Given below are two statements: Statement I: Among Zn, Mn, Sc and...

Statement I:
The third ionization energy ($IE_3$) corresponds to the removal of the third valence electron. Let us look at the electronic configurations of the divalent cations for the given metals:
$Sc^{2+}$: $[Ar] 3d^{1}$
$Mn^{2+}$: $[Ar] 3d^{5}$
$Cu^{2+}$: $[Ar] 3d^{9}$
$Zn^{2+}$: $[Ar] 3d^{10}$

For $Sc$, removing the third electron involves taking it from the $3d^{1}$ state, which results in a highly stable noble gas core ($[Ar]$). Hence, $Sc$ has the lowest $IE_3$.
For $Zn$, the third electron must be removed from a completely filled, highly stable $3d^{10}$ subshell. Combined with its high effective nuclear charge ($Z_{eff}$) across the 3d series, $Zn$ requires the highest energy for the removal of the third electron.
Thus, Statement I is true.

Statement II:
The Crystal Field Stabilization Energy (CFSE) depends on the magnitude of the crystal field splitting energy ($\Delta_o$), which in turn depends on the oxidation state of the central metal ion and the strength of the ligand.
1. Oxidation state: A higher oxidation state on the central metal ion leads to a larger $\Delta_o$. Therefore, the splitting for $Co^{3+}$ complexes is significantly greater than for $Co^{2+}$ complexes. This makes the CFSE of $[Co(H_2O)_6]^{3+}$ greater than that of $[Co(H_2O)_6]^{2+}$.
2. Nature of the ligand: Ethylenediamine ($en$) is a strong field bidentate ligand, whereas $H_2O$ is a weak field ligand. According to the spectrochemical series, $en$ causes a much larger splitting than $H_2O$. Thus, the $\Delta_o$ and CFSE for $[Co(en)_3]^{3+}$ are greater than those for $[Co(H_2O)_6]^{3+}$.

The overall order of CFSE is:
$[Co(H_2O)_6]^{2+} < [Co(H_2O)_6]^{3+} < [Co(en)_3]^{3+}$
Thus, Statement II is true.

Since both statements are correct, the correct option is (1).

Answer: Both Statement I and Statement II are true