pπ–pπ and pπ–dπ Bonding; Changes of Hybridisation

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. Boron trifluoride, boron trichloride and boron tribromide: in all three, boron is short of electrons, so each is a Lewis acid; it wants a lone pair. Now, fluorine is the most electronegative element, pulling electrons from boron the hardest. So boron trifluoride should be the strongest acid of the three, right? Well, it is actually the weakest! How can that be? Today we find out. Questions on this come up often in JEE, so let's begin.

0:36Two kinds of pi bond: p pi-p pi and p pi-d pi

So, first, the names in the heading. Pi is the Greek letter like a little table. p pi, dash, p pi, is a pi bond from the sideways overlap of two p orbitals; p pi, dash, d pi, is the same, but between a p and a d orbital. Okay? Now, carbon, nitrogen and oxygen are second-period atoms. Tell me, does the second shell have any d orbitals? No, it does not! Only two s and two p. So when two second-period atoms make a pi bond, both must use p orbitals. Pi bonds between second-period atoms are p pi, p pi. Right? Now take a third-period atom, like sulphur or phosphorus, or a heavier one, bonded to oxygen, the O on screen. These atoms do have empty d orbitals. So the pi bond is often p pi, d pi. Notice the word, often. Not always! We need it in example two. With me? So who gives what? Oxygen's filled two p orbital overlaps an empty d orbital of sulphur, phosphorus, chlorine or xenon. On screen, C l is chlorine and X e is xenon. So the p comes from oxygen, and the d from the central atom. Clear? Three examples. Sulphate, S O four, two minus. Phosphate, P O four, three minus. The small four below means four oxygens, and the charge sits up top. And xenon trioxide, X e O three. All three have p pi, d pi bonds. Got it?

2:13Back bonding

Now, back bonding. An outer atom has a filled p orbital, and the central atom has an empty orbital. The filled one overlaps the empty one, so electrons flow back, from the outer atom to the central atom, as the arrow shows. Okay? Now our puzzle molecule, B F three; the small three below means three fluorines. Boron is s p two, the little two up top, so one p orbital is left out of the hybrids. Boron's three valence electrons all go into its three bonds with fluorine. So what is in that spare two p orbital? Nothing. It is empty, exactly! And each fluorine has filled two p orbitals. So the filled two p of fluorine overlaps the empty two p of boron. Both are two p orbitals, the same size, so this overlap is really good. Right? So what changes? The boron fluorine bond becomes shorter and stronger. And boron's empty orbital is now partly filled, so boron wants a lone pair less. Its Lewis acidity becomes weaker. Clear? Now, chlorine back bonds with its three p orbital, and bromine with its four p. These are bigger than boron's small two p. So, is their back bonding better or worse? Worse, correct! Bigger orbitals match boron's two p poorly, so back bonding is weaker, and boron stays hungrier. So boron trifluoride is the weakest Lewis acid, then boron trichloride, and boron tribromide is the strongest, as the less than signs show. Puzzle solved! Electronegativity alone gave the wrong answer. With me? One more case, with electrons going the other way. Trisilylamine: one nitrogen holding three S i H three groups, the bracket with a three outside. S i is silicon. Here, nitrogen's lone pair drifts into silicon's empty d orbitals. The double arrow means, so. So the molecule is planar. Got it?

4:14When does hybridisation change?

Alright, the last idea: when does hybridisation change? First, the steric number: sigma bond pairs plus lone pairs on the central atom. Sigma is the Greek letter on screen, like a small o with a tail. Okay? And the rule: hybridisation changes when the steric number changes. So always count both, the bond pairs and the lone pairs. Right? So, aluminium chloride takes a chloride ion, giving an ion with four chlorines and a minus charge; the arrow means, changes to. Aluminium goes from three bond pairs to four: s p two to s p three. Clear? Same with ethene, C two H four, becoming ethane, C two H six. Each carbon goes from three sigma bonds to four, so s p two to s p three. Okay? Now, water takes a proton and becomes H three O plus. Oxygen goes from two bonds to three. So its hybridisation changes too, right? No, no, careful! This is the trap. Water's oxygen has two bond pairs and two lone pairs: four. In H three O plus, one lone pair simply becomes a bond pair: three plus one, still four. So it stays s p three. No change! Got it?

5:31Example 1: hybridisation change, BF3 to BF4- or NH3 to NH4+

Example one. In which change does the hybridisation of the underlined atom change? B F three to B F four minus, with boron underlined, or N H three to N H four plus, with nitrogen underlined? Pause the video, and try it first. Done? Okay, let's check. Boron first. In B F three, boron has three bond pairs and no lone pair. Steric number three, so s p two. Right? In B F four minus, a fluoride ion has joined, so boron has four bond pairs. Steric number four, s p three. It changes! Okay? Now nitrogen. From N H three to N H four plus, nitrogen also gains one more bond. So, does its hybridisation change? No! Remember water? Same story. In N H three, nitrogen has three bond pairs plus one lone pair: four. So it is s p three already. Clear? In N H four plus, that lone pair simply becomes the fourth bond pair. Still four, still s p three. No change. So the answer: B F three to B F four minus, s p two to s p three. Nice and easy, isn't it?

6:46Example 2: p pi-p pi and p pi-d pi bonds in SO3

Now a harder one, example two. Sulphur trioxide, S O three, drawn with three S double bond O bonds. How many of its pi bonds are p pi, p pi, and how many are p pi, d pi? Pause the video and try it. Alright. Three double bonds, so three pi bonds. Sulphur is a third-period atom, bonded to oxygen. So all three pi bonds are p pi, d pi, right? No, no, wait! Remember the word, often? First find sulphur's hybridisation. Three sigma bonds, one to each oxygen, and no lone pair. Steric number three, so s p two. With me? Now, s p two mixes the three s orbital with only two of the three p orbitals. So one three p orbital stays unhybridised. Right? That three p orbital overlaps sideways with a two p orbital of oxygen. p with p, so one bond is p pi, p pi. Okay? Sulphur has no p orbital left for the other two pi bonds, so each of them uses an empty three d orbital of sulphur. That gives two p pi, d pi bonds. Got it? So, one p pi, p pi, and two p pi, d pi. If you said three p pi, d pi, that was the trap. Always check what the hybridisation leaves behind!

8:09Quick revision: p pi-d pi, back bonding, hybridisation

So, quick revision. Pi bonds between second-period atoms are p pi, p pi; a third-period atom bonded to oxygen often forms p pi, d pi. Back bonding makes boron trifluoride the weakest Lewis acid of the three. And hybridisation changes when the steric number changes. Now go and try the practice problems on this page. You can do them. All the best!