The document discusses atomic structure and mass spectrometry. It defines key terms like mass number, atomic number, and isotope. It explains the process of mass spectrometry, including ionization, acceleration, deflection, and detection of ions. Graphs of ionization energies are analyzed to determine electronic configurations and periodic trends. Successive ionization energies are explained by electron shielding effects. Radioactive decay and half-life are also defined.

1. Unit 1
Atomic Structure
(& mass spec)
Periodicity
(Period 3 and IEs)
Quick Peek

2. Define the terms mass number and atomic number. Use these terms to
explain the difference between the terms element and isotope.
Isotope: atoms of an element with the same atomic number
(proton number) but different mass number (due to different
numbers of neutrons).
Mass number = The total number of neutrons and protons in an
atom of an element
Atomic number = The number of protons in an atom of an
element
What does a mass spectrometer do?

3. THE MASS SPECTROMETER
= an instrument which :
(a) converts neutral atoms (or molecules) into
positively charged ions by removing an electron
(b) then separates these ions according to their
relative mass (m) to relative charge (z) ratio, (m/z)
(c) then measures "m/z" and the % abundance
for each ion.
(d) then produces a MASS SPECTRUM = graph of
“m/z” versus % abundance

4. A Mass
Spectrometer
to form
SEPARATE
atoms /
molecules
to remove air
molecules which
would also be
measured
Order of actions
= V I A D D R
= Victory Is A Definite Detectable Result
V
I
A
D
D
R

5. Ionisation
Electron gun fires high-energy
electrons at the minimum energy on
to the sample
 Minimum energy: so no more than 1 electron is knocked out, so
reducing the risk of 2+, 3+ ion formation
A(g) A+(g) + e-
 produces a positive ion of the atom (A+) or molecule (M+)
High energy
electrons
Positive ions
repelled
Electron gun –
electrically
heated coil
Electron trap (+)
Vapourised
sample
Ion repeller (+)
M2+ ions may also be formed if 2 e- knocked off – very rare
Ions of molecules can fragment
M(g) M+(g) + e-

6. Acceleration
High speed
beam of ionised
sample
Ionisation chamber at
+10000 volts
Final plate at
0 voltsIntermediate plate
+ve ions repelled by high +ve potential
 accelerates the +ve ions through a slit
 narrow beam of fast moving +ve ions.
Sample needs to be ionised and fast moving to
make subsequent separation and detection possible

7. Deflection
Electromagnet
Ion stream C
Ion stream A
Ion stream B

Mixed ion stream
from accelerating
unit
High speed beam of +ve ions is deflected by a strong, variable
magnetic field.
Deflection is GREATER for :
(a) lighter (m lower) ions and
(b) more charged (z higher) ions
i.e. ions with LOWER m/z ratio
Does ion stream A, B or C have
lowest m/z ratio? :
A = Lowest m/z
C = Highest m/z
Note : deflection also greater for faster ions
This field strength is steadily increased
causing ions of INCREASING m/z value to be deflected in turn
onto the detector.
Which ion stream, A, B or C, is being detected in the diagram? B

8. Detection and measurement
Ion stream B
Metal box
Wire to
amplifier
Each ion reaching the
detector takes an e-
from the metal box
 tiny current produced
 current measures
abundance of that ion.
Output from recorder is
called a ‘mass
spectrum’
showing abundance (or
detector current) against
m/z ratio for each isotope.
Relative
abundance
m/z

9. Ar(Kr) = 84.06= (82 x 12/100) + (83 x 12/100) + (84 x 55/100) + (86 x 21/100)
Minor peaks are also observed at m/z 41, 41.5, 42 and 43. Why ?
These are caused by Kr2+ particles produced by the rare event of 2
electrons being knocked off the atom during ionisation.
The % distribution of isotopes is the same for the Kr2+ particles as it is for
the Kr+ particles, even though their abundance is considerably lower than
the major m/z peaks

10. Explain the terms ionisation, fragmentation, acceleration, deflection and detection
as used in mass spectrometry.
Ionisation
Fragmentation
Acceleration
Deflection
Detection
A stream of high speed electrons bombards gaseous sample and
knocks off an electron to form a gaseous cation
Bond(s) break in the gaseous cation and creates a smaller
molecule (or atom) cation and a free radical molecule (or atom)
Beam of gaseous cations passes through holes / slits in two
negatively charged plates with a potential difference across them
Beams of cations are deflected by magnetic field.
The smaller the m/z the greater the deflection.
Gaseous cations with a particular m/z ratio hit the detector and
acquire electrons thereby generating a transient electrical current.
The amount of current is proportional to the abundance of cations
colliding at the detector.

11. Define
Ar of element X Mr of molecule Y
Average* mass of one
atom of X
1/12th mass of one atom
of 12C isotope
*weighted average
related to abundance of
its naturally occurring
isotopes
Average mass of one
molecule of Y
1/12th mass of one atom
of 12C isotope

12. Calculate the relative mass of an 16O atom [Ar(O)]
Calculate the relative mass of a 12C16O2 molecule [Mr(CO2)]
= Mass of one O atom
1/12 x Mass of one 12C atom
= 2.656 x 10-26
1/12 x 1.992 x 10-26
kg
kg
= 16 (g/mol)
= Mass of one CO2 molecule
1/12 x Mass of one 12C atom
= 7.304 x 10-26
1/12 x 1.992 x 10-26
kg
kg = 44 (g/mol)

13. Define the term first ionisation energy and write an equation to represent the first ionisation
energy of chlorine.
State whether the process in b) is exothermic or endothermic and explain your choice.
State and explain the trend in first ionisation energies down a group of the Periodic Table.
The minimum energy required to remove one mole of electrons
from one mole of gaseous atoms to form one mole of singly charged
gaseous cations
Cl(g)  Cl+ (g) + e-
Endothermic because energy must be supplied to overcome the
electrostatic force of attraction between negatively charged electron
and the positively charged nucleus
First ionisation energies decrease down a group.
Although nuclear charge increases down a group, atomic radius
increases, as does electron shielding by occupied orbital shells, hence
the outer shell electron(s) are less attracted to the nucleus and require
less energy to remove.
What is the second ionisation energy?

14. (c) Which of these three isotopes of sulphur would you expect to have the:
highest first ionisation energy? .
highest melting point?
greatest chemical reactivity?
(d) Use the following % composition data to calculate the relative atomic mass of sulphur.
Naturally occurring sulphur contains 95.0% sulphur-32, 0.8% sulphur-33 and 4.2% sulphur-34.
All identical because all have same electron configuration
Sulphur-34 because higher mass atoms create stronger van der Waal forces
All identical because all have same electron configuration
Ar = (32 x 95/100) + (33 x 0.8/100) + (34 x4.2/100) = 32.092
Naturally occurring sulphur contains 95.0% sulphur-32, 0.76% sulphur-33 and
4.2% sulphur-34.
S
32
16 S
33
16
S
34
16

15. Electronegative trend is valid for 1st, 2nd, 3rd periods,
but d-block elements fluctuate EN values (effect upon 4th, 5th periods etc.)
FrF Fr 0.7 and F 4.0 electronegativities . What is electronegativity?
Which two elements, if combined in a compound, would produce a compound with
the greatest difference in electronegativity?
What happens to electronegativity, ionization energy and atomic radius
as you go DOWN A GROUP?
Ability of an atom to attract electron
density (or e- or –ve charge) in a
covalent bond or shared pair

16. IncreasingAtomicSizeWhy?
Across a period: effective nuclear charge acting on electrons increases (electrons occupy
existing shells, no additional shielding, however increased proton nuclear charge)
Down a group: effective nuclear charge acting on electrons decreases (increasing electron
shells creates increased electron shielding)
Decreasing Atomic Size

17. 3
5
4
n
log10 In
VARIATIONS IN IONISATION ENERGIES FOR SILICON
Hence, electron arrangement
is :2,8,4
Si
X
X
X
X
XX
X
X
X
X XX
X
X
 4 electrons FURTHEST
from and MOST SHIELDED
from the nucleus  EASIEST
to remove
 2 electrons VERY CLOSE to
and NOT SHIELDED from the
nucleus  MOST DIFFICULT to
remove
 8 electrons at
INTERMEDIATE
DISTANCE, with
INTERMEDIATE
SHIELDING from the
nucleus 
INTERMEDIATE
DIFFICULTY to
remove
This model of atomic
structure is associated
with Niels Bohr (1915)

18. 3
5
4
n
log10 In
VARIATIONS IN IONISATION ENERGIES FOR ELEMENT C
Click on the group
number of element
suggested by these
data
1 2
3 4
5 6
7 0

19. VARIATIONS IN THE SUCCESSIVE
IONISATION ENERGIES
OF THE SODIUM ATOM
2.50
3.00
3.50
4.00
4.50
5.00
5.50
1 2 3 4 5 6 7 8 9 10 11
log In
Number of electrons removed

20. (a) Element X has an atomic number of 9. Figure 1 shows its mass spectrum and Figure 2 is a
graph of its successive lg ionisation energies.
(b) (i) Write down the electronic configuration of the element X.
(ii) To which group of the Periodic Table does X belong?
(iii) What is the relative atomic mass of X?
iv) Describe and explain the trend shown by the ionisation graph in Figure 2
Fig 2
logIn
n
Fig 1
Rel.Abundance
m : z ratio
19 38
x x x x x
x
x
x x
1s2 2s2 2p5 since Fig 2 shows basic e- configuration to be 2,7
Group 7 since Fig 2 shows 7 e- in outer e- shell
19 19=F+ and 38=F2
+

21. As successive electrons are removed, the ion which
remains has an increasing positive charge so more
energy is needed to remove further electrons
The large increase between IE7 and IE8 occurs because
this is the transition between inner shell (closer to the
nucleus, less shieldung, greater effective nuclear
attraction) and outer shell (further from the nucleus,
greater shielding, less effective nuclear attraction)
electrons.
Written answer: explaining the observed trend

22. FIRST IONISATION ENERGY (I1) vs ATOMIC NUMBER
0
500
1000
1500
2000
2500
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Atomic Number (Z)
I1/kJpermole
Na
LARGE DECREASE in IE1 between periods because e-
removed from more distant and more shielded higher
energy electron orbital shells, which overcomes the
increased nuclear attraction due to increased proton
numberExplained by transition between s and
higher energy/better shielded p sub-level
Explained by electron removed
from spin-paired p suborbital
where there is repulsion between
paired e which reduces required IE
Si
Mg
Al
P
S
Cl
Ar
O
B

23. N
N/2
N/4
N/8
0
Radioactive decay (e.g. 14C)
Number
of atoms
TimeSecond Half-lifeFirst Half-life ThirdHalf-life
X
X
X
X
X
X
X
X
X
X
X
X
t½ = the time taken for N atoms to decay to N/2
atoms (e.g. 5730y for 14C)
Nt=No e-λt
ln Nt = -λt
No