Rules for Filling Orbitals

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Electrons like low energy, right? And the lowest energy orbital of all is one s. Now, iron has twenty six electrons. So why don't all twenty six simply crowd into one s? Hmm, something must be stopping them. Today we meet the three rules that decide where every electron goes. And look, in JEE, almost every question on configurations, unpaired electrons or magnetic moment starts from these three rules. So, let's begin.

0:36Rule 1: Aufbau, lowest energy first

So, rule number one: the Aufbau principle. Electrons fill the orbitals in order of increasing energy: the lowest one first, then the next one up. Just like filling a building from the ground floor upwards, okay? But which subshell is lower? Remember the n plus l rule from the last chapter? Here n is the principal quantum number, and l the azimuthal one. Lower n plus l fills first, and when n plus l is equal, the lower n fills first. See, the diagonal arrows in the figure read out exactly this order, and the chain of less-than signs beside them lists it. Clear? Quick check, and no peeping at the figure! Which fills first: four s, or three d? Yes, four s! Four s has n plus l equal to four plus zero, four. Three d has three plus two, five. So four s fills first, even though three is smaller than four. You already know that trap from the last lesson, don't you?

1:37Rule 2: Pauli exclusion principle

Alright, rule number two: the Pauli exclusion principle. No two electrons in an atom can have all four quantum numbers the same. The four are on screen: n, l, then m l, the magnetic quantum number, with a small l below the m, and m s, the spin quantum number, with a small s below. Okay? Now think. Two electrons sit in the same orbital. Same orbital means the same n, the same l and the same m l. So what must be different? Yes, exactly, the spin! One electron has m s equal to plus one half, the other has minus one half: opposite spins. And there are only these two values. So, how many electrons can one orbital hold? Just two, maximum! We draw it like this: the little box is one orbital, and each arrow is one electron. One arrow up, one arrow down: the two opposite spins. And now our puzzle is solved. One s can take only two electrons, so the rest of iron's electrons must go into higher orbitals. Makes perfect sense, doesn't it?

2:41Rule 3: Hund's rule of maximum multiplicity

Now rule number three: Hund's rule of maximum multiplicity. It is about degenerate orbitals, that means orbitals of the same energy, like the p, d or f orbitals of one subshell. With me? So, electrons are entering the p orbitals of one subshell. Tell me, do they pair up in the first orbital, or spread out first? Yes, they spread out! Each orbital first gets one electron, and these single electrons all have parallel spins, pointing the same way. Only after every orbital has one does pairing begin. Just like seats in a bus: everyone first takes an empty seat, and only then does anyone share. Simple, right? Let's see it for nitrogen, N, with seven electrons; that is the seven in the brackets. Aufbau order: one s two, two s two, two p three. The small number up top counts the electrons in that subshell. Two plus two plus three, seven. Okay? Now the three electrons in two p. Three boxes, three orbitals, and Hund says one electron in each, all arrows pointing up, parallel. So nitrogen has three unpaired electrons. Pairing them up too early is a classic JEE mistake. Don't do it!

3:56Shorthand: the noble gas core

Alright, one shortcut before the examples. Writing every subshell from one s onwards gets long. So we write the noble gas just before the element as a core. Argon, written inside square brackets, stands for all eighteen electrons of argon, the argon core, okay? Now back to iron, our puzzle atom, with twenty six electrons. After the argon core, how many are left? Eight, correct! Twenty six minus eighteen. Following Aufbau, four s takes two, and the remaining six go into three d. So iron is argon core, three d six, four s two. Notice, three d is written before four s here, in order of the shell number, but four s still filled first, okay? So that is where iron's twenty six electrons really sit.

4:48Example 1: configurations of K, Ca and Sc

Now, example one. Write the configurations of potassium, K, with nineteen electrons; calcium, with twenty; and scandium, with twenty one. Pause the video, and try it first. Done? Okay, let's check. Potassium: the argon core first. Then one more electron, and it goes into four s, because four s fills before three d. So, argon core, four s one. Calcium has one more. It joins the same four s orbital, with the opposite spin; Pauli allows two. Argon core, four s two. Now four s is full. Now scandium. Four s is full, so the next electron goes to four p, the same shell, right? No, no, careful! Check n plus l. Three d: three plus two, five. Four p: four plus one, also five. Equal! So the lower n wins, and that is three d. So scandium is argon core, three d one, four s two. Scandium is where the three d series starts. Nice and simple, isn't it?

5:56Example 2: which rule each wrong configuration breaks

Now a slightly harder one, JEE style. Nothing to calculate here; you have to catch the mistake. Which rule does each configuration break? First, carbon written as one s two, two s two, two p x two. The small x below the p means one particular p orbital, the p x orbital. Second, lithium as one s three. Third, lithium as one s two, two p one. Pause the video and try it. Done? Okay, let's check. Carbon: both two p electrons are paired up in the p x orbital, while the other p orbitals sit empty. But degenerate orbitals must fill singly first! So the first one breaks Hund's rule. Second, lithium as one s three. Quick, how many electrons can one orbital hold? Only two! Three electrons in one s would force two of them to have all four quantum numbers the same. So this breaks the Pauli principle. It's our iron puzzle in miniature, isn't it? Now the third: one s two, two p one. One s has only two electrons, so Pauli is happy. Two p has a single electron, so Hund has nothing to say. So this one is correct, right? Actually, no! That is the trap. Where is two s? It got skipped! Two s has n plus l equal to two, and two p has three. So two s must fill before two p, and this breaks the Aufbau principle. See, three mistakes, three rules. Got it?

7:30Quick revision: Aufbau, Pauli and Hund's rule

So, quick revision. Aufbau: lower n plus l fills first, so four s before three d. Pauli: at most two electrons per orbital, with opposite spins. Hund: in p, d and f orbitals, fill singly first, with parallel spins. Now go and try the practice problems on this page. Three rules, and you are ready. All the best!