Question

Which of the following carbonyls will have the strongest $\text{ C}-\text{O }$ bond?
A) $\text{ Mn(CO)}_{6}^{+}\text{ }$
B) $\text{ Cr(CO}{{\text{)}}_{\text{6}}}\text{ }$
C) $\text{ V (CO)}_{6}^{-}\text{ }$
D) $\text{ Fe(CO}{{\text{)}}_{\text{5}}}\text{ }$

Answer 1

The ligands such as $\text{ CO }$, $\text{ C}{{\text{N}}^{-}}\text{ }$ and $\text{ N}{{\text{O}}^{\text{+}}}\text{ }$ are capable of accepting an appreciable amount of electron density from the metal atom into their empty $\text{ }\pi \text{ }$ or $\text{ }\pi *\text{ }$ orbitals. These are called $\text{ }\pi -$ acceptors or $\text{ }\pi -$ acid ligands. The strength of the $\text{ M }-\text{ C }$bond depends on the charge of the metal atom. The strength of the $\text{ C}-\text{O }$ bond is inversely related to the strength of the $\text{ M }-\text{ C }$ bond.

Complete step by step answer:
The carbonyl $\text{ CO }$ molecule has the highest occupied molecular orbital contain the lone pair of electrons that is projecting away from the carbon atoms. When $\text{ CO }$it acted as a ligand, it acts as a weak donor of an electron to the metal atoms and forms a $\text{ M }\leftarrow \text{ C }$ bond. The empty antibonding molecular orbital of the $\text{ CO }$overlaps with the d-orbitals of the transition metal and forms a bond.
This results in the bond formation from the filled d-orbitals on the metal into vacant orbitals on$\text{ CO }$ligands. This is called back bonding. The bond is represented as$\text{ M }\to \text{ C }$.
The empty orbital of the carbonyl carbon accepts the electron density from the metal and results in the formation of $\text{ d}\pi -\text{p}\pi \text{ }$ bonding. The extent of back bonding depends on the positive charge on the metal.
As the positive charge on the central metal atom increase, the more readily the metal can donate its electrons to the antibonding p-orbital of carbonyl ligand. This bonding increases the strength of the metal –carbonyl bond and thus weakens the carbon-oxygen bond between the carbonyl ligand.
Let's have a look at the charge on the metal atoms in the given coordination compounds of the carbonyl. The carbonyl$\text{ CO }$ ligand is a neutral ligand. The charge on it is equal to zero. Thus the charge on the central atom is as follows,
A) $\text{ Mn(CO)}_{6}^{+}\text{ }$ :
The central metal $\text{ Mn }$ has a $\text{ + 1 }$ charge on it.

B) $\text{ Cr(CO}{{\text{)}}_{\text{6}}}\text{ }$ :
The central metal atom $\text{ Cr }$ has a zero change on it.

C) $\text{ V (CO)}_{6}^{-}\text{ }$ :
The central metal atom $\text{ V }$ has a negative charge $\text{ }-1\text{ }$ on it.

D) $\text{ Fe(CO}{{\text{)}}_{\text{5}}}\text{ }$ :
The central metal atom $\text{ Fe }$ has a zero charge on it.
The $\text{ Mn }$ have a positive charge compared to other metals. Thus it is not easy for the manganese to donate its electron density to the antibonding orbitals. Therefore, $\text{ Mn }$ would form a weak metal –carbonyl bond.
Therefore, the $\text{ C}-\text{O }$ bond is the strongest $\text{ Mn(CO)}_{6}^{+}\text{ }$.
So, the correct answer is “Option A”.

Note: Note that, if the metal has a negative charge, it can easily donate its electron density to the carbon atom of the carbonyl ligand. This back bonding forms a strong bond between the metal and carbonyl and weakens the $\text{ C}-\text{O }$ bond. The$\text{ Fe(CO}{{\text{)}}_{\text{5}}}\text{ }$ thus greater chances of $\text{ d}\pi -\text{p}\pi \text{ }$ back bonding and thus have the low bond order of $\text{ C}-\text{O }$ bond. This is well studied using $\text{ IR }$ spectroscopy.