Question

Which of the following complexes is diamagnetic?

• A. \( [Fe(H_2O)_6]^{3+} \)
• B. \( [CoF_6]^{3-} \)
• C. \( [Ni(CN)_4]^{2-} \)
• D. \( [MnCl_4]^{2-} \)

Hint:

Strong field ligands like \( CN^- \) often produce low-spin complexes with paired electrons, leading to diamagnetism.

Answer 1

The Correct Option is C

Concept: A diamagnetic complex contains no unpaired electrons. Strong field ligands can pair electrons and produce diamagnetism.

Step 1: Analyzing option (A).
In \( [Fe(H_2O)_6]^{3+} \): \[ Fe^{3+} : 3d^5 \] \( H_2O \) is a weak field ligand. Thus, electrons remain unpaired. Hence, the complex is paramagnetic.

Step 2: Analyzing option (B).
In \( [CoF_6]^{3-} \): \[ Co^{3+} : 3d^6 \] \( F^- \) is a weak field ligand. The complex is high-spin with unpaired electrons. Therefore, it is paramagnetic.

Step 3: Analyzing option (C).
In \( [Ni(CN)_4]^{2-} \): \[ Ni^{2+} : 3d^8 \] \( CN^- \) is a strong field ligand. Electrons pair up and form a square planar complex. All electrons become paired. Therefore, the complex is diamagnetic.

Step 4: Analyzing option (D).
In \( [MnCl_4]^{2-} \): \[ Mn^{2+} : 3d^5 \] \( Cl^- \) is a weak field ligand. The complex contains several unpaired electrons. Hence, it is paramagnetic.

Step 5: Final conclusion.
Only \( [Ni(CN)_4]^{2-} \) is diamagnetic. \[ \boxed{[Ni(CN)_4]^{2-}} \]