Consider the elements: Na, Mg, Al, Si, P.
d. Which element is di...

Question

Consider the elements: Na, Mg, Al, Si, P. 
d. Which element is diamagnetic?

Accepted Answer

hey everyone in this example. We need to select which element below is a diet magnetic element. We should recall that our term di magnetic is referring to an atom with all paired electrons in its configuration and is therefore able to repel magnets. So looking at our atom titanium in part A. We would recognize on our periodic table that titanium is in Group four B. Which we would recognize as our D block or transition metal section. And this is across period four or titanium is across period four of our periodic tables. And so we want to begin with the shorthand noble gas configuration. So we would find the noble gas that is in period three of our periodic tables and that would correspond to noble gas being argon. So we start out our configuration with argon and then we want to go ahead and refer to our principle which states that we would fill in lower energy atomic orbital's first before filling higher energy orbital's. So we would recognize that because our atom titanium again is in the D block. We would recall that on period four of our periodic tables are D block begins at the third energy level. So we would start our configuration with three D for the third energy level of our D block. And within our D block we would count for two units to land on our atom titanium. So we place an exponent of two there. We want to call that R. D orbital or D. Sublevel has a total of five orbital so will represent them by lines here and we would fill in this exponent here of two as two electrons in this D orbital. So we should recall our huns rule where we would fill in each orbital with one electron first before pairing them up. So we would have one electron filled in here and another electron filled in here. And just for color coordination will keep this expanded as to to represent those two electrons continuing on our configuration. We want to recognize that we would also pass through our S sub level, which on the fourth period begins at the fourth energy level. So we would have four S and R. S. Sub level. We would recall we would need to fully fill this up to get to titanium in the D block. So we would Recall that it only has one orbital which can hold a maximum of two electrons. So we would have four S 2. So completing our configuration. This is our full configuration for titanium. And we would recognize that we have to UNP aired electrons in the D block and they're paired because they don't have another opposite spin electron added onto the configuration. So we would say that therefore titanium is not die magnetic. It would therefore be para magnetic. We refer to Adams with unpaid electrons in their configuration as para magnetic. So we would rule out choice A. As an answer choice. It's not di magnetic. Moving on to be we have the atom tantalum. We would recognize that on our periodic tables. Tantalum is located in Group five B. Which we also recall is our d block. It's located in period six. So so far we have group five b. period six And we can also make note of its atomic number, which we recognize atomic number is represented by the symbol Z. So we can say Z is equal to 73 for tantalum on our periodic table. So to begin our configuration because we know tantalum is on the 6th period. We want to find the noble gas on the fifth period and that would correspond to our noble gas being seen on it's on the fifth period of our periodic tables. And again, we want to recall our alfa principle, meaning that we would fill in lower energy level orbital's first. So we would recognize that because tantalum is atomic number 73. We need to get through our atoms between atomic numbers 57 through which covers our F block. And so we would begin our configuration with the F block, which we recall starts out at the fourth energy level, meaning we would have four F. To start off our configuration and we need to fully fill up this sub level. We would recall that R F block has a total of seven orbital's which can hold a maximum of electrons. So we would have four F 14 in the exponents. Moving on in our configuration because we recognize that tantalum is in Group five B, which is our transition metal D block section of our periodic table. We would then move on to that as our next sub level and we would recall that in the sixth period of our periodic tables. The D block begins at the fifth energy level, meaning we would have five D within the D block. We would count for a total of three units which stands for our three electrons that we fill in our D sublevel And then to complete our configuration, we would recognize that because we're in the 6th period. We also fill up our six s to sub level with the two electrons total in that sub level. So drawing out our d orbital again, we recall it has five orbital so 1234 and five. And because we have the exponents three representing the three electrons that we fill in, we would fill them in according to Hunt's rule one by one. So we have one electron here, the second here and the third here. And we would say that we have three unpaid electrons in the D block again. And so therefore tantalum is also not dia magnetic. And just so it's clear those three electrons came from the three exponents that we have in our D block, which we counted for three units to land on the adam tantalum. So moving on to part C, we have the atom lithium. So we would find lithium on our periodic tables in group one a across period two. So we would want to find the noble gas that comes before period too. And that is going to correspond to our noble gas helium which leads us right into our S sub level where we find lithium. And we would account for one unit because it is in group one A. So we would have to us because we are in the second period and that one unit represents one electron filling in our S sublevel. Here and again we recall that R. S. Sublevel only has one orbital which can hold a maximum of two electrons. And right now we have that exponent of one. So we have just one electron filled in this s orbital. And so we would have one unpaid electron in the S block now. And so we would say therefore lithium is not die a magnetic, we can rule it out as an anti choice and move on to D. So we've ruled out A B and C. So far, let's move on to D. We have the atoms in on for D. This is A. D. And for our Adam Xenon we would recognize that on a periodic table. It's in Group eight A. Which we recall is our noble gas group across period five our fifth period on the periodic table. And so we want to find the noble gas that comes on period four. And that would correspond to our noble gas being krypton to start out our configuration. So because we know that Xenon is in period five, we would recall that we would need to fill in our D orbital fully to get to Xenon In period five. And so we would recall that on period five. The D Block Or D. Sublevel begins at the 4th energy level. So we would have four D. And we want to fully fill this in so we can get to the atom xenon. So we would recall that the D block has five orbital's which can hold a maximum of 10 electrons. So we would have four D 10 moving on in our configuration because we're on the fifth period. We also would fill in our s sublevel at five s because it's at the fifth energy level, we would have it fully filled in. So we have five S two. And then once we get to our P sub level, which is also going to be at the fifth energy level on the fifth period of the periodic table, we would count for a total of six units. Two places are exponents in the configuration here, which represents our six electrons in our P block. And we would recall that R P block has a total of three orbital's which can hold a maximum of six electrons. And right now we can fill in those six electrons here because we have that exponent of six in our exponents here. So we would have 123, 4, 5 and six. And so here we would say we have all paired electrons. And so therefore we would say zenon is di magnetic. Sorry about that. It is diane magnetic because it has all these paired electrons in its configuration. So this is a good answer choice. Let's go ahead and consider our last answer choice though, which is D. Or sorry, E for potassium. So we can consider its configuration below. So we have potassium here. We want to recall that on our periodic tables. potassium is in group sorry about that. It's in group one a. Which is in our 4th period of our periodic table. So to begin its configuration, okay, we want to find a noble gas in the third period of our periodic tables and we would say that that corresponds to the noble gas argon. We then want to go into our s sub level where we would count for a total of just one unit to land on our atom potassium. So we would have four S. One and it's for us because we're on the fourth period of our periodic table. So again we have just one orbital in our S sublevel which we would therefore fill in with a total of just one electron because we have that expletive one there for our S sublevel here for our configuration of potassium. So we have one unpaid electron in the S block. And so so therefore potassium is not die a magnetic. So to complete this example, our only correct answer choice is going to be choice D. For xenon being the only atom that was dia magnetic as an element according to its configuration, where we had all paired electrons. So D. Is our final answer to complete this example. I hope that everything I went through is clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.

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