There are two possible [M(OH)4]- complexes of first-series transition metals that have three unpaired electrons.

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

(b) Using orbital diagrams, give a valence bond description of the bonding in each complex.

(c) Based on common oxidation states of first-series transition metals (Figure 21.6), which [M(OH)4]- complex is more likely to exist? 

<QUESTION REFERENCES FIGURE 21.6>-

Video transcript

All right. Hi everyone. So this question says that there are two possible first series transition metals that form NCN 42 negative complexes with two unpaired electrons determine the identity of the metals and their oxidation state. Here we have four different answer choices, proposing different oxidation states as well as different identities for these first series transition metals. Now, our first step is going to be to determine the oxidation state of the metal in the complex. Recall first and foremost that the charge of the overall complex is equal to the total oxidation state of the complex. And so the sum of all the oxidation states in your complex should equal the total oxidation state. So in this case, we don't quite know what the oxidation state of the metal is. So I'm going to treat it as a variable X in my equation here. And so when I add X to the total oxidation state of all of my cyanide ligands, I should get a total of negative two because that is the overall oxidation state. Now recall that cyanide is anionic with an oxidation state equal to its charge of negative one. So in this case, I will substitute the oxidation state of cyanide four negative one, which means that my equation simplifies to X subtracted by four equals negative two. And so when I add both sides by four, my X, which is equal to the oxidation state of the metal is equal to positive two. So now the question is to determine the electron configuration of the metal itself. Recall that the first series transition medals or the transition medals in the first row of the D block ranging from scandium two zinc. This means that the electronic configuration for these metals when they are neutral range from Aragon four S 2 3d 1 up to ARGO four S 2 3d 10. However, recall that these metals have an oxidation state of positive two, which means that two electrons must have been removed from the highest energy orbital, which in this case is the four S orbital. So when removing the two electrons from the four S orbital, the electronic configuration actually ranges from Argon 3d 1, two oops to Argon 3d tent. So now we have to consider the electronic configurations that contain two unpaired electrons and recall that there are going to be three different or two different possibilities for that. When we consider the fight that in the orbital diagram for the 3d orbital, the two possibilities are D eight as well as D two because in or rather for the D eight configuration, all five orbitals in the 3d block receive one unpaired electron before three of them eventually become paired, leaving two unpaired electrons. Whereas in the D two configuration, both electrons simply remain in unpaired orbitals. So this means that we can narrow down our electronic configuration slightly more, right, the possible electronic configurations for our metal with a positive two charge are Argon 3d 2 and Oregon 3d 8. However, it would be easiest to identify these elements in the periodic table using their neutral configurations. So to find their neutral configurations, starting with a positive two configuration, we simply add two electrons to the four S orbital leading to an electronic configuration of Argon, four S 2 3d 2 and Oregon four S 2 3d 8. So now when identifying elements from the periodic table, we can use the electronic configuration of the neutral atom because the number of electrons in a neutral atom is equal to the number of protons, which is indicated by the atomic number. So for the electronic configuration of argon or S 2 3d 2, argon contributes 18 electrons, we have two more from the four S orbital and two more from the 3d orbital giving us a total of 22 which means that the atomic number should also or rather should be 22. Now, for the electronic configuration, Argon four S 2 3d 8, we have 18 electrons from Argon, two more from the four S orbital and eight more from the 3d orbital that yields a total of 28 which is equal to the atomic number of 28. And so the element with an atomic number of 22 is titanium, whereas the element with the atomic number of 28 is nickel and there you have it. So here is our final answer. Specifically, it is going to be option D in the multiple choice because the oxidation state of the metal is positive too. And the identity of the metals are nickel and titanium. So 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.