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

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