Goal: We need to find which complex has the maximum crystal field stabilization.

Key idea: Compare the CFSE (crystal field stabilization energy) for each complex.

For octahedral complexes: $\text{CFSE} = (-0.4\,n_{t_{2g}} + 0.6\,n_{e_g})\Delta_o$

For tetrahedral complexes: $\text{CFSE} = (-0.6\,n_e + 0.4\,n_{t_2})\Delta_t$

Also, strong-field ligands like $CN^-$ can cause pairing and give low-spin complexes, often increasing stabilization.

Option A: $[Ti(H_2O)_6]^{3+}$

• Metal ion: $Ti^{3+}$
• Electronic configuration of $Ti$: $[Ar]3d^24s^2$
• So $Ti^{3+} = 3d^1$
• $H_2O$ is a weak-field ligand, but for $d^1$ spin state does not matter.

Octahedral splitting: $t_{2g}^1 e_g^0$

Therefore, $\text{CFSE} = 1(-0.4\Delta_o) = -0.4\Delta_o$

Option B: $[Co(H_2O)_6]^{2+}$

• Metal ion: $Co^{2+}$
• $Co: [Ar]3d^74s^2$
• So $Co^{2+} = 3d^7$
• $H_2O$ is weak field, so this is high spin octahedral.

Configuration: $t_{2g}^5 e_g^2$

Hence, $\text{CFSE} = 5(-0.4\Delta_o) + 2(0.6\Delta_o)$ $= -2.0\Delta_o + 1.2\Delta_o$ $= -0.8\Delta_o$

Option C: $[Co(CN)_6]^{3-}$

• Ligand charge: $6(CN^-) = -6$
• Overall charge is $-3$
• So oxidation state of Co is $+3$
• Thus $Co^{3+} = 3d^6$
• $CN^-$ is a strong-field ligand, so this is low-spin octahedral.

Configuration: $t_{2g}^6 e_g^0$

Therefore, $\text{CFSE} = 6(-0.4\Delta_o) = -2.4\Delta_o$

This is very large stabilization.

Option D: $[Cu(NH_3)_4]^{2+}$

This is typically treated as a tetra-coordinate $Cu^{2+}$ complex. $Cu^{2+}$ is: $3d^9$

Whether square planar or distorted tetrahedral, the crystal field stabilization is not larger than the very large low-spin octahedral $d^6$ stabilization in option C.

If considered tetrahedral $d^9$: $e^4 t_2^5$ $\text{CFSE} = 4(-0.6\Delta_t) + 5(0.4\Delta_t) = -2.4\Delta_t + 2.0\Delta_t = -0.4\Delta_t$

Since $\Delta_t < \Delta_o$, this is much smaller than option C.

Comparison of stabilizations:

• A: $-0.4\Delta_o$
• B: $-0.8\Delta_o$
• C: $-2.4\Delta_o$
• D: much smaller than C

The maximum stabilization is clearly for: $[Co(CN)_6]^{3-}$