Animated Solution for Chemistry - Coordination Compounds: The correct statement(s) regarding the binary transition metal carbonyl compounds is (are) (Atomic numbers : Fe = 26, Ni = 28)

Let's begin by evaluating the first statement regarding the total number of valence shell electrons. To determine the stability of transition metal complexes, we often rely on the 18-electron rule. This rule states that thermodynamically stable transition metal complexes are formed when the sum of the metal's valence electrons and the electrons donated by the ligands equals $18$.

For Iron pentacarbonyl, ${\text{Fe(CO)}}_5$, the central iron atom has an atomic number of $26$. Its electronic configuration is $[\text{Ar}]3d^{6}4s^2$, giving it $8$ valence electrons. Each of the five carbon monoxide ($\text{CO}$) ligands donates a pair of electrons, contributing $10$ electrons in total. Adding these together, $8+10=18$ valence electrons.

Similarly, for Nickel tetracarbonyl, ${\text{Ni(CO)}}_4$, nickel ($Z=28$) has a configuration of $[\text{Ar}]3d^{8}4s^2$, which means it has $10$ valence electrons. The four $\text{CO}$ ligands donate $8$ electrons. The sum is $10+8=18$ valence electrons. Therefore, both complexes have $18$ valence electrons, not $16$. Statement (A) is incorrect.

Moving to the second statement, we must consider the nature of the $\text{CO}$ ligand. According to the spectrochemical series, carbon monoxide is an exceptionally strong field ligand.

When a strong field ligand approaches a transition metal, it causes a large splitting in the d-orbital energies (${\Delta }_o>P$). This large energy gap forces the metal's d-electrons to pair up in the lower energy orbitals rather than occupying the higher energy orbitals. This pairing of electrons results in what we call low spin complexes. Thus, statement (B) is perfectly correct.

The core of this problem lies in understanding the unique interaction between the metal and the carbonyl ligand, known as synergic bonding. This bond is a two-way street.

First, the $\text{CO}$ molecule donates a lone pair of electrons from its carbon atom into an empty orbital on the metal, forming a standard $\sigma$-bond.

However, the metal doesn't just take; it gives back. The metal atom donates electron density from its filled d-orbitals into the empty ${\pi }^{\ast }$ (antibonding) orbitals of the $\text{CO}$ ligand. This is called $\pi$-back donation. This back-donation creates a partial double bond character between the metal and the carbon, significantly strengthening the $\text{M-C}$ bond. But, because these electrons are entering an antibonding orbital of the $\text{CO}$ molecule, the $\text{C-O}$ bond order decreases, thereby weakening the $\text{C-O}$ bond.

Finally, let's analyze how the oxidation state of the metal affects this delicate balance. If we lower the oxidation state of the metal, we are essentially increasing the electron density on the metal center.

A metal that is richer in electrons will be much more capable of pushing electron density back into the $\text{CO}$ ligand's ${\pi }^{\ast }$ orbitals. This increased $\pi$-back donation further strengthens the $\text{M-C}$ bond. Consequently, statement (C) is correct.

Conversely, if we increase the oxidation state (making the metal more positive), the metal becomes electron-deficient. It holds onto its electrons tightly and engages in less $\pi$-back donation. With fewer electrons entering the $\text{CO}$ antibonding orbital, the $\text{C-O}$ bond is not weakened as much; in fact, relative to a lower oxidation state, the $\text{C-O}$ bond strengthens. Therefore, statement (D), which claims the $\text{C-O}$ bond weakens when the oxidation state increases, is incorrect.