Why is hexaamminecobaltII) is blue, while [Ir(NH3)63+ is colorless? Both are low

Question

Why is hexaamminecobaltII) is blue, while [Ir(NH3)63+ is colorless? Both are low spin d complexes. O A. [Ir(NHa)al3+ is octahedral while hexaamminecobalt(lI) is tetrahedral. O B. Both complexes are octahedral but hexaamminecobalt(lI) has different ligands than [Ir(NH3)613, making it blue. C. Since lr is a third row and cobalt is a first row transition metal , the size of ° is larger for [Ir(NH3)13+ than it is for hexaamminecobalt(lI), making it colorless. 0 D. The charge in the two metals is different, making ° larger for [Ir NH16 3+ than it is for hexaamminecobalt(ll), making it colorless

Explanation / Answer

For complexes of metals of same group with same ligands and charge, the crystal field stabilization increases down the group due to increased size of the valence d- orbitals and hence lower Metal-Ligand repulsion. Ir being in the third row transihion series has larger CFSE than Co. So the iridium complex is colorless.

The answer is option C.

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