Question

Match List I with List II:

List I - Complex		List II - Crystal Field Splitting energy (△0)
A.	[Ti(H2O)6]2+	I.	– 1.2
B.	[V(H2O)6]2+	II.	– 0.6
C.	[Mn(H2O)6]3+	III.	0
D.	[Fe(H2O)6]3+	IV.	– 0.8

Choose the correct answer from the options given below:

• A. A-II, B-IV, C-III, D-I
• B. A-IV, B-I, C-II, D-III
• C. A-II, B-IV, C-I, D-III
• D. A-IV, B-I, C-III, D-II

Answer 1

The Correct Option is B

To solve this, we will calculate the crystal field splitting energy (CFSE) for each complex. For Ti\(^2+\) (A):
Ti\(^2+\) has a \( 3d^2 \) electron configuration. The CFSE for Ti\(^2+\) is: \[ \text{CFSE} = -0.4 \times t_{2g} + 0.6 \times e_g \] Substitute the values: \[ \text{CFSE} = -0.4 \times 2 + 0.6 \times 0 = -0.8 \quad \Rightarrow \quad \text{CFSE} = -0.8 \, \text{eV} \] For V\(^2+\) (B):
V\(^2+\) has a \( 3d^3 \) electron configuration. The CFSE for V\(^2+\) is: \[ \text{CFSE} = -0.4 \times t_{2g} + 0.6 \times e_g \] Substitute the values: \[ \text{CFSE} = -0.4 \times 3 + 0.6 \times 0 = -1.2 \quad \Rightarrow \quad \text{CFSE} = -1.2 \, \text{eV} \] For Mn\(^3+\) (C):
Mn\(^3+\) has a \( 3d^4 \) electron configuration. The CFSE for Mn\(^3+\) is: \[ \text{CFSE} = -0.4 \times t_{2g} + 0.6 \times e_g \] Substitute the values: \[ \text{CFSE} = -0.4 \times 4 + 0.6 \times 1 = -1.6 + 0.6 = -1.0 \quad \Rightarrow \quad \text{CFSE} = 0 \, \text{eV} \] For Fe\(^3+\) (D):
Fe\(^3+\) has a \( 3d^5 \) electron configuration. The CFSE for Fe\(^3+\) is: \[ \text{CFSE} = -0.4 \times t_{2g} + 0.6 \times e_g \] Substitute the values: \[ \text{CFSE} = -0.4 \times 3 + 0.6 \times 2 = -1.2 + 1.2 = 0 \quad \Rightarrow \quad \text{CFSE} = 0 \, \text{eV} \] Thus, the correct matching is:
\( A \) matches with II: \( -0.8 \, \text{eV} \)
\( B \) matches with IV: \( -1.2 \, \text{eV} \)
\( C \) matches with III: \( 0 \, \text{eV} \)
\( D \) matches with I: \( -0.6 \, \text{eV} \)