] Please take into consideration that I am a new to chemistry. Some of the afirmations I wrote here may be wrong, if this is the case I am apologizing for the wrong information and I will please a member to correct me
1. Ionization energy is measured in eV (electron Volts), meaning how much volts (energy) is required to pop off one electron from an atom.
A. Cu with atomic number (protons) = 29 has the following electron configuration: 1s2
This could be written aswell: [Ar]3d9
meaning that it has the electron configuration of Argon plus 9e-
on the d
shell and 2e-
on the s
Considering that the valence number of electrons, of Cu is 2 (2 e-
on the `s' shell) it is stable (on the `s' shell, maximum 2 electrons can fit in, whereas in `p' shell 6 maximum and in `d' shell,maximum 10). You have to note though that, the more electrons, the more protons, the bigger the attraction thus, the electrons are pulled towards the nucleus, in which the atom is getting smaller, but the pulling attractions happening are bigger, so taking one electron away from a stable shell is very hard, you must put a lot of energy to pop that off (which means the ionization energy is big).
B. Potassium with atomic # = 19 has the electron configuration: 1s2
Written as: [Ar]4s1
having the same electron configuration as Argon but with one more electron on the 4s shell.
Having in mind that we acknowledged that the atom is getting smaller when there are more electrons because the number of protons is increasing aswell, however, in the case of K, being a bigger atom an without a stable 4s shell the pulling electron force towards the nucleus is smaller so poping off that electron from the 4s shell is easier (the ionization energy required is lower -- lower electron volts is needed in order to pop the electron off).
C. The situation reverses at the second ionization because, in case of K, if you pop that one electron from the 4s shell, it will have a full stable 3s and 3p shell (most outer shells). Meaning it will have 8 valence electrons. When an atom has 8 valence electrons (8 electrons on its outer most shells) it is a stable atom and the energy required to pop one of them is huge. Whereas in the case of Cu, if you pop one electron off from its outer most shell (4s) it will have 4s1
, being in the situation that K was before, lower attraction towards the nucleus, bigger atom.
2. We already know that Cu is a stable atom because it has 2 electrons on its outer most shell (4s shell) so the energy required to pop one electron off this shell is big.
Zn however requires an even bigger eV energy to ionize it (pop one electron off). Zn electron configuration is: [Ar]3d10
. So notice that it has one more e-
on the 3d shell compared to the 9e-
that Cu has in its 3d shell. What we said before? The more the electrons, the more the protons, the smaller the atom, the bigger the attraction, the higher the energy to ionize.
So to answer your question, Zn requires a bigger energy (with one more eV) than Cu requires.
3. You should watch the following videos (at least this is how I managed to learn all of this especially because it is explained with drawing and a human voice; the explanation makes it easy for the brain to digest the information too):
2. More on Orbitals and electron configuration
3. Electron configuration
4. Even more on electron configuration
5. Valence electrons
6. Table trend: Ionization energy
All of these videos are from Khan Academy.
Btw, I draw this trend when I exercised on those videos a while ago, it may help you (be aware that the image is quite big):