1. IB Chemistry ATOMIC THEORY

2. Atomic Structure

3. 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

4. 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 sub-atomic particles is to use relative masses

5. 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

6. 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

7. Atomic Structure - Questions
What are the three sub atomic particles that make 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?

8. Atomic Structure - Questions
5. Copy and complete the following table:

9. 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.

10. 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:
Cl Cl 35 17 37
17 protons
17 electrons
18 neutrons
17 protons
17 electrons
20 neutrons
The isotopes of chlorine are often referred to as chlorine-35 and chlorine-37

11. 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 • faster rate of diffusion
Lower densities • lower melting and boiling points

12. Isotopes - Questions
Explain what isotopes using hydrogen as an example.
One isotope of the element chlorine, contains 20 neutrons. Which other element also contains 20 neutrons?
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

13. 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.

14. 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)

15. 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-
Acceleration – 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.

16. 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

17. 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.

18. Mass Spectrometer
Detection – ions that reach the detector cause electrons to be released in an ion-current 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.

19. 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.

20. Atomic Structure – Mass Spectrometer
Isotopes of boron
m/z value 11 10
Relative abundance %
18.7
81.3
Ar of boron = (11 x 18.7) + (10 x 81.3)
( ) =
100
= = 10.2

21. Mass Spectrometer – Questions
A mass spec chart for a sample of neon shows that it contains:
90.9% 20Ne
0.17% 21Ne
8.93% 22Ne
Calculate the relative atomic mass of neon
You must show all your working!

22. Mass Spectrometer – Questions
90.9% 20Ne
0.17% 21Ne
8.93% 22Ne
(90.9 x 20) + (0.17 x 21) + (8.93 x 22)
100
Ar= 20.18

23. Mass Spectrometer – Questions
204
206
207
208
52.3
23.6
22.6
1.5
Calculate the relative atomic mass of lead
You must show all your working!

24. 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
100
100
Ar=

25. Principle energy level
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
Number of sub-levels 1 2 3 4
These sub-levels are assigned the letters, s, p, d, f

26. Maximum number of electrons
Energy Levels
Each type of sub-level can hold a different maximum number of electron.
Sub-level
Maximum number of electrons s 2 p 6 d 10 f 14

27. 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?

28. Energy Levels
Let’s take a look at the Periodic Table to see how this fits in.

29. Electronic Structure 1s2 So how do you write it? Example
For magnesium:
1s2, 2s2, 2p6, 3s2
Energy level
Number of
electrons
Sub-level

30. 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.

31. Electronic Structure
The order in this the energy levels are filled is called the Aufbau Principle.
Example (Sodium – 2, 8, 1)

32. 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:
hydrogen c) oxygen e) copper
carbon d) aluminium f) fluorine

33. 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 e-
1s2 2s2 2p6 3s1 
1s2 2s2 2p6
Na atom
Na+ ion
Negative ions: formed by the gain of e-
1s2 2s2 2p4 
1s2 2s2 2p5
O atom
O- ion

34. 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

35. 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) Cl- b) Br- c) P3-

36. Electronic Structure - Questions
Copy and complete the following table:
Atomic no.
Mass no.
No. of protons
No. of neutrons
No. of electrons
Electronic structure Mg 12
1s2 2s2 2p6 3s2
Al3+ 27 10
S2- 16
Sc3+ 21 45
Ni2+ 30 26

37. 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

38. Orbitals The orbitals in different sub-levels have different shapes:
s orbitals
p orbitals

39. 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: 1s 2s 2p    
Electronic structure of carbon, 1s2, 2s2, 2p2  

40. 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. 1s 2s 2p 
Electronic structure of carbon, 1s2, 2s2, 2p2

41. Orbitals
Using boxes to represent orbitals, give the full electronic structure of the following atoms:
a) lithium b) fluorine c) potassium
d) nitrogen e) oxygen 1s 2s 2p

42. Orbitals
Using boxes to represent orbitals, give the full electronic structure of the following atoms:
a) lithium b) fluorine c) potassium
d) nitrogen e) oxygen
Electronic structure of lithium: 1s2, 2s1 1s 2s 2p   

43. Orbitals
Using boxes to represent orbitals, give the full electronic structure of the following atoms:
a) lithium b) fluorine c) potassium
d) nitrogen e) oxygen
Electronic structure of fluorine: 1s2, 2s2, 2p5 1s 2s 2p 

44. Orbitals
Using boxes to represent orbitals, give the full electronic structure of the following atoms:
a) lithium b) fluorine c) potassium
d) nitrogen e) oxygen 1s 2s 2p 3s 3p   4s
Electronic structure of potassium: 1s2, 2s2, 2p6, 3s2, 3p6, 4s1

45. Orbitals
Using boxes to represent orbitals, give the full electronic structure of the following atoms:
a) lithium b) fluorine c) potassium
d) nitrogen e) oxygen
Electronic structure of nitrogen: 1s2, 2s2, 2p3 1s 2s 2p 

46. Orbitals
Using boxes to represent orbitals, give the full electronic structure of the following atoms:
a) lithium b) fluorine c) potassium
d) nitrogen e) oxygen
Electronic structure of oxygen: 1s2, 2s2, 2p4 1s 2s 2p  

47. 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.

48. 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)

49. 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 
Large nuclear charge   
Small force of attraction 
Large force of attraction 
Smaller ionization energy
Greater ionization energy

50. 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 further away from positive nucleus  +
Electrons closer to positive nucleus 
Large force of attraction 
Small force of attraction 
Greater ionization energy
Smaller ionization energy

51. Ionization energy - Questions
Write an equation to represent the first ionization of:
a) aluminium
b) lithium
c) sodium

52. + 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

53. 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

54. 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 sub-level. 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. 3s 3p  
Phosphorus 3s 3p  
Sulphur

55. 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?

56. Trends across a Period - Questions
Now take a look at the graph below: H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca
Explain what the graph shows in as much detail as possible
There is one other break in the general pattern going across a Period. What is it and explain why that is.

57. Trends down a Group Ionization energy decreases going down a 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. + +
Down the Group +

58. Ionization energy - Questions
Explain why sodium has a higher first ionization energy than potassium.
Explain why the first ionization energy of boron is less than that of beryllium.
Why does helium have the highest first ionisation energy of all the elements?
Complete Tasks

59. Successive Ionization energy
Definition: 2nd i.e.
The energy per mole for the process
X+(g) X2+(g) +e-
And so on for further successive ionisation energies

60. 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

61. 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

62. 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

63. 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
The