The Ni²⁺(aq) cation is green, but Zn²⁺(aq) is colorless. Explain.

Video transcript

All right. Hi, everyone. So this question is asking us to explain why palladium two positive and nucleus solution appears brown W cadmium, two positive appears colorless. Now, here we have four different answer choices here. And all of them happened to discuss properties of the electrons found in the D orbitals of either palladium or cadmium. Specifically, they discuss whether or not the D electrons of either cadmium or palladium can be promoted into higher energy levels via the absorption of light. So please keep that in mind as we go along through this question. So if I scroll down here to give myself some more space, recall first and foremost that the presence of color in a given compound refers to its ability to absorb and therefore reflect a visible light. Now, I want to first consider the electronic configurations of both palladium and cadmium. But first I want to start off with palladium. Now recall it in a given neutral atom, the atomic number which corresponds to the number of protons is going to be equal to the number of electrons. So palladium has an atomic number of 46 which corresponds to 46 protons. Now when palladium is neutral, that means palladium has 46 electrons as well. And its electronic configuration is a krypton core followed by or D 10 because I want to point out here that palladium is actually an exception to standard electronic configuration because instead of filling out the S shell before the D shell, the D shell is filled completely, instead resulting in a more stable configuration. Now when drawing the or finding the configuration of palladium two plus a positive charge implies that electrons are being removed from the highest energy level, which in this case is four D, right. So here for palladium two plus, we're going to see a krypton core but only eight electrons in the D shell because two have since been removed, which means that the electronic configuration of palladium ion or palladium two positive is a krypton core followed by four D eight. So at this point, we can go ahead and fill in the D orbitals or palladium two positive recall that in this case, we have five, three of which are considered lower energy and two of which are considered higher energy. Now, the three lower energy orbitals are DXY DXZ and DYZ, whereas the higher energy orbitals are DX to the second and TX to the second minus or subtracted by Y to the second. Now, here we have eight D electrons to distribute according to both the alpha principle and Han's rule. So the abau principle says that the lowest lower energy orbitals are to be filled in first. And Hunt's rule states that orbitals of the same energy are half filled before they are fully filled, which means that I will distribute one electron to each of the three lower energy orbitals before I will give each of them a pair, meaning that six electrons have been distributed into the three lower energy orbitals. And so my remaining two are going to occupy each of my higher energy orbitals. Now, the first thing I want to point out here is that the higher energy orbitals have empty space, so to speak, in the sense that each high energy orbital is only half filled and can therefore accommodate one extra electron each. Now, this is important because recall according to crystal field theory, there can be an electronic transition from those electrons in the lower energy D orbitals to those in the higher energy D orbitals. And that process requires a certain amount of energy referred to as delta. Now delta is a function of the amount of light that was absorbed to gain the necessary energy needed for the electronic transition. So when compounds or complexes absorb light to achieve this transition, a color is observed. So considering the electronic configuration of palladium, two positive electrons, oops electrons can be promoted to the higher energy D orbitals because the higher energy D orbitals have space to accommodate more electrons. And that process can be achieved by absorbing light, which explains why palladium or solution of palladium appears to have a color to it because it does absorb light and therefore reflect light. So now lets compare this to cadmium neutral cadmium has an atomic number of 48 which corresponds to 48 protons. Now assuming cadmium is neutral, it will have 48 electrons to cancel out our protons completely. So the electron configuration of cadmium when it is neutral is a krypton core followed by five s 24 D 10. No. Here, Cadmium happen to have the same charge as palladium, right. It is too positive, which means that two electrons must be removed from the highest energy level, which in this case is going to be five S. So the electronic configuration of cadmium two positive is krypton followed by four D 10. And here we can fill in or D orbitals the same way we did or palladium, right. So here are oops my lower energy orbitals and here are my higher energy orbitals. So now I will start by filling in my lower energy orbitals, starting off by giving each orbital one electron and then pairing them. So my remaining four electrons will be distributed among my two higher energy orbitals. So herein lies the difference between cadmium two positive and palladium two positive because the higher energy orbitals of cadmium two positive do not have any extra space to accommodate electrons from the lower orbitals that may attempt to transition So therefore, right, the D orbitals are completely fill, which means that electrons cannot actually be promoted from the lower orbitals or lower energy orbitals to the higher energy ones, which means that cat beam is not going to absorb light in this case. So if I scroll up one more time or answer, our answer is going to be option a in the multiple choice palladium, two positive in aqueous solution appears brown while cadmium two positive appears colorless because electrons in palladium two positive can be promoted to higher energy D orbitals by absorbing light. And with that being said, if you stuck around to the end of this video, thank you so very much for watching and I hope you found this helpful.