A Four-Step Method, and the Limits of VBT

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Iron three plus and manganese two plus both have five d electrons. Put four chlorides around iron three plus, and all five stay unpaired. Put six cyanides around manganese two plus, and only one stays unpaired! Same five d electrons, so why the difference? Today we build a four step method that answers this quickly, and then we see where valence bond theory falls short. Questions like these come up often in JEE, so let's begin.

0:35Four steps to the bonding in a complex

So, here is the method. Step one: find the oxidation state of the metal, and from it, the number of d electrons. The metal's charge and the ligand charges must add up to the charge on the complex. Okay? Now let me ask you this. Why must we know the charge on the metal before we judge the ligand? Because the same ligand can behave differently with different metal ions, and the d count decides how many electrons we place! Step two: judge the ligand. Strong ligands force the d electrons to pair up, and weak ligands do not. Right? Step three: place the electrons, and count the free inner d orbitals. The n minus one d, with n minus one in round brackets, means the d orbitals of the shell below the outer one, three d for our metals. With me? And step four: read off the hybridisation, geometry, unpaired electrons and the spin only moment, mu s, the Greek letter mu with a small s below it. Clear? That moment comes from the last lesson. mu s equals the square root of n times n plus two, the n plus two in round brackets, in Bohr magnetons, written B M, where this n is a new n, the number of unpaired electrons, not the shell number. Got it?

1:56Step two: check the metal's charge first

So, step two in more detail. Cyanide, C N minus, and carbon monoxide, C O, are strong. Chloride, C l minus, and fluoride, F minus, are weak. The full lists are in Section four point one. Okay? Now a tricky one. What about water? Is it a strong ligand or a weak one? If you said weak, you are only half right, and that is the trap! With a plus two ion, written M two plus, where M is the metal, like manganese, iron or cobalt two plus, water, H two O, the small two below counting hydrogens, and ammonia, N H three, are weak. Right? But with a plus three ion of the three d series, ammonia becomes strong, and with cobalt three plus, even water is strong! So a ligand can change sides with the metal's charge, which is why step one comes first. Clear? The course gives the proof. C o, that is cobalt, with H two O six times, all in square brackets, three plus, is diamagnetic, but the same complex with cobalt two plus has three unpaired electrons. Got it? Why the difference? The higher charge on cobalt three plus pulls the water molecules closer, and that is enough to pair its six d electrons, d with a raised six. Clear? For the full order of ligand strength, you will meet the spectrochemical series in chapter five. With me?

3:28Where valence bond theory falls short

Alright, valence bond theory, VBT for short, gave us shape and magnetism so neatly. So let me ask you this: can it also explain why so many complexes are coloured? Sadly, no, it cannot. VBT says nothing about the colour of complexes, or about their absorption spectra, which means the light they absorb. Okay? Second, it cannot predict which ligands are strong or weak. Remember cobalt three plus with water? We know water pairs the electrons there only because the complex is diamagnetic, which is magnetic data, measured first. Right? Third, it gives no picture of energies: no splitting, and no stabilisation energy. It only says paired or unpaired, nothing in numbers. Clear? Fourth, magnetic moments can depend on temperature, and VBT cannot explain this dependence. Got it? Fifth, it can misplace electrons. Take C u, that is copper, with N H three four times, all in square brackets, two plus, which VBT makes square planar, d s p two, with a raised two. Copper two plus has nine d electrons, so to free a three d orbital, VBT pushes the ninth one up into four p, where it should be lost easily. So VBT predicts easy oxidation to copper three, the Roman three in round brackets. But that is not observed! With me? And finally, it does not clearly distinguish weak and strong ligands. Questions often ask what VBT cannot explain, so keep this list handy. Okay?

5:10Example 1: [FeCl4]- is sp3, tetrahedral, 5.92 BM

Let's try the first example. Apply the four steps to F e C l four, in square brackets, with a minus charge outside. Pause the video and try it first. Done? Okay, let's check. Let x stand for the oxidation state of iron. Four chlorides, four times minus one, give minus four, and the complex is minus one. So x minus four equals minus one, and the double arrow gives x equals plus three. Iron is three d six, four s two, so iron three plus is three d five, the raised five counting d electrons. Okay? Now for step two, tell me: is chloride a strong ligand or a weak one, and so, how many unpaired electrons do we get? It is weak, yes, exactly! Chloride cannot force pairing, so the five d electrons sit one in each three d orbital, and all five are unpaired. Right? Then in step three we ask whether any three d orbital is free. No, all five hold an electron. So the four chloride lone pairs go into the four s orbital and the three four p orbitals, written four s, four p, with a raised three. Clear? And in step four, one s orbital and three p orbitals give s p three, so the complex is tetrahedral. Five unpaired electrons give mu s equal to the square root of five times seven, root thirty five, about five point nine two Bohr magnetons; the wavy equals sign means about. Got it?

6:47Example 2: [Mn(CN)6]4- is d2sp3 with one unpaired electron

Now a harder one, Example two. Apply the four steps to M n, C N six times, in square brackets, four minus. Here is a hint for you: be careful at step two. Pause the video and try it first. Done? Okay, let's check. Let x be the oxidation state of manganese. Six cyanides give minus six, and the complex is minus four. So x minus six equals minus four, and x is plus two. Manganese is three d five, four s two, so manganese two plus is three d five. Okay? Now step two, and here is the trap. Manganese two plus is a plus two ion, and with plus two ions, water and ammonia were weak. So cyanide must be weak here too, right? No, no, wait, that rule is only for water and ammonia! Cyanide is on the strong list with no condition, so it forces the electrons to pair. With me? Now in step three, all five electrons are squeezed into just three orbitals of the three d set, two, two and one, as the brackets show. So two three d orbitals are left free. Clear? And in step four, two inner d orbitals, the four s orbital and the three four p orbitals give d two s p three, the raised numbers counting orbitals. So it is octahedral, an inner orbital complex, with one unpaired electron. Got it? So mu s is the square root of one times three, root three, about one point seven three Bohr magnetons. Okay? Now, remember our puzzle? Iron three plus and manganese two plus both have five d electrons. Chloride left all five unpaired, but cyanide paired four of them. The ligand made the difference, and our puzzle is solved. Right?

8:40Quick revision: four steps, the M2+ trap, limits of VBT

So, quick revision. Four steps: oxidation state and d count, the ligand, the electrons, then read off. Water and ammonia are weak with plus two ions, so check the metal's charge first. And VBT cannot explain colour, energies or temperature effects, nor predict ligand strength in advance. Now go and try the practice problem on this page. All the best!