, 3p4
1000
Cl
1s2, 2s2, 2p6, 3s2, 3p5
1260
Ar
1s2, 2s2, 2p6, 3s2, 3p6
1520
Clue: which sub-level (s, p, d or f is the outer electron in?
Trends across a Period -
Questions
First ionisation
energy/kJ mol-1
Now take a look at the graph below:
3000
2500
2000
1500
1000
500
0
He
Ne
N
H
Li
0
C
Be
B
5
F
O
P
Mg
Na
10
Al
Si
Cl
Ar
Ca
S
15
K
20
25
Atomic number (Z)
a) Explain what the graph shows in as much detail as
possible
b) There is one other break in the general pattern going
across a Period. What is it and explain why that is.
Trends down a Group
+
+
+
Down the Group
Ionization energy decreases going
down a Group.
Going down a Group in the Periodic
Table, the electron removed during
the first ionization is from a higher
energy level and hence it is further
from the nucleus.
The nuclear charge also increases,
but the effect of the increased
nuclear charge is reduced by the
inner electrons which shield the outer
electrons.
+
Ionization energy
- Questions
1. Explain why sodium has a higher first
ionization energy than potassium.
2. Explain why the first ionization energy
of boron is less than that of beryllium.
3. Why does helium have the highest
first ionisation energy of all the
elements?
4. Complete Tasks
Successive Ionization energy
Definition: 2nd i.e.
The energy per mole for the process
X+(g)
X2+(g) +eAnd so on for further successive ionisation energies
Successive Ionization energy
 Successive i.e’s increases because
electrons are being removed from
increasingly positive ions.
 Therefore, nuclear attraction is
greater.
 Large jumps seen when electron is
removed form a new sublevel closer to
the nucleus
Successive Ionization energy
Large increase between 4th
and 3rd shells – electron
closer to nucleus
2nd i.e higher than first
– electron has greater
pull from nucleus
Electron Affinity
 Energy Change per mole for:
X (g) + e-
X-(g)
That is, for the gaseous atoms to gain an
electron to form anions
Electron Affinity
The first e.a is negative (exothermic) because the
electron is attracted to the positive charge on the
atom’s nucleus.
The second e.a is positive (endothermic) because an
electron is being added to an ion which is already
negative : repulsion occurs