1. Determine oxidation state and electronic configuration of Co

In $[CoF_3(H_2O)_3]$, both $F^-$ and $H_2O$ are ligands:

• $3F^- \Rightarrow -3$
• $3H_2O \Rightarrow 0$

Since the complex is neutral, $x-3=0 \Rightarrow x=+3$ So cobalt is $Co^{3+}$.

Atomic number of Co = 27

Neutral Co: $[Ar]3d^74s^2$

Therefore, $Co^{3+}:[Ar]3d^6$

So the metal ion is a $d^6$ system.

2. Decide high spin or low spin

Given in the question: $\Delta_o < P$ This means pairing energy is greater than crystal field splitting, so the complex is high spin.

Thus for octahedral $d^6$ high spin configuration: $t_{2g}^4 e_g^2$

3. Compute CFSE

For octahedral complexes:

• each electron in $t_{2g}$ contributes $-0.4\Delta_o$
• each electron in $e_g$ contributes $+0.6\Delta_o$

So, $\text{CFSE} = 4(-0.4\Delta_o) + 2(+0.6\Delta_o)$ $= -1.6\Delta_o + 1.2\Delta_o$ $= -0.4\Delta_o$

4. Check pairing energy term

A free $d^6$ ion already has one paired electron arrangement, and high-spin octahedral $d^6$ also has one pair. Hence there is no extra pairing energy term to be added in CFSE here.

Therefore, $\boxed{\text{CFSE} = -0.4\Delta_o}$

5. Evaluate options

• A: $-0.8\Delta_o$ ❌
• B: $-0.4\Delta_o$ ✅
• C: $-0.8\Delta_o + 2P$ ❌
• D: $-0.4\Delta_o + P$ ❌

So the correct answer is Option B.

6. Comparison with stored answer

Stored correct answer: B

My derived answer: B

They agree.