Why is the first ionization energy of "O"O lower than for "N"N, but the second ionization energy for "O"O higher?
1 Answer
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"IE1(N)=14.534 eV IE1(O)=13.618 eV "IE"_2 ("N") = "29.601 eV"" "" "" ""IE"_2 ("O") = "35.121 eV"IE2(N)=29.601 eV IE2(O)=35.121 eV
or...
"IE"_1 ("N") ~~ "1402 kJ/mol"" "" ""IE"_1 ("O") ~~ "1314 kJ/mol"IE1(N)≈1402 kJ/mol IE1(O)≈1314 kJ/mol "IE"_2 ("N") ~~ "2856 kJ/mol"" "" ""IE"_2 ("O") ~~ "3389 kJ/mol"IE2(N)≈2856 kJ/mol IE2(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^3N:[He]2s22p3
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
It is because oxygen atom is smaller due to a higher effective nuclear charge
Z_(eff) = Z - S , whereS is approximated to be the number of core electrons andZ 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.
