Tetravalent Carbon and Hybridisation

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Ethane, ethene and ethyne are all just carbon and hydrogen, right? Yet the hydrogen of ethyne is acidic enough to react with sodium amide, and that of ethene is not. So what is different? Hmm, the answer is in how carbon mixes its orbitals, which is called hybridisation. It comes up often in JEE, so let's begin!

0:27Why carbon makes four bonds

So, let us start with carbon, whose configuration is one s two, two s two, two p two. The small raised numbers count the electrons in each subshell, okay? So the outer shell has four valence electrons. Now, to complete its octet, should carbon lose four electrons, or gain four? The answer is neither, because carbon does not lose or gain four electrons easily. So it shares them, forming four covalent bonds, and that is why carbon is tetravalent. Clear? Now, the bond between two carbon atoms is strong, so carbon atoms join into chains, branches and rings; this is catenation. Carbon also forms double bonds, the two lines on screen, and triple bonds, the three lines. Together, these explain the enormous number of organic compounds. With me?

1:16Hybridisation: carbon mixes its orbitals

Now, the big idea. Carbon's two s orbital mixes with one, two or three of its two p orbitals, and gives equivalent new orbitals, called hybrid orbitals. That mixing is hybridisation, okay? These hybrid orbitals form the sigma bonds, written with the Greek letter sigma, and they also hold the lone pairs. Got it? The p orbitals that were not mixed form pi bonds, by sideways overlap; that is the Greek letter pi on screen. So, hybrids make sigma bonds, and leftover p orbitals make pi bonds. Right?

1:49The three hybrid states of carbon

So, three ways to mix. First, s p three: one s mixed with three p orbitals, and the raised three counts those p orbitals. No p is left, so no pi bond. The shape is tetrahedral, one hundred and nine point five degrees, like methane in the picture, where the solid wedge points towards you and the dashed line away from you. Clear? Next, s p two: one s mixed with two p orbitals. One p orbital is left, and it makes one pi bond. The shape is trigonal planar, one hundred and twenty degrees, like each carbon of ethene. Okay? And then s p: one s mixed with just one p. Two p orbitals are left, so two pi bonds. The shape is linear, one hundred and eighty degrees, like ethyne. Got it? Now, s-character. In s p three, four orbitals were mixed, and only one was an s orbital. So what share of each hybrid is s? That is one out of four, twenty five percent, yes! Likewise, s p two is one out of three, thirty three point three percent, and s p is one out of two, fifty percent. With me?

3:00Topper's trick: the one-line rule

Now, a topper's trick. For any atom, add the number of sigma bonds it forms to the number of its lone pairs. Four means s p three, three means s p two, and two means s p. The arrow on screen just means gives, okay? And for a carbon with no charge that forms four bonds, just count its pi bonds: none gives s p three, one gives s p two, and two give s p, really simple, isn't it?

3:27Three common traps

Alright, now the traps. Look at the badge: a middle carbon with a double bond on each side. Double bonds, so s p two, right? No, no, careful, and count the pi bonds first. Two double bonds means two pi bonds on one carbon, so it is s p, not s p two. The same goes for the carbon of carbon dioxide, okay? Trap number two is a lone pair next to a pi bond, as on the nitrogen of an amide, aniline or pyrrole. It is delocalised into the pi system, so the atom is s p two, although the one-line rule gives four. Got it? Trap number three is the charged carbon. The carbon of a carbocation has three sigma bonds and an empty p orbital, so it is s p two. Clear? But a simple alkyl carbanion has three sigma bonds and a lone pair. Three plus one is four, so it is s p three. Do not mix them up, okay?

4:24More s-character: what changes

Now let us see why s-character matters: an s orbital lies closer to the nucleus than a p orbital. And s-character rises from s p three to s p two to s p; the pointed sign means less than. Clear? So as the s-character rises, the hybrid orbital gets shorter, and the bond it forms is shorter and stronger. Right? The carbon also holds the bonding electrons more tightly, so it is more electronegative. Now back to our puzzle. Ethane, ethene, ethyne: whose hydrogen is the most acidic? It is ethyne, yes, exactly! On screen, the small numbers below count the hydrogens, and the sign opening the other way means greater than: ethyne, then ethene, then ethane. That is line three at work: when the hydrogen leaves, the more electronegative s p carbon holds the electron pair left behind most tightly. So ethyne reacts with sodium amide, the N a N H two on screen, and ethene does not. Our puzzle is solved!

5:28Example 1: hybridisation of each carbon in penta-2,3-diene

Time for the first example. Give the hybridisation of each carbon, from left to right, in penta two, three diene: C H three, C H, double bond, C, double bond, C H, C H three. Remember the traps, and pause the video to try it first. Done? Okay, let's check. The two end carbons are C H three groups. There is no pi bond on them, so both are s p three. Okay? Carbon number two and carbon number four each have one double bond, so one pi bond each, and that gives s p two. Right? Now the middle one, carbon number three. Two double bonds, so two pi bonds, and it is s p, not s p two! That is exactly trap number one. Got it? So, from left to right: s p three, s p two, s p, s p two, s p three. Nice and symmetric, and the middle carbon is the one to watch.

6:24Example 2: nitrogen in trimethylamine, acetamide, pyridine

Now a harder one, JEE level. What is the hybridisation of nitrogen in, part a, trimethylamine, part b, acetamide, C H three, C O, N H two, and part c, pyridine? Pause the video and try it first. Done? Okay, let's check. In trimethylamine, the nitrogen forms three sigma bonds and has one lone pair, localised, with no pi bond next to it. Three plus one is four, so s p three. Okay? Now acetamide. The nitrogen again has three sigma bonds and one lone pair. Three plus one is four, so s p three again, right? No, no, wait, because this is trap number two! The nitrogen sits next to the C double bond O group, and its lone pair is delocalised into it. So it is planar, and s p two. Clear? Last, pyridine. Its nitrogen forms two sigma bonds in the ring. It also has one lone pair, lying in the ring plane. Two plus one is three, so s p two. With me? So the answers are s p three, s p two, and s p two. Same nitrogen, three situations, so always check where the lone pair goes!

7:42Quick revision: sigma plus lone pairs, traps, s-character

So, quick revision. Sigma bonds plus lone pairs: four, three and two give s p three, s p two and s p. Two pi bonds on one carbon means s p, and a delocalised lone pair gives s p two. More s-character means shorter bonds, a more electronegative carbon, and a more acidic hydrogen. Now go and try the practice problems on this page. All the best!