Identify the oxidation state and d-electron count for each complex:

For :

Ethylenediamine (en) is neutral.

Oxidation state of cobalt is +3.

Cobalt (atomic number 27) in the +3 state gives a configuration.

With strong/moderate field ligands like en, the complex is low spin, leading to a configuration of .

For :
Each fluoride ion is so, with six of them, Co must be in the +3 state again.

However, since fluoride is a weak field ligand, this complex adopts a high spin configuration
for , resulting in .
For .
Water is neutral.
Manganese in the +2 state gives a configuration.

With weak field water, the configuration is high spin: .

For :
Zinc in the +2 state has a configuration.
In an octahedral field, all 10 electrons fill the orbitals as .

Calculate the crystal field stabilization energy (CFSE) for each case:

The CFSE in an octahedral field can be estimated as:
For (low-spin in :
For (high-spin in ):
For (high-spin in :
For (for in ):
Compare the CFSE values:
gives a CFSE of (most negative, hence highest stabilization).

The others result in less stabilization (or net zero).

Conclusion:

The complex with the highest CFSE is the one with the configuration , which

corresponds to .

Thus, the answer is:

Option A: .