Statement I:
The third ionization energy (IE3) corresponds to the removal of the third valence electron. Let us look at the electronic configurations of the divalent cations for the given metals:
Sc2+: [Ar]3d1
Mn2+: [Ar]3d5
Cu2+: [Ar]3d9
Zn2+: [Ar]3d10
For Sc, removing the third electron involves taking it from the 3d1 state, which results in a highly stable noble gas core ([Ar]). Hence, Sc has the lowest IE3.
For Zn, the third electron must be removed from a completely filled, highly stable 3d10 subshell. Combined with its high effective nuclear charge (Zeff) 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 (Δo), which in turn depends on the oxidation state of the central metal ion and the strength of the ligand.
Oxidation state: A higher oxidation state on the central metal ion leads to a larger Δo. Therefore, the splitting for Co3+ complexes is significantly greater than for Co2+ complexes. This makes the CFSE of [Co(H2O)6]3+ greater than that of [Co(H2O)6]2+.
Nature of the ligand: Ethylenediamine (en) is a strong field bidentate ligand, whereas H2O is a weak field ligand. According to the spectrochemical series, en causes a much larger splitting than H2O. Thus, the Δo and CFSE for [Co(en)3]3+ are greater than those for [Co(H2O)6]3+.
The overall order of CFSE is:
[Co(H2O)6]2+<[Co(H2O)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