Counting Electrons by Quantum Numbers

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Chromium has twenty four electrons, and every one of them carries its own address of four quantum numbers. Now the examiner asks: how many of these twenty four electrons have m sub l equal to zero? That is the magnetic quantum number, okay? Will you write out all twenty four addresses, one by one? No way! In the exam there is no time for that. Today you will learn to count them in under a minute, subshell by subshell. And look, in JEE, these quick counting questions on l and m sub l are quite common. So, let's begin.

0:42Recap: l and ml values of each subshell

So, a quick recap from the last chapter. To count electrons today, we need two quantum numbers. First, small l, the azimuthal quantum number. It names the subshell: l zero is s, one is p, two is d, and three is f. Clear? Second, m sub l: m, with a small l written below it, the magnetic quantum number. Its values run from minus l, through zero, up to plus l; the dots mean every value in between, in steps of one. And each value is one orbital, okay? Alright, tell me. Take a p subshell, so l equal to one. Which values can m sub l take? Minus one, zero and plus one. Exactly! Three values, so three p orbitals. Now notice one small thing, and keep it in your pocket. The range from minus l to plus l always passes through zero. So every subshell, s, p, d or f, has exactly one orbital with m sub l zero. Remember this. Later it will save you from a big trap.

1:49How to count: subshell by subshell

Alright, now the method. Step one: write the full configuration, every subshell from one s onwards, not the short noble gas core form. Why? Because the electrons hidden inside the core also have quantum numbers, and they count too! Step two: go subshell by subshell. Every s electron has l equal to zero, and m sub l equal to zero. An s subshell has only one orbital, right? So there is no other choice. Now a full p subshell, p six; the small six up top is the number of electrons, okay? How many of these six electrons have m sub l equal to zero? Two, yes! In a full subshell, every orbital holds two electrons, one spin up and one spin down. So in a full p, d or f subshell, each m sub l value has exactly two electrons. Really simple. Got it?

2:40Part-filled subshells: Hund's rule

Step four, and this is where students slip. Now, a part-filled subshell. Here Hund's rule decides: the electrons fill the m sub l values one at a time, one in each orbital first, before any pairing. You know this rule from the lesson on filling orbitals, right? Quick check. p three: three electrons in a p subshell. How many of them have m sub l equal to plus one? Just one! Three electrons, three orbitals, one each: one with minus one, one with zero, and one with plus one. With me? But which m sub l gets the first electron? By convention, you start either from plus l or from minus l. Both are used, so in your answer, state which order you followed. When every value gets one electron, like in p three, the order does not change the count anyway. Clear?

3:34Example 1: electrons with l=2 and l=0 in iron

Now, example one. How many electrons in ground-state iron have l equal to two? And how many have l equal to zero? Z is the atomic number, so Z equal to twenty six means iron has twenty six electrons. Pause the video, and try it first. Done? Okay, let's check. Iron, written F e on screen. Step one, the full configuration: one s two, two s two, two p six, three s two, three p six, three d six, four s two. Add them up, and you get twenty six. Right? Now, l equal to two. Which subshell letter is that? d, exactly! And iron has only one d subshell: three d six. So six electrons have l equal to two. Next, l equal to zero, and that means s. There is four s two at the end, so two electrons, right? No, no, careful! Every s subshell has l equal to zero, not just the outermost one. One s two, two s two, three s two, and four s two, okay? Two plus two plus two plus two: eight. So the answers are six, and eight. Now you see why step one says the full configuration, right? The inner electrons count too!

4:54Example 2: electrons with l=2 and ml=0 in chromium

Now a harder one, JEE level, and it is our puzzle from the start. How many electrons in a ground-state chromium atom have, part a, l equal to two, and part b, m sub l equal to zero? Pause and try. Hint: chromium is one of the exceptions, so be careful with its configuration. Done? Okay, let's check. Remember the exception from earlier? Chromium, C r on screen, has a half-filled three d. So: one s two, two s two, two p six, three s two, three p six, three d five, four s one. Quick check: the small numbers up top add up to chromium's atomic number. Right? Part a is quick. l equal to two means d, and the only d subshell is three d five. Five electrons. Now part b, m sub l equal to zero. Start with s: every s electron has m sub l zero. One s two, two s two, three s two, four s one: two plus two plus two plus one, seven. So the answer is seven, right? Many students stop right here. No, no, careful! This is the JEE trap. Remember the fact in your pocket? Every subshell has one orbital with m sub l zero, p and d included. Two p six and three p six are full, so each has two electrons with m sub l zero. Four more. And three d five? How many of its five electrons have m sub l zero? Just one, exactly! It is half-filled, so by Hund's rule there is one electron in each of the five m sub l values. And here the order does not matter: every value gets one anyway. So, seven plus four plus one: twelve. The answers are five, and twelve. See? Our puzzle from the start, solved in under a minute! Not seven, but twelve. Got it?

6:50Quick revision: full configuration first, then count

So, quick revision. First, write the full configuration; the inner electrons count too. Every s electron has l zero and m sub l zero, and a full subshell has two electrons in each m sub l. And a part-filled subshell follows Hund's rule, one per m sub l, and you state the order you used. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!