Question

Ligand's are broadly classified into two classes classical and non-classical ligands, depending on their donor and acceptor ability. Classical ligands form classical complexes while non-classical ligands form non-classical complexes. Bonding mechanism in non-classical is called synergic bonding. In compound $[M(CO)_n]^z$, the correct match for highest 'M-C' bond length for given M, n and z respectively:

• A. Cr, 6, 0
• B. V, 6, -1
• C. Ti, 6, -2
• D. Mn, 6, +1

Answer 1

Answer: (C)

Ti, 6, -2

Answer 2

The M-C bond length in metal carbonyls is influenced by the strength of the M-C bond, which is a result of synergic bonding (sigma donation from CO to metal and pi back-donation from metal to CO). A longer M-C bond length indicates a weaker M-C bond. The strength of the M-C bond is generally inversely related to the oxidation state of the metal. A lower (more negative) oxidation state of the metal leads to a weaker M-C bond and thus a longer M-C bond length.

Let's determine the oxidation state of M in each option, where the charge of CO is 0 and the overall charge of the complex is 'z'. The oxidation state of M (let's call it 'x') is given by $x + n \times (\text{charge of CO}) = z$. Since the charge of CO is 0, $x + n \times 0 = z$, which means $x = z$.

1. For Cr, n=6, z=0: Oxidation state of Cr is 0.
2. For V, n=6, z=-1: Oxidation state of V is -1.
3. For Ti, n=6, z=-2: Oxidation state of Ti is -2.
4. For Mn, n=6, z=+1: Oxidation state of Mn is +1.

The oxidation states are +1, 0, -1, and -2. The lowest oxidation state is -2, corresponding to Ti. Therefore, the M-C bond length will be longest for the complex with the lowest oxidation state.

The order of increasing M-C bond length is: $[Mn(CO)_6]^+ < Cr(CO)_6 < [V(CO)_6]^- < [Ti(CO)_6]^{2-}$

Thus, the highest M-C bond length is observed for $[Ti(CO)_6]^{2-}$, where M=Ti, n=6, and z=-2.