Write oxidation state of Ni in [Ni(CO)4].

Question of 101

Q. Write oxidation state of Ni in [Ni(CO)4].

Rajasthan RbseRajasthan Board Senior Secondary Examination 2020Subjective· 1mImportance★★★★★

0% · 0/101 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Metal Carbonyl Bonding

Metal Carbonyl Bonding: From Intuition to Precise Statement

Imagine you have a metal atom — say, nickel or iron — sitting in a complex. Now imagine a carbon monoxide molecule (CO) approaching it. CO is a small, stable molecule, so why would it stick to a metal at all? The answer lies in a beautiful two-way interaction that chemists call synergic bonding.

The Intuition: A Two-Way Street

Think of CO as having two important features:

A lone pair on the carbon atom (like an outstretched hand offering electrons)
Empty π orbitals* (like open pockets that can accept electrons back)

When CO approaches a metal, the lone pair on carbon donates electrons to the metal. That's the forward donation — CO acts as a Lewis base, the metal as a Lewis acid. But here's the clever part: the metal, now electron-rich from that donation, can push some of its own electrons back into those empty π* orbitals on CO. That's the back-donation.

This isn't a one-time handshake. It's a continuous, mutual reinforcement: the forward donation makes the metal more electron-rich, which strengthens back-donation, which in turn makes the CO more willing to donate further. This synergy is why metal carbonyls are so stable.

Note

The carbon end of CO always binds to the metal — never the oxygen. The lone pair on carbon is the one that donates.

The Precise Statement

Metal carbonyl bonding consists of two simultaneous components:

σ-donation: The carbon atom's lone pair (in a σ-type orbital) donates electron density into an empty d-orbital (or hybrid orbital) on the metal. This creates a σ bond.
π-back-donation: Filled d-orbitals on the metal donate electron density into the empty π* antibonding orbitals on CO. This creates a π bond.

\text{M} \leftarrow \text{C} \equiv \text{O} \quad \text{(σ-donation)}

The net effect is a strong M–C bond, but with an important consequence: the C–O bond weakens. Why? Because back-donation puts electrons into the antibonding π* orbitals of CO, which reduces the bond order of the C–O triple bond.

Important

The C–O bond order in a metal carbonyl is less than 3. The more back-donation, the weaker the C–O bond. This is why the IR stretching frequency of CO drops from ~2143 cm⁻¹ (free CO) to values as low as 1700 cm⁻¹ in some metal carbonyls.

Visualizing the Interaction

Component	Donor	Acceptor	Effect on M–C bond	Effect on C–O bond
σ-donation	C (lone pair)	Metal (empty d-orbital)	Strengthens	No direct effect
π-back-donation	Metal (filled d-orbital)	CO (π* orbital)	Strengthens	Weakens (lowers bond order)

A Concrete Example: Ni(CO)₄ …

Why this formula?

Metal Carbonyl Bonding: Why the Key Concepts Hold

Metal carbonyls are compounds where carbon monoxide (CO) bonds to a metal atom. The bonding is synergistic — two processes reinforce each other. Let's break down why the key ideas work.

1. The Basic Picture: σ-Donation and π-Backdonation

CO bonds to a metal via two simultaneous interactions:

σ-Donation: The lone pair on carbon (from the HOMO of CO, a σ orbital) donates electrons to an empty d-orbital (or hybrid orbital) on the metal.
π-Backdonation: Filled d-orbitals on the metal donate electron density into the empty π* (antibonding) orbital of CO.

Why both are needed?

If only σ-donation occurred, the metal would become too electron-rich (negative charge buildup). π-backdonation relieves this by pushing electrons back onto CO. This mutual reinforcement is called synergistic bonding.

Key result: The M–C bond is strengthened, but the C–O bond is weakened (because electrons go into the antibonding π* orbital of CO).

2. The 18-Electron Rule: Why It Holds

The 18-electron rule states that stable metal carbonyls tend to have 18 valence electrons around the metal. Why 18?

In a transition metal, the valence orbitals are: one s, three p, and five d — total 9 orbitals.
Each orbital can hold 2 electrons (Pauli principle), so maximum capacity = 18 electrons.
This gives a noble gas configuration (like Kr, Xe, etc.), which is energetically stable.

Example:

Ni(CO)₄: Ni has 10 valence electrons (group 10). Each CO donates 2 electrons (σ-donation). Total = 10 + 4×2 = 18 → stable.

Why not 16 or 20?

16-electron complexes exist (e.g., V(CO)₆) but are less stable — they tend to dimerize or react.
20-electron complexes are rare because the extra electrons would occupy antibonding orbitals, destabilizing the molecule.

Exam tip: Count electrons carefully — include metal d-electrons and 2 electrons per CO ligand (as neutral donor).

3. The 18-Electron Rule Formula: Derivation

For a metal carbonyl complex $\ce{M(CO)_x L_y}$ (where L = other ligands), the total electron count is:

\text{Total electrons} = \text{Metal valence electrons} + 2x + 2y

Why 2 per CO?

Each CO donates its lone pair (2 electrons) to the metal. Even though π-backdonation occurs, the net donation from CO is still counted as 2 electrons in the electron-counting formalism (neutral ligand method).

Example:

$\ce{Fe(CO)_5}$ : Fe (group 8) has 8 valence electrons. 5 CO × 2 = 10. Total = 8 + 10 = 18.

4. The 18-Electron Rule: Why It's Not Always Exact

Some stable carbonyls violate the rule (e.g., $\ce{V(CO)_6}$ has 17 electrons). Why?

Steric effects: Large ligands may prevent enough COs from binding to reach 18.
Electronic factors: Some metals prefer 16 electrons (e.g., square planar complexes like $\ce{[Ni(CN)_4]^{2-}}$ ).
Radical stability: $\ce{V(CO)_6}$ is a stable 17-electron radical because the unpaired electron is delocalized over the CO ligands.

Key takeaway: The 18-electron rule is a guideline, not a law. It works best for low-valent, first-row transition metal carbonyls.

5. The Synergistic Effect: Why C–O Bond Weakens

When π-backdonation occurs, electrons fill the π* orbital of CO. This orbital is antibonding with respect to C–O.

Result: The C–O bond order decreases (from 3 in free CO to ~2.5 in a carbonyl complex). …

Start your 14-day free trial to unlock the full solution →

CO is a neutral ligand, and the complex [Ni(CO)4] carries no overall charge, so Ni must be in the zero oxidation state.

In [Ni(CO)4] (nickel tetracarbonyl):

Carbon monoxide (CO) is a neutral ligand — it donates its lone pair on carbon without carrying any charge. …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 16 on this concept.

CBSE 2026Set V11 mark
$[\text{Ni(CO)}_4]$
›Reveal solutionSolution

In $[\text{Ni(CO)}_4]$ carbon monoxide is a neutral ligand and the complex carries no charge, so nickel is in the $0$ oxidation state.

Let the oxidation state of Ni be $x$ . Carbon monoxide (CO) is a neutral ligand (charge $= 0$ ), and the complex $[\text{Ni(CO)}_4]$ is neutral overall: …
Start your 14-day free trial to see the full solution →
CBSE 2026Set A1 markMCQ
$Q. The oxidation state of Ni in Ni(CO)4 is$
(a) +4
(b) +3
(c) +2
(d) 0
›Reveal solutionSolution

Carbonyl (CO) is a neutral ligand and the complex is neutral, so nickel is in the 0 oxidation state.

In tetracarbonylnickel(0), Ni(CO)4, each CO is a neutral ligand (charge 0) and the overall complex is neutral. Let x be the oxidation state of Ni:

x + 4(0) = 0 => x = 0

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 mark
$Q. Write the oxidation state of Nickel in Ni(CO)4.$
›Reveal solutionSolution

Since carbonyl (CO) is a neutral ligand and the complex Ni(CO)4 carries no overall charge, the oxidation number of Ni must be zero.

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set X11 markMCQ
$Q. The structure of pentacarbonyliron(0) is,$
(a) tetrahedral
(b) trigonal bipyramidal
(c) octahedral
(d) square pyramidal
›Reveal solutionSolution

Pentacarbonyliron(0), $[\text{Fe(CO)}_5]$ , has a coordination number of 5 with $sp^3d$ hybridisation, giving a trigonal bipyramidal shape.

Pentacarbonyliron(0) is $[\text{Fe(CO)}_5]$ . Iron is in the zero oxidation state, bonded to five CO ligands. A coordination number of five with $sp^3d$ hybridisation gives a …
Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 mark
$Q. Fill in the blank: Oxidation state of iron in Fe(CO)5 is ________.$
›Reveal solutionSolution

In Fe(CO)5, carbonyl (CO) is a neutral ligand, so the oxidation state of Fe must equal the overall (zero) charge of the complex, i.e. 0.

CO (carbon monoxide) is always a NEUTRAL ligand in coordination chemistry (it donates a lone pair from carbon without carrying any charge itself).

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 mark
$Q. Ni(CO)₄ is diamagnetic. (True / False)$
›Reveal solutionSolution

True — in Ni(CO)₄ nickel is in the zero oxidation state with a paired, sp³-hybridised d¹⁰ configuration, so the complex is diamagnetic.

In Ni(CO)₄, CO has no charge, so nickel is present in the 0 oxidation state, giving the free-atom configuration $3d^8 4s^2 \rightarrow 3d^{10} 4s^0$ (all 10 d-electrons paired up as the 4s electrons are used for hybridisation/bonding). Ni uses one 4s and three 4p orbitals to form four equivalent sp³ hybrid orbitals, giving a tetrahedral geometry, and since all …

Start your 14-day free trial to see the full solution →
CBSE 2024Set D1 markMCQ
$Q. The oxidation state of Ni in Ni(CO)4 is$
(a) 0
(b) 1
(c) 2
(d) 4
›Reveal solutionSolution

In Ni(CO)4 the CO ligands are neutral and the whole molecule is neutral, so Ni must be 0.

Let the oxidation state of Ni be x. Carbonyl (CO) is a neutral ligand (charge 0), and the complex Ni(CO)4 has no overall charge.

x + 4(0) = 0 => x = 0

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$Q. Oxidation state of Fe in [Fe(CO)5] is -$
(a) 0
(b) +2
(c) +3
(d) +4
›Reveal solutionSolution

Since CO is a neutral ligand and the overall complex carries no charge, Fe must be in the zero oxidation state.

Let the oxidation state of Fe be x.

CO (carbonyl) is a neutral ligand, so 5 CO groups contribute a total charge of 0.

[Fe(CO)5] itself is a neutral molecule (no charge written outside the bracket).

So: x + 5(0) = 0, which gives x = 0. …

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$Q. Which of the following is π-acid ligand?$
(a) NH3
(b) CO
(c) F⁻
(d) en
›Reveal solutionSolution

A π-acid ligand has an empty π*/d orbital that can accept back-donated electron density from the metal, in addition to donating a σ lone pair.

CO donates its carbon lone pair to the metal (σ-donor) and simultaneously accepts electron density back from a filled metal d-orbital into its own empty π* antibonding orbital (π-acceptor); this 'synergic' σ-donor + π-acceptor behaviour defines a π-acid ligand, and CO is the textbook example (as in …

Start your 14-day free trial to see the full solution →
CBSE 2023Set ANNUAL1 markMCQ
$Q. The metal carbon bond in metal carbonyls exhibit$
(a) only σ-character
(b) only π-character
(c) both σ and π-character
(d) ionic character
›Reveal solutionSolution

Metal carbonyl bonding is 'synergic': CO donates a σ lone pair to the metal, and the metal back-donates electron density into CO's empty π* orbital.

In a metal carbonyl (e.g. Ni(CO)₄, Fe(CO)₅), the carbon of CO donates its lone pair into a vacant metal d-orbital, forming a σ M←C bond. Simultaneously, a filled metal d-orbital of appropriate symmetry donates (back-donates) electron density into the empty antibonding π* orbital of CO, forming a M→CO π bond. This combination of σ-donati …

Start your 14-day free trial to see the full solution →
CBSE 2022Set ANNUAL1 mark
$Q. Draw the synergic bonding present in carbonyl complex.$
›Reveal solutionSolution
Synergic bonding in metal carbonyls is a two-way (mutually reinforcing) bond: CO donates a sigma lone pair to the metal, while the metal donates back electron density into CO's empty pi* orbital; this simultaneous forward and back donation is what makes the M-C bond in carbonyls unusually strong.

The M-C bond in a metal carbonyl, M-C(triple bond)O, consists of two components that reinforce each other:

Sigma bond (ligand-to-metal donation): The carbon atom of CO has a lone pair of electrons (in an sp hybrid orbital) that is donated into a vacant d-orbital (or hybrid orbital) of the metal atom, forming a coordinate sigma bond: M <- C(triple bond)O

Pi back-bond (metal-to-ligand donation): A filled d-orbital of the metal (of correct symmetry) overlaps with the empty antibonding pi* orbital of CO, donating electron density BACK from the metal into the CO ligand: M -> C(triple bond)O (pi back donation)

Diagram (described): the metal M is joined to C by a straight sigma-bond arrow pointing from C's lone pair into M's empty d-orbital (M <- :C(triple bond)O), and by a second, curved pi-bond arrow pointing from a filled M d-orbital sideways into the empty pi* orbital of the C-O bond (M -> C(triple bond)O, drawn as two lobes overlapping perpendicular to the M-C-O axis).

…
Start your 14-day free trial to see the full solution →
CBSE 2022Set ANNUAL1 markMCQ
$Q. In metal carbonyls the ligand is ______.$
(a) carbon dioxide
(b) carbon monoxide
(c) carbonate
(d) aldehydes and ketones
›Reveal solutionSolution

"Metal carbonyls" are, by definition, coordination complexes of a metal with carbon monoxide as the ligand.

Metal carbonyls are a class of coordination (organometallic) compounds in which carbon monoxide (CO) molecules act as ligands bonded to a central transition-metal atom through the lone pair on the carbon atom, e.g. $Ni(CO)_4$ (tetracarbonylnickel(0)), $Fe(CO)_5$ , $Cr(CO)_6$ . CO is a strong-field, $\pi$ -acceptor ligand that also back-donates electron density from the filled metal d-orbitals into its own e …

Start your 14-day free trial to see the full solution →

←Previous chapterCh 4The d- and f-Block Elements

Q.Write oxidation state of Ni in [Ni(CO)4].

🎓Unlock everything free for 14 days

Unlock everything free for 14 days