Step 1: Understanding the Topic
This question deals with the concept of synergic bonding in metal carbonyl complexes. The length of the metal-carbon (M–C) bond is determined by the interplay of two bonding components: $\sigma$-donation from the carbonyl ligand to the metal and $\pi$-back-donation from the metal's d-orbitals into the carbonyl's empty $\pi^*$ antibonding orbitals.
Step 2: Key Approach - The Role of $\pi$-Back-Donation
The strength of the M–C bond is critically dependent on the extent of $\pi$-back-donation.
Greater $\pi$-back-donation increases the electron density in the M–C bonding region, making the M–C bond stronger and shorter.
The extent of $\pi$-back-donation is determined by the electron density on the central metal atom. A more electron-rich metal (i.e., one with a lower or more negative oxidation state) is a better $\pi$-donor.
Step 3: Detailed Analysis
Let's determine the oxidation state of Mn in the anionic part of each complex and compare the electron density.
For $Na_2[Mn(CO)_4]$, the anion is $[Mn(CO)_4]^{2-}$. Since CO is neutral, the oxidation state of Mn is -2.
For $Na[Mn(CO)_5]$, the anion is $[Mn(CO)_5]^{-}$. The oxidation state of Mn is -1.
Let's assume the third complex for comparison is a neutral or positively charged species, for instance, $[Mn(CO)_6]^{+}$. Here, the oxidation state of Mn would be +1.
Now, let's compare the electron density on the Mn center:
\[
\text{Electron Density: } Mn(-2)>Mn(-1)>Mn(+1)
\]
A higher electron density (more negative charge) on the metal allows for stronger $\pi$-back-donation to the CO ligands.
\[
\text{Strength of Back-Donation: } [Mn(CO)_4]^{2-}>[Mn(CO)_5]^{-}>[Mn(CO)_6]^{+}
\]
Since stronger back-donation leads to a shorter M–C bond, the trend in bond length will be the reverse of the trend in back-donation strength.
Step 4: Final Answer
The M–C bond length increases as the metal center becomes less electron-rich and back-donation weakens.
\[
\text{M–C Bond Length Order: } \boxed{ [Mn(CO)_4]^{2-}<[Mn(CO)_5]^{-}<[Mn(CO)_6]^{+} }
\]