Question

The correct order of the following complexes in terms of their crystal field stabilization energies is:

• A. \( [\text{Co(NH}_3)_6]^{2+} < [\text{Co(NH}_3)_6]^{3+} < [\text{Co(NH}_3)_4]^{2+} < [\text{Co(en)}_3]^{3+} \)
• B. \( [\text{Co(en)}_3]^{3+} < [\text{Co(NH}_3)_6]^{3+} < [\text{Co(NH}_3)_6]^{2+} < [\text{Co(NH}_3)_4]^{2+} \)
• C. \( [\text{Co(NH}_3)_4]^{2+} < [\text{Co(NH}_3)_6]^{2+} < [\text{Co(NH}_3)_6]^{3+} < [\text{Co(en)}_3]^{3+} \)
• D. \( [\text{Co(NH}_3)_4]^{2+} < [\text{Co(NH}_3)_6]^{2+} < [\text{Co(en)}_3]^{3+} < [\text{Co(NH}_3)_6]^{3+} \)

Hint:

For coordination compounds, the crystal field stabilization energy is influenced by the oxidation state of the metal and the ligand field strength. Stronger field ligands and higher oxidation states generally lead to higher CFSE.

Answer 1

The Correct Option is C

To determine the correct order of the complexes in terms of their Crystal Field Stabilization Energies (CFSE), we need to analyze each complex based on the following factors:

1. Nature of the ligand: The ability of the ligand to split the d-orbitals is called the ligand's field strength. Ligands like en (ethylenediamine) are stronger than NH₃ according to the spectrochemical series.
2. Oxidation state of the metal: Higher oxidation states increase the splitting of the d-orbitals.
3. Geometry of the complex: The complexes are given to have octahedral geometry unless mentioned otherwise.

Now, let's consider each complex individually:

• \([\text{Co(NH}_3)_4]^{2+}\): Cobalt is in a +2 oxidation state, and NH₃ is a moderate field ligand.
• \([\text{Co(NH}_3)_6]^{2+}\): Cobalt is in a +2 oxidation state, and with six ammonia molecules, the ligand field slightly increases compared to four ligands due to enhanced coordination.
• \([\text{Co(NH}_3)_6]^{3+}\): Cobalt is in a +3 oxidation state, significantly increasing the field splitting of the d-orbitals because of the higher positive charge.
• \([\text{Co(en)}_3]^{3+}\): Cobalt in a +3 oxidation state with en ligands, which are stronger field ligands than NH₃, results in maximum stability of the complex.

Considering the above analyses and the spectrochemical series, we conclude the increasing order of CFSE:

1. [\text{Co(NH}_3)_4]^{2+}
2. [\text{Co(NH}_3)_6]^{2+}
3. [\text{Co(NH}_3)_6]^{3+}
4. [\text{Co(en)}_3]^{3+}

This is because the combination of the stronger ligand and higher oxidation state results in the greatest CFSE.

Thus, the correct order is:

\( [\text{Co(NH}_3)_4]^{2+} < [\text{Co(NH}_3)_6]^{2+} < [\text{Co(NH}_3)_6]^{3+} < [\text{Co(en)}_3]^{3+} \)

Answer 2

The crystal field stabilization energy (CFSE) depends on the oxidation state and the ligand field strength. Generally: - The higher the oxidation state of the metal, the stronger the ligand field and the higher the CFSE. - \( \text{NH}_3 \) is a stronger field ligand than \( \text{en} \) (ethylenediamine). Thus: - \( [\text{Co(NH}_3)_4]^{2+} \) will have the lowest CFSE, as it is in a lower oxidation state. - \( [\text{Co(NH}_3)_6]^{2+} \) has a higher CFSE compared to \( [\text{Co(NH}_3)_4]^{2+} \). - \( [\text{Co(NH}_3)_6]^{3+} \) has a higher oxidation state, leading to higher CFSE. - \( [\text{Co(en)}_3]^{3+} \) has the highest CFSE due to the strong ligand field of \( \text{en} \). Hence, the correct order is: \[ [\text{Co(NH}_3)_4]^{2+} < [\text{Co(NH}_3)_6]^{2+} < [\text{Co(NH}_3)_6]^{3+} < [\text{Co(en)}_3]^{3+} \]