Question

Which of the following complexes is expected to be diamagnetic?

• A. \( [Fe(H_2O)_6]^{2+} \)
• B. \( [CoF_6]^{3-} \)
• C. \( [Ni(CN)_4]^{2-} \)
• D. \( [Mn(H_2O)_6]^{2+} \)

Hint:

Strong field ligands such as: \[ CN^- ,\ CO,\ NH_3 \] favor electron pairing and may produce diamagnetic complexes.

Answer 1

The Correct Option is C

Concept:
Diamagnetic substances contain no unpaired electrons. In coordination compounds, magnetic behavior depends on:
• Oxidation state of metal ion
• Strength of ligand field
• Pairing of electrons Strong field ligands like \( CN^- \) cause pairing of electrons.

Step 1: Analyze \( [Ni(CN)_4]^{2-} \).
Oxidation state of Ni: \[ x+4(-1)=-2 \] \[ x=+2 \] Electronic configuration: \[ Ni^{2+}: 3d^8 \] Since \( CN^- \) is a strong field ligand, electrons pair up. The complex becomes square planar with all electrons paired. Hence it is diamagnetic.

Step 2: Analyze remaining complexes.
\[ [Fe(H_2O)_6]^{2+} \] contains weak field ligand \( H_2O \), producing unpaired electrons. \[ [CoF_6]^{3-} \] contains weak ligand \( F^- \), giving high-spin complex. \[ [Mn(H_2O)_6]^{2+} \] also contains several unpaired electrons. Hence all are paramagnetic. Therefore: \[ \boxed{[Ni(CN)_4]^{2-}} \] is diamagnetic.