[rayyau35]

1. Explain why the first ionization energy for copper( Z=29) is higher than that for potassium (Z= 19 ),whereas the second ionization potentials are in the reverse order.

2.Which is larger in the ionizayion energy between Zn+ and Cu+ ions?Explain with reasons.Zn (Z=30)


More detailed and dont use 2,8,8,X  as explanations ,use 1s2s2p electronic configurations.Question 1,2  is given for 20 marks totally and dont just type few words to suppose that I know.I hope you will use more "because" this word and not just one sentence to discribe directly.

[argorb]

[!] 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! Thanks!

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: 1s²2s²2p⁶3s²3p⁶3d⁹4s²
This could be written aswell: [Ar]3d⁹4s² meaning that it has the electron configuration of Argon plus 9e^- on the d shell and 2e^- on the s shell.

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: 1s²2s²2p⁶3s²3p⁶4s¹
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]3d¹⁰4s². 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):
1. Orbitals
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):

[wizrak]

Argorb

Coppper has electronic configuration of 1s²2s²2p⁶3s²3p⁶3d¹⁰4s¹ because its more stable than 3d⁹4s²

Anyways if you see the periodic table we can see that copper and potassium is in the same period. Hence its valence electron to be lost (i.e. get ionised) is in the same energy level.
Also, by observing the atomic number we know that copper has got a higher nuclear charge. Hence it pulls the valence electron more closely than potassium. Hence, we need to put more energy in to ionize copper than potassium

HOWEVER, 2nd ionisation energy involves taking electron away from a cation i.e. K⁺ and Cu⁺.
K⁺ has an electron configuration of [Ar]. This shows us that the valence electron is at a lower energy level than Cu⁺. Hence valence electron is closer to the nucleus and nuclear charge is said to be more "effective" i.e. attract the valence electron more strongly. Thus 2nd I.E for potassium is higher than copper.

[argorb]

Argorb

Coppper has electronic configuration of 1s²2s²2p⁶3s²3p⁶3d¹⁰4s¹ because its more stable than 3d⁹4s²

Are you saying that Cu e^- from 4th shell prefers to jump back on the `d' shell to fill it to the maximum so it can be more stable? But this will mean that its metallic nature will increase, i.e. another atom (lets say a hallogen) will take its last e^- very easily; is this the reason that it might want to loose that extra e^- in order to be stable with its other full shells?

[rayyau35]

but I had read the book,the electronic configuration of Cu is ..3d10 4s1 .But the first electron from which orbital(3d or4s??)  is removed by the first ionization energy?I dont know how to express that.After the  first ionization energy, the configuration becomes 3d8  4s2   right?But Both obitals have been included in the change.

Anyone explains that?

[argorb]

but I had read the book,the electronic configuration of Cu is ..3d10 4s1 .But the first electron from which orbital(3d or4s??)  is removed by the first ionization energy?I dont know how to express that.After the  first ionization energy, the configuration becomes 3d8  4s2   right?But Both obitals have been included in the change.

Anyone explains that?

From my understanding of what I've learned, "only" (I edited because probably the better way of saying this is, a big probability represents ->) the `p' and`s' shells are participating in the reactions/ionization. The outer most shells (p and s) e^- are the reactants. Because `d',`f' etc shells requires more energy to fill those in, electrons that fill them are hard to remove. So in case of Cu, which I mentioned that is stable because it has 8 electrons on its outer most shells, 6 on `p' and 2 on`s' = 6 + 2 = 8 e^-. So if you ionize Cu and turn him into a cation, it will have 1 electron remaining on its outer most shell. If Cu would react with K, it will take the single electron on potassium's outer most shell (4s¹) and will fill in the `d' shell to have it fill up to the maximum. So its electron configuration will then be, [Ar] 3d<sup>10</sup>4s<sup>2</sup>, but will turn into a anion, because it will have a negative charge. But if it will react with let's say F, it will probably loose one electron from the `s' shell, and if the case with d¹⁰4s¹ is true this means that it will loose the outer most shell's single electron (the 1 from 4s) and it will turn into a cation, it will have a positive charge but will be a stable atom, which in `his' opinion, this is a good deal. I might be wrong though, I am not sure :-)
I would be glad if someone corrects me and make this more clear (especially the case with 3d¹⁰4s¹).