1. Identify the ligand field

The ligand $\text{phen}$ (1,10-phenanthroline) is a strong-field ligand. Hence these octahedral complexes will generally be low-spin.

So for $[M(\text{phen})_3]^{2+}$ and after oxidation $[M(\text{phen})_3]^{3+}$, we use low-spin octahedral electron configurations.

2. Interpret the question

We need the complex ion which has CFSE in the +2 state but loses CFSE on oxidation to +3 state.

That means after oxidation, the metal ion should become an electronic configuration with zero CFSE in octahedral field.

In octahedral crystal field, the configuration with zero CFSE is:

• high-spin $d^5$, or
• $d^0$, or
• $d^{10}$

Since oxidation is from $M^{2+}$ to $M^{3+}$, we check which metal gives such a situation.

3. Check each option

Option A: $[Fe(\text{phen})_3]^{2+}$

• $Fe^{2+}$ is $3d^6$
• With strong-field ligand, low-spin octahedral: $t_{2g}^6 e_g^0$
• CFSE in +2 state: $6(-0.4\Delta_o)= -2.4\Delta_o$

After oxidation:

• $Fe^{3+}$ is $3d^5$
• With strong-field ligand, low-spin octahedral: $t_{2g}^5 e_g^0$
• CFSE: $5(-0.4\Delta_o)= -2.0\Delta_o$

So CFSE is not lost; it remains nonzero.

Hence A is not correct.

Option B: $[Zn(\text{phen})_3]^{2+}$

• $Zn^{2+}$ is $3d^{10}$
• Octahedral CFSE for $d^{10}$: $0$

After oxidation:

• $Zn^{3+}$ would be $3d^9$ (though very uncommon), not zero CFSE.

But the complex in +2 state already has no CFSE, so it cannot "lose" CFSE upon oxidation.

Hence B is not correct.

Option C: $[Co(\text{phen})_3]^{2+}$

• $Co^{2+}$ is $3d^7$
• Strong-field, low-spin octahedral: $t_{2g}^6 e_g^1$
• CFSE: $6(-0.4\Delta_o)+1(+0.6\Delta_o)= -1.8\Delta_o$

After oxidation:

• $Co^{3+}$ is $3d^6$
• Strong-field, low-spin octahedral: $t_{2g}^6 e_g^0$
• CFSE: $-2.4\Delta_o$

CFSE is not lost; it actually increases in magnitude.

Hence C is not correct.

Option D: $[Ni(\text{phen})_3]^{2+}$

• $Ni^{2+}$ is $3d^8$
• Octahedral configuration: $t_{2g}^6 e_g^2$
• CFSE: $6(-0.4\Delta_o)+2(+0.6\Delta_o)= -1.2\Delta_o$

After oxidation:

• $Ni^{3+}$ is $3d^7$
• With strong-field ligand, low-spin: $t_{2g}^6 e_g^1$
• CFSE: $-1.8\Delta_o$

Again CFSE is not lost.

Hence D is not correct.

4. Conclusion from direct CFSE analysis

None of the options actually lose CFSE to zero on oxidation from $+2$ to $+3$ if we use the correct strong-field treatment for $\text{phen}$.

In particular, option A gives: $Fe^{2+}: d^6 \to t_{2g}^6, \quad Fe^{3+}: d^5 \to t_{2g}^5$ so CFSE remains nonzero after oxidation.

Therefore, the stored answer A does not agree with crystal field analysis.

5. Possible source of the stored answer

If one incorrectly assumes $Fe^{3+}$ becomes high-spin $d^5$, then octahedral CFSE would be zero, and A would be chosen. But with $\text{phen}$, a strong-field ligand, $Fe^{3+}$ should be treated as low-spin, not high-spin.

So under standard coordination chemistry, no option is correct.