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Ch.7 - Periodic Properties of the Elements

Chapter 7, Problem 88

Note from the following table that there is a significant increase in atomic radius upon moving from Y to La, whereas the radii of Zr to Hf are the same. Suggest an explanation for this effect. Atomic Radii (pm) Sc 170 Ti 160 Y 190 Zr 175 La 207 Hf 175

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hey everyone in this example, we need to provide a possible explanation for why the atomic grade I of niobium and tantalum are almost the same even if they're on different periods in the periodic table. So we should recall that for our trend of atomic ready or atomic radius on the periodic table. Sorry, that should have won us. We're going to recall that it's going to be increasing as we move from the left towards the bottom right of our periodic table. However, according to this problem, if we look at niobium compared to tantalum, we would not have this trend apply because niobium and tantalum are an exception to the atomic radius trend since their atomic raid. I are technically close to one another, although they're in different periods, We would recognize that niobium is in period three Or sorry, period five of our periodic tables. And tantalum we would recognize is in period six of our periodic tables. So technically tantalum should have a higher atomic radius since it's lower on the periodic table. Whatever we're going to prove why it does not. So what we want to recall is the concept of effective nuclear charge. And we should recall that it's represented by this symbol here and we would calculate it by taking our symbol for our atomic number, which we recall is easy and we're going to subtract that from our screening constant which is represented by a capital S. We should recall that our screening constant is ultimately referring to our number of core electrons in our configuration for our given atom. So we'll go ahead and represent these values by different colors, so that it's not confusing when we get our or will we plug in the values here? So to find first the effective nuclear charge of niobium, we're going to begin with niobium atomic number. So we should recall that niobium has an atomic number according to the periodic table in the transition metal D block of 41. And we should recall that for neutral atoms only our atomic number is going to equal our number of protons and equal our number of electrons in this prompt, we are not given the niobium atom with a charge, so it is a neutral atom. So we would say that Since we have 41 protons, we therefore have 41 or since we have an atomic number of 41, we therefore have 41 protons And 41 electrons. So to calculate for a screening constant, we would plug in for Z. Which we just stated is going to be a value of and this is going to be subtracted from our screening constant. So we need to determine our number of core electrons and how we would get the number of core electrons is by analyzing our configuration for niobium. So for the shorthand configuration or noble gas configuration of niobium, we would pick the noble gas that comes before niobium on the periodic table. So that would correspond to the noble gas krypton. So we put that in brackets to begin its configuration and then we go through the five S to sublevel because niobium is on the fifth period. And because niobium is in the D block transition metal section, we would recall that in the transition metal section. Niobium begins at the fourth energy level. So we would have four D and we would count in the D block a total of three units to land on our adam niobium. However, we should recall that because of our alphabetic principle, we're going to actually have a configuration where we're going to recall that we should fill in our lower energy orbital's first before filling our higher energy level ones. So we want to fill in this fourth energy level at least fully or halfway. So how we can do that is by exciting one of our electrons from our fifth energy level here. So that we now instead have five S. One And then four D 4 because that electron was excited down to the fourth energy level to make our d orbital a bit more than halfway, a bit less than halfway filled. So we can say almost halfway filled. And so what we can correct this electron configuration to is now krypton and you want to write out the lower energy level first. So we would have four D 45 S one since Niobium has an electron configuration exception since it's in the D block. And so to figure out our core electrons, we want to recall that our valence electrons are going to be our outermost energy level electrons. However our core electrons are going to be all the energy levels that come before that. And so we should recall that krypton according to its atomic number has an atomic number of 36. So we would say that krypton therefore will have 36 electrons. And then because we recognize this X one here of four, we have four more electrons coming from the fourth energy level. So we would add to the 36 electrons for electrons. And this is going to give us our core electrons equal to a value of 40. And we should recall that this is going to equal our screening constant value. So to calculate our effective nuclear charge of niobium, we're going to subtract 40 and this difference gives us an effective nuclear charge equal to a value of one. So we're going to compare our effective nuclear charge of tantalum to this value. So again, we want to recall that tantalum on the periodic table which is located across period six. And the D block transition metal section has an atomic number equal to a value of 73. So we would have 73 And we'll just use purple to keep things consistent. So this is our atomic number 73. Subtracted from our core electrons which we get from our configuration of tantalum and we would recall that tantalum has a configuration of first the noble gas seen on because it comes before tantalum on the periodic table, We would recognize that because tantalum is in the third row of our D block, it's going to have a fully filled in F block. And we would recall that the F block begins at the fourth energy level. So we have four F. And because it's fully filled in, it should have it should be four F 14. So we fully fill this in to get to tantalum configuration and we continue on and move up to the five D three sub level where in the D block we count for three units to get to the atom tantalum and we also recognize that we passed through the six s to sub level. Since we would recall that on the fifth energy level of our D block, we are also at the sixth energy level of our S block. So for our valence electrons, that's going to be the outermost energy level here, being the sixth energy level. And then for our core electrons, We're going to recognize that Xenon on our periodic tables has the atomic number 54, which also stands for 54 electrons. We're going to add the 14 electrons plus the three electrons in our exponents of our Lower energy levels. And this is going to give us a total of 71 core electrons. And so we would plug that in for our calculation of effective nuclear charge. And this difference here gives us a Effective nuclear charge for tantalum equal to a value of two. So what we can conclude is that our effective nuclear charge of niobium is less than the effective nuclear charge of our adam tantalum. We should recall that the adam with the higher effective nuclear charge value is going to have a smaller atomic radius oops so atomic radius. So because we recognize that tantalum here has a higher effective nuclear charge value of two, we're going to recognize that it's going to have a smaller atomic radius and to explain why that is going to be due to the fact that it has a fully filled in F block At the 4th energy level here With its 14 electrons filled in. And we should recall that this is going to not be able to shield our nucleus of our atom of tantalum effectively enough from the outer electrons. We should recall that effective nuclear charge is describing the distance between our protons and our nucleus of our atom from our electrons in focus which are the outer electrons. So we can start out our explanation by saying that tantalum has a fully filled in F block in its configuration. And so therefore our nucleus is not effectively shielded from the outer electrons. And so we would say that therefore the electrons our closer or rather we should say that the protons, the protons in the nucleus pull the electrons closer two, We can just say pull the electrons closer. And so we have a stronger grip of these electrons. The protons have a stronger grip of these electrons And so we will therefore have an effective nuclear charge that is higher. So the electrons being closer increases the effective nuclear charge and the radius of tantalum, we can say specifically atomic radius of tantalum decreases and so and becomes closer in size to the atom ni opium. So this explanation that we've outlined here is going to be our final answer to explain why our atomic radius is of our nay opium and adam tantalum are going to be similar in size instead of us, assuming that according to our trend, tantalum would have a greater nucleus. It's not going to have a greater nucleus, it's going to have or sorry, a greater atomic radius. It's not going to have a greater atomic radius, it's going to have a atomic radius that is either smaller or closer in value to niobium. Due to the fact that again we said that tantalum has a fully filled in f block. The f block does not effectively shield our nucleus from outer electrons. And so the protons in the nucleus have a stronger pull of our electrons pulling them closer here to itself. And this will therefore increase our effective nuclear charge of tantalum and decrease the size of our atomic radius of tantalum. So therefore it's going to have an atomic size closer to nao pia. And this explains why we have this exception to our trend of atomic radius on the periodic table for these two atoms. So I hope that everything I explained was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.
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