So we're going to be going over atomic theory, mass spec, mole conversions, and some applications of that electron configuration, photoelectron spectroscopy, and periodic trends. So the way I kind of see this happening today is we're going to be looking at a few example and practice type problems. And if you would like to interject once again, go ahead and try asking questions either in the ask a question tab or over to the right. So for the first half of the stream, I was thinking that we could look at an example element. And that example element that I wanted to look at was tungsten.
Tungsten is element number 74 on the periodic table. It's also got the element symbol W. With that being said, it has an average atomic mass of 183.84 amu.
And it has a molar mass of 183.84. grams per mole. So let's talk about a couple of these things here. First off, right, the atomic number is the number of protons within an atom. This number for a specific atom cannot change.
So tungsten, every single element of tungsten has 74 protons. The number of neutrons and the number of electrons can't, excuse me, can change. But once again, the number of protons doesn't really, and if it does in some sort of like nuclear process the identity of the atom changes. As far as the average atomic mass goes and the molar mass, when we're looking at the average atomic mass and the molar mass notice that those two numbers are the same.
When we're kind of identifying and looking at these we're really looking at a single atom versus an entire mole of those atoms. We'll talk more about the average atomic mass and molar mass in some of our examples, but once again, if you're looking at the average atomic mass, the unit AMU is measuring the mass of a single atom, whereas molar mass is measuring the mass of one mole of atoms. And don't forget that a mole is an Avogadro's number of atoms, which is 6.02 times 10 to the 23rd. Unstable isotopes of tungsten are 182, 183, 184, and 186. 185 is not stable. So none of these things are stuff that you have to memorize.
They will either be on your periodic table, which hopefully you all have your periodic table sitting right next to you, or it'll be information given to you on the test. So for example, let's go ahead and check out this next question here. How many protons, neutrons, and electrons does a neutral isotope of tungsten-183 have? So what I want you to do take like 10-20 seconds see if you guys can kind of figure out how you would approach this and we'll see kind of where we're going with this.
If you feel if you feel good with this so far awesome. If you have any questions once again please feel free to ask. So how many protons neutrons and electrons does a neutral isotope of tungsten-183 have?
So Hopefully, if you were paying attention before, we know that because the atomic number for tungsten is 74, it has 74 protons. And the next thing I want to say is that because this is a neutral isotope, meaning there is no positive or negative charge, net positive negative charge on this atom, it also has the same number of electrons, 74. Last but not least, 183 is going to be the mass number. 183 is going to be the mass number of tungsten.
Mass number. is not the same thing as mass. Mass number is just the number of protons and neutrons this thing has. So if we go ahead and sum up the numbers of protons and neutrons, it will equal 183. And therefore, to find the number of neutrons, all you have to do is subtract those two numbers. So 183 minus 74 is going to be 109. So tungsten-183 has 109 neutrons.
Now, while I totally wish that, like, these are the types of questions you'd see on the AP test or in your unit test or something like that it's probably not it's a little low level but these things are absolutely necessary in order to understand kind of some of the stuff that's going to be coming up so last but not least one more time one quick review the atomic number is equal to the number of protons the mass number is equal to the protons and neutrons and then of course the average atomic mass is the average mass of an isotope, a single atom of those things. So let's go ahead and move on to mass spectroscopy. So we kind of looked at the individual parts of the atom, right? The atom is made of protons, neutrons, and electrons. the protons neutrons are in the nucleus and the electrons are outside so let's go ahead and take a look at the mass spectroscopy and what we're going to do is we're actually also going to look at this for tungsten so what i want all of you to do i'm going to post up some data for tungsten i want you to look at this data and i want you to find the remainder the remainder of tungsten 186. so here's the data for you When you get that value, I want you to go ahead and comment your answer down below.
I'll do this myself. We know it's 26.5 plus 14.31 plus 30.64. All percentages add to 100%.
So yeah, Connor, it looks like you got it. if we subtract 100 from this you're going to get 28 28.55 um might have just been a decimal thing or sometimes when i calculate mine wrong so 28.55 should be the answer put that there cool So what I wanted to do is this. I'm going to be kind of switching around screens here.
And so hello, big screen again. I'm going to switch over. I have a whiteboard app and I want to kind of look at a couple of things within this whiteboard app. So here we have here we have some data for. our tungsten and what we can do is we can actually use this data to graph the so mass to charge ratio which basically is just saying the isotope we're looking at so which isotope we're looking at and there's four isotopes and then over on the y-axis we have the abundance or the percentage so we can actually take this data and look and use this data for our you know calculating abilities And we find that 182, 182, once again, was 26%.
So it's about a quarter of the way up. 182, 183, 184, and 186. We'll jump up for that one, 186. I'll change my colors up here. So about a quarter of the way up, we're going to have 182. Tungsten, 183 as well. 14% so a little bit lower 30% is tungsten 184 and last but not least about the same high as the first one you're going to go ahead and get tungsten 28 so notice that we can represent this data we can represent this data using differing isotopes and of course the massive charge ratio all righty and once again if i'm moving a little fast Just let me know in the question section. So awesome.
And it looks like Sydney went ahead and actually ended up calculating using the data, the average atomic mass, which, gosh darn it, that was going to be my next question. So let me go ahead and show everyone how Sydney went ahead and approached that problem there. So I'm going to go ahead and switch out my videos here.
And I'll be going once again back in between these videos. So if we have this graph here, we can go ahead and take that data. And the average atomic mass is just going to be a weighted average of the isotopes that occur for that element, along with the percentages that those isotopes appear. Notice that whenever you do this, you have to go ahead and change it.
You have to go ahead and change it into the... Sorry, I went ahead and clicked away. There we go.
Sorry, I'll just fix in my screens here. Alright, so notice that when you have when you use the average atomic mass, you have to change into the decimal form. And yeah, once again, Sydney, when you go ahead and plug all these things in, you find that it's 183.84 atomic mass units. One excellent check to see if you did this right is check the periodic table. If your number is equal to the number on the periodic table, then you know you got it right.
Sometimes the way they make these harder is by having you identify what the element is. So they'll give you the mass data and the percentage data and say, here is here, excuse me, are these values? What is the identity of this element? All right, cool.
So moving on from this, we're going to go ahead and look at some mole questions. So once again, moving on, we're going to look at some mole questions. So which contains more tungsten atoms? We're sticking with our theme here.
One gram of tungsten, let's see, that would be tungsten 6 oxide or tungsten 4 oxide. So W03 or one gram of W02. And as far as this particular problem goes, what you need is the molar masses. So I'm going to go ahead and give those to you there. This is a totally common multiple choice type question.
They give you a certain mass and they say, what do you have more of this element or that element based on the masses? And by the way, you should never compare the masses. And in this case, it's kind of harder to do because you have the same mass. So yeah, any ideas with this one? All right, so I'll go ahead and move on with this.
What I want to kind of take away from this is that there's one atom of tungsten per compound in both WO3 and WO2. Because of that, the only thing that truly matters is the molar masses. You're going to be taking the one gram and dividing it by the molar mass and then setting up a one to one mole ratio because there's one atom of tungsten for every one compound.
By doing so, we find that one gram of W02 has more tungsten atoms within it. One gram of W02 has more tungsten atoms than this. If you have any questions about this, once again, go ahead and write in the question tab down below. or over in the comment section to the right. Alrighty, so enough with the mole stuff, like more direct mole questions.
What I would like to do is kind of transition us into an empirical formula problem. And once again, for this, I am gonna be switching over to the whiteboard. So while I kind of do this, bear with me. But if you have your notebooks, you have your calculators, and you want to go ahead and try this one out, I would totally recommend trying it out. So once again, the empirical formula question is written here.
And as you kind of work your way through this, you can kind of come back to my screen and I will help you through it as well. So let's do this one more time. I'll leave it up for you.
Three point one three grams of tungsten are combusted with oxygen gas. The resulting compound contains tungsten. and oxygen and has a mass of 3.67 grams.
What is the formula for the oxide that formed? What is the formula for the oxide that formed? So I'm going to go ahead and work my way through this on the whiteboard once again.
Let's go ahead and do here. And what I'm going to go ahead and do is just undo the graph that I've drawn. All righty, yes. We know we had 3.13 grams of the tungsten compound. We know we had 3.13 grams of the tungsten compound.
And we know that after it combusted, it got heavier, it went to 3.67 grams of the tungsten oxide compound. So as far as that goes, or is that goes there yeah yeah w x o y we don't know the ratio in which these compounds form excuse me in which these elements form are they coming into a one-to-one ratio one to two ratio a 15 to 13 ratio as long as it's one of small whole numbers you can go ahead and determine that using the empirical formula method now generally speaking the most important thing you should always do is convert to moles of the given substance. Now while we know how much tungsten we have, which is listed right here, we unfortunately don't know how much oxygen we have until we analyze this data.
We know that when a certain amount of this combusted, it got heavier, meaning that 3.13 grams of this compound is tungsten, and therefore the rest is oxygen. All we have to do to determine how much oxygen is in it is subtract. And this should end up getting us 0.54 grams of oxygen.
Cool. So now that we know the mass of tungsten and the mass of oxygen, all we have to go ahead and do is convert each of these values to moles. By doing so, we find the mole ratio in which tungsten and oxygen come together.
So the first thing we're going to do is take the mass of tungsten and convert to moles using the molar mass. Molar mass of tungsten hasn't changed. It's 183. 0.84 grams for every one mole. Calculating this out, gets us 0.0179 or 180 mole of tungsten. And we're going to do the same thing for oxygen.
We're going to go ahead and find out that 0.54 grams of oxygen, converting that to moles gets us 0.03380338 mol of O. So essentially what we just determined is that when tungsten and oxygen form a compound, when tungsten and oxygen form this compound, the ratio in which they come together is 0.018 to 0.0338. 0.018 to 0.0338, which isn't a usable formula, a usable value for us within this in any way, shape or form.
So in order to determine the ratio in which these two things come together, we take the smallest amount of moles and divide both of those values by it. And all this is doing is putting these numbers into a whole number. So if we're comparing these two, it looks to me like the 1.8 is the smallest value. So we're going to take both of these and 0.18, divide both of these here. And by doing so, we should end up with a 1. to roughly two ratio.
I think when we divide the second one here, you end up with 0.01. Yeah, 1.88, which we can go ahead and round that up to two, you really only want to be looking for repeating sixes, or the point five, when you get either of those, you multiply by three or two respectively to go ahead and get that whole number. So what this essentially found for us, this essentially found for us, is that the tungsten and oxygen compound is W O two.
And this is going to be essentially the empirical formula of the substance that we're looking at. Awesome. So with that being said, let me go ahead and switch back over to low. my Chrome tab with our slides and hopefully once again we ended up finding that the tungsten compound was W02.
So if you have any questions with that feel free to once again list in the question section down below and I will definitely take some time at the end of the slides to answer those questions and go over. So the unit one is kind of structured in an odd way such that When we're looking at the atom, you can look at it in a very experimental sort of way where we're reacting with other elements. We're getting masses for these things and we're able to use the experimental data to figure out the formulas of the compounds. I end up doing with my class a hydrate lab. So the hydrates are ionic compounds where water forms bonds with them.
And you can find out how many water compounds, water molecules rather. will coordinate or bond to an ionic compound. We also will look at, you know, a lot of these mole conversions, mass spec. And then halfway through the unit, it kind of flips onto its head and starts talking about quantum mechanics and the electron configuration. Well, we start kind of discussing and identifying why do the elements electrons behave the way that they do.
So that's where we're going to be going next. And once again, at the end, if you have any more questions about moles or anything like that, I'd be more than happy to kind of go over those. So without further ado, let's move on to electron configurations.
So I find that when I teach this section, kind of front loading a lot of vocabulary and understanding this vocabulary is super crucial and critical for this class. So the kind of vocab that we should be looking at is the orbital. Erwin Schrodinger defined these orbitals through his wave equation and it says that in an orbital you can have two electrons and this orbital is a region in space where the electron spends roughly 90% of its time and that means that 10% of the time the electron is literally anywhere else in the universe it's it's probably not it's so probably very close to the atom it's just not within one of those maps that can be generated by the wave equation.
Energy level defines how far the electron is from the nucleus. Now, that's not the 100% definition of it, but at the end of the day, just stating that an electron in a higher energy level is further away from the nucleus than an electron that is in a lower energy level will get you 99.9% of the questions that relate to this on the AP test. Electrons in higher energy levels are...
further from the nucleus. The sublevel just defines the shape of the orbital. There are four types of sublevels which I'm going to go over in the next slide. We have s, p, d, and f, and they have different shapes.
That's pretty much the only distinguishing factor between them. And then last but not least, electron spin. I want to remind everyone that the electrons can occupy one orbital, but in order to do so they have to have opposite spins.
Spin is a physics term and it just states that the electron is either spin up or spin down. And that's all you have to know in the world of chemistry. And the way that once again, I like to tell my students about this is that in order for two electrons to belong in the same orbital, they have to have opposite spins.
Cool. So with that being said, let's check out some of the different shapes and sublevels that there are. I highly recommend you memorizing this chart. As far as this goes, this is data that will not be given to you, and you just need to know the shapes and different sublevels.
Also, the number of orbitals and electrons per energy level is crucial for you to know in order to do electron configurations. So the s orbital, the s orbital is a sphere. It's the most simple orbital, and the electron is somewhere within that sphere. 90% of the time.
There is one orbital of the s sub level per energy level. And because two electrons can occupy that one level, that means that for the s sub level, it can have two electrons per energy level, p gets a little bit more complicated, it is a dumbbell shaped. And as far as this dumbbell shape, it can be oriented on one of the three axes, either the x axis, so it's kind of like this infinity sign on the x.
it's on the y-axis which is up and down so that's gonna be on the y-axis and then of course because atoms are three-dimensional you can go ahead and have that z-axis which is coming in and out of the plane because it's a little bit more complicated that there are more orbitals per energy level they are degenerate orbitals which means they all have equal energy and we can have six electrons per energy level now Hopefully that kind of makes sense because there's still two electrons per orbital. It's just that each of those electrons, excuse me, it's just that each of those belong to one of the three possible p orbitals. So that gives us a total of six. D is even more complicated.
They, I like to describe them as four leaf clovers, even though not all five of them are exactly four leaf clovers. And then F, F, honestly, they're just complicated. But one of my old high school teachers used to say that they look like flowers.
So that's why I like the list though on the way I did. So here's an image of them. Don't bother on the opposite colors, the red and blue, that has nothing to do with the AP curriculum.
It's just the way that mathematically these lobes and orbitals kind of weigh themselves out. It's a result of the wave equation once again. for you. The only thing you really need to know is the shapes of these things. And as you can see, they get significantly more complicated as you go down.
And that's also why I like to think that there's more orientations per energy level. As far as the s orbital goes, it's just a sphere. the next level down you have the three different p orbitals on each of the sub level uh excuse me on each of the axes you jump one more level down you have the five p orbitals notice most of them kind of look like the the four leaf clovers but then you have that one in the middle which is kind of an odd shape there and then down at the bottom you have seven of the f orbitals those seven f orbitals those seven f orbitals have um You will never be expected to draw them.
You know, there's so many of them that it's not expected to know much about them. So I usually just say, eh, they're just complicated. You just need to know there's seven of them.
So the way I teach this to my students is the block method on the periodic table. If you know the block method, what I want for you to do, so hopefully you're listening a little bit here. If you know the block method, I want you to list in the comment section one element from the p block. So if you're listening, I want you to list one element from the p block in the comment section. Yeah, phosphorus is in the p block, and it's the element p.
Neon is also in the p block. Very good. So what we're looking at here is the different blocks of the periodic table.
What's really nice about this is it basically gives us the pattern, the pathway in which we can go ahead and fill these electrons when we're doing the electron configuration. So the off-ball principle says that we have to fill the lowest energy orbitals first. Off-ball is literally German for build up, right? So we're starting at the lowest level and then building our way up as the atoms electrons occupy higher and higher energy levels. The most straightforward element is going to be helium, excuse me, hydrogen, sorry, hydrogen.
has one electron, so when we do the electron configuration for it, we have to define the location of one of one electron. If you see hydrogen, hydrogen's in the upper left-hand corner, it's in the s block, so that means it's an electron. It's one electron that we're describing is orbiting or existing in a spherically shaped orbital. It's in the first energy level, meaning it's closest to the nucleus, and therefore you have the hydrogen electron configuration. Hydrogen is going to be one.
S1 and we go ahead and list the superscript up above. So what I want you all to do really quick on your notebooks, I want you to write the electron configuration for nitrogen. Take a couple seconds here and listen. Electron configuration for nitrogen.
Fortunately, we're moving away from tungsten. Tungsten has an odd electron configuration and it doesn't follow the rules that we've established. Alright, I'll give you five more seconds.
Hopefully using the block method is not too bad. Nice, Connor. It looks like Connor's working his way through it.
1s2, 2s2, and 2p3. So as far as this goes, nitrogen has two electrons in the lowest energy orbital, the 1s orbital. It has two more electrons in the next highest one, which is the 2s orbital.
And then it has three more electrons in three p orbitals and the two in the second energy level. So once again, what this kind of means is that first number there, 1s1, 1s2, excuse me, one is the energy level. It's describing the distance of the two electrons from the nucleus.
S is the shape and the superscripts that kind of look like exponents. the superscripts describe the amount of electrons. So essentially nitrogen has seven electrons, two are in the first energy level, five are in the second energy level, and in that second energy level they all exist in different shaped orbitals. Last but not least I want to remind everyone about the orbital notation. The orbital notation is a notation that scientists and chemists use to identify the spin of the electron within the orbital.
Now based on Hund's rule, Hund's rule says that electrons must fill degenerate orbitals before pairing with other electrons. Because nitrogen has, and rather every element, has three p orbitals per energy level, the three electrons fill the degenerate or equal energy p orbitals before they kind of come back and pair with each other. 1s and 2s are lower in energy, so those electrons get filled first. couple of like language terms that come with this, we know that one s orbital has the paired electrons, the two s has paired electrons. And then if you're looking for nitrogen, nitrogen has 123 unpaired electrons, they have three a total of three.
So something really cool that we can kind of do with this is we can look at the photo electron spectra for an element. Now, the photo electron spectra is basically graphical evidence of the electron configurations. And it is a really straightforward and kind of mellow graph.
It tells you the relative amount of electrons per energy level. And if you have the periodic table blocks down, then you can easily memorize the order in which electrons need to fill the energy levels and sublevels. So I'm going to go ahead and pull up the nitrogen photoelectron spectra. which is right here. And we can see kind of a couple weird things about this graph.
First thing I want to point out is on the x axis, the x axis is backwards. Notice that it starts with 100 over on the left, and then it goes to 10, 1 and point one. So as far as this goes, what it's essentially showing us is a logarithmic increase, or rather a logarithmic decrease. And the electrons that are closest to the y-axis are the ones that are highest in energy to remove. They have the highest ionization.
Also notice about this graph, the first two peaks have the same relative height. When we're looking at that relative height, it identifies the number of electrons. So what is this kind of telling us? Is that those peaks, the first two peaks, are identifying the 1s and the 2s.
S subshell. The 1S had two electrons in it. The 2S had two electrons in it. And that's why they are the same height. As far as that last peak, that third peak, it is roughly one third of the other ones taller.
And that kind of indicates that there are three electrons in there. And once again, this graphically represents nitrogen's electron configuration, 1S2, 2S2, 2P3. There is two peaks, once again.
those two peaks allow for us to identify the number of electrons and they are the same and then that third peak on the right is a little bit taller indicating the third electron there the electrons that are in the 1s orbital are harder to remove from the atom this is because they're significantly closer to the nucleus based on coulomb's law right opposite charges attract and if those opposite charges attract the closer the electrons are to the nucleus harder they are to remove and that's why they have a significantly higher ionization energy that's why they are once again much higher compared to the second energy level over to the right so one more time there is the electron configuration for nitrogen there's the photo electron spectra and what i'm trying to illustrate one last time before we move on from our electron configuration discussion is the fact that those electrons and the electron configuration are just visually represented almost equivalently to those electron configurations. Excellent. All right, cool. So last but not least, and then you guys can all go on with the rest of your Sunday.
Once again, I'll take a few minutes to kind of talk about chemistry, the questions that you have from Unit 1 at the very end. The periodic trends. So as far as the periodic trends go, there are four important trends that you need to know.
We have the atomic radius, which is defined as half the distance between two identical nuclei bonded together. The reason we have this odd definition is because you have to remember atoms don't really have an edge to them. So we have to identify that, hey, this is where the edge is.
And the way they do it is by taking two atoms and the distance between them, halfway the distance between the two nuclei. is going to be the edge of their atom. At the end of the day though, you can go ahead and just really call it the size of the atom. Ionization energy is the amount of energy required to remove an electron from an atom.
If something requires energy, it's usually pretty difficult to do. And that's also true for the ionization energy of atoms. If the atom has a large ionization energy, it is difficult to remove the electron.
Once again, it's the amount of energy required to remove an electron. Electron affinity has always been one of my more challenging concepts to kind of grasp within the world of chemistry. But the first thing you should know about this is that every time an atom gains an electron, it releases some amount of energy. This is true for most of the atoms on the periodic table, not all. But once again, most atoms, when they gain an electron, they release some amount of energy.
And the electron affinity is referring to the amount of energy released when an atom gains an electron. I kind of like to think of this as like a sigh of relief, right? Like, oh, right, I gained an electron.
And depending on how big that sigh of relief is, this changes how much it wanted that electron. Now, if we're kind of considering some of the atoms on the periodic table, you should hopefully realize that larger atoms, when they gain that electron, larger atoms, they're adding that electron to their valence shell. And as far as that goes, it's not as exciting. for a smaller atom, right?
Smaller atoms, the electron when it adds to their valence shell is significantly closer to the nucleus. So the amount of energy that they release when they gain their electron is going to be greater than larger atoms. And then electronegativity.
Hands down, electronegativity, in my opinion, is probably one of the most important vocab words you'll learn all year long. It's referring to the ability for an atom to attract electrons in chemical bonds. These atoms kind of play tug of war over the electrons and this dictates what types of bonds the atoms can form between them. This will be the basis of covalent and ionic bonds when we talk about that within the next unit.
You also talk about electronegativity a lot if you are in the world of APBio. Organic chemistry talks about electronegativity. We love electronegativity. Once again, it's the ability for an atom to attract electrons in a chemical bond. The trend for the atomic radius is the only one that is to the left and down increasing.
And this might be a little bit strange to you. If you're looking at the periodic table, you're like, hmm, lithium is the first element in the second period. And lithium is larger than every other element in that period. So it's larger than beryllium, boron, carbon, fluorine, oxygen, nitrogen and neon. It's the largest in terms of radius.
compared to all of those. And I'll discuss why this is in a second. The trend for ionization energy, electron affinity, and electronegativity is the opposite. It is increasing as you move up and to the right. And hands down, there are two reasons for this.
There are two reasons, and you need to identify the first thing you go into this is, are the atoms that you're comparing in the same group, which is up and down, or are they in the same period, which is left and right? So if the atoms you are comparing are in the same group, once again, up and down, then the the thing that dictates the trend is going to be the energy level the valence electrons occupy. Remember energy level describes the distance of the electron from the nucleus. Energy level is the distance of the electron from the nucleus.
So if the electrons are further away in a higher energy level the electrons, excuse me, the atom is going to be larger. And that's why as you move down the group on the periodic table, the atoms get larger. That follows the trend as we would kind of expect.
And as electrons get further away from the nucleus, they become less attracted to the nucleus as well. If the atom you are comparing are in the same period, that means left and right, then the electrons are in the same energy level. You cannot use that to define your trend.
It is all about the nuclear charge in that case. Nuclear charge is defined as the number of protons within the nucleus. If you are looking at carbon and you're comparing it to nitrogen, carbon has six protons, which once again, you look on the figure out a table, that's the atomic number. Nitrogen has seven protons. So therefore, as you move once again from carbon to nitrogen, nitrogen has more protons in the nucleus, and it's able to kind of suck those electrons in.
closer to the atom, making the atom relatively small. So with that being said, I'm at the end of my unit one review. What I'm going to do is take five minutes and just kind of answer questions freely. I can use my whiteboard.
I can also just answer them using word of mouth. Once again, if you go to the ask us question, please. question tab at the bottom you can ask questions there um otherwise thank you so much for attending and uh you guys can wait a second or two um to see where these questions are going Awesome.
And if you're kind of thinking of questions to discuss or go over, I can tell, you know, just crappy jokes until then. So why can't you trust atoms? Because they make up everything.
So good. More about the photo electron spectroscopy and interpreting data. Cool.
Yeah, all right, cool. So let's do this. I'm gonna jump over to the whiteboard app.
We're gonna erase some things here. Thanks for being patient. All right, so I'm gonna take a few seconds to look at photo electron spectroscopy. There we go. Cool.
Alright, so every photo electron spectroscopy graph that you will receive. is going to be graphed, let me do a change this to my pen, here we go, is going to represent on the x-axis the potential energy or the ionization energy, so I'm just going to write energy, and the y-axis is always going to represent the number of electrons, and this is relative number of electrons. The x-axis always lists the higher energy over on the left-hand side. So we'll go ahead and say like 1,101. And once again, the way that the types of questions that you will kind of get from this are of the following.
They'll give you one of these graphs and they'll say, okay, cool. So here's this graph. You have this here.
this here. And just for your reference, why I'm gonna make it twice as tall, twice as tall. So this would be right there, too. And that would be four. Cool.
So based on the graph that I've just drawn for you, what electron configuration or choosing what element is represented by this, and you can kind of go about this in a couple of ways. If I'm looking at this, I go, okay, so there are two electrons in the first energy level. And the first energy level we know is the one s, there are two electrons and the next energy level, and that's going to be two s. And it looks like there is four more here.
So the electron configuration we're looking at is one s2, two s2, two p4. And basically, this is a workaround or work backwards of looking at the electron configuration for an atom and then identifying what that atom is. So what element, what atom has eight electrons, what neutral atom has eight electrons. I think there's a little bit of a lag.
So I'll wait a second while you guys are looking at this. Oxygen. Yeah, nice.
So oxygen has eight electrons. So what essentially I just drew is the photoelectron spectrograph of oxygen. Another question that I've seen come up with these is comparing the peak placements.
So for example, if I were to look at this one here, so it's going to be, I'm going to draw this one in red, you have this, this, and then this one actually was going to jump up a little higher. So this is one we'll go ahead and say is relative five. So that's 1s2 2s2 2p5. So what element what element would that one be?
That one is fluorine, correct? Fluorine? I don't think it's fluorine. I think it's fluorine. So then the last question I have is this, if you're looking at these two graphs, how come the 1s peak for fluorine is further shifted to the left?
Why is this further left? And if you're checking this out, the answer to this is actually not too bad. The reason that fluorine's 1s peak, and actually most of the rest of the peaks will be further to the left, is because it's harder to remove these electrons.
Why are they harder to remove those electrons? Because fluorine has more protons within the nucleus. As you add protons to the nucleus, the electrons that are in the same energy level, which are these ones right here, get more and more attracted to the nucleus and begin shifting further and further left.
Go ahead and change that there so you can see that. So if we were to draw neon's peak, neon's peak would be further to the left because neon has 10 electrons. If you were to draw sodium's peak, sodium peak is going to be significantly further to the left. And that's because sodium has electrons in a higher energy level. And this is true as we add.
protons to the nucleus, the electrons on the outside get more and more attracted to the atom. Awesome. Artyom.
So without further ado, I think we're kind of coming towards the end of the stream. I am once again, so happy and proud of everybody here. And I do have one more question.
So I'll definitely take a second to answer that. But what I also want to remind everyone is that on September 25th. there's an AMA session with Dylan black and you can do that and work and ask him any questions whether that be related to this or anything else so Amy asks if an element is ISO electronic would it have the same photo electron spectra so let's check that out I think that's a really actually excellent question so I'm gonna go ahead and just clear this here First off, let's just define what isoelectronic means for everyone that is kind of following along. Isoelectronic just means it has the same number of electrons, same electrons.
And the only way that something can have the same number of electrons, if it's the same atom, or if it gains or loses electrons. So if we go ahead and look at fluorine. excuse me, not fluorine, the fluoride ion versus neon, both of these will end up having, so that's F1 minus, and that would be just neon.
Both of these would end up having 10 electrons. So to answer your question, Amy, one more time, the peaks, the relative peak heights would be exactly the same. You would have a two to two to six ratio of the electrons.
So this would essentially look like this, where you'd have your y and x axis, each one would look relatively two to two to six. So I'm going to draw this one up here. So this one will say for neon.
Fluorine, however, while it would still have a 2 to 2 to 6 peak, would be shifted to the right slightly. And that has to do with the same reason as before. Fluorine, or the fluoride ion rather, has less electrons than neon does.
So the less protons, excuse me, in the nucleus. So it's not able to attract those outer electrons as well as neon can. That makes it harder to remove neon's electrons, and therefore it causes neon's peaks to shift slightly to the left, which is harder or requires more energy. Awesome question. All right.
Awesome, everybody.