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

1 Answer
Aug 12, 2017

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.