Energy Order and Electronic Configurations

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. Take oxygen gas, O two. Draw its Lewis structure: a double bond between the two oxygen atoms, and two lone pairs on each. Every electron is paired, right? So oxygen should not be attracted by a magnet. But it is! Oxygen is paramagnetic, and that needs unpaired electrons. So where are they hiding? Molecular orbital theory will tell us. This comes up often in JEE, so let's begin.

0:33The filling order up to 14 electrons

So, in the last lesson we built molecular orbitals. Now, let's see the order in which electrons fill them. On the right is the diagram for N two; we will read it in a moment. Start at the bottom: sigma one s, then sigma star one s. Sigma is the Greek letter on screen, and the little star means antibonding, remember? Then sigma two s, and sigma star two s. The less-than sign means lower in energy. Okay? Next come two pi orbitals: pi two p x and pi two p y. Pi is the other Greek letter, and the small x and y below tell you the axis. The equals sign means they have exactly the same energy. And above them comes sigma two p z, with a small z below. Right? Then the antibonding partners: pi star two p x, equal to pi star two p y, and at the very top, sigma star two p z. In the figure, the two s and two p levels of each nitrogen sit on the left and right, and the middle shows this ladder from sigma two s upwards. Clear? Now, careful: this order is only for up to fourteen electrons. That means B two, C two, N two, and their ions. The small two below each symbol means two atoms in the molecule. With me? And the reason sigma two p z sits above the pi pair here is s p mixing. The two s and two p orbitals mix, and that pushes sigma two p z up. The little dash just joins s and p. Got it?

2:08Bond order, and the KK shorthand

Now, to judge the bond, recall the bond order: half of, N b minus N a. N b, with a small b, counts the electrons in bonding orbitals, and N a, with a small a, counts those in antibonding orbitals, the starred ones. Okay? Now think about the two lowest levels. Sigma one s and sigma star one s are both full: the small two on top means two electrons each. Do they change the bond order? Correct, they don't! Two bonding and two antibonding electrons cancel. So we write these four electrons simply as K K, okay? And K K adds nothing to the bond order. Right? Now let's fill N two. Nitrogen has seven electrons, so N two has fourteen. K K, then sigma two s two, sigma star two s two, pi two p x two, pi two p y two, and the last two go into sigma two p z. In the figure, each arrow is an electron. With me? Now count. Bonding: two in K K, two in sigma two s, four in the pi pair, and two in sigma two p z. That is ten. Antibonding: two in K K and two in sigma star two s, so four. Half of ten minus four: three. So, a triple bond! The figure writes B O for bond order, and counts only the outer electrons: half of eight minus two, the same three. Clear? And look at the arrows: every electron is paired, up and down. So N two is diamagnetic: it is not attracted by a magnet. Okay? One more rule. If the bond order comes out zero, the species does not exist. Remember that.

3:55From O2 on: sigma 2pz drops below pi 2p

Now, from O two onwards, that is O two, F two, and N e two, which is neon, something changes. Sigma two p z drops back below the pi pair, the opposite of nitrogen. Okay? Let's fill O two: eight electrons from each oxygen, sixteen in all. K K, sigma two s two, sigma star two s two, sigma two p z two, and pi two p four, which means both pi orbitals are full. That is fourteen so far, with two more to place. Right? Two electrons are left, and next comes the pi star pair: two orbitals of equal energy. How do these two electrons sit? Together in one orbital, or one in each?

4:38Why oxygen is paramagnetic

One in each, yes, exactly! That is Hund's rule. Pi star two p x one, and pi star two p y one. The figure shows only the two p region: sigma two p z is now the lowest level, and the two single arrows sit in the pi star pair. Got it? Now the bond order. Bonding electrons: ten. Antibonding: two in K K, two in sigma star two s, and two in pi star, so six. Half of ten minus six: two. That is a double bond, just as Lewis said. Okay? And now our puzzle is solved! Those two electrons in pi star are unpaired, so oxygen is paramagnetic. Lewis paired them up; molecular orbital theory shows they are not. Beautiful, isn't it?

5:26Example 1: MO configuration and bond order of F2

Alright, your turn. Example one. Find the bond order and the magnetic behaviour of F two, fluorine. Pause the video, and try it first. Done? Okay, let's check. Fluorine has nine electrons, so F two has eighteen. That is more than fourteen, so we use the oxygen order: sigma two p z fills before the pi pair. Right? Fill it like oxygen: K K, sigma two s two, sigma star two s two, sigma two p z two, pi two p four. That is fourteen. The last four go into the pi star pair, pi star two p four, so pi star is full. Okay? Now count. Bonding: two in K K, then two, two and four, so ten. Antibonding: two in K K, two in sigma star two s, and four in pi star, so eight. Half of ten minus eight: one. So, a single bond between the two fluorine atoms. Clear? Now the magnetism. Every orbital that has electrons is full, even pi star, so every electron is paired. So F two is diamagnetic. Easy, right?

6:38Example 2: why B2 is paramagnetic

Now a harder one. Example two. B two, the boron molecule, is paramagnetic. Explain why. Pause and try. Hint: count the electrons first, then choose the right order. Alright. Boron has five electrons, so B two has ten. K K, sigma two s two, sigma star two s two: that is eight. Two left. Many students put both in sigma two p z, paired up. So, diamagnetic, right? No, no, careful! That is the trap. Ten electrons is up to fourteen, so we use the nitrogen order: the pi pair lies below sigma two p z. Okay? So the last two go into pi two p x and pi two p y, two orbitals of equal energy. And by Hund's rule, one in each: pi two p x one, pi two p y one. With me? Two unpaired electrons! That is why B two is paramagnetic, just like oxygen. Right? And the bond order: bonding six, that is two in K K, two in sigma two s, and two in the pi pair. Antibonding four. Half of six minus four: one. Notice, the wrong order also gives bond order one. Only the magnetism catches the mistake, so always count the electrons first. Got it?

8:02Quick revision: energy order, bond order, magnetism

So, quick revision. Up to fourteen electrons, the pi pair lies below sigma two p z, because of s p mixing. From O two on, sigma two p z drops below the pi pair. Bond order is half of, N b minus N a, and unpaired electrons make a molecule paramagnetic. Now go and try the practice problems on this page. You are ready for them. All the best!