IE

1 of 9 © Boardworks Ltd 2009

Starter: Electron configuration: true or false?


What is ionization energy?

Ionization is a process in which atoms lose or gain electrons and become ions.

The first ionization (I1) energy of an element is the energy required to remove one mole of electrons from one mole of gaseous atoms.

M(g) → M+(g) + e-

(g)

Looking at trends in ionization energies can reveal useful evidence for the arrangement of electrons in atoms and ions.

The second ionization (I2) energy involves the removal of a second electron:

M+(g) → M2+

(g) + e-(g)


Successive ionisation energies

• The energy required to remove successive electrons!• i.e. Second ionisation energy - the energy required to

remove one electron from each atom in one mole of gaseous 1+ ions to form 2+ ions.

• For example, lithium:

1st IE (Li(g) Li+(g) + e-) = +520 KJmol-1

2nd IE (Li+(g) Li2+(g) + e-) = +7298 KJmol-1

3rd IE (Li2+(g) Li3+(g) + e-) = +11815 KJmol-1• Each successive IE is larger due to less electron

repulsion, so electrons are subjected to greater nuclear attraction


Ionization energy definitions


Three factors affect ionisation energy.

1.Atomic radius

1.Nuclear charge

1.Electron shielding


1. Atomic Radius

• Decreases across a period, increases down a group

• The nearer the outer electrons are to the nucleus, the harder they are to remove and therefore the larger the ionisation energy. Hence ionisation energy generally increases across a period


2. Nuclear Charge

• the more positive the nuclear charge, the harder it is to remove electrons, therefore the larger the ionisation energy

• i.e. as you go across a period, the nuclear charge increases by 1 each time, making it harder to lose electrons and therefore an increase in IE.


3. Electron Shielding

• The lower the amount of shielding between the nucleus and outermost electron, the stronger the effective nuclear charge.

• Therefore the harder it is to remove that outer electron so the larger the ionisation energy


Plotting the successive ionization energies of magnesium clearly shows the existence of different energy levels, and the number of electrons at each level.

Successive ionization energies increase as more electrons are removed.

Evidence for energy levels

Large jumps in the ionization energy reveal where electrons are being removed from the next principal energy level, such as between the 2nd and 3rd, and 10th and 11th ionization energies for magnesium.

electron removed

ion

izat

ion

en

erg

y

2

3

4

5

6

12th

11th

10th

9th

8th

7th

6th

5th

4th

3rd

2nd1

st


Example Exam Q

• Element X has the following values (kJmol-1) for successive ionisation energies:

• 1093, 2359, 4627, 6229, 37838, 47285

a)What group in the periodic table is it in?

b)Explain your answer to a)


Answer

• there is a large increase in ionisation energy between the 4th and 5th IE

• This shows that the electron has been removed from a different shell. In this example the element would be in group 4 (the 5th electron to be removed is in a shell closer to the nucleus)

• This increase occurs because the electron is closer to the nucleus so greater nuclear attraction, and because of less electron shielding.


Trends across a period :Evidence for sub-levels

The first ionization energies for the elements in period 3 show a general increase.

The outer electron in Al is in a 3p orbital which has slightly higher energy than an 3s orbital therefore needs less energy to remove it

However, aluminium’s value is below that of magnesium. This suggests that the third principal energy level is not one single energy level.

element

firs

t io

niz

atio

n e

ner

gy

(kJ

mo

l-1)

600

800

1000

1200

1400

Na Mg Al P Ar400

Si S Cl

1600


Evidence for Hund’s rule

The first ionization energies for the elements in period 3 show a general increase.

element

firs

t io

niz

atio

n e

ner

gy

(KJ

mo

l-1)

600

800

1000

1200

1400

Na Mg Al P Ar400

Si S Cl

1600

However, sulfur’s value is below that of phosphorus. The highest energy electrons of both are in the 3p sub-level. In Phosphorus, all the 3p orbitals contain 1 electron. In Sulfur, 1 orbital has 2 electrons. The repulsion between the 2 electrons makes it easier to remove 1, therefore lower IE.


So what would you expect to happen to the ionisation energies down a group?

Although the nuclear charge increases the ionisation energies decrease due to increased atomic radius, and increased electron shielding. The outer electron is in a main level that gets further from the nucleus each time. These two factors outweigh the nuclear charge gain and hence we see the pattern shown. The outermost electron is more easily lost, hence a lower IE


Although the nuclear charge increases the ionisation energies decrease due to increased atomic radius, and increased electron shielding.

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