Question

Compare the M–C bond length in the following complexes:
$Na_2[Mn(CO)_4]$, $Na[Mn(CO)_5]$, $Na[Mn(CO)_6]$.

Hint:

More negative charge on metal carbonyl complex increases $\pi$-back bonding, which shortens the M–C bond.

Answer 1

Step 1: Understand metal–CO bonding.
In metal carbonyl complexes, bonding involves: 
1. $\sigma$-donation from CO to metal. 
2. $\pi$-back bonding from metal to CO. 
Greater $\pi$-back bonding strengthens M–C bond and shortens M–C bond length. 
Step 2: Effect of oxidation state. 
More negative charge on complex → higher electron density on metal. 
Higher electron density → stronger $\pi$-back bonding. 
Stronger back bonding → shorter M–C bond. 
Now compare charges: 
\[ [Mn(CO)_4]^{2-} (\text{in } Na_2[Mn(CO)_4]) \] \[ [Mn(CO)_5]^{-} (\text{in } Na[Mn(CO)_5]) \] \[ [Mn(CO)_6]^0 (\text{neutral}) \] 
Electron density order on metal: 
\[ 2->1->0. \] 
Thus back bonding strength: 
\[ [Mn(CO)_4]^{2-}>[Mn(CO)_5]^{-}>[Mn(CO)_6]. \] 
Step 3: Final order of M–C bond length. 
Stronger back bonding → shorter bond. 
Hence bond length order (shortest to longest): 
\[ \boxed{ [Mn(CO)_4]^{2-} < [Mn(CO)_5]^{-} < [Mn(CO)_6] }. \] 
Or in decreasing bond length: 
\[ \boxed{ [Mn(CO)_6] > [Mn(CO)_5]^{-} > [Mn(CO)_4]^{2-} }. \]