High Spin, Low Spin and CFSE

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. Take a cobalt three plus ion, with six d electrons. Surround it with six fluoride ions, and four electrons stay unpaired. Surround it with six ammonia molecules instead, and every electron is paired. Same ion, same electrons, different result? Something is strange here, right? In JEE, questions on high spin and low spin come up often. So, let's begin.

0:30The fourth electron has a choice

So, here is our splitting diagram. On the left, the free ion has all five d orbitals at one level. In the octahedral part, the dark blue trio at the bottom is t two g, with a small two g below the t, and the orange pair on top is e g, with a small g below; the labels in brackets name the orbitals. The double arrow is delta naught, the Greek capital delta with a small o below. Okay? The raised number on d counts the d electrons. For d one, d two and d three, they simply go into t two g, one in each orbital, no choice yet. Right? But now the fourth electron arrives, and all three t two g orbitals already hold one electron each. So where should it go? It has two options, and both cost energy. Option one is to go up into e g, and that costs delta naught, the gap marked by the double arrow. With me? Option two is to pair up in t two g, which costs the pairing energy, capital P, okay? So it is a contest between delta naught and P, and the cheaper one wins. Clear? If delta naught is less than P, with the less than sign, going up is cheaper, so the electron enters e g. This is high spin, the weak field case: d four becomes t two g three, e g one, and the raised numbers count electrons in each set. Got it? And if delta naught is greater than P, with the greater than sign, pairing is cheaper. That is low spin, the strong field case: d four becomes t two g four, e g zero. Okay?

2:03Crystal field stabilisation energy (CFSE)

Now we put a number on it, the CFSE, or crystal field stabilisation energy. The t two g set sits zero point four delta naught below the barycentre, the grey dashed line in the figure, and e g zero point six above it. So n with a small t two g below, the t two g electron count, gets minus zero point four, and n with a small e g below, plus zero point six. Clear? For low spin we may add one more term in brackets: extra pairings times P, and the cross means times. Extra means beyond those of the high spin form. Right? Let's try one together. High spin d four is t two g three, e g one. What is its CFSE? Three times minus zero point four gives minus one point two, and the single e g electron adds zero point six. So the answer is minus zero point six delta naught. Simple, isn't it? High spin d five, t two g three, e g two, gives minus one point two, plus one point two. They cancel, so the CFSE is zero; remember that. With me? Now low spin d six, t two g six: six times minus zero point four is minus two point four delta naught. All six electrons are paired, three pairs, so we add three P, right? Oh no, careful, that is the trap! High spin d six, t two g four, e g two, already has one pair, as four electrons share three orbitals. So three minus one gives two extra pairings, and the term is plus two P, okay?

3:39Octahedral: only d4 to d7 have a choice

So, let me ask you this. In an octahedral field, which d counts can be high spin or low spin? Only d four, d five, d six and d seven, yes! Only there does an electron, from the fourth to the seventh, have a real choice. Got it? For d one to d three, where the dash means to, electrons just enter t two g; for d eight to d ten, t two g fills and the rest go into e g anyway. So, one configuration, whatever the ligand. Clear? In a tetrahedral field, the right half of the figure, each electron in the lower set, e, counts minus zero point six delta t, and each one in the upper set, t two, with a small two below, counts plus zero point four delta t. Delta t, with a small t below, is the tetrahedral splitting; the wavy equals sign under the figure says it is about four ninths of delta naught. With me?

4:29Example 1: spin and CFSE of [CoF6]3- and [Co(NH3)6]3+

Now let's solve our cobalt puzzle. Give the configuration and CFSE of these two complexes, ignoring P, the pairing energy. First, open square bracket, C o, F six, close bracket, three minus; and second, open square bracket, C o, open round bracket, N H three, close round bracket, six, close square bracket, three plus. The raised signs are the charges. Pause the video and try it first. Done? Okay, let's check. Six fluorides give minus six and the complex is three minus, so cobalt is plus three; ammonia is neutral, so plus three again. Both are cobalt three plus, a d six ion. Right? Now here is the key question. Is fluoride a weak field ligand, or a strong field one? Yes, exactly, it is a weak field ligand! F with a raised minus, fluoride, sits to the left of water in the spectrochemical series, and those ligands are weak field. So it is high spin, t two g four, e g two: four times minus zero point four, plus two times zero point six, gives minus zero point four delta naught. Okay? Ammonia, N H three, is a strong field ligand, so all six electrons pair up in t two g, giving t two g six. Six times minus zero point four gives minus two point four delta naught. Got it? And the puzzle is solved. In t two g four, e g two, one orbital holds a pair, so four electrons are unpaired; in t two g six, none are. Same ion, but the ligand decided! Clear?

6:06Example 2: high or low spin for d4, Delta o 15000, P 20000

Now a harder one. For a d four ion, delta naught is fifteen thousand per centimetre, and P is twenty thousand per centimetre; that is how we read c m with a raised minus one. Which spin state is more stable? Pause the video and try it first. Done? Okay, let's check. Low spin, t two g four, gives minus one point six delta naught, and high spin only minus zero point six. More negative, so low spin wins, right? No, no, wait, we forgot the pairing energy, and that is the trap! High spin, t two g three, e g one, has no pairing: minus zero point six times fifteen thousand gives minus nine thousand per centimetre. Okay? Low spin has one extra pairing, so we add P once. Minus one point six times fifteen thousand is minus twenty four thousand, and adding twenty thousand gives minus four thousand per centimetre. Right? Minus nine thousand is less than minus four thousand, and the more negative value wins. So high spin, t two g three, e g one, is more stable. Got it? And a topper's quick check: delta naught, fifteen thousand, is less than P, twenty thousand, so the electron goes up, and it is high spin in one line. Neat, right?

7:25Quick revision: Delta o vs P, CFSE rule, d4 to d7 choice

So, quick revision. Delta naught less than P gives high spin, weak field, and greater than P gives low spin, strong field. The CFSE is minus zero point four delta naught per t two g electron, plus zero point six per e g electron, and plus P per extra pairing. In an octahedral field, only d four to d seven have a choice, and for high spin d five, the CFSE is zero. Now try the practice problems on this page. All the best!