• Energy?is required to remove an outer shell electron as this involves breaking the?attractive?forces?between the electron and the?positively?charged?nucleus
• There are several factors which affect the magnitude of the ionisation energy:
• Nuclear charge
  • Positive nuclear charge increases with increasing number of protons
  • The greater the positive charge, the greater the attractive forces between the outer electron(s) and the nucleus
  • More energy is required to overcome these forces so ionisation energy?increases?with increasing nuclear charge
• Shielding
  • Electrons?repel?each other and electrons occupying the inner shells repel electrons located in shells further outside the nucleus and prevent them from feeling the?full?effect?of the nuclear charge
  • The greater the shielding effect is, the weaker the attractive forces between the positive nucleus and the negatively charged electrons
  • Less energy is required to overcome the weakened attractive forces so ionisation energy?decreases?with increasing shielding effects

Shielding makes it easier to remove the outermost electrons

• Atomic/ionic radius
  • The larger the radius, the greater the distance between the nucleus and the outer shell electron(s)
  • Increasing distance?weakens?the strength of the attractive forces
  • Larger atoms/ions also result in greater?shielding?due to the presence of more inner electrons
  • Less energy is required to remove the outer shell electron(s) so ionisation energy?decreases?with increasing atomic/ionic radius
• Spin-pair repulsion
  • Spin pair repulsion occurs when the electron being removed is spin paired with another electron in the same orbital
  • The proximity of the like charges of electrons in the orbital results in repulsion
  • Less energy is required to remove one of the electrons so ionisation energy?decreases?when there is spin-pair repulsion

Summary of factors affecting ionisation energies of atoms

• 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

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 subshellNa?????? 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 subshellMg?????? 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
• This is due to?weakened?shielding?effects?through the loss of three electrons
• The large jump corresponds to moving from the?third shell?to the?second shell
• Al???????? 1s²?2s²?2p⁶?3s²?3p¹