Properties and Bonding in Binary Transition Metal Carbonyl Complexes

To determine the correct statement(s), let us analyze each option individually:

1. Option (A): The total number of valence electrons at the metal center is given by: $\text{Total Valence Electrons} = (\text{Valence electrons of metal}) + 2 \times (\text{Number of CO ligands})$

   • For $\text{Fe(CO)}_5$: Iron ($\text{Fe}$, $Z = 26$) has the valence electronic configuration $[\text{Ar}]\,3d^6\,4s^2$, giving $8$ valence electrons. The $5$ carbonyl ligands each contribute $2$ electrons: $\text{Total valence electrons} = 8 + (5 \times 2) = 18$
   • For $\text{Ni(CO)}_4$: Nickel ($\text{Ni}$, $Z = 28$) has the valence electronic configuration $[\text{Ar}]\,3d^8\,4s^2$, giving $10$ valence electrons. The $4$ carbonyl ligands each contribute $2$ electrons: $\text{Total valence electrons} = 10 + (4 \times 2) = 18$ Both complexes obey the $18$-electron rule. Thus, statement (A) is incorrect.

2. Option (B): Carbon monoxide ($\text{CO}$) is a very strong field ligand and a strong $\pi$-acceptor, which produces a large crystal field splitting energy ($\Delta$). This causes extensive electron pairing in the $d$-orbitals, rendering binary transition metal carbonyls predominantly low-spin (often diamagnetic) in nature. Thus, statement (B) is correct.

3. Option (C): Bonding in metal carbonyls involves synergic bonding:

   • A $\sigma$-bond is formed by the donation of a lone pair from the carbon of $\text{CO}$ into an empty $d$-orbital of the metal.
   • A $\pi$-back bond is formed by the back-donation of electron density from filled metal $d$-orbitals into the vacant $\pi^*$ (antibonding) molecular orbitals of $\text{CO}$ ($M(d\pi) \to \text{CO}(\pi^*)$).

   When the oxidation state of the metal is lowered, the electron density on the metal increases, facilitating greater extent of $d\pi \to \pi^*$ back-donation. This increases the metal-carbon ($\text{M}-\text{C}$) bond order, thereby strengthening the metal-carbon bond. Thus, statement (C) is correct.

4. Option (D): When the oxidation state of the metal is increased, the metal becomes more electron-deficient. Consequently, back-donation into the $\pi^*$ antibonding orbital of $\text{CO}$ decreases. With fewer electrons in the antibonding orbital, the $\text{C}-\text{O}$ bond order increases, making the carbonyl $\text{C}-\text{O}$ bond stronger (not weaker). Thus, statement (D) is incorrect.

Conclusion: The correct statements are (B) and (C).