Shapes with Bond Pairs Only

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Draw methane, C H four, on paper: carbon in the middle, four hydrogens around it like a plus sign, every angle ninety degrees. But the real angle in methane is about one hundred and nine and a half degrees. Where did almost twenty extra degrees come from? Strange, right? Today we will find out, with the shapes of molecules whose central atom has only bond pairs. In JEE, questions on shapes come up often. So, let's begin.

0:35No lone pairs: two, three and four pairs

So, the basic idea. Electron pairs round a central atom repel each other, so they spread out as far apart as possible. If every pair is a bond pair, with no lone pairs, the shape is just the arrangement of those pairs. Shape equals electron-pair geometry, okay? The figure shows five basic arrangements: the grey-blue ball in the middle is the central atom, and the orange balls are the bonded atoms. Let's start with two pairs. Two bond pairs on the central atom. Where will they sit, to be as far apart as possible? On opposite sides, yes, exactly! So the shape is linear, with a bond angle of one hundred and eighty degrees. The small circle up top means degrees. In brackets, the example: Be Cl two. The small two below means two chlorine atoms. That's the first shape in the figure. Clear? Now, three pairs. They spread out in a flat triangle, one hundred and twenty degrees apart. Trigonal planar: trigonal for the three corners, planar because it is flat. The example is B F three. Got it? Now, four pairs, and this is our puzzle. Will four pairs sit in a flat plus sign, at ninety degrees, the way we draw methane on paper? No, no! Molecules are not flat drawings; they live in three dimensions. In space, four pairs get further apart than ninety degrees: they point to the corners of a tetrahedron, a pyramid on a triangular base. The angle is one hundred and nine degrees, twenty eight minutes. The small dash, the prime, means minutes. In the figure, the thick line comes towards you, and the dashed line goes behind the page. So methane is tetrahedral. Puzzle solved! Right?

2:24Five and six pairs

Alright, five pairs, as in P Cl five. The shape is trigonal bipyramidal. Bi means two: two triangular pyramids joined base to base. That's the fourth shape in the figure. Okay? Look carefully. Three bonds lie in a flat triangle round the middle, one hundred and twenty degrees apart: the equatorial bonds, like the equator. The other two point straight up and down, along the axis: the axial bonds, each at ninety degrees to the equatorial ones. So two different angles between neighbouring bonds, one hundred and twenty and ninety, in one molecule. With me? Now think. All five bonds are P Cl bonds. So they must all be the same length, right? Oh no, careful! They are not. Each axial bond is at ninety degrees to all three equatorial bonds, but each equatorial bond is at ninety degrees to only the two axial bonds. So the axial bonds suffer more repulsion. They are longer, and weaker, than the equatorial bonds. Got it? And six pairs, as in S F six. Four bonds in a square, one up and one down. The shape is octahedral, and every neighbouring bond angle is ninety degrees. That's the last shape in the figure. Clear?

3:44Seven pairs, ions, and the double-bond trap

Now, seven pairs, as in I F seven. Five bonds lie in a flat pentagon, seventy two degrees apart, and two more stand above and below it, at ninety degrees. So the shape is pentagonal bipyramidal. Okay? And ions follow the same rule. N O two plus: the small plus up top is the charge. Two pairs round nitrogen, so linear. C O three two minus, the carbonate ion: three pairs, trigonal planar. N H four plus, the ammonium ion: four pairs, tetrahedral. Right? Now, here is the trap. Carbon dioxide: O double bond C double bond O, okay? The two lines between the atoms mean a double bond. Two double bonds, so four bonding pairs on carbon. So, tetrahedral, right? No, no, wait! Carbon dioxide is linear. Each double bond is one sigma bond and one pi bond: the two Greek letters on screen. The pi pair sits in the same region as the sigma pair, so it adds no new direction. Only the sigma pairs decide the shape: two on carbon, so linear, one hundred and eighty degrees. Clear? And one important condition. All this works only when the central atom has no lone pair. C H three plus: three bond pairs, no lone pair, so trigonal planar. But C H three minus has a lone pair as well, and it is pyramidal. So always check for lone pairs first. Got it?

5:15Example 1: shapes of SO3 and the sulphate ion

Alright, example one. Give the shape of S O three, and of S O four two minus, the sulphate ion. The small four below means four oxygens, and the two minus up top is the charge. Pause the video, and try it first. Done? Okay, let's check. In S O three, sulphur makes a double bond to each oxygen. Three double bonds, six bonding pairs, so octahedral, right? Oh no, careful! Same trap again. Use the quick count from the VSEPR model video. Sulphur has six valence electrons, and oxygens joined by double bonds add nothing. So the pairs on sulphur are half of six: three. Three oxygens take all three pairs, so there is no lone pair. Okay? And the pi bonds do not count, so only the three sigma pairs decide the shape. Three pairs, no lone pair: S O three is trigonal planar. Easy, right? Now the sulphate ion. Six valence electrons, plus two for the charge of two minus. Half of six plus two: four pairs. Four oxygens take all four, so again, no lone pair. With me? Four pairs, no lone pair: the sulphate ion is tetrahedral, just like methane. That's it!

6:37Example 2: counting 90 degree angles in SF6 and PF5

Now a harder one. How many ninety degree bond angles are there in S F six, and in P F five? Pause and try. Hint: pick one bond, and count the bonds at ninety degrees to it. Done? Okay, let's check. S F six is octahedral. Pick one bond, say the one pointing up. The four bonds in the square are at ninety degrees to it, and only the bottom bond is opposite. So each bond is at ninety degrees to four others. Okay? Six bonds, and each one makes four right angles. Six times four is twenty four. So, twenty four right angles, right? No, no, careful! Every angle has two arms, so each one got counted twice, once from each arm. So divide by two: six times four, over two, equals twelve. Twelve right angles in S F six. Clear? Now P F five: five pairs, so trigonal bipyramidal. The equatorial bonds are one hundred and twenty degrees apart, and the two axial bonds point opposite ways. So the only ninety degree angles are between an axial and an equatorial bond. Okay? Now, count from the axial bonds only. Each axial bond makes a right angle with every equatorial bond. Should we divide by two here as well? No, not this time! We counted only from the axial bonds, so each right angle was counted once. Two axial bonds, three equatorial bonds each: two times three, six. So, twelve in S F six, and six in P F five. Always ask: did I count an angle twice?

8:23Quick revision: shapes for 2 to 7 bond pairs

So, quick revision. No lone pair on the central atom: the shape is the electron-pair geometry, and only sigma pairs count. Two, three and four pairs: linear at one hundred and eighty degrees, trigonal planar at one hundred and twenty, and tetrahedral at one hundred and nine degrees, twenty eight minutes. Five, six and seven pairs: trigonal bipyramidal, octahedral and pentagonal bipyramidal. Now go and try the practice problems on this page. All the best!