, $\text{CN}^-$ is much stronger than $\text{H}_2\text{O}$.

The oxidation state of the metal ion

Higher oxidation state gives larger $\Delta_o$, because the metal ion attracts the ligand electrons more strongly.

Now compare the given complexes.

I: $[\text{Co(H}_2\text{O)}_6]^{2+}$ has $\text{Co}^{2+}$

II: $[\text{Co(H}_2\text{O)}_6]^{3+}$ has $\text{Co}^{3+}$

Both have the same ligand, $\text{H}_2\text{O}$, but II has a higher oxidation state.

So,

$$ \Delta_o(\text{I}) < \Delta_o(\text{II}) $$

Hence, option A is correct.

Now compare I and III:

I has $\text{Co}^{2+}$ with $\text{H}_2\text{O}$.

III has $\text{Fe}^{3+}$ with $\text{H}_2\text{O}$.

Since $\text{Fe}^{3+}$ has a higher oxidation state than $\text{Co}^{2+}$, the splitting will be larger for III.

So,

$$ \Delta_o(\text{I}) < \Delta_o(\text{III}) $$

Hence, option B is also correct.

For IV:

IV is $[\text{Fe(CN)}_6]^{3-}$, containing the strong field ligand $\text{CN}^-$.

Since $\text{CN}^-$ is much stronger than $\text{H}_2\text{O}$,

$$ \Delta_o(\text{IV}) > \Delta_o(\text{II}) $$

So option C, $\text{IV} < \text{II}$, is incorrect.

Option D is just the reverse of option A, so it is also incorrect.

Correct options are:

$$ \boxed{\text{A and B}} $$

Coordination Compounds

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