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Consider this set of ionization energies.
Consider this set of ionization energies.
IE1 = 578 kJ/mol,
IE2 = 1820 kJ/mol,
IE3 = 2750 kJ/mol,
IE4 = 11,600 kJ/mol.
To which third-period element do these ionization energies belong?
♦ Relevant knowledge
Energy of ionization is the energy required to take out the most outer, or loosely bound electron from the electron clouds of the atom from the element of. The most distant electron(s) within any atom can be described as the valence electrons. Since valence electrons lie most distant from the nucleus of the Atom, they are affected by the positively charged atom’s nucleus. They are also the least affected. Thus, valence electrons will be the first ones to be removed by nuclei in another atom which has more positively charged effects on electrons. Furthermore, the electrons of more than one is removed from an atom. As electrons are removed, it becomes more difficult and takes more energy to remove electrons. Ionization energy is measured as Kilojoules per mole.
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Solution:
The ionization energies for Al are indicated by the values of four ionization energy.
Al uses electronic configuration
Al (13) = 1s2 (2p6 3s2 (3s2/3p1)
This means that the first electron will be removed from the 3p subshell with less energy (IE1 = 578kJ/mol).
Due to 3s subshell fullness, the energy required for the removal of the second electron was higher (IE2 = 1820 KJ/mol).
Due to the inert configuration of Ne, the removal of the fourth electron required a large amount of energy. (IE4 = 11600 kj/mol).