1.1 Atomic Structure

Question 1

Time of Flight Mass Spectrometry - function

Answer 1

It is the most useful instrument for accurate determination of the relative atomic mass of an element, based on the abundance and mass of each of its isotopes

Question 2

Time of Flight Mass Spectrometry - condition

Answer 2

Must be kept under a high vacuum to prevent the ions that are produced from colliding with molecules present in the air

Question 3

Stages of Time of Flight Mass Spectrometry

Answer 3

• Ionisation (Electron bombardment / Impact or Electrospray Ionisation)
• Acceleration
• Ion drift
• Ion Detection
• data analysis

Question 4

1) Electrospray Ionisation - use and properties

Answer 4

• This method is used for substances which have a higher molecular mass
• Unlike with electron impact ionisation, fragmentation is unlikely to happen - This is often called a soft ionisation technique

Question 5

1) Electrospray Ionisation - process

Answer 5

1. the sample is dissolved in a volatile & polar solvent
2. the solvent is injected at high pressure through a fine hollow needle connected to the positive terminal of a high voltage supply producing a fine mist
3. Causes the sample molecule (X) to be ionised and to gain a
   proton (H+) from the solvent, forming XH+
4. The solvent evaporates into the vacuum while the XH+ ions are attracted towards a negatively charged plate

X (g) + H+ –> XH+ (g)

Question 6

Electron impact / bombardment Ionisation - use and properties

Answer 6

This method of ionisation is used for elements and substances which have a lower molecular mass.

this technique cannot be used with larger moleculesas it will cause them to fragment and negatively impact the spectra analysis

Question 7

Electron impact / bombardment Ionisation - process

Answer 7

1. The sample is vapourised and
   injected at low pressure
2. An electron gun fires high energy
   electrons at the injected sample
3. This causes an outer electron to be knocked
   out from each particle, forming a 1+ ion

X (g) -> X+ (g) + e-

Question 8

Acceleration

Answer 8

The positive ions formed from either ionisation method are accelerated to a constant kinetic energy (KE) by an electric field towards a negatively charged plate

All ions have constant kinetic energy so lighter ions travel fastest Since their velocity (speed) is dependent on their mass

Question 9

Stage 3: Ion Drift

Answer 9

The 1+ ions will pass through a hole in the negatively charged plate and move into a flight tube
The time of flight of each 1+ ion in this tube depends on their velocity

Question 10

Stage 4: Ion Detection

Answer 10

The ions travel along a tube, called the flight
tube, where they then reach a detector (another negatively charged plate)
The positive ions reach the negatively charged detection plate and gain an electron and produce a current
This size of the current is proportional to the abundance of those ions hitting the plate and gaining an electron

Question 11

data analysis

Answer 11

Two pieces of information are
analysed in combination:
1. The current that is produced
2. The flight tube times (TOF)

This produces a spectra with the relative
abundance produced by ions with varying
mass/charge (m/z) ratios

Question 12

electron configuration

Answer 12

The arrangement of electrons in an atom is called the electron configuration

Question 13

principal energy levels or principal quantum shells

Answer 13

Electrons are arranged around the nucleus in principal energy levels or principal quantum shells
Principal quantum numbers (n) are used to number the energy levels or quantum shells

The lower the principal quantum number, the closer the shell is to the nucleus and the lower the energy
The higher the principal quantum number, the greater the energy of the shell and the further away from the nucleus

Question 14

subshells

Answer 14

The principal quantum shells are split into subshells which are given the letters s, p, d and f
The energy of the electrons in the subshells increases in the order s < p < d < f

All the orbitals in the same subshell have the same energy and are said to be degenerate

Question 15

Orbitals

Answer 15

Subshells contain one or more atomic orbitals
Orbitals exist at specific energy levels and electrons can only be found at these specific levels, not in between them
Each atomic orbital can be occupied by a maximum of two electrons
This means that the number of orbitals in each subshell is as follows:
s : one orbital (1 x 2 = total of 2 electrons)
p : three orbitals ( 3 x 2 = total of 6 electrons)
d : five orbitals (5 x 2 = total of 10 electrons)
f : seven orbitals (7 x 2 = total of 14 electrons)

Question 16

s orbital shape

Answer 16

The s orbitals are spherical
The size of the s orbitals increases with increasing shell number
E.g. the s orbital of the third quantum shell (n = 3) is bigger than the s orbital of the first quantum shell (n = 1)

Question 17

p orbital shape

Answer 17

The p orbitals have a dumbbell shape
Every shell has three p orbitals except for the first one (n = 1)
The p orbitals occupy the x, y and z axes and point at right angles to each other, so are oriented perpendicular to one another
The lobes of the p orbitals become larger and longer with increasing shell number

Question 18

Ground state

Answer 18

The ground state is the most stable electronic configuration of an atom which has the lowest amount of energy
This is achieved by filling the subshells of energy with the lowest energy first (1s)

Question 19

3d vs 4s

Answer 19

The subshells increase in energy as follows: s ＜ p ＜ d ＜ f
The only exception to these rules is the 3d orbital which has slightly higher energy than the 4s orbital
Because of this, the 4s orbital is filled before the 3d orbital

electrons are put in 4s then in 3d
electrons are taken out of 4s then from 3d

Question 20

The electron configuration

Answer 20

The electron configuration gives information about the number of electrons in each shell, subshell and orbital of an atom

Writing out the electron configuration tells us how the electrons in an atom or ion are arranged in their shells, subshells and orbitals

Question 21

The electron configuration - ions

Answer 21

Ions are formed when atoms lose or gain electrons
Negative ions are formed by adding electrons to the outer subshell
Positive ions are formed by removing electrons from the outer subshell

Question 22

methods of writing out the electron configuration

Answer 22

This can be done using the full electron configuration or the shorthand version

The full electron configuration describes the arrangement of all electrons from the 1s subshell up

e.g.
Potassium has 19 electrons
The full electron configuration of potassium is 1s2 2s2 2p6 3s2 3p6 4s1

The shorthand electron configuration includes using the symbol of the nearest preceding noble gas to account for however many electrons are in that noble gas

e.g.

The nearest preceding noble gas to potassium is argon
This accounts for 18 electrons of the 19 electrons that potassium has
The shorthand electron configuration of potassium is [Ar] 4s1

Question 23

exceptions to the general electron configuration rules

Answer 23

Chromium and copper have the following electron configurations, which are different to what you may expect:
Cr is [Ar] 3d5 4s1 not [Ar] 3d4 4s2
Cu is [Ar] 3d10 4s1 not [Ar] 3d9 4s2
This is because the [Ar] 3d5 4s1 and [Ar] 3d10 4s1 configurations are energetically stable

Question 24

Presenting the Electron Configuration

Answer 24

Electrons can be imagined as small spinning charges which rotate around their own axis in either a clockwise or anticlockwise direction
The spin of the electron is represented by its direction

Question 25

Presenting the Electron Configuration - spin-pair repulsion

Answer 25

Electrons with similar spin repel each other which is also called spin-pair repulsion
Electrons will therefore occupy separate orbitals in the same subshell where possible, to minimize this repulsion and have their spin in the same direction
E.g. if there are three electrons in a p subshell, one electron will go into each px, py and pz orbital

Electrons are only paired when there are no more empty orbitals available within a subshell, in which case the spins are the opposite spins to minimize repulsion

Question 26

Box Notation

Answer 26

The electron configuration can be represented using the electrons in boxes notation
Each box represents an atomic orbital
The boxes are arranged in order of increasing energy from bottom to top
The electrons are represented by opposite arrows to show the spin of the electrons

Question 27

What is Ionisation Energy?

Answer 27

The Ionisation Energy (IE) of an element is the amount of energy required to remove one mole of electrons from one mole of gaseous atoms of an element to form one mole of gaseous ions

Question 28

Ionisation Energy properties

Answer 28

Ionisation energies are measured under standard conditions which are 298 K and 101 kPa
The units of IE are kilojoules per mole (kJ mol-1)
General Formula:
X(g) -> X+(g) + e-

E.g. the first ionisation energy of gaseous calcium:
Ca(g) → Ca+ (g) + e- IE1 = +590 kJ mol-1

Question 29

Trends in Ionisation Energies

Answer 29

The size of the first ionisation energy is affected by four factors:
* Size of the nuclear charge (how far across a period the element is)
* Distance of outer electrons from the nucleus (how far down the group the element is)
* Shielding effect of inner electrons
* Spin-pair repulsion

Question 30

Ionisation energy across a period

Answer 30

The ionisation energy across a period generally increases due to the following factors:
* Across a period the nuclear charge increases
* This causes the atomic radius of the atoms to decrease, as the outer shell is pulled closer to the nucleus, so the distance between the nucleus and the outer electrons decreases
* The shielding by inner shell electrons remain reasonably constant as electrons are being added to the same shell
* It becomes harder to remove an electron as you move across a period; more energy is needed
* So, the ionisation energy increases

Nuclear charge increases
Atomic radius decrease
Same shielding

Question 31

Ionisation energy down a group

Answer 31

The ionisation energy down a group decreases due to the following factors:
* The number of protons in the atom is increased, so the nuclear charge increases
* But, the atomic radius of the atoms increases as you are adding more shells of electrons, making the atoms bigger
* So, the distance between the nucleus and outer electron increases as you descend the group
* The shielding by inner shell electrons increases as there are more shells of electrons
* These factors outweigh the increased nuclear charge, meaning it becomes easier to remove the outer electron as you descend a group
* So, the ionisation energy decreases

Nuclear charge increases
Number of electron shells increases
Shielding increases
ionisation energy decreases

Question 32

successive Ionisation Energies: Equations

Answer 32

The second ionisation energy (IE2) is the energy required to remove the second mole of electrons from each +1 ion in a mole of gaseous +1 ions, to form one mole of +2 ions
The third ionisation energy (IE3) is the energy required to remove the third mole of electrons from each +2 ion in a mole of gaseous +2 ions, to form one mole of +3 ions
And so on…
The electrons from an atom can be continued to be removed until only the nucleus is left

Question 33

successive Ionisation Energies- energy needed

Answer 33

Successive ionisation energies of an element
The successive ionisation energies of an element increase
This is because once you have removed the outer electron from an atom, you have formed a positive ion
Removing an electron from a positive ion is more difficult than from a neutral atom
As more electrons are removed, the attractive forces increase due to decreasing shielding and an increase in the proton to electron ratio
The increase in ionisation energy, however, is not constant and is dependent on the atom’s electronic configuration

Question 34

successive Ionisation Energies- energy needed example

Answer 34

The first electron removed has a low IE1 as it is easily removed from the atom due to the spin-pair repulsion of the electrons in the 4s orbital
The second electron is more difficult to remove than the first electron as there is no spin-pair repulsion
The third electron is much more difficult to remove than the second one corresponding to the fact that the third electron is in a principal quantum shell which is closer to the nucleus (3p)

Removal of the fourth electron is more difficult as the orbital is no longer full, and there is less spin-pair repulsion

Question 35

successive Ionisation Energies- energy needed 3 more example

Answer 35

Successive ionisation data can be used to:
Predict or confirm the simple electronic configuration of elements
Confirm the number of electrons in the outer shell of an element
Deduce the Group an element belongs to in the Periodic Table
By analyzing where the large jumps appear and the number of electrons removed when these large jumps occur, the electron configuration of an atom can be determined
Na, Mg and Al will be used as examples to deduce the electronic configuration and positions of elements in the Periodic Table using their successive ionisation energies
Successive Ionisation Energies Table

Atomic Structure First Four Ionisation Energies of Sodium, Magnesium & Aluminium Table, downloadable AS & A Level Chemistry revision notes

Sodium
For sodium, there is a huge jump from the first to the second ionisation energy, indicating that it is much easier to remove the first electron than the second
Therefore, the first electron to be removed must be the last electron in the valence shell thus Na belongs to group I
The large jump corresponds to moving from the 3s to the full 2p subshell
Na 1s2 2s2 2p6 3s1

Magnesium
There is a huge increase from the second to the third ionisation energy, indicating that it is far easier to remove the first two electrons than the third
Therefore the valence shell must contain only two electrons indicating that magnesium belongs to group II
The large jump corresponds to moving from the 3s to the full 2p subshell
Mg 1s2 2s2 2p6 3s2

Aluminium
There is a huge increase from the third to the fourth ionisation energy, indicating that it is far easier to remove the first three electrons than the fourth
The 3p electron and 3s electrons are relatively easy to remove compared with the 2p electrons which are located closer to the nucleus and experience greater nuclear charge
The large jump corresponds to moving from the third shell to the second shell
Al 1s2 2s2 2p6 3s2 3p1

Question 36

mass number

Answer 36

The mass number (or nucleon number) is the total number of protons and neutrons in the nucleus of an atom

Question 37

atomic number

Answer 37

The atomic number (or proton number) is the number of protons in the nucleus of an atom

Question 38

variance in Isotopes chemical properties

Answer 38

Isotopes have the same chemical properties but different physical properties

Chemical properties
* Isotopes of the same element display the same chemical characteristics
* This is because they have the same number of electrons in their outer shells.
* Electrons take part in chemical reactions and therefore determine the chemistry of an atom

Question 39

variance in Isotopes Physical properties

Answer 39

Isotopes have the same chemical properties but different physical properties

Physical properties
* The only difference between isotopes is the number of neutrons
* Since these are neutral subatomic particles, they only add mass to the atom
* As a result of this, isotopes have different physical properties such as small differences in their mass and density

Question 40

RAM from Mass Spectra

Answer 40

Question 41

History of the Atom - 1 - John Dalton

Answer 41

1800’s - John Dalton suggests that each of the elements are made from just one type of atom: tiny spheres that could not be divided

Question 42

History of the Atom - 2 - JJ Thomson

Answer 42

1897 - J Thomson discovers the electron and proposes the plum pudding model: the atom is a ball of positive charge and the negative charge are embedded in it (like blueberries in a blueberry muffin)

Question 43

History of the Atom - 3 - Ernest Rutherford vs thompson

Answer 43

1911 - Ernest Rutherford fired alpha particles at a piece of very thin gold foil(about 10,000 atoms thick).
Thomson’s plum pudding model predicted that…
All alpha particles… passed straight through
However what was observed was:
Most alpha particles… passed straight through
A very few alpha particles… were deflected by more than 90 degrees
Most of the atom is empty space
All the positive charge and most of the mass is concentrated in a small volume (the nucleus)

Question 44

History of the Atom - 4 - Niels Bohr

Answer 44

1913 - Niels Bohr proposed a new model of the atom with four
basic principles:
* Electrons only exist in fixed orbits (shells) and not anywhere in between.
* Each shell has a fixed energy.
* When an electron moves between shells electromagnetic radiation is emitted or absorbed.
* Because the energy of shells is fixed, the radiation will have a fixed frequency - The frequencies of radiation emitted and absorbed by atoms were already known from experiments. The Bohr model fitted these observations.

Question 45

most accurate model from today

Answer 45

The most accurate model we have today is based on quantum
mechanics. The quantum model explains some observations that can’t be accounted for by the Bohr model

Question 46

mass spectrum

Answer 46

A mass spectrum is a type of chart produced by a mass spectrometer. It shows information about the sample that was passed through the mass spectrometer.

If the sample is an element, each line will represent a different isotope of the element. The y-axis gives the relative isotopic abundance of ions, often as a percentage. The x-axis units are given as a ‘mass/charge’ ratio.

The spectrum in Figure 3 was produced using electron impact ionisation. One electron has been knocked off each particle to turn them into +1 ions - so the mass/charge ratio of each peak is the same as the relative mass of that isotope.

(If electrospray ionisation had been used instead, a H+ ion would have been added to each particle to form +1 ions - so the mass/charge ratio of each peak would be one unit greater than the relative mass of each isotope.) All of the spectra shown in this topic have been produced using electron impact ionisation.

Question 47

lonisation trends across periods e.g. The drop between Groups 2 and 3: Mg & Al

Answer 47

The drop between Groups 2 and 3 shows sub-shell structure.
Mg 1s2 2s2 2p6 3s2
AI 1s2 2s2 2p6 3s2 3p1

• Aluminium’s outer electron is in a 3p orbital rather than a 3s. The 3p orbital has a slightly higher energy than the 3s orbital, so the electron is, on average, to be found further from the nucleus.
• The 3p orbital has additional shielding provided by the 3s electrons.
• These two factors together are strong enough to override the effect of the increased nuclear charge, resulting in the ionisation energy dropping slightly.

Question 48

lonisation trends across periods e.g. The drop between Groups 5 and 6: P & S

Answer 48

P 1s2 2s2 2p6 3s2 3p3
S 1s2 2s2 2p6 3s2 3p4

• The shielding is identical in the phosphorus and sulfur atoms, and the electron is being removed from an identical orbital.
• In phosphorus’s case, the electron is being removed from a singly-occupied orbital. But in sulfur, the electron is being removed from an orbital containing two electrons.
• The repulsion between two electrons in an orbital means that electrons are easier to remove from shared orbitals. It’s yet more evidence for the electronic structure model.

Question 49

Relative atomic mass definition

Answer 49

Average mass of one atom of an element compared to 1/12th of the mass of one atom of carbon 12

Question 50

Trends in ionisation energy period 2 - things to remember

Answer 50

Group 1 - 2 ionisation energy increase - eg Be has higher ionisation energy than Li.
Both have same shielding
Both have roughly same distance from nucleus
Be has a higher nuclear charge
Force of attraction between nucleus and outer electron is higher doe Be
So Be has higher ionisation energy than Li.

Question 51

The mass spec of tellurium is shown in the graph.

The mass spectrum contains a small peak at 64. Explain the existence of this peak

Answer 51

128 Te 2+ (which has double the charge of 128 Te 1+ ) will have a mass to charge ratio of 64 as 128/2 is 64.

This isn’t caused by fragmentation as atoms do not fragment.

Question 52

suggest what may cause the RAM of this sample to be different to that shown on the periodic table

Answer 52

There were other isotopes in the sample which were in too small quantities to be recorded

Question 53

Chlorine exists as two isotopes 35 Cl and 37 Cl in the ratio 3:1

Which statement about peaks in the mass spectrum of Cl2 is correct?

Answer 53

B is correct - expand like expanding brackets in algebra