Consider the elements: B, C, N, O, F.
d. Which element has three...

Question

Consider the elements: B, C, N, O, F. 
d. Which element has three unpaired electrons?

Accepted Answer

hey everyone in this example, we need to choose the element with at least one unpaid electron set. We should recall that. We're going to have to go ahead into our orbital diagrams in order to identify this unpaid set of electrons. So beginning with silicon we would find it on a period of tables and group four a. And we would recognize to make this easiest on ourselves that Silicon is in the third energy level. And the lowest sub shell within the third energy level starts out at the S sub shell. We should recall that R S sub shell only has one orbital. And so this would be our start of our orbital diagram. We also want to recall that the next energy level or the next sub shell up from the S sub shell is still going to be our third energy level but it's going to be the P sub shell. And we should recall that within our p sub shell we have three orbital's so we can draw three of these boxes for our three P sub level. So instead of drawing three p there, let's go ahead and just draw it in the middle. And we should recall that when we're filling up electrons in our orbital's in every sub level, we need to go ahead and honor, we would recall as our principal which states that we would have electrons feel the lowest energy sub level before moving on to fill the highest energy sub level. Now, before we start filling in our electrons here, we should recall that silicon is in group for a And therefore has four valence electrons in its outermost energy level. So again, we're in the third energy level for each of our sub shells because we have that three coefficient in front of our sublevel name. So that's telling us that again, we're in the Or our outermost energy level is the 3rd energy level for silicon. So to fill in our four valence electrons, we're going to honor this principle that we recalled and because we can agree that s is the lowest energy sub level, let's go ahead and fully fill that up first. So we would have an upward arrow for our first electron. And then we would fill this orbital in by placing a downward arrow for our second electron. So far we filled in two of our four valence electrons. And so now we can move on up to the next energy level which is going to be or sorry, the next sub level, which is going to be the three piece sub level. And we can focus on filling the electrons here. So to honor the principle in our three p orbital where we have three orbital's represented by our three boxes. We're going to fill in our first electron with an upward arrow and then we're going to finish off our last valence electron with another upward arrow in the second Orbital of our three piece sublevel. And so this would be our orbital diagram for part a silicon. Now we should count how many unpaid electrons we have in this orbital diagram. And as you can see we have one here and then we have a second unpaid electron here. And so we would say that therefore based on this orbital diagram we have to um paired electrons so we can rule out choice A. Because it is not the element with one unpaid electron in its orbital diagram. So let's try again and move onto choice B, which states we have the element aluminum. Now, when we find aluminum on our periodic tables we find it in group three a. On our periodic tables corresponding to therefore three valence electrons. So we can go ahead and begin our orbital diagram through the same shortcut method where we'll start at the outermost energy level. And that would be our three s orbital. Or sorry, sublevel here with one orbital Next we want to go ahead and draw out our three p Sub level with our three orbits which would be the next sub level up from the S sub level. And again we want to go ahead and honor our alpha principle. So because we know we have three valence electrons in our outermost energy level of our aluminum orbital diagram, we can go ahead and fill those electrons in following our alphabet principle. So again with the lowest energy sub level, we're going to go ahead and fill in our first electron with an upward arrow and then fully fill in this energy level or sub level with a downward arrow electron. So we have two electrons filled in there and now for our third valence electron, we're going to go ahead and fill that in our next sub level up. Which is the three P with either we can do an upward arrow or downward arrow but we'll just do the upward arrow as we did before. And this would be our third valence electron for aluminum for its orbital diagram, completing this orbital diagram. And so analyzing this orbital diagram to count the total number of unpaid electrons, we would see that we have a total of one unpaid electron and the first orbital in our three P orbital. And so therefore we would say we have one unpaid electron. So be would be a good prospect for an answer choice. Moving on to choice C because we want to make sure we don't have any other correct choices. We have the atom magnesium. We would recall that magnesium we would find In group two a of our periodic tables therefore corresponding to two valence electrons. So we can go ahead and draw its orbital diagram by drawing out the three s orbital because it's also In the 3rd energy level. And we can make our orbital diagrams simpler by starting out at the outermost sub level and energy level. So what we can do is again honor our off about principle. We have two valence electrons to fill in this orbital diagram. So we'll place the first electron with the upward arrow and the second fully filling up our lowest energy level, the three S sub level with a downward arrow. And so we would say. Therefore based on this orbital diagram, therefore we have zero on paired electrons because we fully filled in our lowest energy sub level in our orbital diagram. And that was the three S sub level leaving us with zero on paired electrons because we did have two valence electrons in magnesium. So we were able to fully fill our three S orbital so we can rule out choice. See it's not a good choice moving on to choice D. We have the element phosphorus and when we look to our periodic tables we would see phosphorus is in group five A. Therefore corresponding to five valence electrons in its outermost energy level. So let's begin by drawing our first sublevel. The three S sub level which we recall has one orbital. Then we go ahead and move on up to the three piece sublevel where again, we still have those three orbital's. However, again we only have five valence electrons in phosphorus. So five outermost electrons in its outermost energy level. And so we'll start off by honoring our principle and filling in the lowest energy sub level with electrons first. So that would be one upward arrow for one electron here and then a second downward arrow for our three S sub level to be fully filled in. And so we filled in a total of two of our five electrons. So now we have three more electrons to fill in. We're going to go ahead and honor the ALfa principle by moving on up to the three piece sub level to fill in the last three electrons. And so we would fill them in orbital by orbital. So one electron here, another electron here and then another electron here. And so we would say that based on this orbital diagram. Therefore we have three UNP aired electrons in this diagram. And so we would rule out choice D because three is too many. We only want an orbital diagram with one unpaid electron. Moving on to our last option here, we have choice E as an option. So for sulfur, we would recognize that sulfur. We would find in group six A of our periodic tables. And so therefore that would correspond to six valence electrons. So to make our orbital diagrams simple, we're going to start off at the outermost energy level. And because we have silicon in period three, we would start off at the three S sub level. Next we also would move on up to a three piece sub level which we recall has these three orbital's. And so let's go ahead and honor our alphabetic principle by filling in our first few valence electrons in the lowest energy sub level first. So we would have one upward arrow for the first electron and then another downward arrow for the second electron so far. We've used two of our six valence electrons. Now we're going to go ahead and fill in the last four electrons in our three P Sublevel. So we would have the third valence electron field in the first orbital, the fourth filled in the second orbital and the fifth filled in the third orbital. And then now we can start pairing up our arrows here By filling in the 6th valence electron of sulfur in the first orbital there. So we have a pair of electrons in the first orbital of our three piece of level. And we would say that therefore as a whole based on this diagram, we have to um paired electrons And our three p orbital or Sublevel. So let's go ahead. And again we can rule out choice because we want one pair of electrons. So we can go ahead and confirm that choice. B was the most correct choice as the element with one pair of one paired electron in its three piece sub level. So aluminum is the final answer. To complete this example, which corresponds to choice. Be in our multiple choice. If you have any questions please leave them down below. And I'll see everyone in the next practice video

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