Part B: Harder Problems

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Water has a dipole moment of one point eight four debye, and it has two O H bonds. So each O H bond gives half of it: zero point nine two debye, right? Hmm, too easy, isn't it? By the end of this video, you will find the real answer, and it is much bigger. Today is Part B of our problem bank: harder problems, of the kind JEE asks, and each one mixes two or three ideas of the unit. We pick up three tools first, then solve two problems. So, let's begin.

0:35Tool 1: dipole moments are arrows

So, tool one. The dipole moment of a molecule, written with the Greek letter mu, is the vector sum of its bond moments. Each bond moment is an arrow, so direction matters, not just size. Okay? Take B F three, boron trifluoride; the small three below is a subscript, for three fluorines. It is flat, with three equal B F moments at one hundred and twenty degrees; the small circle is the degree sign. Three equal arrows, spread evenly, cancel completely. So mu equals zero. Right? Now a Part B question. Para dinitrobenzene, with the small p for para: two N O two groups on a benzene ring, opposite each other. The N O two groups are planar and lie along the ring axis, so their moments are exactly opposite and cancel: mu is zero. Clear? But hydroquinone, para dihydroxybenzene, has two O H groups, also opposite each other on the ring. Para again, so its dipole moment is zero too, right? No, no, careful! Look at the COH group; the dashes on screen are bonds. This group is bent at oxygen. So the two hydrogen atoms can point in different directions, and the group moments need not cancel. The equals sign with a slash means not equal: mu is not zero. Para alone is not enough; each group moment must lie along the axis. Got it?

1:58Tool 2: lone pairs squeeze angles, stretch bonds

Tool two: lone pairs. On the right are the five basic arrangements of electron pairs. Look at the fourth one, the trigonal bipyramid: five pairs, with angles of ninety and one hundred and twenty degrees. S F four, sulphur tetrafluoride, has five pairs round sulphur. One of them is a lone pair, and it sits in the equatorial plane, round the middle. With me? Now the two equatorial fluorines. In a perfect trigonal bipyramid, the equatorial angle is one hundred and twenty degrees. So the F, S, F angle is one twenty, right? Oh no, careful! It is only about one hundred and two degrees; the wavy equals sign means approximately. Why? The lone pair repels the two equatorial bond pairs strongly, and squeezes their angle far below one twenty. Okay? Now C l F three, chlorine trifluoride. Five pairs again, but two lone pairs this time, and both sit in the equatorial plane. So it has two kinds of chlorine fluorine bond: axial, up and down, and equatorial, round the middle. Right? So tell me: which kind of bond is longer, axial or equatorial? The axial ones, yes, exactly! Each axial bond is at ninety degrees to both equatorial lone pairs, and suffers both repulsions. So it is pushed out: one hundred and seventy picometres, written p m, against one hundred and sixty for the equatorial bond. Clear?

3:28Tool 3: BF3 to BF4-, three ideas

Tool three, and this is what Part B does: one change, three ideas. B F three becomes B F four minus; the minus sign up top is its negative charge. Boron goes from s p two, planar, to s p three, tetrahedral; the two and the three are written up top. Okay? Now the bond length. The B F bond grows from one hundred and thirty to one hundred and forty five picometres; the arrow means changes to. Why should the same bond get longer? Back bonding! In B F three, boron has an empty p orbital, and the fluorines give some electron density back into it. That is p pi, p pi back bonding, with the Greek letter pi. So each B F bond gets partial double-bond character, and stays short. With me? But in B F four minus, boron is s p three, and there is no vacant p orbital on boron. No back bonding, so a longer bond. Shape, hybridisation and bond length, all from one change. Got it?

4:29Example 1: partial charge from dipole moment and bond length

Now, example one. Here is a linear molecule, A B, with a bond moment of one point five debye, the capital D, and a bond length of one hundred and fifty picometres. Find the charge on each atom, in units of e, the size of the charge on one electron. Pause the video, and try it first. Done? Okay, let's check. Dipole moment is charge times distance: mu equals q times d. So q equals mu over d. Simple, isn't it? So q is one point five over one fifty, right? Can we just divide these two numbers? No, no, wait! Debye and picometre are not S I units, so change both first. One debye is three point three three six times ten to the power minus thirty coulomb metre, the C and the m on screen. So mu is one point five times that. And one fifty picometres is one point five times ten to the power minus ten metre. Okay? Now divide. The one point fives cancel. So q is three point three four times ten to the power minus twenty coulomb. Right? Last step: in units of e, divide by one point six times ten to the power minus nineteen. That is about zero point two one. So each atom carries about zero point two one e, just a fraction of a full charge. Clear?

5:50Example 2: O-H bond moment of water from 1.84 D

Now our puzzle from the start. The dipole moment of water, H two O, is one point eight four debye, and the HOH angle is one hundred and four point five degrees. Find the O H bond moment. Pause and try. Hint: the moments are arrows. Done? Okay. Two O H bonds, one point eight four in total. So each bond is half of it, zero point nine two debye, right? No, no, careful! This is a common JEE trap. Halving works only if both arrows point the same way. In water they are at one hundred and four point five degrees. The resultant runs exactly between them, so each O H moment makes half that angle with it: fifty two point two five degrees. Okay? Now split each bond moment into two parts. The sideways parts are equal and opposite, so they cancel. The parts along the resultant add. Right? Each bond gives mu O H times cosine of fifty two point two five degrees; the small O H below mu names the bond. Two bonds, so one point eight four equals two times mu O H times cosine of fifty two point two five degrees. Two times that cosine is one point two two four. Clear? So the bond moment is one point eight four over one point two two four: about one point five zero debye. Puzzle solved! Each bond is one point five debye, not zero point nine two. The two arrows partly cancel, so together they give only one point eight four. With me?

7:25Quick revision: dipole arrows, lone pairs, back bonding

So, quick revision. One: add bond moments as arrows, and symmetry can cancel them. Two: lone pairs squeeze bond angles, and lengthen the bonds at ninety degrees to them. Three: back bonding shortens the B F bond: one thirty picometres in B F three, one forty five in B F four minus. Four: q equals mu over d, but change debye and picometres to S I units first. Now go and try the two practice problems on this page. All the best!