Electron Configuration

chemistry lecture number 22

electron

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configuration

electron

s occupy energy levels energy levels have SPD sublevels and the sublevels have orbitals the S Su has one PS3 orbital DS orbitals five orbitals and F has seven now suppose that an atom was stripped of its electrons and then suppose we add the uh I guess I'm sorry, the electrons are added back one at a time, what suble and orbital energy level would the electrons occupy? Now the electrons fill the orbitals according to a pattern, so here's the question, we have all these different energy levels, we have different sublevels. you know s sp sp p d s p d and f and then within these Su levels we have all these different orbitals and the S has one orbital here and the P has the three here and the D has the five, so where would the electrons go if you had to? add one at a time, would you go to this energy level or this energy level or this energy level? uh, if the electron went to this energy level, what would its LEL go to? and if it was at a certain level Su, let's say it was at the P Su LEL the electrons you know would go here or they would go here, so we have to figure out where in this whole mess the electrons would go and the distribution of electrons follows a pattern.

Now we are going to use a diagram to find. the pattern and first we need to use the chart below now this is a chart that you saw in the previous electricity you should have it memorized the only addition we have added is that in the fifth energy level there is spdf and I think G comes after F, but it usually doesn't go beyond F, so remember that these are the Su lels within each of the energy levels and we could take this to 6, seven and 8 and still write spdf, but we just need to go up. to five, so what we're going to do is draw a diagram based on this information right here, so this just shows the energy levels and the sublevels that are in each energy level, so the first energy level has a S, the second has an S and A P, the third energy level has s, p and d, the fourth has SPD and f and then all the rest we just write SPD and f, so the fifth has SP D and F, okay, like this that this is what we are going to do.

I'm going to start drawing arrows through each of these terms and the arrows will start from the top left and go down, sorry, top right and go down to the bottom left and we'll cut the arrows and each one. Every time we draw an arrow, it has to go through the first term of the next horizontal row anyway, so the first arrow goes like this, okay, the next arrow I draw has to go through the first term of this one of the next row. so the next row is this row here and there is the first term, the next arrow says like this here is the first term of the next row, so the next arrow has to go like this and press that here is the first term of the next row, then the next goes through here here is the first term of the next row, so the next arrow goes through here is now possible.

I could have written a row six and a seven and kept that pattern, but we only need to get up to five. now look at the order in which we draw the arrows the first arrow goes through 1s so I'm going to write 1s the next arrow goes through 2s so I'm going to write 2s the next arrow goes through 2p and 3s and that's what I'm going to write in that order. I'm going to write 2p first and then 3s, so I'm going to write 2p and then 3s. The next arrow goes through 3p and 4S, so I'm going to write 3 p and then 4S and then the next error goes through 3D 4 P 5S 3D 4 p 5 S, okay, so these are written in the order that the arrows go through them, the next thing we need to do is, I'm going to draw horizontal lines over the top of each of these terms and the number of lines I draw depends on the subble S has one orbital P has three orbitals D has five orbitals and F oh, we didn't go as far as F Well, yeah, so we don't have it, but we won't need to use it, so this was in the previous lesson, you need to memorize this, okay, so one s s has an orbital.

I'm going to draw a single line above. it's like this 2 s once again s has one 2 p p p has three orbitals so I'm going to draw three 1 2 horizontal lines on it s has one 3 p p has three one 2 three 4 s s has 1 3 d d has five so Let's draw five lines above the 3D 1 2 3 4 and five 4 p p has three above 1 2 3 then 5 S S has 1 and that's it, so this setup here is this little diagram which is what we're going to use to calculate figure out where to place electrons and this shows that as you go from left to right you get further and further away from the nucleus and the energy increases as you go from left to right so you can imagine that the nucleus would be right here and all the electrons They're going to want to get closer to the core, so they're going to fill in these little spaces from left to right, so I didn't draw the 5S here because the source wouldn't have been visible, we're just going to go up to 4p, so as you go from left to right, the energy increases and it basically means that when we say that the energy increases it means that you are moving further and further away from the core, you need to memorize this diagram or memorize the procedure to do.

This diagram is fine now. The next thing we're going to do is draw up and down arrows on the horizontal line to indicate that an electron is in an orbital, and generally we're going to fill the lines from left to right. right and the electrons want to get closer to the nucleus so the left side fills first so when we complete this diagram we are going to start on this side we are going to cross nicely and when I complete the diagram from left to right that will also be This is known as the arc principle and it just means that electrons fill the lower energy levels first, before filling the higher energy levels.

It's just a fancy way of saying that we complete the diagram from left to right, so here they are. The rules that we're going to use to complete the orbital diagram uh are okay, so first you choose the lowest energy level, which means you're going to start as far to the left as you can to choose the lowest SPD or F of suev s one more time. It just means that we're going to go from left to right, fill the orbital with a maximum of two electrons and have them spin in opposite directions and I'll draw a picture.

I'll show you what that means in a moment and before a second. The electron can be placed in any orbital. All orbitals of that suble must contain at least one electron and spin in the same direction. That's a lot of words. We'll show you what rule number four means when we draw the diagram right now. It probably doesn't mean much, it will mean something when you see that it's really right, so to do this you need to use a periodic table to get the atomic number and remember that the atomic number tells you how many electrons the atom has.

Okay, so hydrogen has one electron, where does the electron go in all this mess? Well, we start from left to right, right, the first electron goes here, further to the left, we are going to draw an arrow to indicate that it is an electron, so the arrow represents an electron and that is the electron spinning in one direction , okay, that's pretty simple, that's the electron

configuration

of a hydrogen, let's do the next element, draw the electron configuration for helium, two electrons, one, two, and notice I put two arrows here, why? Because each line or each orbital can contain a maximum of two electrons, also note that one arrow is up and one arrow is down.

When two electrons occupy the same orbital, they have to rotate in opposite directions. That is the poly exclusion principle. Let's draw the electronic configuration for the next element lithium three electrons here we go 1 2 3 and we have to put the third electron here because this one already has two it can't contain more than two, so if we run out of space here we go to the next one, let's draw the electronic configuration of burum four electrons 1 2 3 four good two electrons in that rotating in the opposite direction let's draw the electronic configuration of Boron 1 2 3 4 five good then we go from left to right we are putting two arrows each and the arrows are rotating in directions opposites, now we're going to make one that's going to deviate a little bit from the pattern, so I'm going to throw a little curve ball to draw the electron configuration of the carbon. six 1 2 3 4 five six that's where you put the electron six now why didn't I put this electron on this line backwards?

It's because of rule number four and rule number four says that before you can put a second electron in this one all of these others have to have at least one arrow, so I can't put a second electron in this one until the rest of these They also have one electron and when you place them they have to face each other in the same direction, okay, let's make nitrogen. I think you can still see pattern seven for nitrogen. 1 2 3 4 5 6 7 once again, why didn't I put this one in this one or this one before any of these lines can have a second electron, the rest of the lines have to have an arrow, okay, let's make oxygen, eight electrons 1 2 3 4 5 6 7 8 now I can start putting a second electron in this one because these others now have one electron in them so I can't put second electrons in here until all the others have at least one and they do, that's why I was able to take the eighth electron and put it right there. let's do Florine 1 2 3 4 5 6 7 8 nine okay and I was able to put the second electron here because all these others had at least one electron, okay let's do 10 electrons 1 2 3 4 5 6 7 8 9 10 Okay, so I was able to put the last electron here because all of these others had at least one electron.

Okay, let's do it for vadium atomic number 23, so I think you should be able to do this since you've seen. pattern 23 here we go 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 and 23 and then 24 and 25 would go here and here so 23 24 25 and then 26 go here and then you would fill it here and then you would start doing this, so before you can put second arrows on any of these, the rest of these have to have one as soon as everyone has one, so you go back and start filling them in now instead of drawing arrows, a shorthand form of the electron configuration uses superscripts and the superscript number is the number of arrows and for example if the 3 p orbitals contain five arrows you would write 3 P5 so for example let's write the configuration electrons of chlorine which has an atomic number of 17 so here we go 17 electrons 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 and the way we would abbreviate this is 1 s you have two electrons here so you have two there two electrons in the 2 so we have two here 2 4 six electrons in the 2 p orbital so six would go there two electrons here two there five electrons here five there and then you would just rewrite this like this so that's the electron configuration abbreviated for chlorine and, in fact, we can abbreviate it even further.

We can abbreviate the electron configurations even further by using the configuration of the elements in the rightmost vertical column of the periodic table and the rightmost column has, excuse me, helium, neon, argon. Krypton Xenon and Ron, so basically here's your periodic table and we're going to use the electron configurations of these elements in the far right column. Well, you're here, so let me give you an example. In the first part, we are going to abbreviate chlorine. and the first part of the chlorine configuration matches the neon configuration, so this is the electronic configuration of chlorine that we just drew.

If you look at the first part of the neon configuration, you have 10 electrons here and neon has 10 electrons, so here is neon's atomic number of 10, so if we were to abbreviate it using these superscripts, we would let you know 1 S2 2 S2 2p6 3 S2 and 3p5 this electron configuration matches the abbreviated configuration of neon, so instead of writing everything This is why we don't substitute instead of 1 S2 2 S2 and 2p6 we will simply substitute the symbol uh for neon, which will be this instead of all this and we won't have to write as much instead. After writing all this, we will simply write the letters NE e instead of this part here so that the abbreviated configuration of chlorine can be written as the first part of neon plus 3 S2 3p5 for a PDF transcript of this lecture.

Go to Richard Lou.com. This has been the 22nd chemistry lecture, Electronic Configuration.