A-level Chemistry/AQA/Module 1/Periodicity

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Ionisation Energy[edit | edit source]

Ionisation Energy is the energy needed to dislodge an electron from the attraction which holds it in an atom/ion/molecule. The first ionisation energy is the energy needed to dislodge the least powerfully held electron. Second ionization energy is the energy needed to dislodge the second least powerfully held electron etc.

The size of the ionisation energy for an electron is related to the size of the attraction between that electron and the nucleus to which it is bound. 4 factors effect the size of this attraction.

The number of protons in the nucleus.
The higher the number of protons in the nucleus the higher its total positive charge and the higher the electrostatic attraction between it and the negatively charged electron.

The distance between the electron and the nucleus it is bound to.
Attraction decreases with distance.

The number of intervening electrons between the electron and the nucleus - 'Screening'
Intervening electrons have a negative charge which repulses other electrons and reduces the attraction between electron and nucleus. This is referred to as screening. Note that intervening electrons do not 'block' attraction as the name suggests rather they reduce attraction by providing an opposing repulsion.

Whether or not the electron under consideration is alone in its orbital or paired with another electron
Paired electrons repulse each other and reduce attraction between the electron and the nucleus.

Changes in first Ionisation Energy within groups and periods[edit | edit source]

The First Ionisation Energy decreases down Group II. This is because each element has a new shell, increasing the amount of shielding and the atomic radius.

The First Ionisation Energy increases across Period III. This is because:

  • The effective nuclear charge increases, due to increase in the number of protons.
  • The radius of atom decreases with increasing atomic number, hence more energy is required to remove outer electrons.
  • The number of shells remains the same; leading to the same amount of shielding.

There is a drop from Mg to Al because there is a new subshell which is higher in energy (more energy is required to remove the outer electrons). There is another drop from P to S because both electrons are now filled in the 3p subshell, leading to some repulsion.

Changes in Atomic Radius[edit | edit source]

Across Period 3 (Na to Ar) as the number of protons in the nucleus increases, the effective nuclear charge increases but shielding by inner electrons is kept unchanged as new electrons enter the same shell as they are attracted by the protons. This in turn will decrease the atomic radius.