Why isn't the scandium(2+) cation isoelectronic with potassium?

Question

I saw that I got this question wrong on a test, and was very confused. Why aren't they isoelectronic? $\ce{Sc^2+}$ with atomic number 21 should have 19 electrons since 21 - 2 = 19, which is the same as the number of electrons as Potassium with 19 protons and electrons. Does it have something to do with being in different sublevels, one being s and the other d and all?

Answer 1

The neutral $\ce{Sc}$ atom has the electron configuration of $\ce{1s^2 2s^2 2p^6 3s^2 3p^6 3d^1 4s^2}$, and the $\ce{Sc^2+}$ ion has electron configuration $\ce{1s^2 2s^2 2p^6 3s^2 3p^6 3d^1}$. The neutral potassium atom has the electron configuration of $\ce{1s^2 2s^2 2p^6 3s^2 3p^6 4s^1}$. While both $\ce{Sc^2+}$ and $\ce{K}$ have the same number of electrons, their electron configurations are different and therefore they are not isoelectronic to one another. (Isoelectronic implies same number of electrons as well as the same structure)

Answer 2

As has already been stated, the two are not isoelectronic because their electronic structures are different: potassium’s single electron being in an s orbital while scandium(II)’s is in a d orbital.

The reason for this difference is the only other thing that is different: the difference in nuclear charge by 2. A higher nuclear charge causes the orbitals to contract and reduces their energy. However, not all orbitals are equally affected by a change in nuclear charge: the lower an orbital’s principal quantum number is, the more it as affected as a general rule. So the 3d orbitals, belonging to the third shell, feel the charge difference more strongly and adapt more than the 4s orbital which already belongs to the fourth shell.

While for neutral potassium and calcium(I) it is very clear that the 4s orbital has a lower energy than the 3d orbitals, the picture reverses itself once you reach scandium — the first d block element. Because the 3d orbitals’ energies have been reduced more than the 4s energy, it suddenly becomes more stable and the single valence electron of scandium(II) will reside in 3d rather than 4s.

In scandium(I), we also need to take electron-electron interaction into account to fully explain the concept. Basically, each electron you remove from an atom will stabilise all orbitals and vice-versa. Again, this effect affects different orbitals differently, mainly depending on the quantum numbers $l$ and $m_l$. So reducing scandium(II) to scandium(I) will shift the orbital energies around to the very interesting result that one electron will occupy a 3d orbital while one will occupy the 4s orbital. (I have glossed over it, but of course spin and spin pairing also play a role.)

In my humble opinion, chemguide.co.uk has done an excellent job at explaining this problem.

Answer 3

This is a very interesting question, I have also experienced this mistake, so should be careful while answering in such questions. You have accept the @ Allison W's answer he has explained it well and I try to give you a technique to answer such question very carefully without mistaking in the future.

1.First you should write the electron configuration of the element which gives you.**

Sc - 1s2 2s2 2p6 3s2 3p6 3d1 4s2

2.Then you should understand how many electrons are left for the given ion. According to this question it should be 2. Then you should leave them from the outer energy level. That means they are in 4s.

Don't misunderstand this with the way of electrons are filling. When the atom is building according to the aufbau principle the first filling electrons are the 4s electrons and then the 3d. So the last electron which is filling is the 3d electron. But you should understand it is not the electron which is most outer of the atom. When the ions are making the leaving electrons which have least attraction of the nucleus. That means the most outer electrons.

3.Then you should see the stability of the made configuration.

This is not affect for this question but you can face such questions. For example,

Sc+-1s2 2s2 2p6 3s2 3p6 3d1 4s1

But this not happening while there is a more stability configuration of below.

Sc+-1s2 2s2 2p6 3s2 3p6 3d2 4s0

So, you should give your attention in such situation to change such like that. (In primary chemistry if you not changed it is not a case, but in advance you should be careful either in this point.)

4.After writing the new electron configuration of the ion now compare it with the given element's configuration.

K- 1s2 2s2 2p6 3s2 3p6 4s1

Sc2+ 1s2 2s2 2p6 3s2 3p6 3d1

Generalize this steps and you will not mistake such question again.