Diborane and Borazine

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle first. Boron has three valence electrons, so its hydride should be B H three, right? But B H three does not exist on its own; two units join to give B two H six, diborane. And diborane holds eight atoms together with only twelve valence electrons. How can that work? Today we find the answer, the banana bond, then meet borazine, the inorganic benzene. This comes up often in JEE, so let's begin.

0:33Electron deficient: too few electrons

B H three, with the small three below counting the hydrogens, does not exist alone. It dimerises, meaning two units join, to give B two H six. Now, each boron brings three valence electrons and each hydrogen brings one. So how many does diborane have in all? Two borons times three is six, and six hydrogens times one is six more. So twelve valence electrons, just six pairs. Right? But in an ordinary structure, each bond is one pair shared by two atoms, and eight atoms need seven bonds. That is fourteen electrons, and we have only twelve; see the less-than sign. So diborane is electron deficient, as the heading says. Clear?

1:16Not planar: terminal bonds and banana bridges

Now look at the figure. On the left is diborane: two borons, each with two outer hydrogens, and two bridging hydrogens on curved orange bonds. On the right, for comparison, is the aluminium chloride dimer; its arrows show bridging chlorines giving lone pairs. Each boron is roughly s p three hybridised, with a small raised three on the p. Okay? The four outer bonds are terminal B H bonds: normal two-centre two-electron bonds, written two c, two e. Each is one hundred and nineteen picometres, written p m, and all four lie in one plane, using eight of our twelve electrons. Got it? That leaves four electrons, one pair per bridge. So each bridge, BHB, holds three atoms with one pair: a three-centre two-electron bond, three c, two e, the banana bond. With one bridge hydrogen above the plane and one below, the molecule is not planar. Bridge bonds are one hundred and thirty four picometres, longer and weaker. With me? Look at the angles. H t, B, H t, the small t for terminal, is about one hundred and twenty degrees, the small raised circle meaning degrees; the wavy equals sign means approximately. H b, B, H b, the small b for bridging, is about ninety seven degrees. Okay? Many students get this next one backwards. Which bonds have more p character, the terminal ones or the bridge ones? Yes, it is the bridge bonds! The terminal bonds make the larger angle, so they have more s character. The bridge bonds, with the smaller angle, have more p character. Clear?

2:57Making diborane

So how do we make diborane? Boron trifluoride, B F three, reacts with lithium aluminium hydride, L i A l H four, in ether. Four B F three plus three L i A l H four give two B two H six, three L i F and three A l F three; the arrow means gives. Okay? In the lab, iodine oxidises sodium borohydride, N a B H four. Two N a B H four plus I two give B two H six, two N a I and H two. Right? In industry, two B F three and six N a H give B two H six and six N a F, at four hundred and fifty kelvin; that is the K over the arrow. Got it?

3:36Reactions of diborane

Next, its reactions. Diborane burns in oxygen to B two O three and water, and even catches fire in air on its own. Delta c H standard, with capital delta the Greek triangle, the small c for combustion and the small circle for standard, is minus one thousand nine hundred and seventy six kilojoules per mole; the raised minus one on mol means per mole. With me? Trimethylamine, N M e three, a Lewis base with M e for methyl, splits each bridge symmetrically into two adducts, B H three dot N M e three. The dot joins the two parts, and boron becomes tetrahedral. Okay? Ammonia, N H three, is also a Lewis base. So it should give two B H three dot N H three, the same kind of product, right? No, we have to be careful here, because that is the trap! Ammonia splits diborane unsymmetrically: one boron takes both ammonias and loses a hydride to the other. The salt is B H two with two N H three, plus charge, and B H four, minus charge, each in square brackets. Clear?

4:42Borazine, the inorganic benzene

That salt from ammonia leads to borazine. Three B two H six plus six N H three give three units of it. Heating, the capital delta over the second arrow, then gives two B three N three H six, borazine, plus twelve H two. Right? Borazine is a planar six-membered ring of alternating boron and nitrogen atoms. It is isoelectronic with benzene, with the same number of electrons, so we call it inorganic benzene. Got it? It has delocalised pi electrons, the Greek letter pi on screen, donated by nitrogen. And it has no bridge bonds, no banana bonds at all. Okay? Now let me ask you this. If borazine is isoelectronic with benzene, is it just as unreactive as benzene? No, it is not, and that is the key difference! The B N bonds are polar, so borazine is more reactive. It adds three H C l, chlorine to boron and hydrogen to nitrogen.

5:41Example 1: moles of H2 from hydrolysis of 1 mol B2H6

Clear? Now, Example one, an easy warm-up. How many moles of hydrogen are liberated when one mole of diborane is completely hydrolysed? Pause the video and try it first. Done? Okay, let's check. Diborane has six hydrogen atoms, and two atoms make one H two molecule. So the answer is three moles, right? No, no, we must be careful, because water brings hydrogen too. B two H six plus six H two O give two BOH three, boron with three O H groups, plus six H two. Okay? Now count the hydrogen atoms. Six from diborane plus twelve from water gives eighteen on the left. On the right, the two BOH three hold six. Right? So the other twelve form six H two: every hydrogen of diborane ends up in H two. The answer is six moles, not three; the m o l means moles. Got it?

6:40Example 2: mass of diborane from 10 g BF3 and LiAlH4

Now, Example two, a little harder. Ten point zero grams of B F three is treated with excess lithium aluminium hydride in ether. What mass of diborane can be formed? Pause the video and try it, starting from the balanced equation. Done? Okay, let's check. Here is our equation: four B F three give two B two H six. Lithium aluminium hydride is in excess, so the B F three decides the amount. Okay? First, n, the amount of B F three in moles. Ten point zero over sixty seven point eight, the molar mass, gives zero point one four seven five moles. Right? Now, four moles of B F three give two moles of diborane. So do we double our moles, or take half? We take half, yes, exactly! Doubling is the trap here. Half of zero point one four seven five is about zero point zero seven three seven moles of diborane. With me? Next, the molar mass of diborane, capital M, is twenty seven point seven grams per mole. So the mass is zero point zero seven three seven times twenty seven point seven, about two point zero four grams. Clear?

7:52Quick revision: banana bonds, NH3 cleavage, borazine

So, quick revision. Diborane has four terminal bonds, two c, two e, and two banana bridges, three c, two e. Trimethylamine splits it symmetrically, ammonia unsymmetrically. Borazine is isoelectronic with benzene, but more reactive, because its bonds are polar. Now try the practice problems on this page. All the best!