Size of an Atom or Ion
Now moving on with our discussion the first property that we will discuss is the atomic radius. Now, if I ask you what do you think would be atomic radius, your obvious answer would be ‘Sir, it’s the radius of an atom.’ But, if I ask you to define it, then the proper definition for atomic radius would be that it is the distance from the center of the nucleus to the outer most shell. Right. So you just take the distance and that is the atomic radius. Now, the atomic radius is not so simple. Why? Because, you are dealing with very, very small particle over here which has a size of 10 to the power minus 10. Now for such small atom it would be very, very difficult for us to find the atomic radius accurately. The second reason because of which finding out the atomic radius becomes very difficult is the probability density picture. If you recall the probability density picture, you see that the probability of finding an electron even at a very, very large distance from the nucleus cannot be zero, except some places. Similar to that, if we take the probability picture of the electron, in the surrounding or if there are neighboring atoms then those neighboring atoms would also cause some sort of a distortion to the electron cloud and that would also result in the inaccuracies of our atomic radius. And, the last reason because of which the atomic radius will not be found out accurately is: the size of the atom changes with the change in its bonding state. So if it is covalently bonding, it might have a different radius. If it is metallic bonding, it might have something different. So we’ll take up all these, we’ll consider these all factors and what we are left with is called as the effective atomic radii. So wherever we are going to study, it’s the actual effective atomic radii and not the atomic radius which is from the center to the outer most shell. Right?
Now the first atomic radius that we are going to study is the covalent radius. Now what exactly is a covalent radius? Now suppose I have one atom over here and another atom over here. Now if I just bring them close and I make them touch each other, i.e. the electron clouds are touching each other, now how do I find the radius? If I take this distance from the center of the nucleus, the distance from the center of both the nucleus, that would be termed as the inter-nuclear distance. And if I take the half of it, that should give me the atomic radius. Right? But when a covalent bond is formed, what happens is they don’t touch – they overlap. Right? When they overlap, do you think the inter-nuclear distance would be the same as before? No, because the space over here becomes common to both of them and has to be subtracted from the inter-nuclear distance. So the inter-nuclear distance becomes shorter. Now if you take the half, this would be called as your covalent radii. Right? So, if you imagine, atomic radius and covalent radius, which one should be greater? Yes, it should be the atomic radius which is greater and the covalent radii would always be smaller than the atomic radius. Similarly if we talk about metallic radius – now metallic radius is again taking the inter-nuclear distance, dividing it by half in the metallic lattices. When the metal atoms combine to form the lattice, there you take the inter-nuclear distance divide it by two and that will be your metallic radius. And, again metallic radius would be smaller as compared to atomic radius. So, atomic radius is the one which is the largest, followed by covalent and metallic would always be smaller than the atomic radius. Now, let’s talk about the periodicity that I was saying, that how does the atomic radius vary when we go along a period or through a group. Now, if I bring back the periodic table, in the second period when we are moving from left to right in a period the atomic radius always decreases. Right? You must be wondering why. Because, the electrons are being added. So if the electrons are added the atomic radius should increase, but that’s not the case.
Now let’s take a look why. Let’s assume the protons to be ropes. Right? And the electrons to be hooks. So, what exactly is happening? If I take the example of lithium and beryllium, the electronic configuration of lithium is this and for beryllium is this. Right? Now in lithium the outermost shell contains only one electron and in beryllium the outermost shell contains two electrons. the number of proton for lithium is 3, the number of proton for beryllium is 4. Now what happens? There are two ropes which are clinging on to the hooks in lithium in the first shell and same is happening for beryllium. Right? So the protons are pulling the first shell towards itself with the same force. Now, the next proton is pulling the next electron which is placed in the second shell and in beryllium also the two ropes are now pulling the shell. So, where do you think the pull should be more? Yes, in beryllium the pull would be more because one object is being pulled by two ropes and lithium only one rope is pulling it up. So, out of lithium and beryllium which should be bigger in size? It is lithium. So using this concept you can easily predict that what would be the atomic radius. So when the electron are being added instead of the atomic radius being increasing it decreases because of this phenomena. Now, the period that we took from lithium to neon, if I draw a graph based on the atomic radius and the atomic number, how should my graph look like? It would be decreasing throughout, right? But if I compare it to the actual graph do you see some discrepancies? Now let us understand why this discrepancy is occurring. Now if I take the example of nitrogen and oxygen – the electronic configuration of nitrogen would be this and for oxygen would be this. Now if you look very carefully the proton subshell in nitrogen is having a half filled configuration but in the case of oxygen one of the proton orbitals has a fully filled orbital. Now what exactly happens? Over here in oxygen when this electron is being added the inter-electronic repulsions increase and the effect of inter-electronic repulsion outweighs the effect of the added, or the increased nuclear charge. So that is why the size of oxygen increases as compared to nitrogen, because in nitrogen no such inter-electronic repulsions are happening. So it follows the general trend that we have been discussing but in oxygen because of this repulsion the discrepancy occurs. Now if I move ahead with fluorine, now for fluorine the electronic configuration is this. Now over here if you look carefully the proton orbitals, there are two proton orbitals which are fully filled. So that means there are two inter-electronic repulsions happening over here. But interestingly the two inter-electronic repulsions, the effect of that does not outweigh the effect of the increased nuclear charge because of which the size of fluorine decreases. So that is how it should have been nitrogen, oxygen and fluorine in the decreasing order but now it becomes this. Oxygen gets a higher atomic radius. So this is the discrepancy. Now in every other period you will have the same trend where I explained to you that from left to right, the atomic radius will decrease.
Now if we discuss what will happen to the atomic radius in the group, just by applying a little bit of common sense we can find out the answer. We don’t need to do anything, because we know when we are going down the group one shell is increasing. Now if one shell is increasing the atomic radius has to increase. So that is why when we are going down the group the atomic radius always increases. Now, one more thing – ne, what would you say? Neon should have the smallest size or it should have a larger size as compared to all other? Yes, neon will have a larger size and not a smaller size because for neon you cannot find the covalent or the metallic radius. You have to actually find out a radius which is called as the van der Waal radii. Because neon will not form covalent bonds nor metallic bonds. So what is a van der Waal radii? You have 1 atom of neon, another atom of neon. They are held together by weak van der Waal forces. So they will not be touching each other, they will not be overlapping – nothing of that sort will happen. They would just be present along each other, alongside each other and you would have the van der Waal forces of attraction. Now over here you just take the inter-nuclear distance and you divide it by 2. You would get the radius of neon. But because there is some space between the two atoms, that is why the atomic radius or the van der Waal radii of neon would be a larger value. So when we are going from left to right in a period I will restate from left to right in a period the atomic radius would decrease till group 17 and then for the noble gases it would increase. That is the periodic trend for atomic radius.
Now let us discuss the ionic radius. Now before that let’s say what ions are. Now if I remove an electron from the atom what charge would it have? It would have a positive charge. So that is what is called as a cation and if I add an electron it becomes an ion. So what we will be discussing over here is the cationic radius and the ionic radius. Now hypothetically speaking let’s take sodium, let me add one electron to sodium and let me remove one electron from sodium. Right? So Na- is practically impossible but hypothetically I am taking it just to explain. Right? So sodium has an electronic configuration (2, 8, 1). Na- would have an electronic configuration of (2, 8, 2). And Na+ would have an electronic configuration of (2, 8). Right? So now out of Na and Na+ which would have a smaller size? Obviously it would be Na+ because Na has 3 shells and Na+ has only 2 shells. So since one shell has completely been removed the size of Na+ becomes smaller. Right? Now what happens with Na-? Now for sodium minus the number of electrons is (2, 8, 2) and it is (2, 8, 1). Now if you recall, what I said in atomic radius when the electrons are being added then the atomic radius should decrease. So that means with sodium minus the atomic radius should be smaller as compared to sodium, but it’s not actually true. Because, over there, in atomic radius, by the increase of one electron at the same time one p was also added. If you take a look over here sodium has 11 protons and sodium minus also has 11 protons. So what happens? Now let us say that there are 11 ropes holding onto 11 hooks for sodium. What will happen in the case of Na-? There will be 11 ropes which are holding on to the 11 electrons and the one electron that I am adding, how will it be attracted to the nucleus? What will the nucleus do? Nucleus will take some of the power from all the ropes and make another rope and give it to that electron. so, the force of attraction between the nucleus and all the electrons now have reduced with a small value. So, the size increases. So that means whenever an electron is added to the atom, i.e., the anionic radius would always have a larger value, then it would be the neutral atom and finally it would be the cationic radius. So cations would always be smaller than the atom and the anions would always be larger than the atom. Right? So this is the crux of the ionic radius.
Now you must be wondering, I made the cationic radius look so simple. Now let’s complicate it. Right? Now if I take the example of magnesium, Mg+ and Mg2+ and I ask you to arrange them in increasing order of ionic radius. How will you do that? Now for magnesium the number of electrons or the electronic configuration that you would have is (2, 8, 2). For Mg2+ it would be (2, 8, 1), and Mg2+ it would be (2, 8). Now easily I can say in all the 3, since Mg2+ has only 2 shells, so Mg2+ would be the smallest. So I can say that it is the smallest. Now I am left with Mg and Mg+. Now Mg+ is a cation but not in the complete oxidation state, right? So now for Mg the number of protons is 12 and for Mg+ the number of protons is 12. But the number of electrons for Mg is 12 and the number of electrons for Mg+ now is 11. Right? So what happens? We arrange the 12 electrons like this and the 11 electrons like this. This is Mg and this is Mg+. Now over here the 12 ropes are holding on to the 12 electrons but over here there are 12 ropes to hold onto only 11 electrons. So what happens? They generally have the ropes holding on to each of the electrons but now since there is one more rope extra. So what happens? That rope gets divided equally into 11 parts and add up to all the ropes that are actually there. So what will happen? These ropes which are adding up, this would increase the attractive pull on the electrons and when the attractive pull on the electrons increase the size will become smaller. So that is how the cationic radius becomes smaller as compared to the atoms. Now if I talk about the periodicity, then as you go down the group, what will happen? Since the shell would keep on increasing in the same group and if you remove the electrons the effect would be common in all of them. So as you go down the group the ionic radii would always increase. Right?
Now let’s complicate it further. Now I will talk about the isoelectronic species. Now for isoelectronic species, there the number of electrons would be equal but the elements would be different. Now let’s take an example where I have N3-, Al3+, Mg2+, O2-, F- and Na+ and if I ask you to arrange them in increasing order of the ionic radii, what shall your answer be? Think. Now for such questions, your approach should be that whatever atoms are given to you or whatever ions are given to you just write the number of electrons and protons below that. So, for N3- the number of electrons would be 10, that is 7 + 3. For Al3+ would be 10, for Mg2+ its 10 and for all of them it would be 10. That is why they are isoelectronic species. but at the same time if you take a look at the number of protons, for nitrogen, the number of protons would be 7, for aluminum it would be 13, for magnesium it would be 12, for oxygen it would be 8, for fluorine it would be 9 and for sodium it would be 11. Now you have to figure out that which would have more attractions and which would have lesser attractions. Now for the approach of this answer you can even think that cations would always have a smaller size. Right? So let’s talk about Al3+, Mg2+ and Na+ separately and figure out how they will be arranged. Now sodium has 11 protons pulling onto 10 electrons. Right? So that means there are 11 ropes pulling 10 electrons. In the case of Mg, what happens? There are 12 ropes pulling on to 10 electrons and in the case of aluminum there are 13 ropes pulling onto the 10 electrons. So where do you think the pull will be more? That is in the case of Al3+. So Al3+ would be the smallest, followed by Mg2+ and Na+. So this is how you arrange the cations. Right? Now let us talk about the O2-, F- and N3-. For N3- you have 7 ropes pulling onto the 10 electrons, for O2- you have 8 ropes pulling onto 10 electrons and for F- you have 9 ropes pulling onto the 10 electrons. So in which case would the attractive force be more? Yes, the attractive force would be more in the case of F- so F- would be smaller followed by O2- and then finally N3-. So this should be your increasing order of the ionic radius. Now if you take a look very carefully what happens? Can I generalize a statement for isoelectronic species – more the cationic charge, smaller would be the radius, and more the ionic charge, larger would be the radius. So if you keep this in mind, you don’t have to do all these calculations. You can just simply find out
Now let me give you a very complicated question and if you do this question within 2 minutes then probably you have understood what atomic radius is and what ionic radius is in complete details. So remember, 2 minutes. Right? So this is the question that you need to solve.
Now the first thing you have to do is, you find out the number of electrons and protons in each of the species. So the number of electrons and protons is this, right? Now if you notice very carefully there are 3 sets of isoelectronic species over here. Where C4+ has only 2 electrons so that has to be the smallest, so that is the smallest. Now we are left with 2 sets where isoelectronic number is 10 or 18. So we know that the isoelectronic species with electrons as 10 would be smaller as compared to the ones with 18 because they would have 3 shells, and the ones with 10 would have 2 shells. So we will arrange the isoelectronic species which have 10 electrons separately and then the numbers with 18. So for the 10, in the previous question we already discussed that the cations which have the largest charge would be the smallest so you will have Al3+ placed first, then you would have Mg2+, Na+ followed by F-, O2- and N3-. Right? So this is the 10 isoelectronic electrons. Now you have Sulphur, chlorine, calcium and potassium, so over there also if you check the charges then you can say that the cations would be placed first so that means Ca2+ because it has a 2+ charge would be smaller as compared to K+ and when you talk about the anions then Cl- would be smaller as compared to S2-. Now if I arrange them all in the increasing order of their ionic radius, so this is how it should be arranged. So that is the answer.
Now the next property that we have is the ionization enthalpy. It is also referred as the ionization potential or the ionization energy. All the 3 terms are for the same thing. Now, in brief if I just tell you what ionization enthalpy is, it is the amount of energy to remove an electron and form the cation of that particular atom. Now if you go by the scientific definition, it says, that it is the amount of energy required to remove the most loosely bound election from the outermost shell of an isolated gaseous atom. Now 3 points have to be taken care of over here, that it is the most loosely bound electron from the outermost shell and an isolate gaseous atom. So these 3 words should be there in your definition all the time because these are the 3 key points. Now why the most loosely bound? Because that is the first electron that we can remove easily. From the outermost shell – why? Because the first electron that we can remove would be the last e that is there in the atom. And why an isolated gaseous atom? Because, if suppose, you have a covalent compound and if you remove one electron from a covalent compound, what will happen? You are distorting the octets of both the atoms. so that is why it is always the isolated gaseous atom that you make a gas, isolate an atom from there and that is the time you can remove an electron so that is what is ionization enthalpy.
Now if we take an example of say aluminum. Now aluminum has the electronic configuration as (2, 8) and 3. Now if I have to form (2, 8, 2), i.e., I need to form Al+ I am removing one electron. so the amount of energy that would be required to remove that electron is what is called as the ionization enthalpy. Right? Now, let me form one more cation that is Al2+. Right? So what will happen? In Al2+ I would remove one electron from Al+. So my electronic configuration would now be (2, 8 and 1), instead of being (2, 8, 2). Right? And then if I remove one more electron from the outermost shell my electronic configuration comes out to be (2, 8). Right? Now, can I form, or can I remove one more electron from the outermost shell of (2, 8)? It’s not possible. Because the octet has been attained and if the octet is achieved no matter how high energy you supply it would not remove its electron. So the maximum ionization that we can do for aluminum is 3. Right? So the energy required from Al to Al+ is what is called as the ionization enthalpy I. From Al+ to Al2+ the amount of energy required is called as the ionization enthalpy II, and from Al2+ to Al3+ the amount of energy required is called as the ionization enthalpy III. So this can be denoted by IE1, IE2 and IE3 and IE4 for aluminum would not be possible because the octet has been achieved. Now if I ask you that at which point is the ionization enthalpy the highest? Will it be IE1, or IE2 or IE3? Yes, so IE1 would be the lowest and IE3 would be the highest. Why? Because it’s basically what you are trying to do is you are trying to remove the electron by breaking the force of attraction in between the nucleus and the electron. So if the force of attraction is very high between the nucleus and the electron then what will happen? A higher amount of energy is required. Right? So now in aluminum and comparing it to Al+ where do you think the size is smaller? The size is smaller in the case of Al+ because that is what we have started from the ionic radii. So if the size is smaller, then the amount of attraction on the outermost electron in Al+ would be greater as compared to Al. so that is why the ionization enthalpy I and ionization enthalpy II, ionization enthalpy II is greater. And similarly when you move on to Al2+, or Al3+ the size keeps on decreasing and as the size keeps on decreasing the force of attraction between the nucleus and the electrons keeps on increasing. So since the force of attraction is increasing the ionization enthalpy would also increase. So there are few factors upon which the ionization enthalpy depends. The first factor is the size of the atom. Now the size of the atom, we have just discussed, smaller the size more would be the ionization enthalpy. The second factor on which ionization enthalpy depends is the nuclear charge. So higher the amount of nuclear charge, i.e., higher the number of protons, what will happen? There would be a higher attractive force for the outermost electron and we would require more energy to remove that electron. Right? So these are the two factors which are important.