Step 1: Understanding the Concept:
The crystal field splitting energy \(\Delta_o\) is the gap between the \(t_{2g}\) and \(e_g\) levels in an octahedral complex. A larger gap means a stronger crystal field. The gap grows with (i) a stronger ligand and (ii) a higher charge on the metal ion.
Step 2: Key Formula or Approach:
We compare the complexes using two rules.
1. For the same metal and ligand, a higher oxidation state gives a larger \(\Delta_o\), because the ligands sit closer to a more highly charged ion.
2. For the same metal ion, the spectrochemical series puts \(\text{CN}^- \gg \text{H}_2\text{O}\).
Step 3: Compare I and II.
Both have the same metal, cobalt, and the same ligand, water. Complex I has \(\text{Co}^{2+}\) and complex II has \(\text{Co}^{3+}\).
The \(\text{Co}^{3+}\) ion is smaller and more highly charged, so water ligands are pulled in closer and split the d orbitals more.
So \(\Delta_o\) of II is larger than \(\Delta_o\) of I. That means I \(<\) II, which is option (A).
Step 4: Check the remaining options.
Option (D) says II \(<\) I. This is the reverse of what we just found, so it is wrong.
Option (C) says IV \(<\) II. Complex IV has \(\text{CN}^-\), a very strong field ligand, on \(\text{Fe}^{3+}\), and it has a much larger splitting than the water complex of \(\text{Co}^{3+}\). So IV is greater than II, and (C) is wrong.
Final Answer:
The comparison I \(<\) II is correct, because \(\text{Co}^{3+}\) gives a larger \(\Delta_o\) than \(\text{Co}^{2+}\) with the same ligand.
\[ \boxed{\text{(A) I} < \text{II}} \]