Given below are two statements: Assertion A: In the complex Ni(CO)₄ and Fe(CO)₅, the metals have zero oxidation state.…

🧠 Verify A and R Independently

Assertion A: In $\mathrm{Ni(CO)_4}$ and $\mathrm{Fe(CO)_5}$, the metals have zero oxidation state.

CO is a neutral ligand. The complexes are also neutral overall. So:

• $\mathrm{Ni(CO)_4}$: Ni + 4(0) = 0 → Ni(0). ✓
• $\mathrm{Fe(CO)_5}$: Fe + 5(0) = 0 → Fe(0). ✓

A is correct.

Reason R: Low oxidation states are found when the ligands have π-donor character.

This is wrong. CO is a π-acceptor (back-bonding from filled metal d-orbitals into CO's empty $\pi^*$). Other low-OS-stabilising ligands (CN⁻, NO, PR₃, alkenes) are also π-acceptors, not π-donors.

The actual mechanism: low-OS metals have lots of d-electron density. π-acceptor ligands relieve that excess by drawing electrons into their empty $\pi^*$ orbitals, stabilising the low OS.

π-donor ligands (F⁻, OH⁻, O²⁻) do the opposite — they pump electrons into the metal, destabilising low OS and favouring high OS instead (which is why CrO₄²⁻ has Cr(VI), MnO₄⁻ has Mn(VII), etc.).

R is incorrect.

🗺️ The Synergic Bonding Picture

In M←CO bonding:

1. $\sigma$ donation: CO lone pair → empty metal orbital.
2. $\pi$ back-donation: filled metal $t_{2g}$ → empty CO $\pi^*$ (π-acceptor behaviour).

The synergy stabilises the M–CO bond and is the entire reason that $\mathrm{Ni(CO)_4}$, $\mathrm{Fe(CO)_5}$, $\mathrm{Cr(CO)_6}$, $\mathrm{Mo(CO)_6}$ etc. exist as neutral M(0) carbonyls.

⚡ Acceptor vs Donor π-Behaviour

| Type | Examples | Stabilises | |---|---|---| | π-acceptor | CO, CN⁻, NO, PR₃, alkenes | low OS | | π-donor | F⁻, OH⁻, O²⁻ | high OS |

⚠️ The Word "Donor" Is the Trap

R sounds plausible because CO is a (σ-)donor — but it's a σ-donor + π-acceptor, not a π-donor. The π behaviour is the determining factor for OS stabilisation.

$\boxed{\text{Answer: (1) A is correct but R is not correct}}$