Given below are two statements :
Statement I : Among Zn, Mn, Sc and Cu, the energy required to remove the third valence electron is highest for Zn and lowest for Sc.
Statement II : The correct order of the following complexes in terms of CFSE is $$[Co(H_2O)_6]^{2+} < [Co(H_2O)_6]^{3+} < [Co(en)_3]^{3+}$$.
In the light of the above statements, choose the correct answer from the options given below:

Checking Statement I

The third ionisation enthalpy corresponds to removing the third valence electron, i.e. the electron that is lost when the neutral atom has already lost its two 4s electrons.

The electronic configurations (ground state) are:

Sc : $$[Ar]\,3d^{1}\,4s^{2}$$
Mn : $$[Ar]\,3d^{5}\,4s^{2}$$
Cu : $$[Ar]\,3d^{10}\,4s^{1}$$
Zn : $$[Ar]\,3d^{10}\,4s^{2}$$

After two electrons are removed (first and second ionisations) the situations are:

Sc $$\rightarrow$$ $$[Ar]\,3d^{1}$$ (third electron will be a 3d electron)
Mn $$\rightarrow$$ $$[Ar]\,3d^{5}$$ (half-filled d5, stable)
Cu $$\rightarrow$$ $$[Ar]\,3d^{10}$$ (completely filled d10, very stable)
Zn $$\rightarrow$$ $$[Ar]\,3d^{10}$$ (completely filled d10, very stable)

The stability of a completely filled d10 subshell is greater than that of a half-filled d5 subshell, which in turn is greater than that of a single 3d electron. Hence

Energy required (IE3): $$\text{Zn} \gt \text{Cu} \gt \text{Mn} \gt \text{Sc}$$

Thus, the highest third ionisation enthalpy is for Zn and the lowest is for Sc. Statement I is true.

Checking Statement II

We compare the crystal-field stabilisation energy (CFSE) of the three octahedral complexes.

1. $$[Co(H_2O)_6]^{2+}$$ : Co(II) is $$3d^{7}$$. With the weak-field ligand H2O, the complex is high-spin:
configuration $$t_{2g}^{5}\,e_g^{2}$$
CFSE $$= (5 \times -0.4\Delta_o) + (2 \times +0.6\Delta_o) = -2.0\Delta_o +1.2\Delta_o = -0.8\Delta_o$$

2. $$[Co(H_2O)_6]^{3+}$$ : Co(III) is $$3d^{6}$$. The oxidation state +3 greatly enlarges $$\Delta_o$$, so the complex becomes low-spin:
configuration $$t_{2g}^{6}\,e_g^{0}$$
CFSE $$= 6 \times -0.4\Delta_o = -2.4\Delta_o$$

3. $$[Co(en)_3]^{3+}$$ : Co(III) is again $$3d^{6}$$, but ethylenediamine (en) is a stronger field ligand than H2O, giving an even larger $$\Delta_o$$ while the electron distribution remains low-spin:
configuration $$t_{2g}^{6}\,e_g^{0}$$
CFSE $$= 6 \times -0.4(\Delta_o') = -2.4\Delta_o'$$ where $$\Delta_o' \gt \Delta_o$$

Because $$\lvert\Delta_o'\rvert \gt \lvert\Delta_o\rvert$$, the numerical magnitude of CFSE is

$$[Co(H_2O)_6]^{2+}\;(\;0.8\Delta_o\;) \lt [Co(H_2O)_6]^{3+}\;(\;2.4\Delta_o\;) \lt [Co(en)_3]^{3+}\;(\;2.4\Delta_o'\;)$$

Therefore Statement II is also true.

Conclusion

Both Statement I and Statement II are true.

Option A which is: Both Statement I and Statement II are true