Question

CoCl₄^2-forms a tetrahedral complex ion and forms Co(CN)₆^3-an octahedral complex ion. What is wrong about the following statements concerning each complex ion and the d orbital splitting diagrams?

a. CoCl₄^2- is an example of a strong field case having two unpaired electrons.

b. Because CN^- is a weak-field ligand, Co(CN)₆^3- will be a low-spin case having four unpaired electrons.

Short answer

CoCl₄^2- is an example of a high spin complex and Cl is a weak field ligand. is Co(CN)₆^3-an example of a low spin complex and CN is strong field ligand.

Step-by-step solution

a) Oxidation state of Co in CoCl42- 

Co is in +2 oxidation state inCoCl₄^2-

Cl is a weak field ligand.

There are three unpaired electrons inCoCl₄^2-

b) Nature of ligands in Co(CN)63-

CN is a strong field ligand, hence Co(CN)₆^3- is low-spin complex.

There are no unpaired electrons in Co(CN)₆^3-