Crystal Field Splitting

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. Six ligands, acting like negative charges, come close to a metal ion, and they push up the energy of all five of its d orbitals. Every single one goes up. And yet, chemists say that three of these orbitals are lowered, or stabilised. Pushed up and lowered, both at once? Something is strange here, right? Today we settle it with crystal field splitting, the basis for spin and colour in complexes. Questions on it come up often in JEE papers, so let's begin.

0:35Crystal field theory: ligands as point charges

So, how does crystal field theory see a complex? It treats every ligand as a point negative charge, or as the negative end of a dipole. And it takes the metal to ligand bond as purely ionic. Okay? Now look at the metal ion alone, the free ion. Its five d orbitals all have the same energy, and we call them degenerate. Clear? Now bring the ligands close. The ligands are negative, and so are the d electrons, so they repel, and the energy of all five orbitals goes up. Right? But let me ask you this. Do all five orbitals rise by the same amount? No, they do not, and that is the key idea! An orbital that points straight at a ligand gets a bigger push, so it rises more. This unequal push splits the five orbitals into groups, and this is called crystal field splitting. With me?

1:28Octahedral field: eg up, t2g down

Let's start with the most common case, octahedral, with six ligands, the left half of the figure. The ligands sit on the three axes: plus and minus x, plus and minus y, and plus and minus z. The plus or minus sign means one ligand on each side. Okay? You know the shapes of the d orbitals. Which of them point straight along the axes, right at the ligands? Yes, exactly, d x squared minus y squared, and d z squared! The small letters below each d, with their raised twos, tell you which way it points. Facing the ligands head on, these two rise more. They form the e g set, with a small g below, the orange pair at the top of the figure. They sit at plus zero point six delta naught. Delta naught, the Greek capital delta with a small o below, is the whole gap, marked by the double arrow. Right? The other three, d x y, d y z and d z x, point between the axes, in between the ligands, so they rise less. They form the t two g set, the dark blue trio, at minus zero point four delta naught. With me? Now wait a second. The screen says minus zero point four, but we said every orbital was pushed up! So how can it be a minus? Here is the answer. Both numbers are measured from the barycentre, the flat grey dashed line labelled barycentre, which is the average energy of the five orbitals in the field. This energy weighted centre does not move: two times plus zero point six, plus three times minus zero point four, gives zero. So stabilised only means below this average. Every orbital was still pushed up, and our puzzle is solved! Clear?

3:10Tetrahedral field: the order flips

Now the right half of the figure, a tetrahedral field, with four ligands. Here, no orbital points straight at a ligand. Okay? So tell me, which set is higher now, the pair, or the trio? Many students say the pair, just like the e g set in the octahedral case. Oh no, careful, because here the order flips! The trio, d x y, d y z and d z x, is now the higher set, called t two, with no small g this time. It sits at plus zero point four delta t, the delta with a small t below, for tetrahedral. Right? And the pair, d x squared minus y squared and d z squared, is the lower set, called e, at minus zero point six delta t. Got it? Now a topper's trick. In both fields, the pair moves by zero point six, and the trio by zero point four; only the signs flip. So remember one pattern, and you have both. Clear? And how big is delta t? For the same metal and ligand, delta t is about four ninths of delta naught; the wavy equals sign means about. Fewer ligands, none pointing straight at an orbital, so a smaller split: four ninths, not nine fourths, okay? Since delta t is so small, it is nearly always less than P, the pairing energy, the cost of pairing two electrons; the sign on screen means less than. So it costs less for an electron to go up into the higher set than to pair, and tetrahedral complexes are practically always high spin. We will study spin fully in a later lesson. With me?

4:44Square planar: four ligands in one plane

Now the third pattern, square planar. Four ligands in one flat plane, sitting on the x and y axes only. Okay? Which ions take this shape? Look for a d eight ion, the raised eight counting its d electrons, with a strong ligand. Or a d eight ion of a four d or five d metal; the slash on screen means or. Right? They are square planar and diamagnetic, with no unpaired electrons. Two examples: open square bracket, N i, open round bracket, C N, close round bracket, four, close square bracket, two minus; and P t, C l four, in square brackets, two minus. The raised two minus is the charge. But why no unpaired electrons? That is our second example. Okay?

5:35Example 1: estimate Delta t when Delta o is 18000 per cm

Let's try the first example. For a metal ion and ligand with delta naught equal to eighteen thousand per centimetre, estimate delta t. The c m with the raised minus one on screen is read per centimetre. Pause the video and try it first. Done? Okay, let's check. Same metal, same ligand, so we use the relation we just learnt: delta t is about four ninths of delta naught. With me? Now the numbers, and the cross means times. Eighteen thousand over nine gives two thousand, and four times two thousand gives eight thousand. So delta t is about eight thousand per centimetre. And if you got forty thousand five hundred? Then you multiplied by nine over four, and that is exactly the trap! The tetrahedral splitting is the smaller one. Clear?

6:24Example 2: empty d orbital in a square planar d8 complex

Now a slightly harder one. Which d orbital is left empty in a square planar d eight complex, and why? Hint: think about where the four ligands sit. Pause the video and try it first. Done? Okay, let's check. The four ligands sit on the x and y axes. So which d orbital points straight at all four of them? It is d x squared minus y squared, yes, exactly! It points along the x and y axes, right at the ligands. So it is raised far above the other four orbitals. Right? Now place the eight d electrons. They fill the four lower orbitals, two in each, and four times two gives eight. Got it? So nothing is left for d x squared minus y squared, and it stays empty. That is our answer. And every electron is paired, so the complex is diamagnetic. That is exactly why the nickel and platinum complexes we saw have no unpaired electrons. See how it fits?

7:25Quick revision: octahedral, tetrahedral, square planar

So, quick revision. Ligands repel the d electrons, and the orbitals facing them rise more. Octahedral: e g goes up by zero point six delta naught, and t two g down by zero point four delta naught; in a tetrahedral field the order flips. Delta t is about four ninths of delta naught, and in square planar d eight, d x squared minus y squared is left empty. Now go and try the practice problem on this page. You can do it. All the best!