Magnetic Properties

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. Take two ions of iron, iron two plus and iron three plus. Iron three plus has lost one more electron, so it has fewer left. So which one has the bigger magnetic moment? Surely the one with more electrons, iron two plus, right? Hmm, think again. By the end of this video, you will calculate it yourself. Today we study the magnetic properties of transition metal ions. This comes up often in JEE, and it is quick marks once you know one formula. So, let's begin.

0:38Paramagnetic, diamagnetic, ferromagnetic

Let me ask you this. Which electrons decide whether a substance is drawn into a magnetic field, the paired ones or the unpaired ones? Yes, exactly, the unpaired electrons! Any substance with unpaired electrons is paramagnetic, which means it is drawn into a magnetic field. Okay? And if every electron is paired, the substance is diamagnetic. Right? Now a special group. Iron, cobalt and nickel, written F e, C o and N i, and the oxide C r O two, where the small two below counts the oxygen atoms, are ferromagnetic. They show strong, permanent magnetisation. Clear?

1:17The spin-only formula

Now the one formula of this lesson. Mu, the Greek letter you see on screen, is the magnetic moment. Mu equals the square root of the whole product, n times n plus two, with n plus two in brackets. The tick shaped sign over that product is the square root, and the B M after it stands for Bohr magnetons, the unit. Okay? And n here is the number of unpaired electrons. Only the unpaired ones count, not the total. That is the whole trick of this topic. With me? Why is it called spin-only? For three d ions, the orbital contribution is ignored, and only the spins of unpaired electrons count. That is the condition behind this formula. Got it?

1:59Five values worth remembering

Let us put in numbers. For one unpaired electron, n is one, so mu is the square root of one times three, root three. The wavy equals sign means approximately equal, so that is about one point seven three Bohr magnetons. Right? Now you try one. Two unpaired electrons, so n is two. What is mu? Yes, exactly! Two times four is eight, and root eight is about two point eight three Bohr magnetons. Okay? In the same way, three, four and five unpaired electrons give root fifteen, root twenty four and root thirty five. That is three point eight seven, four point nine zero, and five point nine two Bohr magnetons. Learn these five values by heart. Clear? And here is a topper's trick for going backwards. If a question gives you mu, square it; the small two raised on mu means squared. Mu squared comes out very close to n times n plus two, that is three, eight, fifteen, twenty four or thirty five, for n from one to five. Got it?

3:03Counting unpaired electrons: Hund's rule

Now, how do we count the unpaired electrons? We use Hund's rule. There are five d orbitals, and the electrons go in singly first, before any pairing starts. Okay? Look at d five, with the five written small and raised, meaning five d electrons. Each box on screen is one d orbital, and each up arrow is one electron. The double arrow means this gives n equals five. Right? Now add a sixth electron. All five boxes already have one. So how many unpaired electrons are left? Four, yes exactly! The sixth electron has to pair up in the first box. An up and a down arrow in one box are a pair of electrons. So d six has only four unpaired electrons, not six. One more electron, but one fewer unpaired electron. With me? So the pattern is this. From d one to d five, n goes one, two, three, four, five. Then from d six to d nine, each new electron pairs up, so n comes down, four, three, two, one. Okay? Now a trap. Zinc two plus is three d ten, ten d electrons. So it must have the biggest moment of all, right? Oh no, careful, ten electrons fill all five boxes completely, two in each, so every electron is paired! Then n is zero, and zinc two plus is diamagnetic. And d zero, with no d electrons at all, is diamagnetic too. Clear? One condition to remember. This counting holds for free three d ions, or ions with weak-field ligands, which we call high spin. Got it?

4:44Example 1: spin-only moments of Fe2+ and Fe3+

Now, Example one, and it is our puzzle from the start. Find the spin-only magnetic moments of iron two plus and iron three plus. The two plus and three plus, written small and raised, are the charges, okay? Pause the video and try it first. Done? Okay, let's check. Iron is argon, three d six, four s two; the argon in square brackets is the argon core. Remember from the last chapter, when an ion forms, the four s electrons leave first. Right? So iron two plus loses its two four s electrons, and it is three d six. Six d electrons, so four unpaired, as in our boxes. Mu is root twenty four, about four point nine zero Bohr magnetons. Okay? Iron three plus loses one more electron, from the three d, so it is three d five. How many unpaired electrons does it have? Yes, it has five unpaired electrons, one in each box! So mu is root thirty five, about five point nine two Bohr magnetons. So our puzzle is solved: iron three plus has fewer electrons but more unpaired ones, so it has the bigger moment. With me?

5:57Example 2: which ions have a moment of 3.87 BM

Now, Example two, a little harder. Which of vanadium two plus, chromium three plus, manganese two plus and cobalt two plus have a magnetic moment of three point eight seven Bohr magnetons? Pause the video and try it first. Done? Okay, let's check. Use the topper's trick and go backwards. Three point eight seven squared is about fifteen, and fifteen is three times five. That is n times n plus two with n equal to three. So we need exactly three unpaired electrons. Okay? Vanadium is three d three, four s two. Remove the two four s electrons, and vanadium two plus is three d three; the arrow means becomes. Chromium is three d five, four s one. Remove the four s electron and two d electrons, and chromium three plus is also three d three. Three unpaired electrons each, so both fit. Right? Manganese two plus is three d five, just like iron three plus. That is five unpaired electrons, not three, so manganese two plus is out. Clear? Now cobalt two plus. It has seven d electrons. Seven is far too many for three unpaired, so it is out as well, right? Oh no, careful, that is the trap! Cobalt is three d seven, four s two, so cobalt two plus is three d seven. Now fill the boxes. The first five electrons go in singly, and the sixth and seventh pair up. So two boxes are paired and three are single, and for high spin cobalt two plus, n is three. With me? So the answer is vanadium two plus, chromium three plus and cobalt two plus. Different d counts, but the same three unpaired electrons. Got it?

7:38Quick revision: paramagnetism, the formula, d1 to d10

So, quick revision. First, unpaired electrons make a substance paramagnetic, and when all are paired, it is diamagnetic. Second, mu equals the square root of n times n plus two Bohr magnetons, where n is the number of unpaired electrons. And third, n rises from d one to d five, falls from d six to d nine, and d zero and d ten have none. Now go and try the two practice problems on this page. You will find them easy. All the best!