Question

Which of the following complexes shows maximum number of unpaired electrons?

• A. \([Fe(CN)_6]^{4-}\)
• B. \([FeF_6]^{3-}\)
• C. \([Co(NH_3)_6]^{3+}\)
• D. \([Ni(CO)_4]\)

Hint:

Weak field ligands such as: \[ F^-,\ Cl^-,\ Br^- \] usually form high spin complexes with maximum unpaired electrons.

Answer 1

The Correct Option is B

Concept: Number of unpaired electrons depends on:
Oxidation state of metal
Nature of ligand
Crystal field splitting
Weak field or strong field ligands
Weak field ligands produce high spin complexes with more unpaired electrons. Step 1: {Analyse \([Fe(CN)_6]^{4-}\).} Oxidation state of Fe: \[ x+6(-1)=-4 \] \[ x=+2 \] Electronic configuration: \[ Fe^{2+}=3d^6 \] \(CN^-\) is strong field ligand. Thus low spin complex forms. Number of unpaired electrons: \[ 0 \] Step 2: {Analyse \([FeF_6]^{3-}\).} Oxidation state: \[ x+6(-1)=-3 \] \[ x=+3 \] Thus: \[ Fe^{3+}=3d^5 \] \(F^-\) is weak field ligand. Hence high spin complex forms. Electronic arrangement gives: \[ 5 \text{ unpaired electrons} \] Step 3: {Analyse \([Co(NH_3)_6]^{3+}\).} \[ Co^{3+}=3d^6 \] \(NH_3\) acts as strong field ligand for \(Co^{3+}\). Low spin complex forms. Number of unpaired electrons: \[ 0 \] Step 4: {Analyse \([Ni(CO)_4]\).} Nickel oxidation state: \[ 0 \] CO is strong field ligand. All electrons become paired. Thus: \[ 0 \text{ unpaired electrons} \] Step 5: {Identify the complex with maximum unpaired electrons.} Maximum unpaired electrons: \[ 5 \] present in: \[ \boxed{[FeF_6]^{3-}} \] Therefore, the correct option is: \[ \boxed{(B)} \]