Quantum Numbers - The Easy Way!

2p sublevel this is 3p this is 4p bromine is somewhere around here bromine ends in 4p5 this is 1 2 3 4 so four p5 so if I want to count the number of p electrons that bromine has this is what we can do count the number of elements in the p block up to bromine this is one two three 4 5 6 7 8 9 10 11 12 up to 17 so bromine has 17 p electrons this is how you can get the answer in the quickest way now you can always do it the old fashioned way so if we write the configuration down to 35 once we get to 2p6 that's 10 electrons then 3s2 3p6 4s2 that's 20. after 4s2 it's 3d10 so we have a total of 30 and then comes 4p5, which adds up to 35. so we have six plus six plus five, which tells us we have 17 p electrons.

Now, how many s electrons does bromine have? You could see that's two four six eight, so bromine has eight s electrons and also ten d electrons, that's how you can count the different types of electrons in a particular element, so let's go over one more example of how many electrons, how many electrons s has. Strontium has atomic number 38. So how can we solve this? Using the periodic table, this is what you can do. This is the s block and helium is also an s electron. So this is helium. Here we have hydrogen under the hydrogen. it has lithium sodium potassium rubidium and on the right it has beryllium magnesium calcium and strontium hydrogen represents 1s1 helium is 1s2 lithium is 2s1 2s2 this is 3s1 3s2 4s1 4s2 5s1 5s2 so the quick way to find the number of electrons s is to count the elements in the block s so helium is part of the block s so that is 2 3 4 5 6 7 8 9 10 so strontium has 10 electrons s let's say if you get a test question that asks you what is the maximum number of electrons that the n is equal to 10 energy levels, how would you answer that?

If you're not sure, let's talk about how it works. First, how many electrons are there in the first energy level. Now remember that in the first energy level you only have sublevel one and s. It can only have two electrons, so there is a maximum of two electrons that can occupy the first energy level. Now, what happens at the second energy level? How many electrons can there be in the second level? In the second level you only have 2s and 2p. We know that they can hold two p can hold six two plus six is ​​eight, so there are eight electrons that can occupy the second energy level.

Now, what about the third energy level? So we have 3s, 3p and 3d, so it will be 2 6 and d can hold up to 10. So there are up to 18 electrons in the third energy level, in the fourth you have 4.4p 4d and 4f, so it will be 2 plus 6 plus 10, which is 18 plus 14 f can hold up to 14 electrons, it has 7 orbitals so now we have 32. Now what about the fifth energy level? There are 5s 5p 5d 5f. after 5f comes 5g and if n is 6, after 6g comes 6h, so it is 2 6 10 14 and after 14 it is 18. As you can see, for each new sublevel you add, you always add four electrons or two orbitals, so 2 6 10 14 18 you still add four so 32 plus 18 is 50.

So is there an equation that relates the energy level to the maximum number of electrons that can occupy the energy level? It turns out that there is the equation that is 2n2 2n2 represents the maximum. number of electrons at a given energy level, so when n is 1 it is 2 times 1 squared 1 squared is 1 times 2 gives 2. when n is 2 2 squared is 4 times 2 that gives you 8. when n is 3 3 squared is 9 times 2 is 18 when n is 4 4 squared is 16 times 2 is 32 and one is 5 5 times 5 is 25 times 2 gives 50. So how many electrons or what is the maximum number of electrons who can occupy the tenth? energy level, so use an equation 2n squared, it will be 2 times 10 squared, which is 10 squared, that is, 10 times 10, that is, 100 times 2, which corresponds to 200 electrons.

Now, how many orbitals are there in the tenth energy level? The number of orbitals is equal to n squared remember that each orbital can hold two electrons so if there are 200 electrons you will have a maximum of 100 orbitals so it is simply 10 squared which will give you 100 orbitals so which the maximum number of electrons is equal to 2n squared and the maximum number of orbitals in an energy level is simply equal to n squared, so be sure to add these equations to your list in addition to the other equations already mentioned. Now let's talk more about orbitals and energy levels, so let's say if n is 3, we know the number of orbitals. will be n squared, which is equal to 9.

So we have the equation, but sometimes in a test you may forget the equation, but if you understand it you will know what steps to take to find the correct answer, so let's understand why the number maximum of orbitals is n squared in the third energy level we know that there are three sublevels 3s 3p and 3d now we know that the s sublevel only has one orbital it has one box p has three orbitals three boxes and d has five orbitals or five boxes So, If you add up the total number of boxes, it's one plus three plus five, which equals nine, so in the third energy level there are a total of nine orbitals.

Let's try one more example. In the fourth energy level we know that the number of orbitals is. n squared or four squared which will give us 16. now in a fourth energy level there are four sublevels 4s 4p 4d and 4f so s has one orbital p has three d has five and f has seven so if you add the

numbers

this is one three five and seven one plus three is four four plus five is nine nine plus seven is sixteen so you can draw it and you can literally count how many orbitals there are at a given energy level now let's say if they give you two quads and

numbers

let's say if you give the value of n and l so let's say if n is three and l is zero and let's say n is four and l is one n is three l is 2 and n is 5 when l is 4 or 3 So for these different cases, identify the sublevel, calculate the maximum number of electrons that these two

quantum

numbers can have and also determine the number of orbitals that those two numbers can have, so let's start with the sublevel when n is 3 and when l. is 0 what sublevel does it correspond to when l is 0 corresponds to s then this is the 3s sublevel s only has one orbital so that's it for the number of orbitals and for the 3s sublevel since there is only one orbital the largest number of electrons What you can have is two electrons, so the answer is 3.

That's the type of sublevel. We have one orbital and a maximum of two electrons. Now what about the next one? Which sublevel corresponds to a

quantum

number of n equal to four and a value of l of one? so l is one for the p sublevel, so this is the four p sublevel and p has three orbitals and there is a maximum of six electrons that can occupy those three orbitals, so how many electrons can occupy or have the quantum numbers 3 and 2 ? for n and l, then when l is 2 we have the d sublevel, so it will be 3d and for d there are five orbitals, two electrons per orbital, so the largest number of electrons that can have these two quantum numbers is ten and what happens with the last one when n is five and l is equal to three l is three for the f sublevel, so this is five f and we know that f can hold up to 14 electrons and there are seven orbitals, so that's it for that one, now how many electrons can have the following three quantum numbers, let's say when n is three, l is two and when ml is one, now that we have a specific type of orbital within the sublevel, the answer has to be two because this identifies only one five o'clock box. boxes that are in the d sublevel and you can only have two electrons in a given orbital, so if you have nl and ml the answer is always three, I mean, it's always two electrons, so for the two when ls2 we have the d sublevel, so this is 3d and there are five boxes one two three four five ml can vary between negative two and two, so there are two electrons that have an ml value of one, so the answer is two.

Now, how many electrons can the following four quantum numbers have according to Paulie? Exclusion Principle: There is only one electron that can have four quantum numbers, so for four and three we have the four f sublevels and we know that f has seven orbitals that will vary between three negative and three positive, so the electron of interest is in this orbital and the spin tells us that it is the one with the downward arrow and it is only one electron that fits this criterion. Now let's say if you have the value of n and the spin, what is the maximum number of electrons that these two quantum numbers can have? is your answer now remember that in each orbital the spin could be plus half or minus half which means that half of the electrons will have a down arrow and the other half can have an up arrow now we know the maximum number of electrons in the third energy level is equal to 2n squared, which is 2 times 3 squared and that is 18.

So, since there are a total or maximum of 18 electrons in the third energy level, nine of those electrons they will have a half negative spin so if I have 3s 3p and 3d here is the 3s block this is 3p and this is 3d there is no 3f for the third level so each orbital can have a down arrow each orbital can only have one arrow down not two because the other one has to be up As you can see, there are nine down arrows, so it will be half of this number, so there are nine electrons that can have these two quantum numbers.

Now, how many electrons can the following quantum numbers have? So if we focus on n and l, l is 2 for d sublevel, then we have 4d sublevel and we know that d can hold up to 10 electrons because d has 5 orbitals, so if d can hold up to 10 electrons, half of those 10 electrons will have the up arrow of the positive half, so the answer should be five, so it's going to be one, two, three, four, five, so once you add more to the equation, if you're looking The maximum number of electrons with those quantum numbers will always be half of the two previous numbers.

Let's say, for example, if we have the quantum numbers and a ml, let's say n is 3 and ml is 1. How many electrons will these two quantum numbers have? So annistry for an industry, you can have the sublevel 3s, 3p and 3d for 3s there is an orbital for 3p this current and for 3d there are 5. for s l is 0 so ml has no choice but to be 0. for p l is 1, so ml can be negative one to one and for d l is two, so ml varies between negatives. two and two now notice that there are only two boxes with an ml value of one so there are only four electrons that can be in these two boxes so the answer is four there are four electrons that have the quantum is 3 and 1, where n is 3 and ml. is 1.

Now let's say that if we want to find the number of electrons that these two quantum numbers have plus a negative half spin, then the answer will be half of four, it will be two, which is these two that it has. to be a downward arrow, let's try another problem like that, so let's say n is five and ml is equal to zero, how many electrons will these two quantum numbers have? So, in the fifth level we have 5s 5p 5d 5f and also 5g, so for the s the sublevel l is 0 which means ml0 for the sublevel p l is 1 which means that ml will vary between negative one and one and for the sublevel d l is two so ml will vary from negative two to two now for sublevel f l is three and f has seven orbitals so ml will vary from negative three to three and for sublevel g, l is four so ml can vary from minus four to four.

Now we need to identify each orbital that has a zero value for ml, so this is one two. three four five, so there are five orbitals with these two quantum numbers and each of those orbitals can hold up to two electrons, so there are a total of 10 electrons with quantum numbers n equals five and ml equals zero. Now let's say if we add the spin to let's say if the spin is half negative, then the answer will be half of what we have here, so half of the 10 electrons will have a down arrow, so the answer final is 5 which has these three quantum numbers, but 10 electrons have the The first two quantum numbers now let's say that if you have a multiple choice question and it asks you which of the following configurations corresponds to a halogen, you would say it is ns1 ns2 ns2 and p4 and s2 and p5, which one is it?

The halogens correspond to ns2 sp5. chalcogens correspond to ns2 and p4, alkali metals correspond to ns1 and alkaline earth metals and s2, so let's talk about alkali metals, so we have lithium, sodium, potassium, rubidium, cesium, etc., all these elements have the ns1 configuration for lithium, is 1s2. 2s1 for sodium ends in 3s1 potassium force1, so all alkali metals will end in ns1 and then you have the next group, the alkaline earth metals, which include beryllium, magnesium, calcium, strontium and barium, so these elements end in ns2. Beryllium is 2s2. calcium 4s2 strontium-5s2 and then you have the chalcogens like oxygen sulfur selenium these have the ns2 and p4 configuration if you write the electron configuration for oxygen is 1s2 2s2 2p4 sulfur ends in 3p4 for sulfur it will be 3s2 3p4 so all the chalcogens It has the configuration ns2 mp4 halogens such as fluorine chlorine bromineIodine is ns2 mp5 now for the noble gases like neon argon krypton is ns2 and p6 the exception is helium which is simply 1s2 but the other noble gases will be ns2 mp6 so if you see this this corresponds to the column or group of elements of the periodic table so the elements in the same column share the same or similar chemical reactivity oxygen sulfur selenium have similar configurations ns2 and p4 and behave in a similar chemical way since they have the same number of electrons, well, what I wanted That is, it is the same number of valence electrons.

Now it turns out that there is a way to know which column it corresponds to, for example, if you have ns2 and p4, simply add the exponents two plus four, which gives you are six so this corresponds to the elements in group six to that is oxygen like this that for selenium those two calculations ns2 np5 if you add 2 and 5 you get 7 which corresponds to group 7a which would be the halogens such as chlorine bromine iodine fluorine ns1 corresponds to group 1a, which are the alkali metals, ns2, corresponds to group 2a, which They are the alkaline earth metals, magnesium, calcium, strontium, things like that, so let's say if you have some elements like manganese, aluminum, germanium, phosphorus and xenon, which element is or corresponds to the group that is ns2. and p1 ns2 mp1 if you add the exponents 2 plus 1 is 3 then this will be group 3a so if you don't see a d somewhere it probably won't be a transition metal so what element is in the group 3rd or?

What element has 3 valence electrons, there will be no xenon, which has eight valence electrons, it is not phosphorus, it has five, phosphorus is in group five eight and it is not germanium because it is in group four. The answer is aluminum, it has three valence electrons. and it is in group 3a so let's try another example let's say that if you want to find the element that corresponds to group ns2 and d7 then if it ends in d you know that it is a transition metal so let's say that the options are vanadium manganese fe c o and n i then two plus seven is nine so it has to be group nine which is basically the ninth column so it will be cobo cobalt has the configuration 4s2 3d7 if you count it starting from the left it is in the ninth column so is in group nine of the periodic table now the last topic of interest is determining whether a set of four fourth numbers is allowed or not, so let's say that if n is 3 l is 2 ml is negative 1 and ms is plus half, Are these four quantum numbers?

They allowed it or not, is there something wrong with these numbers? Now there are a few things you need to remember. l is always less than or equal to n minus one, so if l is equal to or greater than 3 it won't work, so l is less than n, which is fine, ml has to be between negative l and l, so if l is 2, ml has to be between 2 and negative 2, which in this example is fine and ms can only be a fraction, so now there is nothing wrong with these four quantum numbers. What's up with this one?

Let's say if n is 4 and if l is 4 and this is negative 2 and if it's negative half, is there anything wrong here? l has to be less than n it can't be equal to n so it's wrong now let's say if n is 5 l is 2 ml is negative 3 and ms is plus half. Is there something wrong? So first compare n and l if l is less than n then ok. Now look at ml and l if l is two ml it has to be between negative two and two minus three is outside that range so this doesn't work so let's say this is four i.e. one zero let's say negative half, so l is less than n which is good and if l is one ml has to be between negative one and one so zero is within that range so that's fine and ms can be a fraction so It works here, let's try another one, let's say. if n is 3 l is 2 ml is negative 2 and ms is one is it ok or is there an error?

So let's compare l and n l is less than n so that's fine and if l is two ml could be negative 2 to 2 which negative 2 is in that range so that's fine the only problem we have here is that ms is not a fraction. ms can only be plus half or minus 1 over 2. So this doesn't work, so basically that's it for this video. We've covered almost every topic you'll see regarding electron configuration and quantum numbers, so thanks for watching and have a great day.