The correct order of the following complexes in terms of their crystal field stabilization energies is:
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- \( [\text{Co(NH}_3)_6]^{2+}<[\text{Co(NH}_3)_6]^{3+}<[\text{Co(NH}_3)_4]^{2+}<[\text{Co(en)}_3]^{3+} \)
- \( [\text{Co(en)}_3]^{3+}<[\text{Co(NH}_3)_6]^{3+}<[\text{Co(NH}_3)_6]^{2+}<[\text{Co(NH}_3)_4]^{2+} \)
- \( [\text{Co(NH}_3)_4]^{2+}<[\text{Co(NH}_3)_6]^{2+}<[\text{Co(NH}_3)_6]^{3+}<[\text{Co(en)}_3]^{3+} \)
- \( [\text{Co(NH}_3)_4]^{2+}<[\text{Co(NH}_3)_6]^{2+}<[\text{Co(en)}_3]^{3+}<[\text{Co(NH}_3)_6]^{3+} \)
The Correct Option is C
Solution and Explanation
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:
\[[\text{Co(NH}_3)_4]^{2+} < [\text{Co(NH}_3)_6]^{2+}<[\text{Co(NH}_3)_6]^{3+} < [\text{Co(en)}_3]^{3+}\]Top JEE Main Chemistry Questions
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