Question

Why is the first ionization energy of `"O"` lower than for `"N"`, but the second ionization energy for `"O"` higher?

Answer 1

Well, I would use NIST to check these. I looked up the first two ionization energies (by typing the atomic symbol) and got...

• `"IE"_1 ("N") = "14.534 eV"" "" "" ""IE"_1 ("O") = "13.618 eV"`
• `"IE"_2 ("N") = "29.601 eV"" "" "" ""IE"_2 ("O") = "35.121 eV"`

or...

• `"IE"_1 ("N") ~~ "1402 kJ/mol"" "" ""IE"_1 ("O") ~~ "1314 kJ/mol"`
• `"IE"_2 ("N") ~~ "2856 kJ/mol"" "" ""IE"_2 ("O") ~~ "3389 kJ/mol"`

Here, you can see that the first ionization energy of oxygen atom is lower. Consider the electron configurations.

`"N": [He]2s^2 2p^3`

`underbrace(ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr)))`
`" "" "" "color(white)(.)2p`
`ul(uarr darr)`
`color(white)(.)2s`

`"O": [He]2s^2 2p^4`

`underbrace(ul(uarr darr)" "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr)))`
`" "" "" "color(white)(.)2p`
`ul(uarr darr)`
`color(white)(.)2s`

Since oxygen has a paired `2p` electron, that one will repel the others, making it easier to remove than any of the others.

Easier to remove `->` lower ionization energy.

The second ionization energy of oxygen is higher than for nitrogen. In principle, they should be the same for two seemingly identical `2p` electrons, BUT we would neglect an important factor...

It is because oxygen atom is smaller due to a higher effective nuclear charge `Z_(eff) = Z - S`, where `S` is approximated to be the number of core electrons and `Z` is the atomic number.

`Z_(eff,N) ~~ "7 protons" - "2 core e"^(-) ~~ 5`
`Z_(eff,O) ~~ "8 protons" - "2 core e"^(-) ~~ 6`

So, it is more difficult to remove the electron from the smaller atom, whose electrons are more tightly bound to the nucleus.

Harder to remove `->` higher ionization energy.