Analysis of a metal complex shows Fe(ll) is bound to an unknown ligand in a octa

Question

Analysis of a metal complex shows Fe(ll) is bound to an unknown ligand in a octahedral arrangement. The complex is red and exhibits no magnetic properties. Further analysis shows that the possible ligand bound to the Fe(ll) could be chloride or cyanide. Which ligand is most likely to be the correct and explain your reasoning. A second Fe(ll) compound is analyzed and found to be bound to a ligand in a tetrahedral geometry. The compound is orange and does exhibit magnetic properties. Analysis shows the ligand is either chloride or cyanide. Which ligand is most likely to be correct and explain your reasoning. What properties would the Fe(ll) complex exhibit if it was bound to the ligands (CI^1- and CN^1-) in a square planar geometry?

Explanation / Answer

a. Since the Fe(II) complex does not show any magnetic properties, it is diamagnetic in nature. The ligand would be a strong field ligand, that is :cyanide (CN-)". The strong field ligand causes the electrons to pair up in the d-orbitals.

b. Another complex of Fe(II) exhibit magnetic properties. Thus the electrons in d-orbitals are unpaired. The ligand is a weak field ligand and here it is "chloride (Cl-)".

c. Square planar complexes are usually low spin complexes. Thus, in both CN- and Cl- we would observe complexes which would not be magnetically active for Fe(II).

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