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

A) [Co(NH₃)₄]²⁺ < [Co(NH₃)₆]²⁺ < [Co(en)₃]³⁺ < [Co(NH₃)₆]³⁺
B) [Co(NH₃)₆]²⁺ < [Co(NH₃)₆]³⁺ < [Co(NH₃)₄]²⁺ < [Co(en)₃]³⁺
C) [Co(en)₃]³⁺ < [Co(NH₃)₆]³⁺ < [Co(NH₃)₆]²⁺ < [Co(NH₃)₄]²⁺
D) [Co(NH₃)₄]²⁺ < [Co(NH₃)₆]²⁺ < [Co(NH₃)₆]³⁺ < [Co(en)₃]³⁺
Explanation

[Solution by Dr. Abubacker] The tetrahedral Co²⁺ complex has the smallest CFSE ($\Delta_t=\frac49\Delta_o$, and weak). High-spin octahedral Co²⁺ ($d^7$) has $-0.8\Delta_o$. Co³⁺ ($d^6$) forms low-spin complexes with a much larger $\Delta_o$ and CFSE ($-2.4\Delta_o$ plus pairing); en is a stronger-field ligand than NH₃, so [Co(en)₃]³⁺ is the highest. Answer (4).

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

Want to attempt the full paper?
View Full Paper Take Full Paper Online