In which of the following octahedral complex species the magnitude of $$\Delta_0$$ will be maximum?

To determine which octahedral complex has the maximum magnitude of $$\Delta_0$$ (crystal field splitting energy), we need to consider two main factors: the oxidation state of the central metal ion and the nature of the ligands. A higher oxidation state increases $$\Delta_0$$ because the metal ion has a greater charge density, leading to stronger interactions with the ligands. Additionally, ligands can be ranked by their field strength using the spectrochemical series: $$\text{I}^- < \text{Br}^- < \text{Cl}^- < \text{F}^- < \text{OH}^- < \text{H}_2\text{O} < \text{NH}_3 < \text{en} < \text{NO}_2^- < \text{CN}^- < \text{CO}$$. Strong field ligands (like $$\text{CN}^-$$) cause a larger splitting, while weak field ligands (like $$\text{H}_2\text{O}$$) cause a smaller splitting.

Now, let's analyze each option step by step:

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

The central metal ion is cobalt. To find its oxidation state, let the oxidation number of Co be $$x$$. Water $$(\text{H}_2\text{O})$$ is neutral, so the total charge from ligands is 0. The complex has a +2 charge. Therefore, $$x + 0 = +2$$, which gives $$x = +2$$. So, cobalt is in the +2 oxidation state. The ligand is $$\text{H}_2\text{O}$$, which is a weak field ligand. Thus, this complex will have a relatively small $$\Delta_0$$.

Option B: $$[\text{Co}(\text{CN})_6]^{3-}$$

Cyanide $$(\text{CN}^-)$$ has a charge of -1 per ligand. Let the oxidation number of Co be $$x$$. The complex has a -3 charge. There are six ligands, so $$x + 6 \times (-1) = -3$$. Solving: $$x - 6 = -3$$, which gives $$x = +3$$. So, cobalt is in the +3 oxidation state. The ligand is $$\text{CN}^-$$, which is a strong field ligand (among the strongest in the spectrochemical series). Therefore, this complex should have a large $$\Delta_0$$.

Option C: $$[\text{Co}(\text{C}_2\text{O}_4)_3]^{3-}$$

Oxalate $$(\text{C}_2\text{O}_4^{2-})$$ has a charge of -2 per ligand. Let the oxidation number of Co be $$x$$. The complex has a -3 charge. There are three ligands, so $$x + 3 \times (-2) = -3$$. Solving: $$x - 6 = -3$$, which gives $$x = +3$$. So, cobalt is in the +3 oxidation state. The ligand is oxalate, which is a moderately strong field ligand (weaker than $$\text{CN}^-$$ and $$\text{NH}_3$$ in the spectrochemical series). Thus, this complex will have a moderate $$\Delta_0$$, but less than that of strong field ligands.

Option D: $$[\text{Co}(\text{NH}_3)_6]^{3+}$$

Ammonia $$(\text{NH}_3)$$ is neutral. Let the oxidation number of Co be $$x$$. The complex has a +3 charge. There are six ligands, so $$x + 6 \times 0 = +3$$, which gives $$x = +3$$. So, cobalt is in the +3 oxidation state. The ligand is $$\text{NH}_3$$, which is a strong field ligand, but weaker than $$\text{CN}^-$$. Therefore, this complex will have a large $$\Delta_0$$, but less than that of the cyanide complex.

Comparing the complexes:

• Option A has Co(II) and a weak field ligand, so it has the smallest $$\Delta_0$$.
• Options B, C, and D have Co(III), which has a higher oxidation state than Co(II), leading to larger $$\Delta_0$$ values.
• Among the Co(III) complexes, the ligand field strength determines the order of $$\Delta_0$$: $$\text{CN}^- > \text{NH}_3 > \text{C}_2\text{O}_4^{2-}$$ (based on the spectrochemical series).

Therefore, $$[\text{Co}(\text{CN})_6]^{3-}$$ (option B) has both the highest oxidation state of cobalt and the strongest field ligand, resulting in the maximum $$\Delta_0$$.

Hence, the correct answer is Option B.