Ionisation Energy



Ionisation Energy

Ionisation Energy

1: Define And Use The Term First Ionisation Energy, IE:

> Ionisation energy, ΔHi, the energy needed to remove 1 mole of electrons from 1 mole of atoms of an element in the gaseous state to form 1 mole of gaseous ions.

>This is represented by the following equation:


Note: Ionisation energies are measured (calculated) under standard conditions. The general symbol for ionisation energy is ΔHi. Its units are kJ mol–1. The symbol used for the 1st ionisation energy is ΔHi1.

2: How To Construct Equations To Represent First, Second And Subsequent Ionisation Energies?

> Using calcium as an example: 

1st ionisation energy: Ca(g) → Ca+(g) + e– 

ΔHi1 = 590 kJmol–1 

> If a second electron is removed from one ion in a mole of gaseous 1+ ions, it is known as the 2nd ionisation energy, ΔHi2. Again, we use calcium here as an example: 

2nd ionisation energy: Ca+(g) → Ca2+(g) + e 

ΔHi2 = 1150 kJmol–1 

> When a third electron is removed from one ion in a mole of gaseous 2+ ions is called the 3rd ionisation energy. Again, we use calcium here as an example: 

3rd ionisation energy: Ca2+(g) →  Ca3+(g) + e 

ΔHi3 = 4940 kJmol–1 

Electrons can be removed continuously until only the (positive) nucleus is left. We call this sequence of ionization energies, successive (subsequent) ionisation energies.

Note: All values of ionisations are positive and therefore endothermic. This is because energy must be provided so that the electrostatic attractive forces between the nucleus and the electron can be overcome.

3: Identify And Explain The Trends In Ionisation Energies Across A Period And Down A Group Of The Periodic Table:

>Across a Period:

a) The (first) ionization energy Increases across a period.

b) This increase in the (first) ionization energy happens due to the number of shells across the period remains the same. The proton number increases hence the effective nuclear charge increases resulting in the atomic radius decreasing. 

>Down a Group:

a) The (first) ionization energy Decreases down a group.

b) The (first) ionization energy decreases down a group because new shells are added down the group, the attraction of the nucleus to the outer shell (valence) electrons decreases followed by the shielding effect which also increases. 

Note: Shielding effect: the ability of inner shells of electrons to reduce the effective nuclear charge on electrons in the outer shell

4: Identify And Explain The Variation In Successive Ionisation Energies Of An Element:

> Each successive I.E (ionisation energy) is higher than the previous one because as electrons are removed, the attraction between protons and remaining electrons (Valence and those in the inner shells) increases.

> The second ionisation energy belonging to the element always has a greater value than the first ionisation energy. It should be noted that whenever the first electron of an atom is removed a positive ion is formed. The ion causes an increase in the attraction on the remaining electrons that are present and so the energy that is required to remove the next electron is larger. Every successive (subsequent) ionisation energy is bigger than the previous one for the same reason. 

>Some of the increases are much bigger, however, and these big jumps illustrate evidence for the main principle electron shells.

5: Why Are Ionisation Energies Present?

Ionisation energies are present due to the attraction between the nucleus and the outer electron

6: Explain The Factors Influencing The Ionisation Energies Of Elements In Terms Of Nuclear Charge, Atomic/Ionic Radius, Shielding By Inner Shells And Sub-Shells And Spin-Pair Repulsion.

a) Nuclear Charge: The nuclear charge is positive (+) due to the protons in the nucleus. The greater the Nuclear charge, the greater the ionisation energy.

b) Atomic/Ionic Radius: Atomic or ionic radius is the distance from the center of the nucleus to the outermost orbit. The greater the atomic or ionic radius, the lower I.E (ionisation energy) would be. This is because the distance of outermost electrons to nucleus is large and less energy is needed to remove the electron.

Ionic Radius: 

Positive Ion: Smaller radius than the original neutral atom because shell number decreases, shielding effect also decreases but the attraction of nucleus increases. So, more ionisation energy is required.

Negative Ion: Larger ionic radius than neutral atom because electrons are added while nuclear charge remains the same. Lesser ionisation energy is required.

c) Shielding By Inner Shells And Sub-Shells: The inner shells and subshells of electrons repel outermost (outer shell or valence) electrons, thus shielding them from the positive nucleus. The more electron shells, the greater is the shielding effect. The greater the effect, the lower the I.E (ionisation energy) because there are lesser attractive forces between the nucleus & outer shell (valence) electrons.

d) Spin-Pair Repulsion: 
  Cambridge International Examinations Lawrie Ryan, Roger Norris

> A noticeable decrease in ΔHi1 between nitrogen and oxygen can be observed. Oxygen (2s22p4) has one more proton than nitrogen (2s22p3) and the electron removed is in the same 2p subshell. However, the spin-pairing of the electrons is to a large extent involved here. 

>The diagram above shows that the electron removed from the N is from an orbital that contains an unpaired electron. The electron removed from the O is from the orbital that contains a pair of electrons. The extra repulsion between the pair of (negatively charged) electrons in this orbital result in less ionisation energy being needed to remove an electron. So, ΔHi1 for O is lower than N, because of spin-pair repulsion

Note: These patterns repeat themselves across the third period. However, the presence of the d-block elements in Period 4 interrupts in the pattern, as d-block elements have first ionisation energies that are relatively similar and fairly high.

7 and 8: Deduce The Electronic Configurations And Positions Of Elements Using Successive Ionisation Energy Data.


     Cambridge International Examinations Lawrie Ryan, Roger Norris
    Note: the above values are given in the data booklet.

>In the table above there is a big difference between some successive ionisation energies. 

>For nitrogen this occurs between the 5th and 6th ionisation energies. The difference in Sodium can be observed between the 1st and 2nd ionisation energies. These large changes indicate that for the second of these two ionisation energies the electron being removed is from a principal quantum shell closer to the nucleus and also reveals the group the element is in for Nitrogen since it has five outer shell electrons are in the second quantum shell it is in Group 5. Similarly, with sodium, it has one outer shell electron in the third quantum shell. Hence it is in Group 1.

>Similarly configuration from the total number of electrons removed would be of Nitrogen and Sodium (read the table backward to figure out the configuration):




 > The element illustrated above is “Carbon” which is in Group 14 (4). There is a sudden increase in ionisation energy after the 4th electron which shows it has 4 outer shell electrons in the second quantum shell. Revealing the group it is in. Since the total number of electrons is 6 the configuration is: 1s22s22p2 


> In the table above there is a big jump between the 2nd and 3rd ionisations energies indicating that this element must be in group 2 of the periodic table as the 3rd electron is removed from an electron shell that is seen to be much closer to the nucleus with less shielding and so has larger ionisation energy.

Note: in the diagram below of the trend of ionisation energy across Period 3

> I.E (ionisation energy) of Al is lower than Mg because the electron removed in Al is from a (higher energy level) 3p orbital which is further away from the nucleus than an electron in the 3s  being removed from Mg. The Nuclear attraction is less for 3p than 3s hence I.E (ionisation energy) of Al is lower than Mg.

> I.E (ionisation energy) of S is lower than P because electrons being removed in P is in a half-filled, more stable 3p orbital whereas in S, the pairing of electrons in 3p results in increased repulsion (spin-pair repulsion as we saw in period 2 above) hence less energy is needed to remove an electron.
   Cambridge International Examinations Lawrie Ryan, Roger Norris
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