Two first-series transition metals have three unpaired electrons in complex ions of the type [MCl4]2-.

(a) What are the oxidation state and the identity of M in these complexes?

(b) Draw valence bond orbital diagrams for the two possible ions.

(c) Based on common oxidation states of first-series transition metals (Figure 21.6), which ion is more likely to exist?

<QUESTION REFERENCES FIGURE 21.6>

Video transcript

All right. Hi, everyone. So this question says that two second series transition metals can form MBR four negative complex ions with three unpaired electrons determine the oxidation state of the metals and their identities. Now, here we have four different answer choices, proposing an oxidation state of either positive two or positive three as well as different transition metals, including technetium, silver, molybdenum, palladium, manganese copper, chromium and nickel. So the first step is going to be to find the oxidation state of the metal in this complex aisle, right. So here we've got M BR four negative and recall first and foremost that the charge of the complex as a whole is equal to the total oxidation state of the complex. And the reason why I say this is because recall also that the sum of the oxidation states of the metal and all ligands must equal the overall oxidation state. Now here notice how we have four bromo ligands in brick that bromide is an anion with a charge of negative one. Therefore, right, each bromo ligand that we have in the complex is going to contribute an oxidation number of negative one giving me a total of negative four. So if the oxidation state of the metal is X given that, that's what I'm solving for, then X subtracted by four equals negative one, which is the total oxidation state of the complex. And so when I add both sides by four, I get that X is equal to positive three, meaning that the oxidation state of the metal is going to be positive three, which means that I can eliminate options. A and C because they propose an oxidation state of positive two. So now the question becomes, what is the electronic configuration of the possible metals? And to find this, we have to reference our second series transition metals which refer to all transition metals on the second row of the D block, right? Those range from atrium all the way down to Cadia and their electronic configurations range from crypton five S 241 all the way down to Krypton 52 oops four D 10, right? Because as you proceed along the second series, you continue to add electrons to the four D orbital. However, right, what I want to point out here is that this is, or these are the electronic configurations of the neutral atoms. But in this case, we need an oxidation state of positive three. Now, because the oxidation state is positive, this implies that electrons have to be removed from the highest energy orbitals. So this means that of or rather we'll take the neutral electronic configurations and remove three electrons from the highest energy orbitals. In this case, two are removed from the five s orbital and one more is removed from the four D orbital. So when I take this into consideration, the electronic configurations turn into krypton four D zero all the way down to krypton four D nine, taking into account the electron that I've since removed. So now it's these new electronic configurations that we have to take into account when deciding which configuration gives us three unpaired electrons. Like the question mentioned in the beginning now to obtain three unpaired electrons, there are two distinct possibilities. There is D three and D seven because if I draw an orbital diagram for the four deep orbital like so the D three configuration is going to place all three electrons in their own degenerate orbital, meaning that all of them are going to be unpaired. Now, for D seven, the first five electrons go into each degenerate orbital before the remaining two are paired, leaving us with three unpaired electrons still. So to find metals that have the appropriate number of unpaired electrons, they have to have the electronic configurations of either krypton four D three or krypton four D seven with an oxidation state of positive three. Now, as a neutral metal to find the the electronic configuration of the neutral metal, we're simply going to add the three electrons that we took away earlier, right? So Krypton four D three becomes krypton five S two four D four, whereas krypton four D seven becomes krypton five S 24 D eight. And just to reiterate the reason why I'm looking for the neutral electronic configuration is so that I can locate these transition metals on the periodic table. Right? So in reference to the electronic configuration, krypton five S 244 recall that krypton in the neutral state contributes a total of 36 electrons which matches its proton or its atomic number. Therefore, a number of protons, right. So it's krypton 52 44 that's 36 electrons from krypton two more from the five S orbital and four more from the four D orbital giving me a total of 42. Whereas for my second electronic configuration, thats krypton 52 48 Ive got 36 electrons from Krypton, two electrons in the five S orbital and eight from the four D orbital. And this gives me a total of 46. No, we've just calculated the number of electrons, right? But when an atom is neutral, the number of electrons is equal to the number of protons, right. So by locating or calculating the number of electrons present in a neutral atom, we have found the atomic number of the respective element because the atomic number is equal to the number of protons. So an atomic number of 42 corresponds to molybdenum, whereas an atomic number of 46 corresponds to palladium. And there you have it. This is our final answer, right? If I scroll up here to the answer choices, then the oxidation state is going to be positive three. And the identity of the transition metals is molybdenum and palladium. So with all that being said, if you stuck around to the end, thank you so very much for watching and I hope you found this helpful.