Use the periodic table to give the electron configuration for each of the following atoms and ions. 

(c) Co(V) in CoO43=

(d) Co(IV) in CoF62-

Video transcript

All right. Hello everyone. So for this question, let's go ahead and determine the electron configurations of the nickel ion in the following compounds. Using the periodic table. Part one, we have nino 32. And for part two, we have N IC 43 negative, we also have four different entry choices labeled A through D. So let's go ahead and get started by recalling a few things about each of the electronic orbitals. S orbitals can hold a maximum of two electrons, P orbitals, a maximum of six electrons D orbitals, a maximum of 10 electrons and F orbitals, a maximum of 14 electrons. Now, using this information, we can actually use the periodic table and divide it into S blocks, P blocks D blocks and F blocks. This makes it easier to find the electronic configuration of a given element. The idea is to start with hydrogen at one S one and continue down each of the rows until you have reached the element that you are considering. Now in both of the compounds provided nickel is going to exist as an ion with some sort of charge. However, before we can consider the electron configuration of an ion, it's easiest to first find the electron configuration of the neutral element. So starting off here with nickel, right, nickel is a transition metal with an atomic number of 28. This means that it has 28 protons inside of its nucleus. However, recall that when elements are neutral, the number of protons and the number of electrons are equal. So in the neutral state, it's going to have 28 electrons. Now, a simpler way of writing an electronic configuration is by first referencing the noble gas that goes directly before the element that you are considering after the noble gas core is when you begin to fill out subsequent orbitals until youve reached your target element. In the case of nickel, the noble gas that goes directly before it is argon. So the electronic configuration of nickel is going to have an argon core. Afterwards, I proceed through the four S orbital and into the 3d orbital. So that gives me four s 2 3d 8. Now this is the electronic configuration of the neutral nickel atom. So now let's go ahead and find the oxidation states of nickel in both of the compounds provided starting off with an eye no 32. Now, because nickel is a transition metal, it's going to have a variable oxidation state because of that, we can treat it as an unknown variable that we can solve for notably, this compound is neutral. So the oxidation states or charges must cancel out. So if the oxidation state of Nicholl is eggs that are unknown, then X added to the oxidation states of both nitrate ions must equal to zero and nitrate. If you recall has a charge of negative one, since there are two nitrate ions here, I would multiply negative one by two. So my expression simplifies to X subtracted by two equals zero, which means that X and therefore the oxidation state of nickel in part one must be equal to positive two. So in order to obtain an oxidation state of positive two nickel must have lost two electrons. And so those electrons must have been removed from the highest energy orbital. In this case, the four S orbital. So for part one here, the electronic configuration of nickel two positive is going to be equal to argon 3d 8, the eight should be a superscript. So just to reiterate the two electrons from the four as orbital were removed to achieve that oxidation state. So now lets go ahead and talk about the oxidation state of nickel. In the second part, that's N IC L 43 negative. Now, in this case, the oxidation state of nickel and the charge of each chloride ion must equal to the overall charge of the compound which is negative three. So if I add X to the total charge coming from each chloride ion that should equal negative three, now, chloride has a charge of negative one and because there are four. All right here, I would multiply that by four. So X subtracted by four equals negative three. And therefore X which is the oxidation state of nickel is equal to positive one. So this time only one electron get removed from the four S orbital. So for part two, that is Argon four S 1 3d 8 and there you have it. So the electronic configurations here coincide with option B in the multiple choice. For part one, it is Argon 3d 8 and for part two, it is Argon four S 1 3d 8. So with that being said, thank you so very much for watching and I hope you found this helpful.