Unpaired Electrons and Magnetic Moment

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Suppose I give you just one number. An ion of the first transition series has a magnetic moment of three point eight seven B M, okay? From this one number, can you tell me how many unpaired electrons it has? And even name the ion? Sounds like detective work, right? By the end of this lesson, you will do it in ten seconds. And look, in JEE, unpaired electrons and magnetic moment come up often: one count, one formula, and the marks are yours. So, let's begin.

0:38Step 1: count the unpaired electrons

So, everything today starts with one count: the number of unpaired electrons. Remember the little box from the last lessons? The box is one orbital, and the arrow is one electron. An unpaired electron sits alone in its orbital, just one arrow in the box. Clear? And when two electrons share one orbital, one arrow up and one arrow down, opposite spins, they are paired. And remember how the electrons fill, by Hund's rule: in the orbitals of one subshell, they first go in singly, and only then do they pair up, okay? So, let's count. Take d five: the small five up top means five electrons in the d subshell. There are five d orbitals, five boxes, so all five electrons sit alone, one in each box. Five unpaired electrons. As simple as that. Now d eight. Eight electrons, more than five. So more unpaired electrons than d five, right? Maybe even eight unpaired? No, no, careful! There are only five boxes. The first five electrons go in singly, and the next three must pair up, in three of the boxes. So only two boxes still have a single arrow: two unpaired electrons. More electrons does not mean more unpaired electrons. This is a classic JEE trap! And d ten? Ten electrons, five boxes, every box full, one up and one down. Zero unpaired electrons. Remember this one, we will need it soon.

2:06Step 2: the spin-only magnetic moment

Now, the formula. The spin-only magnetic moment is mu, the Greek letter you see on screen. Mu equals the square root of n into n plus two. And the unit is B M, short for Bohr magneton, okay? Notice, here n is the number of unpaired electrons. Only the unpaired ones, not all the electrons. So first you count, just as we did, and then you put n into the formula. Got it? Quick check. Take two unpaired electrons, so n equals two. Then n plus two is four, and two into four is eight. So mu is eight B M, right? No, no! You forgot the square root. Mu is the root of two into four, that is, root eight. Two point eight three B M, not eight! Under exam pressure, students drop this root, so be careful, okay?

2:57The five values to remember

Now, these five values come again and again, so learn them by heart. n equals one: root three, one point seven three. n equals two: root eight, two point eight three, we just did it. n equals three: root fifteen, three point eight seven. Then n equals four: root twenty four, four point nine zero. And n equals five: root thirty five, five point nine two. All in B M, okay? Five unpaired electrons, like d five, give the biggest value on this list. Now a topper's trick. Learn these five numbers once. Then, when a question gives you the magnetic moment, just read the table backwards. Four point nine zero? That is n equals four. No algebra at all. That's it, done!

3:47Diamagnetic or paramagnetic?

Now, two names you must know. If a species has no unpaired electrons at all, then n is zero, and mu is root zero, that is, zero. Such a species is diamagnetic: it is weakly repelled by a magnetic field. And if there is even one unpaired electron, the species is paramagnetic: it is attracted into a magnetic field. So, dia means zero unpaired, para means at least one. Simple, right? Now think. Take a d ten ion: ten d electrons, so many electrons! Surely it is paramagnetic? Oh no, careful! Remember our boxes: in d ten, every electron is paired. So n is zero, and the ion is diamagnetic. Dia or para depends only on the unpaired electrons, never on the total. With me?

4:36Example 1: unpaired electrons from 3.87 BM, and one such ion

Now, remember our puzzle from the start? Let's crack it. An ion of the first transition series has a magnetic moment of three point eight seven B M, okay? How many unpaired electrons does it have? And name one such ion. Pause the video, and try it first. Done? Okay, let's check. Mu is the root of n into n plus two. To remove the root, square both sides. Three point eight seven squared is very nearly fifteen; the wavy equals sign means approximately. So, n into n plus two equals fifteen. Now, which whole number n gives fifteen? Three, yes! Three into five is fifteen. The double arrow means so: three unpaired electrons. And with our topper's trick, you didn't even need this step. Three point eight seven is right there in the table: n equals three. Ten seconds, as I promised! Now the second part: name one such ion. Which ion of the first transition series has three unpaired electrons? Chromium three plus is one! Chromium, C r, is the argon core in square brackets, then three d five, four s one, our famous exception. The three plus up top means it has lost three electrons, and the arrow shows what it becomes. Ions lose the four s electron first, then two from three d. So chromium three plus is three d three: three electrons in five boxes, all alone. Three unpaired. Puzzle solved!

6:12Example 2: largest moment of Fe2+, Mn2+, Co2+, Cr3+

Now a slightly harder one, JEE level. Which ion has the largest spin-only magnetic moment: iron two plus, manganese two plus, cobalt two plus, or chromium three plus? Pause and try. Hint: count the unpaired electrons first. Done? Now, many students think like this. Cobalt two plus has the most d electrons of the four. So cobalt two plus has the largest moment, right? No, no, careful! It is the same trap as d eight. The moment depends only on unpaired electrons, so let's count. Iron two plus, F e two plus: ions lose their four s electrons first, so it is three d six. Five go in singly, the sixth one pairs up. Four unpaired. Manganese two plus: three d five. Five electrons, five boxes, all alone. So all five are unpaired! Cobalt two plus: three d seven. Five go in singly, and then two of them pair up. So only three unpaired. See? The most d electrons, but not the most unpaired! And chromium three plus: three d three, our answer from Example one. Three unpaired, the same as cobalt two plus. So, four, five, three and three. Which ion wins? Manganese two plus, exactly! Five unpaired electrons, so mu is root thirty five, five point nine two B M, the biggest value on our table. See, a half-filled three d gives the most unpaired electrons here.

7:51Quick revision: root n(n+2) BM, para or diamagnetic

So, quick revision. One: count the unpaired electrons with Hund's rule, and remember that ions lose their four s electrons first. Two: mu equals root of n into n plus two, in B M, and learn the five values: one point seven three, two point eight three, three point eight seven, four point nine zero, and five point nine two. Three: zero unpaired electrons means diamagnetic; n greater than or equal to one, that is, one or more, means paramagnetic. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!