The Lanthanoids: Configurations and Oxidation States

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Europium and gadolinium are next-door neighbours in the periodic table, with atomic numbers sixty three and sixty four. Yet pulling out the third electron is hard for europium, and much easier for gadolinium. So why such a big difference? We will settle it in Example two. Today we learn the configurations and oxidation states of the lanthanoids; they come up often in JEE, so let's begin.

0:30Who are the lanthanoids?

So, who are the lanthanoids? They run from cerium, written C e, to lutetium, written L u. Z is the atomic number: fifty eight for cerium, and seventy one for lutetium. That makes fourteen elements. Okay? Now the general configuration. The square brackets with X e inside mean the xenon core. Then comes four f, with a raised one to fourteen, then five d, with a raised zero to one, and then six s two. The raised numbers count the electrons in each subshell, and the dash means a range. Right? Now, many students write every lanthanoid as four f n, five d one, six s two, where n is the number of f electrons. Is that right? No, it is not, and this is a common trap! Most lanthanoids have no five d electron at all; they are simply four f n, six s two. Europium, written E u, is four f seven, six s two, and ytterbium, written Y b, is four f fourteen, six s two. Got it? So which atoms do have a five d electron? Only four of them. Lanthanum, L a, the element just before cerium, is five d one, six s two, and cerium is four f one, five d one, six s two. With me? The other two are gadolinium, G d, four f seven, five d one, six s two, and lutetium, four f fourteen, five d one, six s two. Here is a topper's trick: gadolinium has seven f electrons, half filled, and lutetium has fourteen, completely filled. So remember lanthanum, cerium, the half-filled one, and the full one.

2:20The +3 ions, Ln³⁺

Clear? Now the ions. L n stands for any lanthanoid, and the raised three plus is its charge. Every L n three plus ion is simply xenon, four f n, with nothing left in five d or six s. Okay? So, here is a question for you. When a lanthanoid atom loses electrons, which go first: the four f electrons, or the six s electrons? The six s electrons go first, yes! Remove six s first, then the five d electron if there is one, and only then four f. Remember this order. Right? Let us try it. Lanthanum loses all three outer electrons, so L a three plus is four f zero. Cerium loses its two six s electrons and its five d electron, so C e three plus is four f one. With me? In the same way, gadolinium three plus is four f seven, and lutetium three plus is four f fourteen. Half filled and completely filled again, so keep seven and fourteen in mind. Clear?

3:27Oxidation states: +3, and sometimes +2 or +4

Now, oxidation states. For the lanthanoids, plus three is the characteristic state, the one to expect first. Okay? But plus two and plus four also occur, and here is the pattern. They occur mostly when they give f zero, f seven or f fourteen: an empty, a half-filled, or a completely filled f subshell. It is a guide, not a strict law, as we will see in a moment. Right? So, europium two plus is four f seven, half filled. But plus three is still the stable state for the lanthanoids. So will europium two plus act as an oxidant, or as a reductant? Yes, exactly, a reductant! To reach plus three, europium two plus must lose an electron, so it reduces something else. Ytterbium two plus, which is f fourteen, does the same; both are strong reductants. Samarium two plus, written S m two plus, is also a reductant, even though it is f six, so the f zero, f seven, f fourteen rule is only a guide. Got it? Now flip it around. Cerium four plus is f zero, and terbium four plus, written T b, is f seven. To reach plus three, each must gain an electron, so they oxidise other things. Both are strong oxidants. Clear? And here is a number to back that up. E with a small circle, read E naught, is the standard electrode potential; the slash separates cerium four plus from cerium three plus. Its value is plus one point seven four volts, large and positive, so cerium four plus is a strong oxidant. With me?

5:05Example 1: configurations of Ce3+, Ce4+, Eu2+ and Yb3+

Now, Example one. Write the configurations of cerium three plus, cerium four plus, europium two plus and ytterbium three plus. Pause the video and try it first. Done? Okay, let's check. Cerium is four f one, five d one, six s two. Remove the two six s electrons and the five d electron, and cerium three plus is four f one. Cerium four plus loses one more, from four f, so it is four f zero. Okay? Europium is four f seven, six s two. Two electrons leave, both from six s, so europium two plus is four f seven, half filled. Right? Now ytterbium three plus, and remember, ytterbium has no five d electron. Two electrons come from six s. So where does the third one come from? It comes from four f, yes, exactly! Ytterbium is four f fourteen, six s two, so ytterbium three plus is four f thirteen. If you wrote four f fourteen, that is the trap; it is the two plus ion that keeps f fourteen. Got it? So, the answer, after the xenon core, is four f one, four f zero, four f seven, and four f thirteen. Clear?

6:29Example 2: third ionisation enthalpy of Eu high, Gd low

Now, Example two, our puzzle from the start. Why is the third ionisation enthalpy of europium high, but that of gadolinium low? Pause the video and try it first. Done? Okay, let's check. The third electron comes out of the two plus ion. So what is europium two plus? We found it in Example one. Yes, it is four f seven! Europium two plus is xenon, four f seven, half filled. So the third electron must come out of that stable four f seven and break it. That is why the third ionisation enthalpy of europium is high. Okay? Now gadolinium. It loses its two six s electrons first, so gadolinium two plus is xenon, four f seven, five d one. So the third electron is the five d electron, and the four f seven is not touched. Right? When that electron leaves, gadolinium three plus is four f seven, the stable half-filled subshell. Losing the five d electron reaches four f seven, so the third ionisation enthalpy of gadolinium is low. With me? So the answer is the stability of four f seven. Europium has to break it, while gadolinium reaches it. And that solves our puzzle: next-door neighbours, on opposite sides of the same half-filled subshell. Clear? The same idea works for f fourteen. The third ionisation enthalpy of ytterbium is high, because the electron comes out of f fourteen, while that of lutetium is low, because losing it reaches f fourteen. Got it?

8:11Quick revision: configurations, +3 state, order of removal

So, quick revision. First, the lanthanoids are xenon, four f one to fourteen, five d zero to one, six s two, and only lanthanum, cerium, gadolinium and lutetium have the five d electron. Second, plus three is characteristic, with plus two or plus four mostly when that gives f zero, f seven or f fourteen. And third, for any ion, remove six s first, then five d, then four f. Now go and try the two practice problems on this page. All the best!