Why is hexaamminecobalt(lI) is blue, while [Ir(NH3)63* is colorless? Both are lo

Question

Why is hexaamminecobalt(lI) is blue, while [Ir(NH3)63* is colorless? Both are low spin d complexes. OA. Both complexes are octahedral but hexaamminecobalt(Ill) has different ligands than [Ir(NH3)6)3, making it blue. B. Since Ir is a third row and colbalt is a first row transition metal the size of ? is larger for IrNH3 6 3 than it is for hexaamminecobalt(lll O c. The charge in the two metals is different, making ? larger for [Ir(NH3)6]3t than it is for hexaamminecobalt(?), making it colorless. making it colorless. D. [r(NH3)sl3+ is octahedral while hexaamminecobalt(Il) is tetrahedral

Explanation / Answer

option B is the answer.

oxidation number of both metal are same. But the size of Ir is greater than Co. so energy gap between t2g and eg is greater for Ir in compare to Co.

so the enrgy gap of Ir corresponds to in UV region and that of Co come in visible region.

so Ir complex is color less and Ir complex is blue colored.

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