What Metal Carbonyls Are
A metal carbonyl is a coordination compound in which a transition metal is bonded to carbon monoxide molecules acting as ligands. When the metal binds only CO groups, the compound is called a homoleptic carbonyl. Most transition metals form such carbonyls, and they are notable for having simple, well-defined structures with the metal almost always in a low or zero oxidation state.
What makes carbonyls conceptually rich is the ligand itself. Carbon monoxide is a neutral, two-electron donor that binds through its carbon atom, never the oxygen. On its own CO is a weak base, yet it forms remarkably strong bonds to low-valent metals. Resolving that apparent contradiction is the entire point of the synergic bonding model developed below.
Carbonyls also sit at the boundary between coordination chemistry and organometallic chemistry. The M–C linkage to a carbon-donor ligand is the defining feature of organometallic compounds, which is why NEET frequently pairs carbonyl questions with the broader category of metal–carbon bonded species.
Mononuclear Carbonyls and Their Geometries
Mononuclear carbonyls contain a single metal atom. Their shapes follow directly from the coordination number, and three are essential for NEET:
| Carbonyl | IUPAC name | Coordination No. | Geometry | Oxidation state of M |
|---|---|---|---|---|
| $\ce{Ni(CO)4}$ | Tetracarbonylnickel(0) | 4 | Tetrahedral | 0 |
| $\ce{Fe(CO)5}$ | Pentacarbonyliron(0) | 5 | Trigonal bipyramidal | 0 |
| $\ce{Cr(CO)6}$ | Hexacarbonylchromium(0) | 6 | Octahedral | 0 |
Note that the metal is in the zero oxidation state in every case. This is unusual and is itself a clue: a neutral metal centre is electron-rich, which (as we will see) is exactly what is needed to push electron density back onto the CO ligands.
The three benchmark mononuclear homoleptic carbonyls and the polyhedron each adopts.
Why CO Is a Special Ligand
To understand the M–CO bond, look first at the free CO molecule. It has a triple bond, $\ce{C#O}$, and a lone pair on each atom. Two molecular-orbital features matter:
- A filled $\sigma$ orbital concentrated on carbon — this is the carbon lone pair that CO donates to the metal.
- A pair of empty antibonding $\pi^*$ orbitals — these are vacant, low-lying, and able to accept electron density from the metal.
This combination — a $\sigma$ donor that also carries empty $\pi^*$ acceptor orbitals — is what allows the two-way bonding. A ligand that can both donate a $\sigma$ pair and accept metal $\pi$ electron density is called a $\pi$-acceptor (or $\pi$-acid) ligand. CO is the textbook example.
The Synergic M–CO Bond
The metal–carbon bond in metal carbonyls possesses both $\sigma$ and $\pi$ character. Following NCERT §5.6, it is built from two donations going in opposite directions:
- M–C $\sigma$ bond (ligand → metal): the lone pair on the carbonyl carbon is donated into a vacant orbital of the metal. This is ordinary dative bonding, ligand to metal.
- M–C $\pi$ bond (metal → ligand): a pair of electrons from a filled $d$ orbital of the metal is donated into the vacant antibonding $\pi^*$ orbital of CO. This is the reverse direction — metal to ligand — and is called back-donation or back-bonding.
The crucial point is that the two donations reinforce each other. As CO donates its $\sigma$ pair to the metal, the metal becomes electron-rich and is better able to back-donate; as the metal back-donates into $\pi^*$, the C–O linkage loosens and CO accepts the $\sigma$ donation more readily. This mutual amplification is the synergic effect, and it strengthens the bond between CO and the metal.
Top arrow: CO carbon lone pair donates into a vacant metal orbital ($\sigma$ donation). Bottom arrows: filled metal $d$ orbital overlaps the empty CO $\pi^*$ orbital ($\pi$ back-donation).
Because the $\pi$ contribution arrives from the metal, it is the metal-to-ligand bonding that creates the synergic effect — the ligand-to-metal $\sigma$ alone could not. This is why CO bonds so strongly to low-valent metals despite being a poor classical base.
The $\sigma$-donation half of this picture is the same dative-bond idea used to count electrons in Valence Bond Theory of coordination compounds — revise it alongside back-bonding.
Evidence: Bond Lengths and IR Frequency
The synergic model is not just a story — it makes two measurable predictions, both of which are confirmed experimentally and both of which NEET tests.
Bond-length evidence
Pushing electron density into the antibonding $\pi^*$ orbital of CO reduces the C–O bond order. A lower bond order means a longer C–O bond. Simultaneously, the new metal–carbon $\pi$ overlap adds to the existing $\sigma$ bond, raising the M–C bond order and giving a shorter M–C bond. So the two bond lengths move in opposite directions, exactly as the synergic picture demands.
Infrared (IR) evidence
The C–O stretching vibration is sensitive to bond order. Free carbon monoxide absorbs near $2143\ \text{cm}^{-1}$. When CO coordinates to a metal and receives back-donation, the weakened C–O bond vibrates at a lower frequency. The more back-donation, the lower the C–O stretching frequency. This makes the IR stretch a direct, quantitative reporter of how much $\pi$ electron density the metal has transferred.
| As back-donation increases… | C–O bond order | C–O bond length | C–O IR frequency | M–C bond strength |
|---|---|---|---|---|
| More electron density into CO π* | Decreases | Increases (longer) | Decreases (lower) | Increases (stronger) |
Which bond gets stronger and which gets weaker?
Students often flip the two outcomes. Back-bonding strengthens M–C (the bond being formed) and weakens C–O (because $\pi^*$ is antibonding for C–O). The two always move in opposite directions.
More back-donation → shorter, stronger M–C → longer, weaker C–O → lower C–O IR frequency.
Metal Charge and the Strength of Back-Bonding
How much a metal back-donates depends on how electron-rich it is. A metal carrying extra negative charge has more $d$-electron density to give away, so it back-donates more strongly into CO $\pi^*$. The consequence: across a series of carbonyls differing only in the charge on the metal, the most negative metal produces the longest, weakest C–O bond.
Order the following by increasing C–O bond length: $\ce{Ni(CO)4}$, $\ce{[Mn(CO)6]+}$, $\ce{[Co(CO)4]-}$, $\ce{[Fe(CO)4]^2-}$.
First assign the metal charge in each: Mn is $+1$, Ni is $0$, Co is $-1$, Fe is $-2$. The more negative the metal, the greater the back-donation into $\pi^*$, hence the longer the C–O bond.
Increasing C–O bond length: $\ce{[Mn(CO)6]+} < \ce{Ni(CO)4} < \ce{[Co(CO)4]-} < \ce{[Fe(CO)4]^2-}$. The doubly negative $\ce{[Fe(CO)4]^2-}$ has the maximum back-donation and the longest C–O bond — this is the 2016 NEET answer.
The EAN Rule
The Effective Atomic Number (EAN) rule is a bookkeeping guide to why particular carbonyls are stable. It states that a metal in a stable carbonyl tends to surround itself with enough electrons to reach the configuration of the next noble gas. The EAN is calculated as:
EAN = (atomic number of metal) − (electrons lost to reach its oxidation state) + (electrons donated by all ligands).
Each CO donates two electrons (its $\sigma$ lone pair). For the three benchmark carbonyls, all with the metal in oxidation state $0$, the count works out neatly:
| Carbonyl | Z of metal | e⁻ from CO ligands | EAN | Noble gas |
|---|---|---|---|---|
| $\ce{Ni(CO)4}$ | 28 | 4 × 2 = 8 | 36 | Kr |
| $\ce{Fe(CO)5}$ | 26 | 5 × 2 = 10 | 36 | Kr |
| $\ce{Cr(CO)6}$ | 24 | 6 × 2 = 12 | 36 | Kr |
All three reach $36$, the electron count of krypton, which is why each is stable and why the number of CO groups is fixed at 4, 5 and 6 respectively. The EAN rule also rationalises the diamagnetism of $\ce{Ni(CO)4}$: in the presence of the strong-field CO ligand the metal electrons pair up, leaving no unpaired electrons.
CO is a strong-field ligand
Because CO is such a good $\pi$-acceptor, it sits at the strong-field end of the spectrochemical series. In $\ce{Ni(CO)4}$ the $d$ electrons pair up, so the complex is diamagnetic with $sp^3$ hybridisation and a tetrahedral shape — contrast $\ce{[NiCl4]^2-}$, which is tetrahedral but paramagnetic because $\ce{Cl-}$ is weak-field.
$\ce{Ni(CO)4}$: tetrahedral, $sp^3$, zero unpaired electrons, diamagnetic.
Polynuclear Carbonyls in Brief
Some carbonyls contain more than one metal atom and are called polynuclear carbonyls. Two are named in NCERT:
- Decacarbonyldimanganese(0), $\ce{Mn2(CO)10}$, is made of two square-pyramidal $\ce{Mn(CO)5}$ units joined directly by a Mn–Mn metal–metal bond.
- Octacarbonyldicobalt(0), $\ce{Co2(CO)8}$, has a Co–Co bond bridged by two CO groups. These bridging carbonyls span both metals rather than binding a single one.
Polynuclear carbonyls thus introduce two features absent in mononuclear ones: direct metal–metal bonds and bridging CO ligands. For NEET, the key recall is that $\ce{Ni(CO)4}$ and $\ce{Fe(CO)5}$ are mononuclear, whereas $\ce{Mn2(CO)10}$ and $\ce{Co2(CO)8}$ are dinuclear.
Bonding in Metal Carbonyls in one screen
- CO binds through carbon; the M–C bond has both $\sigma$ and $\pi$ character.
- $\sigma$ donation: carbon lone pair → vacant metal orbital (ligand to metal).
- $\pi$ back-donation: filled metal $d$ orbital → empty CO $\pi^*$ (metal to ligand).
- The two reinforce each other — the synergic effect — strengthening M–C and weakening C–O.
- Evidence: back-bonding lengthens C–O, shortens M–C, and lowers the C–O IR stretch below $2143\ \text{cm}^{-1}$.
- More negative metal → more back-donation → longest C–O bond (e.g. $\ce{[Fe(CO)4]^2-}$).
- Geometries: $\ce{Ni(CO)4}$ tetrahedral, $\ce{Fe(CO)5}$ trigonal bipyramidal, $\ce{Cr(CO)6}$ octahedral; all reach EAN 36 (Kr).
- $\ce{Ni(CO)4}$ is diamagnetic ($sp^3$); polynuclear $\ce{Mn2(CO)10}$ and $\ce{Co2(CO)8}$ have M–M bonds.