Question

How do you estimate the successive ionization energies of oxygen?

Answer 1

Hint: Ionization energy depends on the magnitude of the attraction between the nucleus and the electrons. The more tightly held an electron is the more energy will be needed to remove it from the atom.

Complete step by step answer:
Ionization energy – It is simply the energy needed to remove one or more electrons from a mole of isolated atoms in a gaseous state the ionization process produces positive ions or cations.
The attraction between the nucleus and an electron. That surrounds it oxygen, which is located in group \[16\] and period \[2\] of the periodic table, has an atomic number equal to \[8\] out these \[8\] electrons, only \[2\] are considers are valence electrons electronic configuration of oxygen \[O:1{{S}^{2}}2{{S}^{2}}2{{P}^{4}}\] its ionization energy will increase, the electron-electron repulsion that exists between then will impact the successive ionization energies.
Let us assume that we know oxygen’s first ionization energy which is equal to \[1314KJ/mol.\]
Let us say that we need \[1500KJ/mol\] more for the second electron. This will happen.
\[\to {{2}^{nd}}I.E=1314+1500\]
\[=2814KJ/mol\]
The third electron will require little more energy \[1750KJ\]
\[\to {{3}^{nd}}I.E=2814+1750\]
\[=4564KJ/mol\]
The \[{{4}^{th}}\] (fourth will need a little more, so extra \[2150KJ\]).
\[\to {{4}^{th}}I.E=4564+2150\]
\[=6714KJ/mol\]
The fifth will need a little more add an extra \[2900KJ\]
\[\to {{5}^{th}}I.E=6714+2900\]
\[=9614KJ/mol\]
The sixth will need a little more, so add an extra \[3300KJ\]
\[\to {{6}^{th}}I.E=9614+3500\]
\[=13914KJ/mol\]
We are down to the last two electrons, the one located in the first energy level.

Note: If we get the trend right, then we can say that we estimated the successive ionization energies of oxygen correctly. This is important because the ionization energy for electrons is located on the same energy level. That will be relatively similar in magnitude.