The Kossel–Lewis Approach and the Octet Rule

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Neon has eight electrons in its outer shell, and it hardly reacts at all. Chlorine has seven, just one fewer, and it reacts very readily. So what is so special about the number eight? And here is a twist. In sulphur hexafluoride, S F six, the sulphur atom sits with twelve electrons around it, and the molecule is perfectly stable. So is eight really magic, or not? Today we settle it: the octet rule and its exceptions, which come up often in JEE bonding questions. So, let's begin.

0:38The octet rule: aim for eight

So, start with the noble gases. Neon, argon and the heavier ones each have eight electrons in the valence shell, the outermost shell. That arrangement is very stable, okay? Now Kossel and Lewis had one simple idea. Other atoms combine so that they also reach eight valence electrons, the noble gas arrangement. This is the octet rule; octet just means a set of eight. Now tell me: chlorine has seven. How can it get to eight? Two ways! Kossel said: by transfer. Chlorine simply takes one electron from another atom. Lewis said: by sharing. Two atoms share a pair of electrons, and both count that pair. Transfer or sharing, the target is the same: eight. With me? But careful with the small ones. Hydrogen, and the ions of lithium and beryllium, L i plus and B e two plus on screen, do not end up with eight. They end up with two, a duplet, just like helium, H e, okay? And to keep count, we use Lewis symbols. The element's symbol, with one dot for each valence electron. Sodium, N a, gets one dot; oxygen, O, gets six; chlorine, C l, gets seven. Just the valence electrons from its group. Simple, right?

2:03Exception 1: the incomplete octet

Now, this rule has three exceptions. Exception one: the incomplete octet. The central atom ends up with fewer than eight electrons around it. Look at beryllium chloride, B e C l two. The small two written below means two chlorine atoms, okay? Beryllium has only four electrons around it. Boron trifluoride, B F three, and the monomer of aluminium chloride, a single A l C l three molecule: only six around the central atom. Got it? Such molecules are short of electrons, so we call them electron deficient. And they act as Lewis acids: they can accept a pair of electrons. Clear? Now, a quick question. Take the ammonium ion, N H four plus. The small plus at the top right is its positive charge. Positive means short of electrons, so it must be electron deficient, right? No, no, careful! This is a common trap. Electron deficient is about the count of electrons around an atom, not about the charge. Look at B F three. It is completely neutral, no charge at all, but boron has only six electrons. So it is electron deficient. And N H four plus? Positive, yes. But nitrogen has four bonds, and four times two is eight: a complete octet. So it is not electron deficient at all. Charge and electron count are two different things. Okay?

3:28Exceptions 2 and 3: odd and expanded

Exception two: odd-electron molecules. Nitric oxide, N O; nitrogen dioxide, N O two; and chlorine dioxide, C l O two. Let's count for nitric oxide. Nitrogen brings five valence electrons, and oxygen brings six. So how many in total? And can both atoms get eight? Eleven! An odd number. Electrons sit in pairs, in bonds and in lone pairs. So with an odd total, one electron is always left over, and one atom is left with only seven. No way around it, right? Now exception three goes the other way: the expanded octet. Here the central atom holds more than eight: ten, twelve or even fourteen electrons. How? It uses its d orbitals as well. Okay? Phosphorus pentachloride, P C l five: five bonds, ten electrons. S F six: six bonds, twelve. Iodine heptafluoride, I F seven: seven bonds, fourteen. Sulphuric acid, H two S O four, too. Remember our puzzle? S F six, twelve electrons, perfectly stable. Now you know why: sulphur expanded its octet. But can every atom do this? No! Only atoms of the third period and heavier. The second-period atoms, carbon, nitrogen, oxygen and fluorine, cannot expand. Their second shell has only s and p orbitals; there is no two d. So for them, eight is the absolute limit. Remember this one; we need it in example two.

5:05Other limits of the octet rule

Now two more limits. We started with the noble gases as our model of stability. But they do form compounds! Xenon difluoride, X e F two, and X e O F four. So even noble gases do react, sometimes. And the second limit: the octet rule tells us nothing about the shape of a molecule, and nothing about its energy, that is, how stable it is. For those, we need other ideas, which come later. Clear?

5:34Example 1: which of BeCl2, BCl3, SiCl4, PCl5 obey octet

Alright, now let's use all this. Example one. Count the electrons around the central atom in beryllium chloride, boron trichloride, silicon tetrachloride and phosphorus pentachloride. And say which of them obey the octet rule. Pause the video, and try it first. Done? Okay, let's check. One simple tool: each single bond puts two electrons around the central atom. One condition: no lone pairs on the central atom. Here every valence electron of the central atom is in a bond, so that holds. Okay? So, beryllium chloride: two bonds, two times two, four electrons. Boron trichloride: three times two, six. Both fall short of eight: incomplete octets. Silicon tetrachloride: four times two, eight. That is a perfect octet! Phosphorus pentachloride: five times two, ten. More than eight. Now, P C l five breaks the octet rule. So it cannot exist, right? No, no, wait! P C l five is real; we met it a minute ago as an expanded octet. Breaking the octet rule does not mean a molecule cannot exist. So the answer: four, six, eight and ten, and only silicon tetrachloride obeys the octet rule. Got it?

6:55Example 2: why NCl5 and OF4 cannot exist

Now a slightly harder one. Example two. Which of these cannot exist, and why? N C l five, P C l five, S F six, O F four, and C l F three. Pause and try. Hint: look at the period of the central atom first. Done? Okay, let's check. Phosphorus, sulphur and chlorine are all in the third period. They have vacant three d orbitals, so they can expand their octets. Clear? So P C l five has ten, and S F six has twelve. C l F three: three bonds make six, plus the four electrons chlorine keeps as two lone pairs. Ten in all. All three exist. Now, oxygen sits just above sulphur, in the same group. Sulphur happily makes S F six. So O F four should be fine too, right? No, no, careful! This is the trap. Here the group does not decide it; the period does. Nitrogen and oxygen are second-period atoms: no d orbitals, so never more than eight electrons. Remember? N C l five: five bonds, so nitrogen would need ten electrons. O F four: four bonds put eight around oxygen, and oxygen still keeps two electrons of its own as a lone pair. Eight plus two, ten. Both are impossible. So, the answer: N C l five and O F four cannot exist. See how quickly the period settles it?

8:25Quick revision: octet rule and its three exceptions

So, quick revision. Octet rule: atoms reach eight, by transfer, as Kossel said, or by sharing, as Lewis said. Three exceptions: incomplete octet, odd-electron molecules, and the expanded octet, only for the third period and heavier. And electron deficient means too few electrons, not a positive charge. Now go and try the practice problems on this page. All the best!