Comparison of Third Ionization Energy and CFSE of Cobalt Complexes

To determine the correctness of the given statements, let us analyze each statement individually:

Analysis of Statement I:

The third ionization energy ($IE_3$) corresponds to the energy required to remove the third electron from a doubly charged gaseous cation: $\text{M}^{2+}(g) \longrightarrow \text{M}^{3+}(g) + e^-$

Let us examine the electronic configurations of the $+2$ oxidation states for the given $3d$-series transition elements:

1. Zinc ($\text{Zn}$, $Z = 30$):

   • Ground state configuration: $[\text{Ar}] 3d^{10} 4s^2$
   • $\text{Zn}^{2+}$ configuration: $[\text{Ar}] 3d^{10}$
   • Removing the 3rd electron requires breaking a completely filled, extra-stable $3d^{10}$ subshell. Hence, $\text{Zn}$ has the highest $IE_3$ among all $3d$-series transition elements ($IE_3 \approx 3833\text{ kJ/mol}$).

2. Scandium ($\text{Sc}$, $Z = 21$):

   • Ground state configuration: $[\text{Ar}] 3d^1 4s^2$
   • $\text{Sc}^{2+}$ configuration: $[\text{Ar}] 3d^1$
   • Removing the 3rd electron yields the extremely stable noble gas configuration $[\text{Ar}]$ ($3d^0$). Therefore, the energy required is the lowest ($IE_3 \approx 1814\text{ kJ/mol}$).

3. Manganese ($\text{Mn}$, $Z = 25$) and Copper ($\text{Cu}$, $Z = 29$):

   • $\text{Mn}^{2+}$ configuration: $[\text{Ar}] 3d^5$ ($IE_3 \approx 3252\text{ kJ/mol}$)
   • $\text{Cu}^{2+}$ configuration: $[\text{Ar}] 3d^9$ ($IE_3 \approx 3554\text{ kJ/mol}$)

Comparing the third ionization energies: $IE_3(\text{Sc}) < IE_3(\text{Mn}) < IE_3(\text{Cu}) < IE_3(\text{Zn})$

Thus, among $\text{Zn}, \text{Mn}, \text{Sc},$ and $\text{Cu}$, $IE_3$ is highest for $\text{Zn}$ and lowest for $\text{Sc}$.

Therefore, Statement I is TRUE.

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Analysis of Statement II:

The magnitude of the Crystal Field Splitting Energy (CFSE), represented by the octahedral splitting parameter $\Delta_o$, depends on:

1. Oxidation state of the central metal ion: Higher oxidation state results in greater effective nuclear charge ($Z_{\text{eff}}$), drawing ligands closer and causing greater splitting ($\Delta_o$).

   • Comparing $[\text{Co}(\text{H}_2\text{O})_6]^{2+}$ ($\text{Co}^{2+}$) and $[\text{Co}(\text{H}_2\text{O})_6]^{3+}$ ($\text{Co}^{3+}$): $\Delta_o([\text{Co}(\text{H}_2\text{O})_6]^{2+}) < \Delta_o([\text{Co}(\text{H}_2\text{O})_6]^{3+})$

2. Nature of the ligand (Spectrochemical Series): Stronger field ligands produce larger crystal field splitting energy ($\Delta_o$).

   • According to the spectrochemical series: $\text{H}_2\text{O} < \text{en}$ (ethylenediamine is a bidentate strong-field ligand).
   • Comparing $[\text{Co}(\text{H}_2\text{O})_6]^{3+}$ and $[\text{Co}(\text{en})_3]^{3+}$: $\Delta_o([\text{Co}(\text{H}_2\text{O})_6]^{3+}) < \Delta_o([\text{Co}(\text{en})_3]^{3+})$

Combining these trends, the correct increasing order of CFSE ($\Delta_o$) is: $[\text{Co}(\text{H}_2\text{O})_6]^{2+} < [\text{Co}(\text{H}_2\text{O})_6]^{3+} < [\text{Co}(\text{en})_3]^{3+}$

Therefore, Statement II is TRUE.

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Conclusion:

Since both Statement I and Statement II are true, the correct option is A.