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: \[ \boxed{(B)} \]