Match List-I with List-II.

	List-I Complex		List-II Crystal Field splitting energy ($$\Delta_o$$)
A.	$$[Ti(H_2O)_6]^{2+}$$	I.	-1.2
B.	$$[V(H_2O)_6]^{2+}$$	II.	-0.6
C.	$$[Mn(H_2O)_6]^{3+}$$	III.	0
D.	$$[Fe(H_2O)_6]^{3+}$$	IV.	-0.8

Choose the correct answer from the options given below:

The formula for CFSE in an octahedral field is:

• A. $$[Ti(H_2O)_6]^{2+}$$
  • Titanium oxidation state: Ti2+ (3d2 configuration)
  • Electron distribution: t2g2 eg0
  • $$\text{CFSE} = [2 \times (-0.4) + 0 \times 0.6]\Delta_o = \mathbf{-0.8\Delta_o}$$
  A matches with IV


• B. $$[V(H_2O)_6]^{2+}$$
  • Vanadium oxidation state: V2+ (3d3 configuration)
  • Electron distribution: t2g3 eg0
  •$$\text{CFSE} = [3 \times (-0.4) + 0 \times 0.6]\Delta_o = \mathbf{-1.2\Delta_o}$$
B matches with I


• C. $$[Mn(H_2O)_6]^{3+}$$
  • Manganese oxidation state: Mn3+ (3d4 high-spin configuration)
  • Electron distribution: t2g3 eg1
  • $$\text{CFSE} = [3 \times (-0.4) + 1 \times 0.6]\Delta_o = [-1.2 + 0.6]\Delta_o = \mathbf{-0.6\Delta_o}$$
  C matches with II


• D. $$[Fe(H_{2}O)_{6}]^{3+}$$
  • Iron oxidation state: Fe3+ (3d5 high-spin configuration)
  • Electron distribution: t2g3 eg2
  • $$\text{CFSE} = [3 \times (-0.4) + 2 \times 0.6]\Delta_o = [-1.2 + 1.2]\Delta_o = \mathbf{0\Delta_o}$$
  D matches with III

Summary Mapping:

Conclusion:

The correct option order is A-IV, B-I, C-II, D-III.

Correct Option: C