Question

Which of the following comparisons is correct for \(\Delta _0\) of the following complexes?
I- \([\text{Co(H}_2\text{O)}_6]^{2+}\)
II- \([\text{Co(H}_2\text{O)}_6]^{3+}\)
III- \([\text{Fe(H}_2\text{O)}_6]^{3+}\)
IV- \([\text{Fe(CN)}_6]^{3-}\)

• A. \(\text{I} < \text{II}\)
• B. \(\text{I} < \text{III}\)
• C. \(\text{IV} < \text{II}\)
• D. \(\text{II} < \text{I}\)

Hint:

With the same ligand (water), the metal ion with the higher charge gives the larger splitting, so the \(\text{Co}^{3+}\) complex is above the \(\text{Co}^{2+}\) complex.

Answer 1

The Correct Option is A

Step 1: What decides the size of the splitting.
Two things set $\Delta_o$ in an octahedral complex: how strong the ligand is, and how highly charged the central metal ion is. In the spectrochemical series water is a weaker ligand than cyanide.

Step 2: Look at the pair with identical ligands.
Complexes I and II are both hexaaqua cobalt complexes, so the ligand is not a factor. Only the oxidation state differs, +2 in I and +3 in II.
A +3 ion attracts the water molecules more strongly and the metal-ligand distance is shorter. This raises the splitting.

Step 3: Conclude.
So $\Delta_o(\text{I}) < \Delta_o(\text{II})$. This matches option (A).

Step 4: Why (C) and (D) fail.
In (D) the order of I and II is flipped, which goes against the charge effect. In (C), complex IV has the cyanide ligand, which gives a splitting far above that of any aqua complex, so IV cannot be smaller than II.

Final Answer:
Option (A) is correct. \[ \boxed{\text{A}} \]