The Spectrochemical Series and the Size of Δ

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Fluoride, the F minus ion, carries a full negative charge, and water is neutral. If ligands acted just as point charges, the charged fluoride should split the d orbitals more than the neutral water, right? But the real order of ligands says the opposite: water is the stronger ligand. So how can a neutral molecule beat a negative ion? Hold that thought; we will solve it today, along with what makes the splitting, the Greek capital delta in the title, large or small. This comes up often in JEE, so let's begin.

0:38Spectrochemical series: increasing field strength

So, here is the spectrochemical series: ligands in order of increasing field strength, which means increasing splitting. Each less than sign means weaker than the next, the small minus up top is the charge, and the small numbers below count the atoms, okay? From the weakest: iodide, I minus, then B r minus, SCN minus, C l minus, S two minus, F minus, O H minus, oxalate, C two O four two minus, and then H two O, water. The second line carries on: NCS minus, e d t a four minus, N H three, then e n, ethylenediamine, then C N minus, and at the very end C O, carbon monoxide, the strongest of all. Got it? Water is the dividing line. Everything to the left of water is weak field, while N H three, e n, C N minus and C O are strong field. Remember these four, okay? Now look carefully. SCN minus sits on the first line, and NCS minus on the second. Same three atoms, two different places. How can that be? It depends on which atom bonds to the metal. Thiocyanate bonded through sulfur, SCN minus, is weaker than when it is bonded through nitrogen, NCS minus. Clear?

1:55What makes the splitting Delta large?

Alright, the little triangle in the heading, delta, is the size of the splitting, and the course gives four factors that make it large. Factor one is the ligand: the stronger the ligand, further right in the series, the larger the delta. Right? Factor two is the oxidation state of the metal, written M on screen. Delta naught, with a small o below, is the octahedral splitting. Delta naught for M three plus, in brackets, is roughly one point five to two times that for M two plus; the wavy equals sign means roughly. With me? So let me ask you this. Take F e, H two O six, in square brackets, with charge two plus, and the same complex with three plus. Which has the larger delta naught? Yes, exactly, the three plus complex beats the two plus one! Higher charge, and shorter metal oxygen bonds, written M dash O, so a larger splitting. Okay? Factor three is going down a group. Three d, less than four d, less than five d, and each step raises the splitting by about thirty to fifty percent. Got it? So four d and five d complexes are low spin: the gap is so large that electrons pair up in the lower set instead of jumping across.

3:09Factor four: geometry

Clear? Factor four is the geometry. Remember this picture from the last lesson? The five d orbitals of the free ion split. In the octahedral field, the t two g set goes down, the e g pair goes up, and the gap is delta naught. On the right, the tetrahedral pattern is reversed, and its gap, delta t, with a small t below, is much smaller. For the same metal and ligand, delta t is about four ninths of delta naught, less than half. With me? And for square planar complexes, the course says the splitting is large. So always check the geometry too, right?

3:45Back to the puzzle: why water beats fluoride

Now, back to our puzzle. Crystal field theory, written CFT on screen, treats the ligands as point charges, and the bonding as purely ionic. Okay? So, if CFT were the full story, which should split the d orbitals more, the charged fluoride ion or the neutral water molecule? Fluoride, because of its charge, so CFT predicts F minus greater than H two O, with the greater than sign. But the series says F minus is less than H two O, the opposite! Right? The reason is that the order depends on covalent sigma and pi bonding, the two Greek letters on screen, which CFT ignores. So it cannot explain the order, a real difficulty for the theory. Got it? Same story with C O, carbon monoxide: a neutral molecule, yet the strongest ligand of all, because of pi back bonding. So never judge a ligand by its charge; always use the series. Clear?

4:41Example 1: rank Cr3+ complexes of F-, H2O, NH3, CN-

Now, Example one, from the course. Arrange in increasing delta naught: open square bracket, C r F six, close bracket, three minus; then C r, N H three six, three plus; C r, C N six, three minus; and C r, H two O six, three plus. Pause the video and try it first. Done? Okay, let's check. Two complexes carry three minus, and two carry three plus. So is chromium in different oxidation states here? If you said yes, careful, that is the trap! Let x stand for the oxidation state of chromium in the fluoride complex. Then x plus six times minus one, in brackets, equals minus three, so x is plus three. Right? Cyanide is also minus one, so plus three again, and ammonia and water are neutral, so plus three straight away. Same metal, same oxidation state, six ligands each, so only the ligand changes. That is the topper's trick: first check what stays the same. Okay? Now the series does the job: F minus, less than H two O, less than N H three, less than C N minus. Got it? So, in increasing delta naught: the fluoride complex, the water complex, the ammonia complex, and then the cyanide complex, with the largest splitting. Clear?

6:06Example 2: larger splitting, [NiCl4]2- or [Ni(H2O)6]2+

Now a harder one, Example two. Which has the larger splitting: N i C l four, in square brackets, two minus, or N i, H two O six, two plus? Pause the video and try it first. Done? Okay, let's first see what stays the same. In the chloride complex, with x again for the oxidation state of nickel, x plus four times minus one, in brackets, equals minus two, so nickel is plus two, and water is neutral, so the aqua complex has nickel plus two as well. With me? So the metal and its oxidation state match. But two things are different here. Can you spot both? The first is the geometry, very good! N i C l four two minus is tetrahedral, the aqua complex is octahedral, and delta t is only about four ninths of delta naught. Right? The second difference is the ligand. Chloride is a negative ion and water is neutral, so chloride should be the stronger ligand, right? No, no, wait, remember our puzzle! Charge does not decide: in the series, C l minus comes before H two O, so chloride is the weaker ligand. Okay? So both factors, the geometry and the weaker ligand, shrink the splitting of the chloride complex, and the answer is N i, H two O six, two plus, the aqua complex. See, the fluoride puzzle saved us from that trap. Got it?

7:35Quick revision: the series, four factors, why charge fails

So, quick revision. The series runs from iodide, the weakest, to carbon monoxide, the strongest, and everything left of water is weak field. Delta grows with a stronger ligand, a higher oxidation state, and going from three d to four d to five d. Delta t is about four ninths of delta naught, and charge does not decide the order, because CFT ignores covalent bonding. Now go and try the practice problem on this page. All the best!