Question

How would you write a chemical equation representing the third ionization energy for lithium ?

Answer 1

We realize that ionization energy is the energy needed to eliminate the outermost valence electron from the neutral neutral gaseous atom to frame a cation. The units of ionization energy is $kJ{\text{ }}mo{l^{ - 1}}$ .

Complete step by step answer:
Let's talk about progressive ionization energies. In the event that a gaseous atom is to lose more than one electron, they can be eliminated in a steady progression, that is, in succession and not at the same time. In reality when the gaseous atom loses one electron to frame a monovalent cation, the number of electrons in the cation diminishes by one and these are held by the core of the cation of more prominent power. In this manner, with the equivalent energy, the subsequent electron can't be eliminated also, additional energy is expected to frame a divalent cation. In a similar way, energy needed to eliminate the third electron is required to be higher. This plainly shows that electrons are eliminated in progression or consistently.
Presently go to the inquiry.
Ionization energy – The energy required to remove an electron from an atom.

{Li + I.{E_{1}} - - - - - - > {\text{ }}L{i^ + } + {\text{ }}{e^ - }} \\\ {^{}L{i^ + }{ + ^{}}I.{E_{2}} - - - - - - > L{i^{ + + }} + {\text{ }}{e^ - }} \\\ {^{}L{i^{ + + }}{ + ^{}}I.{E_3} - - - - - - { > ^{}}L{i^{ + + {\text{ }} + }} + {\text{ }}{e^ - }} \end{array}$$ Therefore , the chemical equation representing the third ionization energy for lithium atoms is shown below. $$L{i^{ + + }}{_{(s)}^{}}{ + ^{}}I.{E_3} - - - - - - - - { > ^{}}L{i^{ + + {\text{ }} + }}_{(g)} + {\text{ }}{e^ - }$$ Clearly , this reaction is an endothermic reaction . **Note:** Never forget that the ionization energy to eliminate the subsequent electron is named as second ionization energy, to eliminate third electron the ionization energy is called third ionization energy, etc .