Understanding Periodicity in Chemistry

Hello everybody and welcome to this A-Level Chemistry explanation video about the periodicity topic. In this video I'll take a look at what periodicity is and then we'll explore periodicity through four properties that elements can have. Atomic radius, electronegativity, ionisation energy, melting and boiling points. What I'll do is I'll make sure you understand what these properties are and what their patterns are as we work our way across period three but then... I will make sure that you know what you need to write for exam questions that ask you things about each of these properties and their patterns. Periodicity is the study of the repeating patterns or trends that occur in physical or chemical properties as we move around in the periodic table. And these physical and chemical properties are typically linked to whereabouts in the periodic table you find particular elements. And so we need a language to use to talk about elements position in the periodic table. Now, we've obviously got groups, which are the up and down, the columns and periods, which are the rows. But we also have regions of the periodic table that we call blocks. And the block that an element is in corresponds to the subshell that you find the outer electrons in. So, for instance, the group one and two elements have their outer electrons. in the S subshell. And so we call that the S block. And then the elements that are over to the right-hand side of the periodic table, these final six groups, they have their outer electron in a P subshell. So those are called the P block elements. And last of all that you'll encounter at A-level is the middle of the periodic table that you will have known as the transition elements and they are referred to as the D block elements because their outer electrons are in the D subshell and the position of these elements in the periodic table affects four properties that you need to know about and we're going to go on to look at now. There are four properties that we're going to consider and all of them look at the pattern for these properties as you work your way across period three. So that's this row. of the periodic table. The first property that we're going to take a look at is a really fundamental property that underpins a lot of the others, and it is atomic radius, which is very, very literal as names go. It is the distance between the centre of the atom, called the nucleus, and the electrons in the outermost energy level. And so that is the radius of the atom, and we can express that by the circle that I've got here, and as you would expect, the radius, just like the radius of a circle, is this distance that I've measured like. So it follows logically from the definition that the atomic radius of an atom will be larger if the atom has got more occupied shells of electrons, and then smaller if there are fewer occupied electron shells. And so for period three, the electrons are all in the same energy level, the third energy level. And so what that means is... When we're deciding what the pattern will be for atomic radius, the electrons are all in the same electron shell. And as a consequence of that, the electron shielding will be the same for each of them. And so therefore, what that means is the pattern for atomic radius only depends on how closely those electrons sit to the nucleus. The other feature that changes as we work from left to right across period three is the nucleus gains charge. because each of these atoms has got an extra proton. And so as a result of that, the positive charge of the nucleus is larger, so the attraction for those outer shell electrons increases. And so this electrostatic attraction pulls those electrons in closer to the nucleus. And so therefore, the atomic radius decreases across period three. And in fact, across any of the periods, the atomic radius will decrease. So you might be required to give a three-mark answer for what the pattern is for atomic radius across the period, and a two-mark follow-up explanation for that pattern, as I've bullet-pointed here. Or you could be asked to sketch the pattern, and when you sketch it, you're not required to remember any of the numbers, just the general pattern, which is a decrease as we work our way across the period. Potentially, you won't... be asked directly about the atomic radius of an atom or two atoms. It might be that you're asked to use your knowledge of the atomic radius that atoms possess to explain patterns in other properties. And that's what we're going to look at next. Atomic radius has a direct impact on the electronegativity of a particular atom. You could actually be asked a two-mark question asking you to define what electronegativity is, and it's the ability of an atom to attract electron density, or simply electrons, in a covalent bond. So for instance, if you've got a diatomic molecule such as bromine, Br2, the electron pair that holds... holds those two atoms together will be shared exactly equally. Whereas if you were to have a different molecule, for instance hydrogen bonded to chlorine, those two elements will have different electronegativities. And in this instance the electrons will be slightly closer to the chlorine because the electronegativity of the chlorine atom is greater. So as we work our way across the period, the pattern of electronegativity is a increase. Because the electrons are more attracted to the period three elements as we work our way to the right hand side. And the reason for this is directly related to the atomic radius that we've just discussed. And we could use very much the same language. The nuclear charge or the number of protons is increasing as we work our way to the right hand side. And the electrons are in the same shell or there is similar shielding in each case. And so because of those things, the attraction that the electrons in the covalent bond experience from the nucleus of a particular element will increase as that atom gets smaller and as that atom has a greater charge in its nucleus. And so therefore, this could be a three-mark question. And again, it's the same rules. What's the pattern? What's the two-mark follow-up explanation? Or... you could be asked to draw the electronegativity pattern, and it is an increase as we work our way across the period. It's worth noting that they're likely to say from sodium through to chlorine rather than continuing on to argon, because as you will remember, the group zero elements, the noble gases, they don't form covalent bonds very readily, so they can't really be considered to have an electronegativity. because they don't attract electron density in a covalent bond, because they don't form covalent bonds readily. It should be noted that while I've been talking about across a period, there are patterns in electronegativity within a group as well. And the pattern is that the electronegativity increases as you go up a group. And this is because the atoms get smaller the higher up you go, and therefore the nucleus is... closer to that bonding pair of electrons and therefore there will be a stronger attraction between the nucleus and the bonding pair of electrons and so the electrons will be pulled closer to that particular atom. For instance within group 7, the halogens, fluorine at the top of the group is the most electronegative element and in fact it's the most electronegative element of all. The top right of the periodic table is where you find the most electronegative elements, and the bottom left of the periodic table is where you find the least electronegative elements. The next property that you need to know the pattern for as we work our way across period 3 is something called ionisation energy. Now, ionisation energy is the enthalpy or energy required to remove a mole of electrons from a mole of gaseous atoms. Now, for instance, what that means is if we have sodium and we want to remove a mole of electrons to make a mole of sodium ions, first of all, it needs to be in the gaseous state. So we've got gaseous sodium turning into gaseous sodium ions. So that means it's unipositive. It's lost one mole of electrons. And here are the electrons that have been lost. So the energy change needed for this reaction. would be exactly equal to the ionization energy for sodium. Now, it follows that obviously the electrons are in shells around sodium because they're experiencing attraction from the sodium's nucleus. Now, the stronger the attraction that electron is experiencing, the harder it will be to remove that electron, and therefore the larger the ionization energy will be. So... that means that the ionisation energy for sodium will be different to the ionisation energy for chlorine, etc. This is a definition that you need to remember, just like electronegativity, and it could also be a two-mark question with one mark for each of these two rows. There are definitions that don't have the a mole of, it just says remove an electron from a gaseous atom. And for year 12 chemistry that's absolutely fine, but I would encourage you to get into the habit of saying a mole of electrons from a mole of gaseous atoms. because that's what I would expect you to be doing on the final exams at the end of year 13. I've just said that the ionisation energy will be larger if it is harder to remove the electron from the outer energy level. Well, it will be harder to remove that electron the closer that electron is to the nucleus, and the stronger the attraction, the bigger the positive charge in the nucleus. And so therefore the pattern across period 3... Since we know that as we go across period three, the outer shell electron becomes closer to the nucleus, in other words, the atomic radius decreases. And we also know that the cause for that is that the nuclear charge increases as we go left to right. Well, it stands to reason that the ionisation energy will increase. In other words, it will be harder to remove that outer shell electron for the simple reason. that the nucleus has got an extra proton each step we take towards the right-hand side, and so the nuclear charge will be larger, and the outer shell electron is actually slightly closer to the nucleus as well. So that's the double whammy. We're having both of those reasons being ticked here. The outer shell electrons are closer to the nucleus, and the nucleus has got a larger positive charge. And so that means that the attraction from the nucleus to the outer shell electron is going to be larger as we work our way left to right across this period. And so most typically this will be a three or possibly a four mark answer, as I've drafted down the right hand side. Or you could be asked to recognise the pattern across the period in terms of a graph form. In general, as I've said, the ionisation energy increases across a period. But that's not a pattern that occurs for each individual step as we work our way across a period. In fact, what we get is a pattern like this. So you can see that we have got the general increase as we work our way from left to right across the period, from sodium all the way through to argon. But we get two dips, and this is characteristic for whatever period we're looking at. So you need to learn about period three. but they could sneakily ask you about period two or period four, and the pattern will be the same, and the reason for the pattern will be the same. So what you can see is we've got a region here with the two S block elements, sodium and magnesium, and then there is a dip. And the reason for this dip in the position here is because the outer shell electrons that are being removed are in a P block rather than an S block. You could be asked to explain the shape of... this graph or even remember where the dips occur. So if you're asked to explain why we've got this dip between magnesium and aluminium, my advice would be to think first, well what is the electron arrangement for magnesium? So it's 1s2, 2s2, 2p6, 3s2. Whereas for aluminium it's the same thing but at the end it is 3s2. p1. And so what that means is we now know that the outer electron from aluminium is being removed from a p subshell, whereas for magnesium it's being removed from s. And p subshells have slightly higher energy and are further from the nucleus and are so easier to remove. You also need to be able to explain this second dip that appears between phosphorus and sulfur. And This time it's a little bit harder to explain because all of these elements have their outer electrons in the 3p subshell. If we take a look at phosphorus, the outer electron goes 3s2 3p3. For sulfur it goes 3s2 3p4, so only a subtle difference. The reason for the dip comes from the fact that there are only three orbitals in a p subshell. So that means for phosphorus, each of those electrons that are in the p subshell occupy the orbitals singly. So there's one in each of those three, as I've shown here. Whereas for sulfur, one of those orbitals has to have two electrons in it. And so what that means is that those two electrons, which are both negatively charged, will repel each other. and so one of them will have slightly more energy than you would expect from a 3p electron, and so therefore it will be slightly easier to remove due to that repulsion between the electrons. So to sum up on this graph, you might be asked to explain the shape, whether that's the general increase or those two dips, or you might be asked to finish the graph off. So you could be given the beginning four points on the graph, and they could ask you to add the final four points. And so therefore it's important to remember that you've got an increase and then a dip, and that that dip falls between the previous two. It doesn't matter precisely where. It's just it's got to be lower than the previous but above the one at the very beginning. And then we've got our increase twice and then we've got the dip and again the same thing between the previous two points and then two last increases. The pattern that we've just looked at would continue for all of the periods. So we've got our characteristic increase, dip, increase, increase, dip. increase, increase, as you work your way across a period. Now, as you go down a group, it follows that the electrons are going to be further away from the nucleus because there is an extra occupied energy level. And so what that means is if this was the pattern for period three, well, this would be the pattern for period four with the same shape, but obviously a smaller first ionisation energy because the outer electrons are... further from the nucleus and that means that they are therefore easier to lose because the attraction between the nucleus and the outer electrons is smaller. The final property that we're going to look at is the melting and boiling point for the elements of period three. Before we take a look at that let's remind ourselves about the three states of matter or the three phases that elements can exist in. We've obviously got solids on the left hand side, liquids in the middle and gases on the right. As a solid you can see that we've got the nice regular layers of the atoms. The atoms are not moving, they're just vibrating about fixed points because they don't have very much energy. And there are forces between the particles that holds them in this rigid lattice structure. Liquids are similarly quite close compared to solids. But their layered structure has been disturbed and they are slightly more random and they're moving around more. The reason they're moving around more is they've got more energy. And then for gases, they have got lots of space between their particles. The particles have got lots of energy and they're moving around rapidly as a result. And the forces between the individual particles is particularly low for the gases. Now, melting point. is obviously the temperature at which a solid turns into a liquid. And in fact, also the temperature where a liquid turns into a solid. Now, this will happen when the solid particles have got enough energy to move slightly away from each other, to overcome those forces that exist between the particles in the solid structure. So these forces will influence...... how much energy is needed to separate the particles. The stronger those forces are, the more energy will be needed to move these particles away from each other. And similarly, the weaker those attractions are, the less energy will be required to separate the particles from each other. And as a result of that, the melting point and boiling point will change as you would expect. So strong forces between... particles means that the energy required to make them separate will be significantly high, and that means that the melting point will also be high. And the pattern for boiling point is exactly the same. Everything I've just said holds true for that. So boiling point is obviously the temperature at which a liquid turns into a gas, and if the liquid particles have got strong forces between each other... That means they will be hard to separate, lots of energy will be required, and so the boiling point will be higher. And of course, boiling points are always going to be higher than melting points. If we take a look now at the melting point pattern as we work our way across period three, we need to keep in mind that stronger forces between particles means more energy is going to be required to separate those particles and therefore the melting point will be higher. The pattern for period three melting points looks like this. Now you'll note that I haven't got a scale on my left hand side for my y-axis, I'm just simply saying the melting point of the particular element and showing it as we go across the period. What you should note is that there are three distinct regions. The region on the left hand side for sodium and magnesium and aluminium, that is separate from the other two regions and it makes sense that it would be so because these are the three metallic elements. And then we've got silicon which is a giant covalent element. And then we've got the other four elements, which are all nonmetals. And so they have a different type of force between the particles. So if we start with the metals, obviously metals have got metallic bonding. In metallic bonding, the outer electrons of each of the atoms are delocalised and free to move through the structure. And that means that the metal atoms are in fact positively charged ions. And the... metallic bonding itself that holds these atoms together in their metallic lattice is an electrostatic attraction between the metal positively charged and the negative delocalized electrons. If we take a look at the pattern on the graph itself, we can see that sodium has the lowest of the three melting points, then magnesium, and then actually really closely followed aluminium. So in general, as we work left to right, there is an increase. And that tells us that the metallic bonding must be getting stronger. There must be stronger attractions between those electrons in the lattice for magnesium than it is for sodium, and stronger for aluminium than it is for the other two. And the reason for this is as we go from left to right across those three metallic elements, the metals nucleus... gets more positive. Sodium forms 1 plus ions. Magnesium, 2 plus ions. Aluminium, 3 plus ions. Not only that, but these positive metal ions actually get smaller as we go across the period as well. So aluminium 3 plus will be a smaller ion than sodium 1 plus. And so that has a doubling effect because what we can say is we can wrap those two properties together in something called charge density. So the charge density of the aluminium ion is significantly greater than the sodium ion and so therefore the delocalised electrons in aluminium structure will experience far greater attraction to the aluminium ions than they would in the sodium positive lattice structure. And so what that means is it's far easier to separate sodium ions from each other than it is. magnesium ions. And so the melting point, as we compare those three metals, increases across that region. And so you could be required to justify why one of those metals has got a different melting point to another and back it up with an explanation. When we get on to silicon, silicon has got the greatest melting point of any elements in the period, and that's because silicon is max. macromolecular. It is a giant covalent structure. And what it does is very similar to the structure of carbon, basically identical, is we have a tetrahedral structure which goes on and on and on with thousands and thousands, millions of silicon atoms held together by four strong covalent bonds per atom. These obviously require a huge amount of energy to break for each of them. And there are lots of these covalent bonds. And so that means that the melting point of silicon is going to be absolutely huge, bigger than those three metals that we've just been looking at. The final region of the periodic table that we're looking at here is the four elements from phosphorus through to argon. So these aren't metallic, these aren't giant covalents, these first three are simple molecular or simple covalent. So simple molecular substances obviously exist as molecules and these molecules, when they are experiencing a temperature increase, what happens is that energy goes into breaking the forces between these molecules. Obviously, these are called intermolecular forces. Now, because they are elements, the only intermolecular force that they experience is the force called van der Waals forces. It's the weakest of all the intermolecular forces, but it's the force that exists in everything. Now, van der Waals forces are larger for larger molecules. or they are larger for particles with a greater number of electrons. Phosphorus exists as P4, sulfur as S8, and chlorine as Cl2. And so it's not a big leap to say that sulfur is the biggest of those three, and it will have the greatest number of electrons, and that's why its melting point is the highest of that trio. And phosphorus is the second largest, P4. It's got the second highest melting point. And chlorine is the smallest of those three molecules, and therefore it has got the smallest melting point, the lowest melting point, because it is very easy to separate one molecule of chlorine from another. Last of all, we've got argon. Argon is a noble gas. Noble gases are what's called monoatomic, or monatomic for short. And what that means is they only go around as single atoms. And so when we are melting something like argon, one of the noble gases, what we're doing is we're separating individual atoms of argon, which aren't very strongly attracted to each other. In fact, they only have van der Waals forces between each other as well. And since they are single atoms, these van der Waals forces are really, really weak. And so that's why the melting point is lowest of... all for argon. And so you need to be able to explain the shape of this graph. You could be asked to fill it in if you got some of the points but not the others, or you could be asked to explain or predict what the melting points of pairs of these elements would be. So you could be asked to say why phosphorus has a very low melting point, whereas sodium has a very high melting point, for instance. And you would be exploring what I've just said. You'd start by talking about the type of bonding that exists between the two different elements that you're considering. And then you'd explain why one of them has stronger forces than the other. And therefore, the one with the strongest forces has the higher melting point. Or you could be asked, why does phosphorus have a higher melting point than chlorine? And then you would be saying, well, phosphorus is a larger molecule. Therefore, it's got stronger van der Waals forces. which require more energy to overcome. And you would expect to see the same pattern in boiling points as you work your way across period three for the same sorts of reasons. Obviously, boiling point values are greater than melting point values, but the justification for the patterns follows the same line of reasoning. Okay, that's the end of this video. Hopefully now you feel more confident in the areas of this topic. You should be able to define periodicity, electronegativity and ionisation energy. You should be able to state the block an element is in based on its position in the periodic table. You should be able to be confident working with graphs, completing them for atomic radius, electronegativity, first ionisation energy and melting point. You should be able to explain patterns in the properties I've just mentioned as we work our way across period three, whether that's from a graph. or whether that's completely coming to it cold and you have to remember it with no prompting. You need to be able to explain why two elements have different particular properties, whether that's why does sodium have a higher melting point than chlorine, or why one thing has a different ionisation energy to another. And you need to be able to think on your feet and apply these rules to different periods that you haven't studied in class, and I've not covered here. but the rules that we've said for period three correspond to other periods that we've not covered. Okay, that's the end of this video. I hope it was useful. I'll see you again soon.

What I'll do is I'll make sure you understand what these properties are and what their patterns are as we work our way across period three but then... I will make sure that you know what you need to write for exam questions that ask you things about each of these properties and their patterns. Periodicity is the study of the repeating patterns or trends that occur in physical or chemical properties as we move around in the periodic table.

And these physical and chemical properties are typically linked to whereabouts in the periodic table you find particular elements. And so we need a language to use to talk about elements position in the periodic table. Now, we've obviously got groups, which are the up and down, the columns and periods, which are the rows. But we also have regions of the periodic table that we call blocks. And the block that an element is in corresponds to the subshell that you find the outer electrons in.

So, for instance, the group one and two elements have their outer electrons. in the S subshell. And so we call that the S block. And then the elements that are over to the right-hand side of the periodic table, these final six groups, they have their outer electron in a P subshell.

So those are called the P block elements. And last of all that you'll encounter at A-level is the middle of the periodic table that you will have known as the transition elements and they are referred to as the D block elements because their outer electrons are in the D subshell and the position of these elements in the periodic table affects four properties that you need to know about and we're going to go on to look at now. There are four properties that we're going to consider and all of them look at the pattern for these properties as you work your way across period three.

So that's this row. of the periodic table. The first property that we're going to take a look at is a really fundamental property that underpins a lot of the others, and it is atomic radius, which is very, very literal as names go.

It is the distance between the centre of the atom, called the nucleus, and the electrons in the outermost energy level. And so that is the radius of the atom, and we can express that by the circle that I've got here, and as you would expect, the radius, just like the radius of a circle, is this distance that I've measured like. So it follows logically from the definition that the atomic radius of an atom will be larger if the atom has got more occupied shells of electrons, and then smaller if there are fewer occupied electron shells.

And so for period three, the electrons are all in the same energy level, the third energy level. And so what that means is... When we're deciding what the pattern will be for atomic radius, the electrons are all in the same electron shell. And as a consequence of that, the electron shielding will be the same for each of them. And so therefore, what that means is the pattern for atomic radius only depends on how closely those electrons sit to the nucleus.

The other feature that changes as we work from left to right across period three is the nucleus gains charge. because each of these atoms has got an extra proton. And so as a result of that, the positive charge of the nucleus is larger, so the attraction for those outer shell electrons increases.

And so this electrostatic attraction pulls those electrons in closer to the nucleus. And so therefore, the atomic radius decreases across period three. And in fact, across any of the periods, the atomic radius will decrease.

So you might be required to give a three-mark answer for what the pattern is for atomic radius across the period, and a two-mark follow-up explanation for that pattern, as I've bullet-pointed here. Or you could be asked to sketch the pattern, and when you sketch it, you're not required to remember any of the numbers, just the general pattern, which is a decrease as we work our way across the period. Potentially, you won't... be asked directly about the atomic radius of an atom or two atoms.

It might be that you're asked to use your knowledge of the atomic radius that atoms possess to explain patterns in other properties. And that's what we're going to look at next. Atomic radius has a direct impact on the electronegativity of a particular atom. You could actually be asked a two-mark question asking you to define what electronegativity is, and it's the ability of an atom to attract electron density, or simply electrons, in a covalent bond.

So for instance, if you've got a diatomic molecule such as bromine, Br2, the electron pair that holds... holds those two atoms together will be shared exactly equally. Whereas if you were to have a different molecule, for instance hydrogen bonded to chlorine, those two elements will have different electronegativities.

And in this instance the electrons will be slightly closer to the chlorine because the electronegativity of the chlorine atom is greater. So as we work our way across the period, the pattern of electronegativity is a increase. Because the electrons are more attracted to the period three elements as we work our way to the right hand side. And the reason for this is directly related to the atomic radius that we've just discussed.

And we could use very much the same language. The nuclear charge or the number of protons is increasing as we work our way to the right hand side. And the electrons are in the same shell or there is similar shielding in each case. And so because of those things, the attraction that the electrons in the covalent bond experience from the nucleus of a particular element will increase as that atom gets smaller and as that atom has a greater charge in its nucleus. And so therefore, this could be a three-mark question.

And again, it's the same rules. What's the pattern? What's the two-mark follow-up explanation?

Or... you could be asked to draw the electronegativity pattern, and it is an increase as we work our way across the period. It's worth noting that they're likely to say from sodium through to chlorine rather than continuing on to argon, because as you will remember, the group zero elements, the noble gases, they don't form covalent bonds very readily, so they can't really be considered to have an electronegativity. because they don't attract electron density in a covalent bond, because they don't form covalent bonds readily. It should be noted that while I've been talking about across a period, there are patterns in electronegativity within a group as well.

And the pattern is that the electronegativity increases as you go up a group. And this is because the atoms get smaller the higher up you go, and therefore the nucleus is... closer to that bonding pair of electrons and therefore there will be a stronger attraction between the nucleus and the bonding pair of electrons and so the electrons will be pulled closer to that particular atom. For instance within group 7, the halogens, fluorine at the top of the group is the most electronegative element and in fact it's the most electronegative element of all. The top right of the periodic table is where you find the most electronegative elements, and the bottom left of the periodic table is where you find the least electronegative elements.

The next property that you need to know the pattern for as we work our way across period 3 is something called ionisation energy. Now, ionisation energy is the enthalpy or energy required to remove a mole of electrons from a mole of gaseous atoms. Now, for instance, what that means is if we have sodium and we want to remove a mole of electrons to make a mole of sodium ions, first of all, it needs to be in the gaseous state.

So we've got gaseous sodium turning into gaseous sodium ions. So that means it's unipositive. It's lost one mole of electrons.

And here are the electrons that have been lost. So the energy change needed for this reaction. would be exactly equal to the ionization energy for sodium.

Now, it follows that obviously the electrons are in shells around sodium because they're experiencing attraction from the sodium's nucleus. Now, the stronger the attraction that electron is experiencing, the harder it will be to remove that electron, and therefore the larger the ionization energy will be. So... that means that the ionisation energy for sodium will be different to the ionisation energy for chlorine, etc. This is a definition that you need to remember, just like electronegativity, and it could also be a two-mark question with one mark for each of these two rows.

There are definitions that don't have the a mole of, it just says remove an electron from a gaseous atom. And for year 12 chemistry that's absolutely fine, but I would encourage you to get into the habit of saying a mole of electrons from a mole of gaseous atoms. because that's what I would expect you to be doing on the final exams at the end of year 13. I've just said that the ionisation energy will be larger if it is harder to remove the electron from the outer energy level. Well, it will be harder to remove that electron the closer that electron is to the nucleus, and the stronger the attraction, the bigger the positive charge in the nucleus.

And so therefore the pattern across period 3... Since we know that as we go across period three, the outer shell electron becomes closer to the nucleus, in other words, the atomic radius decreases. And we also know that the cause for that is that the nuclear charge increases as we go left to right.

Well, it stands to reason that the ionisation energy will increase. In other words, it will be harder to remove that outer shell electron for the simple reason. that the nucleus has got an extra proton each step we take towards the right-hand side, and so the nuclear charge will be larger, and the outer shell electron is actually slightly closer to the nucleus as well.

So that's the double whammy. We're having both of those reasons being ticked here. The outer shell electrons are closer to the nucleus, and the nucleus has got a larger positive charge. And so that means that the attraction from the nucleus to the outer shell electron is going to be larger as we work our way left to right across this period.

And so most typically this will be a three or possibly a four mark answer, as I've drafted down the right hand side. Or you could be asked to recognise the pattern across the period in terms of a graph form. In general, as I've said, the ionisation energy increases across a period.

But that's not a pattern that occurs for each individual step as we work our way across a period. In fact, what we get is a pattern like this. So you can see that we have got the general increase as we work our way from left to right across the period, from sodium all the way through to argon.

But we get two dips, and this is characteristic for whatever period we're looking at. So you need to learn about period three. but they could sneakily ask you about period two or period four, and the pattern will be the same, and the reason for the pattern will be the same. So what you can see is we've got a region here with the two S block elements, sodium and magnesium, and then there is a dip. And the reason for this dip in the position here is because the outer shell electrons that are being removed are in a P block rather than an S block.

You could be asked to explain the shape of... this graph or even remember where the dips occur. So if you're asked to explain why we've got this dip between magnesium and aluminium, my advice would be to think first, well what is the electron arrangement for magnesium?

So it's 1s2, 2s2, 2p6, 3s2. Whereas for aluminium it's the same thing but at the end it is 3s2. p1.

And so what that means is we now know that the outer electron from aluminium is being removed from a p subshell, whereas for magnesium it's being removed from s. And p subshells have slightly higher energy and are further from the nucleus and are so easier to remove. You also need to be able to explain this second dip that appears between phosphorus and sulfur. And This time it's a little bit harder to explain because all of these elements have their outer electrons in the 3p subshell. If we take a look at phosphorus, the outer electron goes 3s2 3p3.

For sulfur it goes 3s2 3p4, so only a subtle difference. The reason for the dip comes from the fact that there are only three orbitals in a p subshell. So that means for phosphorus, each of those electrons that are in the p subshell occupy the orbitals singly. So there's one in each of those three, as I've shown here. Whereas for sulfur, one of those orbitals has to have two electrons in it.

And so what that means is that those two electrons, which are both negatively charged, will repel each other. and so one of them will have slightly more energy than you would expect from a 3p electron, and so therefore it will be slightly easier to remove due to that repulsion between the electrons. So to sum up on this graph, you might be asked to explain the shape, whether that's the general increase or those two dips, or you might be asked to finish the graph off.

So you could be given the beginning four points on the graph, and they could ask you to add the final four points. And so therefore it's important to remember that you've got an increase and then a dip, and that that dip falls between the previous two. It doesn't matter precisely where. It's just it's got to be lower than the previous but above the one at the very beginning. And then we've got our increase twice and then we've got the dip and again the same thing between the previous two points and then two last increases.

The pattern that we've just looked at would continue for all of the periods. So we've got our characteristic increase, dip, increase, increase, dip. increase, increase, as you work your way across a period.

Now, as you go down a group, it follows that the electrons are going to be further away from the nucleus because there is an extra occupied energy level. And so what that means is if this was the pattern for period three, well, this would be the pattern for period four with the same shape, but obviously a smaller first ionisation energy because the outer electrons are... further from the nucleus and that means that they are therefore easier to lose because the attraction between the nucleus and the outer electrons is smaller. The final property that we're going to look at is the melting and boiling point for the elements of period three.

Before we take a look at that let's remind ourselves about the three states of matter or the three phases that elements can exist in. We've obviously got solids on the left hand side, liquids in the middle and gases on the right. As a solid you can see that we've got the nice regular layers of the atoms.

The atoms are not moving, they're just vibrating about fixed points because they don't have very much energy. And there are forces between the particles that holds them in this rigid lattice structure. Liquids are similarly quite close compared to solids. But their layered structure has been disturbed and they are slightly more random and they're moving around more.

The reason they're moving around more is they've got more energy. And then for gases, they have got lots of space between their particles. The particles have got lots of energy and they're moving around rapidly as a result.

And the forces between the individual particles is particularly low for the gases. Now, melting point. is obviously the temperature at which a solid turns into a liquid. And in fact, also the temperature where a liquid turns into a solid.

Now, this will happen when the solid particles have got enough energy to move slightly away from each other, to overcome those forces that exist between the particles in the solid structure. So these forces will influence...... how much energy is needed to separate the particles.

The stronger those forces are, the more energy will be needed to move these particles away from each other. And similarly, the weaker those attractions are, the less energy will be required to separate the particles from each other. And as a result of that, the melting point and boiling point will change as you would expect. So strong forces between...

particles means that the energy required to make them separate will be significantly high, and that means that the melting point will also be high. And the pattern for boiling point is exactly the same. Everything I've just said holds true for that. So boiling point is obviously the temperature at which a liquid turns into a gas, and if the liquid particles have got strong forces between each other...

That means they will be hard to separate, lots of energy will be required, and so the boiling point will be higher. And of course, boiling points are always going to be higher than melting points. If we take a look now at the melting point pattern as we work our way across period three, we need to keep in mind that stronger forces between particles means more energy is going to be required to separate those particles and therefore the melting point will be higher. The pattern for period three melting points looks like this. Now you'll note that I haven't got a scale on my left hand side for my y-axis, I'm just simply saying the melting point of the particular element and showing it as we go across the period.

What you should note is that there are three distinct regions. The region on the left hand side for sodium and magnesium and aluminium, that is separate from the other two regions and it makes sense that it would be so because these are the three metallic elements. And then we've got silicon which is a giant covalent element.

And then we've got the other four elements, which are all nonmetals. And so they have a different type of force between the particles. So if we start with the metals, obviously metals have got metallic bonding. In metallic bonding, the outer electrons of each of the atoms are delocalised and free to move through the structure.

And that means that the metal atoms are in fact positively charged ions. And the... metallic bonding itself that holds these atoms together in their metallic lattice is an electrostatic attraction between the metal positively charged and the negative delocalized electrons. If we take a look at the pattern on the graph itself, we can see that sodium has the lowest of the three melting points, then magnesium, and then actually really closely followed aluminium. So in general, as we work left to right, there is an increase.

And that tells us that the metallic bonding must be getting stronger. There must be stronger attractions between those electrons in the lattice for magnesium than it is for sodium, and stronger for aluminium than it is for the other two. And the reason for this is as we go from left to right across those three metallic elements, the metals nucleus...

gets more positive. Sodium forms 1 plus ions. Magnesium, 2 plus ions.

Aluminium, 3 plus ions. Not only that, but these positive metal ions actually get smaller as we go across the period as well. So aluminium 3 plus will be a smaller ion than sodium 1 plus. And so that has a doubling effect because what we can say is we can wrap those two properties together in something called charge density. So the charge density of the aluminium ion is significantly greater than the sodium ion and so therefore the delocalised electrons in aluminium structure will experience far greater attraction to the aluminium ions than they would in the sodium positive lattice structure.

And so what that means is it's far easier to separate sodium ions from each other than it is. magnesium ions. And so the melting point, as we compare those three metals, increases across that region. And so you could be required to justify why one of those metals has got a different melting point to another and back it up with an explanation.

When we get on to silicon, silicon has got the greatest melting point of any elements in the period, and that's because silicon is max. macromolecular. It is a giant covalent structure. And what it does is very similar to the structure of carbon, basically identical, is we have a tetrahedral structure which goes on and on and on with thousands and thousands, millions of silicon atoms held together by four strong covalent bonds per atom.

These obviously require a huge amount of energy to break for each of them. And there are lots of these covalent bonds. And so that means that the melting point of silicon is going to be absolutely huge, bigger than those three metals that we've just been looking at. The final region of the periodic table that we're looking at here is the four elements from phosphorus through to argon. So these aren't metallic, these aren't giant covalents, these first three are simple molecular or simple covalent.

So simple molecular substances obviously exist as molecules and these molecules, when they are experiencing a temperature increase, what happens is that energy goes into breaking the forces between these molecules. Obviously, these are called intermolecular forces. Now, because they are elements, the only intermolecular force that they experience is the force called van der Waals forces. It's the weakest of all the intermolecular forces, but it's the force that exists in everything. Now, van der Waals forces are larger for larger molecules.

or they are larger for particles with a greater number of electrons. Phosphorus exists as P4, sulfur as S8, and chlorine as Cl2. And so it's not a big leap to say that sulfur is the biggest of those three, and it will have the greatest number of electrons, and that's why its melting point is the highest of that trio.

And phosphorus is the second largest, P4. It's got the second highest melting point. And chlorine is the smallest of those three molecules, and therefore it has got the smallest melting point, the lowest melting point, because it is very easy to separate one molecule of chlorine from another.

Last of all, we've got argon. Argon is a noble gas. Noble gases are what's called monoatomic, or monatomic for short.

And what that means is they only go around as single atoms. And so when we are melting something like argon, one of the noble gases, what we're doing is we're separating individual atoms of argon, which aren't very strongly attracted to each other. In fact, they only have van der Waals forces between each other as well.

And since they are single atoms, these van der Waals forces are really, really weak. And so that's why the melting point is lowest of... all for argon.

And so you need to be able to explain the shape of this graph. You could be asked to fill it in if you got some of the points but not the others, or you could be asked to explain or predict what the melting points of pairs of these elements would be. So you could be asked to say why phosphorus has a very low melting point, whereas sodium has a very high melting point, for instance. And you would be exploring what I've just said.

You'd start by talking about the type of bonding that exists between the two different elements that you're considering. And then you'd explain why one of them has stronger forces than the other. And therefore, the one with the strongest forces has the higher melting point. Or you could be asked, why does phosphorus have a higher melting point than chlorine? And then you would be saying, well, phosphorus is a larger molecule.

Therefore, it's got stronger van der Waals forces. which require more energy to overcome. And you would expect to see the same pattern in boiling points as you work your way across period three for the same sorts of reasons.

Obviously, boiling point values are greater than melting point values, but the justification for the patterns follows the same line of reasoning. Okay, that's the end of this video. Hopefully now you feel more confident in the areas of this topic. You should be able to define periodicity, electronegativity and ionisation energy. You should be able to state the block an element is in based on its position in the periodic table.

You should be able to be confident working with graphs, completing them for atomic radius, electronegativity, first ionisation energy and melting point. You should be able to explain patterns in the properties I've just mentioned as we work our way across period three, whether that's from a graph. or whether that's completely coming to it cold and you have to remember it with no prompting. You need to be able to explain why two elements have different particular properties, whether that's why does sodium have a higher melting point than chlorine, or why one thing has a different ionisation energy to another.

And you need to be able to think on your feet and apply these rules to different periods that you haven't studied in class, and I've not covered here. but the rules that we've said for period three correspond to other periods that we've not covered. Okay, that's the end of this video. I hope it was useful.

I'll see you again soon.

Transcript for:Understanding Periodicity in Chemistry

Transcript for:
Understanding Periodicity in Chemistry