Electronegativity, Polarity and Dipole Moment

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take water and carbon dioxide. In both, oxygen pulls the shared electrons towards itself, so every bond has a positive end and a negative end. Water has a dipole moment, right? But carbon dioxide, with two such bonds, has a dipole moment of exactly zero. Polar bonds, and still a non-polar molecule? How? Strange, isn't it? Today one number, the dipole moment, will show you exactly why. In JEE, questions on dipole moments come up often. So, let's begin.

0:40Polar bonds: who pulls harder?

So, electronegativity is how strongly an atom in a bond pulls the shared electrons towards itself. When the two atoms have different electronegativities, the pull is unequal, and we call the bond polar. Okay? Now take hydrogen chloride, H C l. Chlorine is more electronegative than hydrogen. So tell me, which end of the bond becomes slightly negative? Chlorine, yes, exactly! The more electronegative atom carries delta minus: the small Greek letter delta, with a minus sign, means a small negative charge, not a full one. So in H C l, hydrogen gets delta plus, and chlorine gets delta minus. Right?

1:23Dipole moment: charge times distance

Now, how do we measure polarity with a number? Two opposite charges, some distance apart. Which two things should decide how big the effect is? The charge, and the distance, very good! So the dipole moment, written with the Greek letter mu, is q times d, where q is the charge at each end and d is the distance between them. Bigger charge or longer distance, bigger dipole moment. With me? The unit is the debye, and its symbol is a capital D, okay? One debye equals three point three three six times ten to the power minus thirty coulomb metre: the C and the m on screen. We need this number soon. Right? And chemists draw a dipole moment as an arrow, from the positive end to the negative end. So in H C l, the arrow points from hydrogen, the delta plus end, towards chlorine, the delta minus end. Got it?

2:17Percentage ionic character

Now, delta plus and delta minus are only partial charges. If the bond were completely ionic, each end would carry one full electronic charge, e. So a fully ionic bond would have a dipole moment of e times d. Clear? Now compare the real bond with this ideal one. Percentage ionic character equals mu observed, over e times d, times one hundred. The small word observed, below mu, is a subscript: it means the measured dipole moment. Okay?

2:48Molecules: add the moments as vectors

Alright, whole molecules. Each polar bond has its own moment, a bond moment, and lone pairs have moments too. The dipole moment of the molecule is the vector sum of all of them. So direction matters, not just size. Right? For two bonds at an angle theta, it is plain vector addition. Mu equals the square root of mu one squared, plus mu two squared, plus two times mu one times mu two times cos theta. Mu one and mu two, with the small subscripts, are the two bond moments, and theta, the Greek letter, is the angle between them. With me? Now, back to our puzzle. Carbon dioxide, C O two, first in the figure, is linear; the small two below is a subscript for two oxygens. So what is the angle between its two bond moments? One hundred and eighty degrees, yes! The little circle is the degree sign. The two orange arrows are equal and exactly opposite, so they cancel. Put theta equal to one eighty, with mu one equal to mu two, in the formula, and mu comes out zero. Clear? But water, H two O, is bent. In the figure, both orange bond moments lean towards oxygen, so they cannot cancel. The thick dark blue arrow is their resultant. Puzzle solved! Polar bonds are not enough. The shape decides. Got it? Now, one trap. Look at the last two molecules: ammonia, N H three, and N F three. Fluorine is far more electronegative than hydrogen. So N F three must have the bigger dipole moment, right? No, no, careful! Ammonia has one point four seven debye, and N F three only zero point two three. Why? The lone pair on nitrogen, the two dots on top. In ammonia, the nitrogen hydrogen bond moments point towards nitrogen, the same way as the lone-pair moment, so they add: the big dark blue arrow. In N F three, the nitrogen fluorine bond moments point away from nitrogen, against the lone pair, so they largely cancel: the tiny arrow. So lone pairs count too. Okay?

4:52Symmetric molecules: the moments cancel

So, in symmetric molecules, the bond moments cancel exactly. Carbon dioxide, B F three, methane, C H four, and carbon tetrachloride, C C l four: all have mu equal to zero. Yes, even with four polar bonds in C C l four! Right? The same for P C l five, S F six, and X e F four. All symmetric, all zero. Clear? Now, put two identical groups on a benzene ring. Ortho, meta or para: which one gives zero? Para, exactly! Ortho gives root three times mu, meta gives just mu, and para gives zero, where mu is the moment of one group. So para dichlorobenzene, with the small p for para, has zero dipole moment. Got it?

5:39Example 1: which of CO2, SO2, BF3 and NF3 are polar

Now, the first example. Which of these four are polar: C O two, S O two, B F three, and N F three? Hint: think about the shape first. Pause the video and try it. Done? Okay, let's check. Carbon dioxide is linear, and B F three is trigonal planar, a flat triangle with three equal bond moments. In both, the moments cancel, so both are non-polar. Right? Now S O two. On paper it looks just like C O two: one atom in the middle, two oxygens. So non-polar as well, right? Oh no, careful! S O two is bent, not linear. Just like water, the two moments cannot cancel. So it is polar. Clear? And N F three is pyramidal, with the lone pair on top. The moments do not cancel, so it is polar. With me? So the answer: S O two and N F three. Similar formulas, different shapes, different answers. Okay?

6:42Example 2: percentage ionic character of HCl

Now a harder one, with numbers. Hydrogen chloride has a dipole moment of one point zero three debye, and a bond length of one hundred and twenty seven picometres, written p m. Find its percentage ionic character. Pause and try it. Hint: first find its dipole moment if it were fully ionic. Done? Okay, let's check. First, the bond length in metres. One picometre is ten to the power minus twelve metre, so d is one point two seven times ten to the power minus ten metre. Got it? If the bond were fully ionic, the charge would be e: one point six times ten to the power minus nineteen coulomb. Times d, that gives two point zero three times ten to the power minus twenty nine coulomb metre. Right? So now we just divide one point zero three by this number, right? No, no, wait! One point zero three is in debye, but this is in coulomb metre. The units must match. So divide by three point three three six times ten to the power minus thirty, the value of one debye. Six point zero nine debye: the fully ionic value. Clear? Now it is easy. One point zero three over six point zero nine, times one hundred. The wavy equals sign means approximately: about seventeen percent. So H C l is polar, but only about seventeen percent ionic. With me?

8:07Quick revision: polarity, dipole moment, ionic character

So, quick revision. One: in a polar bond, the more electronegative atom gets delta minus, and mu is q times d, in debye. Two: percentage ionic character is mu observed, over e times d, times one hundred. Three: the dipole moment of a molecule is the vector sum of its bond and lone-pair moments, and symmetric molecules give zero. Now go and try the practice problems on this page. All the best!