Metal Carbonyls

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Carbon monoxide, C O, is a very weak base, a poor donor of its lone pair. And yet, in the spectrochemical series, it is the strongest ligand of all. So how can a very weak base be the strongest ligand? Metal carbonyls will solve this, and they come up often in JEE papers. So let's begin.

0:24Metal carbonyls: the simplest organometallic compounds

So, a metal carbonyl has only C O ligands; we call these homoleptic carbonyls. C O binds through its carbon, and these metal to carbon bonds make them the simplest organometallic compounds. The metal is in a low oxidation state, often zero. Okay? Now, three carbonyls to know. First, N i, C O four, in square brackets: nickel with four C O groups, in round brackets, counted by the small four below. It is tetrahedral, s p three, the raised three on the p, and used in the Mond process. Right? Next, F e, C O five, is trigonal bipyramidal, d s p three. And C r, C O six, is octahedral, d two s p three, with a raised two on the d. Got it? Now look at the numbers: four C O groups on nickel, five on iron, and six on chromium. Why does each metal take a different number? Here is the secret: each C O gives two electrons. Nickel brings ten valence electrons, plus eight from four C O groups. Iron brings eight, plus ten. Chromium brings six, plus twelve. The total is always eighteen! That is the eighteen-electron rule. With me?

1:44Odd electron counts: 17, or a metal–metal bond

Now take V, C O six, in square brackets. Vanadium brings five valence electrons, and six C O groups give six times two, which is twelve; the cross means times. Five plus twelve gives seventeen, an odd number, so one electron is unpaired, making it paramagnetic. Clear? So vanadium carbonyl is one electron short. How can it reach eighteen? Yes, exactly, by gaining one electron! It is easily reduced to V, C O six, minus, the raised minus its charge, with the full eighteen. Okay? Manganese brings seven valence electrons and cobalt nine, both odd. So two metal atoms join by a metal to metal bond, written M, dash, M, on screen, and each metal counts it as one electron. Right? Take C o two, C O eight, in square brackets: two cobalt atoms and eight C O groups. In the solid, it has one cobalt to cobalt bond, two bridging C O groups between both cobalts, and six terminal ones, each on one metal. Got it?

2:52Synergic bonding: give and take

Now, how does C O hold on to the metal? First comes sigma donation, sigma being the Greek letter: the carbon lone pair goes into an empty metal orbital, the arrow meaning goes into. Okay? But C O is a very weak base, so this sigma bond alone is weak. And the metal, in a low oxidation state, has filled d orbitals. What can it do with them? It gives some back, yes! The metal donates from a filled d orbital into the empty pi star orbital of C O, pi with a raised star. Pi star is antibonding; this is pi back donation. Right? And each effect strengthens the other: the more the carbon gives, the more the metal can give back. This teamwork is called synergic bonding. With me? Now, the electrons given back enter the pi star of C O, an antibonding orbital. So what happens to the carbon to oxygen bond?

3:51Back donation: weaker C–O, stronger M–C

Yes, it gets weaker! Here is the whole picture in one figure: the grey circle M is the metal, and C O, with its triple bond, is on the right. The orange arrow on top is sigma donation, and the green arrow below is pi back donation. Electrons in pi star lower the C O bond order, so the C O bond gets longer, just as the lower label on the right says. Clear? So the stretching frequency, nu C O, the Greek letter nu with C O small below, falls too. Free carbon monoxide has two thousand one hundred and forty three per centimetre, the c m with a raised minus one. Right? In carbonyls, terminal C O is lower, about one thousand eight hundred and fifty up to two thousand one hundred and twenty per centimetre. Bridging C O, with back donation from two metals, is lower still, about one thousand seven hundred and fifty up to one thousand eight hundred and fifty. Got it? And the metal to carbon bond, written M, dash, C, has both sigma and pi character, so it is stronger and shorter than a single bond, as the upper label on the right says. With me?

5:00More electron density on the metal, more back donation

Now, more electron density on the metal, from a lower oxidation state or a negative charge, means more back donation. Okay? Look at this isoelectronic set, with equal electron counts: V, C O six, minus; C r, C O six; and M n, C O six, plus. The positive charge holds the d electrons tightly, so along this order back donation falls, the C O bond order and nu C O rise, and the M C bond weakens. Right? Now back to our puzzle. C O is a very weak base, so why is it the strongest ligand in the spectrochemical series? It is all because of back donation, yes! Back donation lowers the metal t two g orbitals, the t with a small two g below, and raises delta naught, the Greek capital delta with a small o below. So C O, a pi acceptor, is the strongest ligand, and our puzzle is solved. Clear?

5:58Example 1: 18 electron rule for each Mn in Mn2(CO)10

Let's try the first example. Show that each manganese in M n two, C O ten, in square brackets, obeys the eighteen-electron rule. That is two manganese atoms and ten C O groups. Pause the video and try it first. Done? Okay, let's check. C O is neutral, so manganese is in the zero oxidation state, the zero in round brackets, and it brings seven valence electrons. Okay? Now careful, here is the trap: ten C O groups in all, not ten on each manganese. With no bridging C O, each carries five, and five times two gives ten. The M n to M n bond adds one more. Right? So each manganese has seven plus ten plus one, which gives eighteen, and obeys the rule. If you got twenty eight, you gave all ten C O groups to one manganese. Clear?

6:54Example 2: C-O bond length order of Ni, Co, Fe carbonyls

Now a harder one. Arrange in increasing C O bond length: N i, C O four; C o, C O four, minus; and F e, C O four, two minus, in square brackets, the raised signs being charges. Hint: think of back donation. Pause the video and try it first. Done? Okay, let's check. All three are isoelectronic, d ten, the raised ten counting the d electrons. Nickel is zero, cobalt minus one, iron minus two, as the brackets show. With me? Many students reason like this: iron carries the most negative charge, so it has the most electron density, so its C O bond must be the strongest and the shortest. Is that right? No, no, wait, that is the trap! That extra electron density does not go into the C O bond. The more negative the metal, the more it pushes back into the pi star of C O, which lowers the C O bond order and makes the bond longer, not shorter. Okay? So, in increasing C O bond length, the order is nickel carbonyl, then the cobalt anion, then the iron anion; the sign on screen means less than. See how back donation decides it?

8:09Quick revision: 18 electrons, back bonding, C-O length

So, quick revision. Carbonyls have only C O ligands, a low oxidation state, and usually eighteen electrons. Synergic bonding is sigma donation to the metal plus pi back donation into pi star. More back donation means a stronger M C bond, a longer C O bond, and a lower nu C O frequency. Now try the practice problems on this page. All the best!