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) is influenced by the metal's oxidation state and the ligand's field strength.

General principles dictate: 
- A higher metal oxidation state correlates with a stronger ligand field and consequently a higher CFSE. 
- \( \text{NH}_3 \) functions as a stronger field ligand than \( \text{en} \) (ethylenediamine). 

Consequently, the following relationships apply: - \( [\text{Co(NH}_3)_4]^{2+} \) exhibits the lowest CFSE due to its lower oxidation state. 
- Compared to \( [\text{Co(NH}_3)_4]^{2+} \), \( [\text{Co(NH}_3)_6]^{2+} \) displays a higher CFSE. 
- \( [\text{Co(NH}_3)_6]^{3+} \) has a higher CFSE than \( [\text{Co(NH}_3)_6]^{2+} \) due to its elevated oxidation state. 
- \( [\text{Co(en)}_3]^{3+} \) possesses the highest CFSE, attributed to the potent ligand field exerted by \( \text{en} \). 

Therefore, the order of CFSE is: 

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