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A Cook 2015
Chemistry Led Specification Revision:
Topic 1 - The Atomic Structure and Periodic Table
(a) Know the structure of an atom in terms of electrons, protons and neutrons
• The structure of the atom can be generalised into the three subatomic particles:

• The protons are found within the centre of the
nucleus and consist of a positive charge
• The neutrons are also found within the centre of
the nucleus with no charge
• The electrons are found in the surrounding
energy levels, around the nucleus

(b) Know the relative mass and relative charge of protons, neutrons and electrons
Particle

Symbol

Relative Mass

Relative Charge

Position

Proton

p

1

+1

Nucleus

Neutron

n

1

0

Nucleus

Electron

e-

1/1840

-1

Energy levels
surrounding the
nucleus

(c) Know what is meant by the terms ‘atomic (protons) number’ and ‘mass number’
• Atomic number = Proton number = (number of electrons)
• Mass number = Proton number + number of neutrons

Mass Number

Atomic Number

(d) Be able to determine the number of each type of sub-atomic particle in an atom, molecule or ion from the
atomic (proton) number and mass number
• This chlorine atom contains:
• Protons = Atomic number = 17 protons
• Neutrons = Mass number - Atomic number = 35 - 17 = 18
• Electrons = Proton number = 17 Electrons

A Cook 2015
(e) Understand the term ‘isotope’
• An isotope is the same element but with different numbers of neutrons
• They have the same number of electrons and so as a result have the same chemical structure and
properties (react in the same way) because it is the electrons which give elements their properties
• Chlorine - 35 Isotope = Protons = 17 // Neutrons = 18 // Electrons = 17
• Chlorine - 37 Isotope = Protons = 17 // Neutrons = 20 // Electrons = 17
(f) Be able to define the terms ‘relative isotopic mass’ and ‘relative atomic mass’, based on the 12C scale
...

• Relative Atomic Mass = The weighted mean mass of an element compared to 1/12 the mass of a
Carbon-12 atom, which has a mass of 12
(g) understand the terms ‘relative molecular mass’ and ‘relative formula mass’ including calculating these
values from relative atomic mass
• Relative molecular Mass = Add up all the molecular masses of all the atoms in the molecule
• i
...
Mr(C2H6O) = ((2 x 12
...
0) + 16
...
0
• Relative Formula Mass = Add up all the relative atomic masses of all the ions or atoms in the formula
• i
...
Mr(CaF2) = 40
...
0) = 78
...

• You need to know how to work out the relative atomic mass (Ar) of an element from its isotopic masses
1) Different isotopes of an element occur in different quantities, or isotopic abundances
2) Work out the average mass of all the atoms to find the relative atomic mass
3) if you're given the isotopic abundances in percentages, follow these steps:
1) Multiply each relative isotopic mass by its % relative isotopic abundance, and then add up the
results
2) Divide by 100
Q) Find the relative atomic mass of boron, given that 20
...
0, while 80
...
0
...
((20
...
0) + (80
...
0)) = 1080
2
...
8

A Cook 2015
(i) Be able to predict the mass spectra, including relative peak heights, for diatomic molecules, including
chlorine
• Some elements contain two or more atoms covalently bonded together, if these substances are analyses
by mass spec, you can obtain the relative molecular mass of the element or compound by observing the
peaks with the largest m/z ratios
...
Multiplying all the relative abundances by its relative isotopic abundance, and add up all the results
2
...
(3
...
0 x 37
...
142 / (3+1) = 35
...
92
...
67% of silicon is Si-29
...
1, calculate the abundance and isotopic mass of the third isotope
...
23 + 4
...
90
100- 96
...
10% (3rd isotope)
Now work out isotopic mass of third isotope:
Ar = 28
...
1 = ((92
...
0) + (4
...
0) + (Y x 3
...
1 = 2717
...
10) / 100
2810 = 2717
...
10)
2810 - 2717
...
10
92
...
10
92
...
10 = Y
29
...
Cl-35
75% and Cl-37 with an abundance of 25%
...
75% = 0
...
25
Make a table showing all the different Cl2 molecules
...
75 x 0
...
5625

Cl-35 - Cl-37: 0
...
25 = 0
...
25 x 0
...
1875

Cl-37 - Cl-37: 0
...
25 = 0
...
1875 + 0
...
375
Divide all of the abundances by the smallest abundance to get the smallest whole number ratio and by
working out the mass of each molecule you can predict the mass spectrum of Chlorine diatomic molecule
...
5625 / 0
...
375 / 0
...
0625 / 0
...
This is
the peak with the highest m/z value
...


• The highest m/z peak here is 46
...
0 + 16
...
0 = 46
...

A(g) —> A+(g) + e• Second ionisation energy - The energy required to remove 1 electron from each atom of 1 mole of
gaseous 1+ positive ions to form 1 mole of 2+ gaseous ions
...

• The big jumps is electrons being removed from a different quantum shell to the last (closer to the nucleus)
• This is once piece of evidence for the quantum shells existing

Successive ionisation energies of Na :
1st

2nd
496

4563

3rd
6913

4th
9544

5th
13352

6th
16611

7th
20115

8th
25491

9th
28943

10th

11th

141367

159079

A Cook 2015
Removed from the outside in, so 1st electron removed is in the outermost quantum shell (with the largest
energy) and then the smallest electron energy is seen in the last electron to be removed (thus highest
ionisation energy required to remove)
Description:
• The first electron is considerably easier to remove than the second
...

• The last 2 electrons are much harder to remove than the last 8
...

• The next eight electrons are in the second quantum shell,m which is of lower energy than the third
...

• So the basic electronic configuration for sodium is 2,8,1
...

• The energy of the electron has to be increased to a particular value for it to be removed
• For any atom, the energy value that the electron must have (to be removed) will be the same despite
where the electron is in the atom

High-energy Quantum Shell electron:
• If an electron already has a high energy, then the energy needed to remove it will be small
Low-energy Quantum Shell electron:
• If the electron is in an orbital of a low-energy quantum shell, i
...
1s orbital, then it will need to gain a
considerable amount of energy to be removed
The difference in the energy between the electron when it has been removed and the energy it has when it is
in its original orbital in the quantum shell, is known as the ionisation energy
...

• However they both have different amount of energies present
Shielding:
• Because He’s 1s orbital contains 2 electrons, as opposed to 1 in H, there is greater electron-electron
repulsion between the two electrons within the orbital: each electron is said to ‘SHIED’ the other from the
nuclear charge
• The affect of this factor alone is to incase the energy of the electrons (thus, reducing the ionisation energy)
Nuclear Charges:
However, the nuclear charge of helium is double that of Hydrogen:
• This is because helium contains 2 protons, whereas hydrogen contains only 1 protons
• The effect of the extra proton is a greater nuclear charge
• The effect of the greater nuclear charge means that the electrons are attracted more strongly to the
nucleus and so the energy of the 1s electrons is reduced
• The effect of the increased nuclear charge is greater than that of the shielding and so the first ionisation
energy of He (2370 kJ mol-1) is great than that of Hydrogen (1310 kJ mol-1)
...

• Added to this, the electron is in the 2s orbital experiences electron-electron repulsion from the lower 2
electrons in the 1s orbital (shielding it from the nuclear charge)
• The latter two effects are more significant than the increased nuclear charge
...
The orbital in which the electron exists
2
...
The repulsion (shielding) experienced by the electron from all other electrons present
(m) Understand reasons for the general increase in first ionisation energy across a period
Across a period trends:
Consider Period 2; and the elements Li to Ne
Nucelar Charge Increases:
• Nuclear charge increase, because extra protons are added from left to right
• This will lead to an increased attraction of the nucleus and the electron, therefore leading to a decrease in
the energy of the outermost electron, and increase the first ionisation energy
...

• The increase in nuclear charge is more significant than the increase in electron-electron repulsion
...



Summary of increase of IE across a group:
1
...
Increase of electron-electron repulsion
3
...

Quantum Shell Addition:
• The addition of a quantum shell increases the energy of the outermost electrons (decreasing IE)
• Because:
• The third quantum shell has a higher energy value than the second; the fourth quantum shell is higher in
energy than the third
• This trend continues down the group
• As each new quantum shell is added, the outer electrons experience more electron-electron repulsion
(shielding) from the nuclear charge, increasing the energy of the outer electrons
• This leads to the IE decreasing
On this occasion, down the group, the combined effect of adding extra quantum shells, which decreases the
IE by increasing the energy of the outermost electrons through shielding, is more significant the increase in
nuclear charge
...

• Only specific frequencies are emitted, and these are unique to an individual element
• All particular atoms of an element emit the same EM radiation
• The emission spectrum produced is called a Line Spectrum
• The fact that only certain frequencies of EM radiation are emitted, rather than continuous spectrum, is
compelling evidence that the energy of electrons in atoms can only have certain, fixed values and not a
continuous range of values
...

Successive ionisation energies provide evidence for quantum shells:
• As the successive ionisation energies are seen in order, there is big jumps
• These big jumps are where the electrons are now being taken from a lower quantum shell than the one
before

A Cook 2015
• And thus have a significantly lower energy level (thus much higher IE)
The first ionisation energy provides evidence for the group of element
• The location of these ‘big jumps’ can be used to identify which group the element is in
• Looking at either a continuous chart of successive IE or a logarithm of ionisation energies it is possible to
count the amount of electrons removed before these big jumps
• The electron removed first is the outermost electron, the second is the next electron in line, and the third is
the next, if the jump appears here, then it is evidence that the element is within group two
...

• This provides evidence that the element is in group 2, as each group defines the amount of electrons
within the outermost quantum shell
• If it takes two electrons to be removed before a jump in IE then this shows that there is only 2 electrons in
the outermost quantum shell and thus proof that it is a group 2 element
...
g
...


A Cook 2015
(r) Know the shape of an s-orbital and a p-orbital
S Orbitals
• Consists of a sphere around the nucleus, the single electron in Hydrogen is present in the 1s orbital
...


(s) Know the number of electrons that can occupy s, p and d- subshells
There can be up to two electrons within each orbital
...

• This is shown below in the electronic configurations

Atomic
Number

Symbol

1s

2s

2p

3s

3p

1 H

2

1

4 Be

2

2

5 B

2

2

1

6 C

2

2

2

7 N

2

2

3

8 O

2

2

4

9 F

2

2

5

10 Ne

2

2

6

11 Na

2

2

6

1

12 Mg

2

2

6

4p

2

3 Li

4s

1

2 He

3d

2

(u) Be able to predict the electronic configurations, using 1s notation and electrons-in-boxes notation, of:
Atoms, given the atomic number, Z, up to Z = 36
Ions given the atomic number, Z, and the ionic charge, for s and p block ions only, up to Z = 36
Atomic
Number

Symbol

Electrical Configuration

1 H

1s1

2 He

1s2

3 Li

1s2 2s1

4 Be

1s2 2s2

5 Be

1s2 2s2 2p1

6 C

1s2 2s2 2p2

7 N

1s2 2s2 2p3

8 O

1s2 2s2 2p4

9 F

1s2 2s2 2p5

10 Ne

1s2 2s2 2p6

11 Na

1s2 2s2 2p6 3s1

A Cook 2015
Atomic
Number

Symbol

Electrical Configuration

12 Mg

1s2 2s2 2p6 3s2

13 Al

1s2 2s2 2p6 3s2 3p2

14 Si

1s2 2s2 2p6 3s2 3p3

15 P

1s2 2s2 2p6 3s2 3p3

16 S

1s2 2s2 2p6 3s2 3p4

17 Cl

1s2 2s2 2p6 3s2 3p5

18 Ar

1s2 2s2 2p6 3s2 3p6

19 K

1s2 2s2 2p6 3s2 3p6 4s1

20 Ca

1s2 2s2 2p6 3s2 3p6 4s2

21 Sc

1s2 2s2 2p6 3s2 3p6 3d1 4s2

22 Ti

1s2 2s2 2p6 3s2 3p6 3d2 4s2

23 V

1s2 2s2 2p6 3s2 3p6 3d3 4s2

Electron-box-notation
Hund’s rule: States that electrons will occupy orbitals singly before pairing takes place
The Pauli Exclusion Principle: states that two electrons cannot occupy the same orbital unless they have
opposite spins
...

• Oxygen = 1s2 2s2 2p4
• Notice how in the last 3 boxes the lowest energies will always be filled before the others

(v) Know that elements can be classified as s, p and d-block elements

A Cook 2015
(w) Understand that electronic configuration determines the chemical properties of an element
The electronic configuration determines how many electrons will be in the valence shell, and so as a result,
because the outermost electrons will always attract/repel making the chemical properties of this element, it
directly affects what the properties will be
...

(x) Understand periodicity in terms of a repeating pattern across different periods
Periodicity:
• The elements in a Period, exhibit periodicity
• Elements in periods 2 and 3 show this the best
• Examples of periodicity is a regular repeating electronic configuration, boiling point, melting point, atomic
radii and first ionisation energies
(y) Understand reasons for the trends in the following properties of the elements from periods 2 and 3 of the
Periodic table
• The melting and boiling temperatures of the elements, based on given data, in terms of structure and
bonding
• ionisation energy based on given data or recall of the plots of ionisation energy versus atomic number
Atomic Radii:
• The atomic radii is a measure of the size of the atoms
• It is the distance from the centre of the nucleus to the boundary of the electron clouds
• Since the electron clouds are not well defined, we can find the atomic radii by determining the distance
between two nucleus and dividing it by two

the covalent radii (where the distance
• There is three types of atomic radii,
between two nuclei of covalently bonded atoms is measured) the van der Waals radii (where the distance
between two nuclei of two atoms bonded only by van der Waals forces (intermolecular forces) is
measured) and finally metallic radii (used for metals)
...

• This is because as the number of protons increases the nuclear charge does so too
• This means that there is an increase in the attractive force between the nucleus and the electrons pulling
these electrons in closer towards the nucleus
• This increase in attractive force offsets the increase in electron-electron repulsion as the number of
electrons in the outer quantum shell increases
Melting Point and Boiling Temperatures
Period 2
Li

Be

B

C
(diamond)

N

O

F

Melting
Temperature
/oC

181

1278

2300

3550

-210

-218

-220

Boiling
temperature
/oC

1342

2970

3927

4827

-196

-183

-188

Type of
bonding

Metallic

Metallic

Covalent

Covalent

Covalent

Covalent

Covalent

Structure

Giant
lattice

Giant
lattice

Giant
lattice

Giant
lattice

Simple
Molecular

Simple
Molecular

Simple
Molecular

Na

Mg

Al

Si

P

S

Cl

Melting
Temperature
/oC

98

649

660

1410

44

113

-101

Boiling
temperature
/oC

883

1107

2467

2355

280

445

-35

Type of
bonding

Metallic

Metallic

Metallic

Covalent

Covalent

Covalent

Covalent

Structure

Giant
lattice

Giant
lattice

Giant
lattice

Giant
lattice

Simple
Molecular

Simple
Molecular

Simple
Molecular

Period 3

• Elements with giant lattices have higher melting and boiling points
• Those with simple molecular structures have low melting and boiling points

A Cook 2015
First Ionisation Energies:

Hydrogen and Helium
• The explanation for the increase in the first ionisation energy comes from how there is an increase in
nuclear charge from 1 and 2 as an extra proton is added
...

Anomalies:
The first ionisation energy of (Group 3) Boron is less than that of the (group 2) Beryllium element, and the
first ionisation energy of the (group 6) Oxygen is less than that of the (group 5) Nitrogen
...

In addition the 2p electron experiences greater electron-electron repulsion from the 2s lower orbital
This reduces the effects of the nuclear attraction, increasing the energy which the electron has
As a result, the ionisation energy decreases because the energy which the electron already has due to its
increased energy is near the amount it needs to be ionised
...




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