CIE A Level Chemistry复习笔记1.1.12 Ionisation Energies & Electronic Configurations

• 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