Synergic bonding in metal carbonyls: why does back-donation weaken the C-O bond but strengthen the M-C bond?

Second attempt, so I should know this by now, but metal carbonyls were never properly clear in my head last year. I memorised "synergic" and moved on. This time I want to actually understand it.

What I think I know: CO donates a lone pair from carbon to an empty orbital on the metal (a sigma bond). The metal, which is in zero oxidation state with lots of d electrons, donates electron density back from a filled d orbital into the antibonding $\pi^*$ orbital of CO. That's the back-bonding.

What I don't get: if the metal pushes electrons into an antibonding orbital of CO, I can see the C–O bond should weaken. But NCERT says this back-donation strengthens the M–C bond. How? And how is it "synergic" if both processes help each other?

Also, why does CO bind through carbon when oxygen is more electronegative? And shapes: I have $Ni(CO)_4$ tetrahedral, $Fe(CO)_5$ trigonal bipyramidal and $Cr(CO)_6$ octahedral, but I can't explain them beyond rote. Last, what are the uses (Mond process) that are asked in NEET?

8 answers & replies

[+]Rahul DubeyDropperSample· 16d ago

Two bonds between metal and carbon now, a sigma bond and a pi bond (from back-donation), so the M–C bond gets partial double-bond character, shorter and stronger. And the electrons that went into $\pi^*$ of CO lower the C–O bond order, so it gets slightly longer and weaker.

Synergic means the two effects feed each other. Sigma donation from CO makes the metal more electron-rich, which makes it a better pi donor. Back-donation relieves the build-up of negative charge on the metal, which makes CO a better sigma donor. Each strengthens the other.

[+]Kavya NaikDropperSample· 16d ago

About carbon: the donor lone pair in CO sits on carbon and is the higher-energy, more available one. Oxygen is more electronegative so it holds its lone pairs tightly. Carbon's lone pair is the one that is given away. That's all you need at NEET level.

[+][–]Siddharth RaoDropperSample· 16d agoAuthor
So the oxygen lone pair isn't involved in bonding to the metal at all?
- [+][–]Kavya NaikDropperSample· 16d ago
  Right, in the normal terminal carbonyl it's the carbon end only. Oxygen's lone pairs stay put.

[+]Tushar PawarDropperSample· 16d ago

Shapes: I think $Ni(CO)_4$ would be square planar, since four-coordinate Ni complexes are square planar when the ligand is strong field and CO is about as strong as it gets.

[+]Arjun Nair12thSample· 16d ago

Tushar, the square planar idea is for $Ni^{2+}$ ( $d^8$ ) with strong ligands like CN⁻. In $Ni(CO)_4$ nickel is in zero oxidation state, $3d^{8}4s^2$ rearranged to $3d^{10}$ , and it uses 4s and 4p orbitals, so $sp^3$ and tetrahedral. Diamagnetic.

$Fe(CO)_5$ is $dsp^3$ , trigonal bipyramidal. $Cr(CO)_6$ is $d^2sp^3$ , octahedral. All diamagnetic. Nice check: each obeys the 18-electron count, Cr: 6 + 12 = 18, Fe: 8 + 10 = 18, Ni: 10 + 8 = 18.

[+][–]Tushar PawarDropperSample· 16d ago
ah d10 not d8, my bad. thanks arjun, 18 electron check is slick

[+]Mehak AroraDropperSample· 16d ago

Uses, as the sweet little summary i made:

Mond process: impure nickel is heated with CO at about 330–350 K to form volatile $Ni(CO)_4$ , then the gas is heated to about 450–470 K and it decomposes to give very pure nickel and CO.
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