Set 3 (Molecular Orbitals and Intermolecular Forces)

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. Water, H F and ammonia all form hydrogen bonds, and the single strongest one is between two H F molecules. So which liquid has the highest boiling point? Most students say H F straight away. Strongest bond, highest boiling point. Simple, right? Hold on to that answer; we will check it in this class. This is our third mixed problem set for JEE Main: molecular orbitals and intermolecular forces. These quick questions come up often, and they are easy marks, if you know the traps. So, let's begin.

0:40Tool 1: the energy order and magnetism

So, first tool: the molecular orbital diagram. On the right is the diagram of N two, with the two s and two p orbitals of each atom on the sides. Each short line in the middle is a molecular orbital, and each little arrow is one electron, okay? Sigma and pi are the Greek letters for the two kinds of orbitals, and the little star means antibonding. With me? Now, the order. Up to fourteen electrons, that is B two, C two and N two, s p mixing pushes sigma two p z, with a small z below, above the pi two p pair. So pi two p fills first. You can see it in the diagram. Clear? Now, O two has sixteen electrons. Does it follow the same order? No! From O two on, that is O two, F two and N e two, sigma two p z drops below pi two p. Remember this switch; our second example is built on it. Got it? And magnetism is really simple. Any unpaired electron makes a species paramagnetic. If every electron is paired, it is diamagnetic. Okay?

1:47Tool 2: bond order and bond length

Now, the second tool: bond order. B O, bond order, equals one half times, in brackets, N b minus N a. N with a small b below counts the electrons in bonding orbitals, and N with a small a below counts those in antibonding orbitals. With me? And a higher bond order means a more stable molecule, with a shorter bond. Right? Quick one from our set. O two two plus, O two plus and O two. The two plus up top is a charge of plus two. Which one has the shortest bond? Count first. O two has sixteen electrons. One plus charge means one electron less: fifteen. Two plus: fourteen. Clear? Now fill them in. All three have ten electrons in bonding orbitals. Only the antibonding count changes: four, five and six. So the bond orders are one half times ten minus four, that is three, then two point five, and two. Highest bond order, shortest bond. So O two two plus is the shortest, then O two plus, then O two; these less than signs show the order. Easy marks!

2:58Tool 3: forces between molecules

Now, the third tool: forces between molecules. First, hydrogen bonds; H-bond on screen is short for hydrogen bond. In each pair, the dash is the covalent bond, and the three dots are the hydrogen bond, okay? F H dot dot dot F is the strongest, then O H dot dot dot O, and N H dot dot dot N is the weakest; the greater than signs show the order. Right? Now, back to our puzzle. Water, H F and ammonia, N H three. H F has the strongest single hydrogen bond, so liquid H F boils highest, right? No, no, careful! The number of bonds matters too. Water has two hydrogens and two lone pairs, so each molecule forms up to four hydrogen bonds. H F has only one hydrogen to give. So water boils highest. Puzzle solved! Now, methane, ethane and propane: C H four, C two H six and C three H eight. No hydrogen bonds here. Only London forces, the dispersion forces, which grow with the size of the molecule. So the boiling point rises in this order. Got it? And one more from our set: why does potassium chloride, K C l, dissolve in water? The ions attract the water dipoles. This is ion dipole attraction, also called hydration. With me?

4:23Example 1: count paramagnetic species from H2+ to B2

Alright, example one. How many of these are paramagnetic: H two plus, H e two plus, L i two, B e two and B two? Pause the video, and try it first. Done? Okay, let's check. Count the electrons first. H two plus: one. H e two plus: two plus two, minus one, that is three. L i two: six. B e two: eight. And B two: ten. Right? An odd count always leaves one electron alone. So H two plus and H e two plus have one unpaired electron each: both paramagnetic. Now the other three: six, eight and ten, all even numbers. So everything is paired, and all three are diamagnetic, right? For L i two and B e two, yes. Their electrons fill the sigma orbitals in pairs, so every electron is paired. But wait, look at B two. Ten electrons, that is fourteen or fewer, so pi two p fills before sigma two p z. The last two electrons go into pi two p x and pi two p y, one each, by Hund's rule. Two unpaired electrons! So B two is paramagnetic. Oh, careful with this one. The double arrow on screen means therefore: three are paramagnetic, H two plus, H e two plus and B two. Three, not two. Got it?

5:52Example 2: s-p mixing flips the magnetism of B2 and C2

Now the harder one, JEE Main level. If s p mixing is ignored, how many of B two, C two and N two change their magnetic behaviour? Pause and try. Hint: fill each molecule in both orders. Done? Okay, let's check. Ignoring s p mixing means using the O two order, with sigma two p z before pi two p. The first eight electrons fill everything up to sigma star two s, the same in both orders. So only the electrons after that matter. Okay? B two has ten, so two more. With mixing, they go one each into pi two p x and pi two p y; the small number up top counts the electrons in that orbital. Para, short for paramagnetic. The arrow shows the switch to no mixing: now both electrons pair up in sigma two p z. Dia, short for diamagnetic. So B two changes! Now you try C two, with four more electrons. Does it change too? Yes, exactly! With mixing, the four electrons fill both pi two p orbitals: all paired, diamagnetic. Without mixing, two pair up in sigma two p z, and the last two go one each into the pi orbitals: paramagnetic! So C two flips the other way. Two out of two have changed. So N two must change as well, and the answer is three, right? No, no, wait! N two has six more electrons: enough to fill both pi orbitals and sigma two p z completely, in either order. Every electron paired, diamagnetic both ways. Unchanged! So the answer is two: B two and C two. And now you see why that switch at fourteen electrons really matters.

7:41Quick revision: MO energy order, bond order, H-bonds

Clear? So, quick revision. Up to fourteen electrons, pi two p fills before sigma two p z; from O two on, the reverse. Bond order is one half times N b minus N a, and a higher bond order means a shorter bond. And for hydrogen bonds, both the strength and the number count, while London forces grow with size. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!