Magnetic Moments

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. Nobody can see electrons, right? And yet a chemist can put an iron complex in a magnetic field, measure just one number, five point nine two, and then tell you whether the iron inside is iron two or iron three. How is that even possible? Today you will learn to do it yourself. Questions on magnetic moments come up often in JEE, so let's begin.

0:29Paramagnetic or diamagnetic?

So, first, a basic question. Electrons in a complex can sit in pairs, or alone, unpaired. Which ones make the complex respond to a magnet, the paired ones or the unpaired ones? Yes, exactly, it is the unpaired ones! Any substance with unpaired electrons is called paramagnetic, and it is attracted into a magnetic field. Even one unpaired electron is enough. Okay? And when all the electrons are paired, the substance is diamagnetic, which means it is weakly repelled by the field. So, paramagnetic is pulled in, and diamagnetic is pushed away, weakly. Right? Now one more point from the course. For complexes of the three d metals, the orbital contribution to the moment is largely quenched, which means mostly switched off. So the measured moment is close to the spin only value, the part from the unpaired spins. Clear?

1:24The spin-only formula

Now, the formula. The Greek letter mu, with a small s below it for spin only, is the moment. mu s equals the square root of n times n plus two, with n plus two in brackets, in Bohr magnetons, written B M on screen. Okay? And here, n is the number of unpaired electrons. Only the unpaired ones, remember, not all the d electrons. Got it? So let us try the smallest case. If a complex has just one unpaired electron, what is its spin only moment? Very good if you said root three! One times three is three, and the square root of three is one point seven three Bohr magnetons. With me? In the same way, two, three, four and five unpaired electrons give two point eight three, three point eight seven, four point nine zero, and five point nine two. Learn these five numbers by heart. Right? Now let us go backwards, because that is how questions usually come. The spin only moment is four point nine zero Bohr magnetons. How many unpaired electrons are there? Yes, the answer is four, correct! Here is a quick trick: square the moment. The wavy equals sign means approximately, so four point nine zero squared is about twenty four. And four times six is twenty four, so n is four. Clear?

2:52High spin or low spin: only d⁴ to d⁷

Now, in an octahedral complex, the same metal ion can end up with more unpaired electrons, or with fewer. We call these high spin and low spin, and the course says only d four to d seven can be either one. Here d four means four electrons in the d orbitals, the small four written up top. Okay? So who decides? The ligand decides. With a weak field ligand, the electrons are not forced to pair, so they spread out before any pairing, and we get high spin, with more unpaired electrons. Right? And a strong field ligand forces the electrons to pair up, so we get low spin, with fewer unpaired electrons. Got it? Now the numbers from the course. The first number is for high spin, and the second, after the slash, is for low spin. For d four, high spin has four unpaired and low spin has two. For d five, five and one. With me? For d six, high spin has four unpaired, and low spin has zero. And for d seven, high spin has three, and low spin has one. Clear? Now be careful here. Suppose a d six complex is low spin. What is its spin only moment? Many students put n equal to six here, and that is the trap! Low spin d six has zero unpaired electrons, so n is zero, the moment is zero, and the complex is diamagnetic. Always count unpaired electrons only. Okay?

4:21Example 1: Fe(II) or Fe(III) from a moment of 5.92 BM

Now, remember our puzzle from the start? Example one. An iron complex has mu equal to five point nine two Bohr magnetons. Is it iron two or iron three, written as F e with two or three in Roman numerals? And is the complex high spin or low spin? Pause the video and try it first. Done? Okay, let's check. Five point nine two squared is about thirty five, which is n times n plus two when n is five, because five times seven is thirty five. So there are five unpaired electrons. Right? Now, after the argon core in square brackets, iron has six electrons in three d and two in four s, counted by the small numbers up top. Ions lose the four s electrons first. So iron two has six d electrons, and iron three has five. With me? So let me ask you this. Can iron two, with six d electrons, ever give five unpaired electrons? The answer is no, it cannot. Remember, d six has at most four unpaired electrons, and that is in high spin. But we need five, so it is not iron two. Okay? So it must be iron three, with five d electrons. And d five gives five unpaired electrons only in high spin. So the answer is iron three, high spin. See? One number from a magnet told us the oxidation state and the spin, and that solves our puzzle. Got it?

5:55Example 2: find the d count from 2.83 and 4.90 BM

Now a harder one, Example two. An octahedral complex of a three d metal in the plus three state has a spin only moment of two point eight three Bohr magnetons with a strong field ligand, and four point nine zero with a weak field ligand. Find the number of d electrons, and name one such metal. Pause the video and try it first. Done? Okay, let's check. Strong field means low spin. Two point eight three squared is about eight, and two times four is eight, so n is two. Okay? Weak field means high spin. And four point nine zero, as we already know, gives n equal to four. Right? Now remember our table from before. Four unpaired in high spin happens twice, for d four and for d six. So which one is it? If you said d six, careful, that is the trap! We need four in high spin and two in low spin, four slash two. But d six gives four slash zero, so its low spin form has zero unpaired, not two. With me? Only d four fits both: four unpaired in high spin, and two in low spin. Exactly the pair from the course table. Got it? Last step. Which three d metal, in the plus three state, has four d electrons? The answer is manganese. Manganese, M n, is three d five, four s two. Remove the two four s electrons and one three d electron, and M n three plus, with the charge written up top, is three d four. Clear? So the answer is d four, and one such metal ion is M n three plus. Two moments together pinned down the number of d electrons. Okay?

7:48Quick revision: para vs dia, root n(n+2), high and low spin

So, quick revision. Unpaired electrons make a complex paramagnetic; all paired makes it diamagnetic. The spin only moment is the square root of n times n plus two Bohr magnetons, from one point seven three for one unpaired electron up to five point nine two for five. And from d four to d seven in an octahedral complex, high or low spin changes n, so always count the unpaired electrons. Now go and try the practice problems on this page. All the best!