In the ground-state electron configuration of Fe³⁺,...

Question

In the ground-state electron configuration of Fe³⁺, how many unpaired electrons are present?

Accepted Answer

Welcome back, everyone. For the following ions provide their orbital diagrams and identify whether it is diamagnetic or paramagnetic. Let's recall that a substance that is diamagnetic will have no unpaired electrons in its electron configuration. And this will correspond to the substance being slightly repelled by a magnet. Now, let's recall next, that a substance that is paramagnetic will have unpaired electrons in its electron configuration. And this will correspond to a weak attraction to a magnet or a magnetic field. So now with this outlines, let's begin with our first ion which we're given for part one given as our paladium two plus ion recall that for neutral paladium on our periodic table, we have an atomic number equal to 46 corresponding to 46 protons and for again, neutral paladium 46 electrons. So beginning by writing out our electron configuration for neutral paladium, we would start in our s block across our first period of our periodic table where we would pass through our one s two subs shell because in our exponent we have two, since we would fill in two electrons and then moving forward, we would pass through our second energy level hitting our two S two subs shell where we would fill in two electrons. Then moving forward in our configuration, we would hit our two P six subs shell filling in six electrons in our P block. Moving forward. In our configuration, we would hit our three S two subs shell filling in two electrons. And then moving forward in our configuration, we would hit our three P six subs shell filling in six electrons. Then moving forward, we would hit our four S two subs shell filling in two electrons and then moving to our D block, which at our fourth energy level, we should recall, begins at the third energy level. So we would have 3d 10 filling in 10 electrons. Then moving forward, we would hit our four P six subs shell filling in six electrons and then continuing on, we would hit our five S two subs shell filling in two electrons where ultimately, we hit our D block again, which at the fifth period of our periodic table begins at the fourth energy level. So we would have four D and we would count for a total of eight units to land on our atom paladium, meaning we would fill in eight electrons in our exponent. Now, we want to outline that our valence shell is going to consist of our five S and four D subs shell consisting of our valence electrons. So sorry, this should say valence shell where the exponents represent our valence electrons. Now, as we stated, we're dealing with a plus two ca of paladium. So we would lose two electrons which we want to remove from our highest energy orbital. And so we would remove the two electrons from our five S two subs shell. And so now we would have a electron configuration for our paladium two plus ion in its valence shell. So we're going to write out our valence electron configuration specifically being five S 04 D eight. And this electron configuration for our paladium two plus ion would makes sense, especially considering we removed two electrons from our highest energy orbital and filled, kept our lower energy orbital, which is our four D subs shell slash orbital filled with its eight electrons and so or rather half filled. Since again, we should recall that our D block can hold a maximum of electrons. So we've applied successfully our alpha principle and huns rule and we can note that down. So now with this electron configuration for our valence shell of our palladium two plus ion, we want to draw its orbital diagram again for paladium two plus. So again, we have our five S subs shell and this should say five s here, which consists of just one orbital which we will not fill in any electrons in. And then we have our four D subs shell which we would recall that our D block contains a total of five orbitals which I'm representing as boxes in which we fill in only eight electrons. And recall that we would single singly fill in our electrons with an upward arrow or downward arrow, whatever your choice is in each orbital. So we would have five filled in so far and we need three more electrons filled in. We would make a downward arrow in the first orbital pairing up our first orbital with electrons, a downward arrow in our second orbital giving us seven electrons so far. And then a downward arrow in our third orbital, which would leave us with a total of two unpaired electrons in our D subs shell. And therefore, we would say that paladium two plus is pera magnetic since it contains two unpaired electrons. And this would be our first answer so far. Moving on to our second ion we're given which is our manganese seven plus cat ion. First, considering neutral manganese recall that on our periodic table, it will correspond to an atomic number or Z value equal to 25 telling us that we have 25 protons for neutral manganese and 25 electrons for neutral manganese. And now writing out its electron configuration, we would begin with going through our one S two subs shell moving forward to S two subs shell. Then our two P six subs shell, we would then have our three S two subs shell. Next our three P six subs shell moving forward in our configuration. So that we land on manganese, we would then hit our four S two subs shell and recall that at the fourth period of our period table, our D block begins at the third energy level. So we would have 3d where we would count for a total of five units to fill in five electrons so that we land on our atom manganese. And so for our valence shell, we would have our four S and 3d subs shell. And based on our seven plus charge for our manganese ion, we would lose seven electrons. So beginning by removing from our valence shell, we would remove from our higher energy orbital, which is our four S subs shell, the two electrons. And then from our second highest subs shell, our 3d subs shell, we would move five electrons and that would remove our total of seven valence electrons where now we can state that our valence electron configuration for manganese seven plus is going to be now four S zero 3d zero. And that would leave us with ultimately nothing to show. And so we would actually interpret this as now our next highest filled orbital from our configuration of neutral manganese so that we would have three S 23 P six as our vans electron configuration for manganese seven plus. So now drawing out our orbitals, we begin with our S3 S subs shell with represented as one orbital with just one box here. And then we have our three P six orbital, which we should recall our P subs shell contains a total of three orbitals which I'm representing as boxes. And now, as we have in our valence electron configuration of manganese seven plus, we have two electrons filled in our S subs shell as arrows with opposite spins. And then in our three P subs shell, we have six bounce electrons filled in so filling in an electron as one arrow in any direction singly in each orbital first and then pairing them up for our last three electrons. In our three piece subs shell, we would have our total of six valence electrons filled in. We can see that we have no unpaired electrons in our valance orbital diagram of manganese two plus. And so therefore, manganese is going to be dia magnetic as a seven plus ion. And this would be our second answer so far. Next, we would move on to our third ion given which is our zinc two plus ion. So first considering neutral sync on our periodic table, we would find sync with a corresponding Z value or atomic number of 30 corresponding to 30 protons and 30 electrons for neutral zinc. So writing out its electron configuration in order to land on zinc on our periodic table, we would go through our one S two subs shell, our two S two subs shell, then our two P six subs shell followed by our three S two subs shell followed by our three p six subs shell followed by our four S two subs shell. And then again, as we stated before, at the fourth energy level or fourth period of our periodic table, our D block begins at the third energy level giving us 3d where we would count for a total of 10 units to land on our atom zinc. So we would fill in 10 electrons in our 3d subs shell. And so for our valance, our valance subs shell, we would have four s two 3d 10. And going back to our ion of zinc two plus, we would based on that plus two charge lose two electrons in which following our outbow principle, we want to keep our lower energy orbital 3d 10 fully filled in with its 10 electrons. And so we would remove the two electrons from our highest energy orbital, which is our four S subs shell where we remove those two electrons. And that would leave us now with our valence electron configuration for zinc two plus as four S zero 3d 10. So showing that as an orbital diagram for zinc two plus, we have our four S subs shell represented by one orbital as a box. And then we have our 3d subs shell, which we should recall again, our D block consists of a total of five orbitals which I'm representing as five boxes. And then our four S subs shell has zero electrons filled in. However, our 3d subs shell given by the exponent has 10 electrons filled in. And so we would begin filling in our orbital singly first with any direction arrow representing our electrons. So we would have five electrons filled in as single arrows in each orbital. And then now pairing up our electrons for our last five valence electrons in our 3d subs shell, we would have a downward arrow or in any direction that you chose initially the opposite arrow. Now completing our 10 electrons filled in our 3d subs shell. And given by this electron configuration, we can see that we have zero or we can say no unpaired electrons in our orbital diagram. And therefore, zinc two plus is going to be diamagnetic as our next answer. So this is our third answer so far. And now we have our fourth ion given in our prompt, which considers our molybdenum M 03 plus ion. Now, beginning with considering neutral molybdenum, which corresponds to a Z value or atomic number of 42 telling us that we have 42 protons and 42 electrons in neutral molybdenum. Writing out its electron configuration. We'll write that next to our symbol for neutral me molybdenum. So we would have a electron configuration of one S 22, S two, two P 63, S 23 P 64 S two, recall that at our fourth period, our D block begins at the third energy level. So we would have 3d 10, then four P six, then hitting our fifth period, we would have five S two and then hitting our D block where molybdenum is located in our transition metal section of our periodic table, we begin our D block at the fifth period being at the fourth energy level. So we would have four D and we would count for a total of four units to land on our atom molybdenum, meaning we would fill in four electrons. And so our valence shell being our outermost shell would consist of our five S two and four D subs shell, meaning that our or based on our three plus ion of molybdenum, we would lose three electrons from our valence shell. So beginning with our higher subs shell in our valence shell and removing the first two electrons in our five S subs shell and then removing a single electron from our four D subs shell, we would have minus one. And that would give us a valance electron configuration, four molybdenum as five S 04 D three. So writing out our orbital diagram, 4 are Molybdenum 3 Plus Ion. We would begin with our single orbital for our 5 s subs shell in which we have no electrons filled in for our molybdenum three plus ion. And then we would have for our four D subs shell, a total of five orbitals in which we need to only fill in three electrons based on our exponent of three and so single filling in those electrons as any direction arrow, I'll use an upward arrow. We have one in our first orbital, two in our second orbital and three in our third orbital. And this would give us a total of three unpaired electrons in our four D subs shell for molybdenum three plus. And so therefore, we would say Molina three plus is para magnetic. And this would be our fourth final answer to complete this example. And so all of our final answers will correspond to choice B in the multiple choice which states that our paladium two plus ion is paramagnetic with its two unpaired valence electrons. As we showed above our manganese seven plus Kion is diamagnetic with no unpaired electrons based on its orbital diagram. Our zinc two plus ion is going to be diamagnetic with also zero unpaired electrons in its orbital diagram. And then our molybdenum three plus ion is going to be paramagnetic with its total of three unpaired electrons in its orbital diagram. B is our final answer. I hope this made sense. And I'll see everyone in the next video.

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