Unit 1

Atomic Structure (& mass spec)

Periodicity (Period 3 and IEs)

Quick Peek

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?

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

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

IA

D

D

R

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-

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

DeflectionElectromagnet

Ion stream C

Ion stream AIon 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) ionsi.e. ions with LOWER m/z ratio

Does ion stream A, B or C have lowest m/z ratio? :

A = Lowest m/zC = Highest m/z

Note : deflection also greater for faster ions

This field strength is steadily increasedcausing 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

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

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

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.

DefineAr 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

Calculate the relative mass of an 16O atom [Ar(O)]

Calculate the relative mass of a 12C16O2 molecule [Mr(CO2)]

= Mass of one O atom1/12 x Mass of one 12C atom

= 2.656 x 10-26 1/12 x 1.992 x 10-26

kgkg = 16 (g/mol)

= Mass of one CO2 molecule1/12 x Mass of one 12C atom

= 7.304 x 10-26 1/12 x 1.992 x 10-26

kgkg = 44 (g/mol)

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 electronsfrom 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?

(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.

S32

16 S33

16S

34

16

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

Incr

easi

ng A

tom

ic S

ize

Why?

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

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

XX

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)

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 23 45 6

7 0

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

(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

l

og I n

n

Fig 1

Rel

. Abu

ndan

ce

m : z ratio

19 38x 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+

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

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)

I 1 /k

J p

er m

ole

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

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

XX

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