Step 1: Understanding the Concept:
A complex is diamagnetic if all of its electrons are paired. If even one unpaired electron is present, the complex is paramagnetic.
Magnetic behavior is explained by Crystal Field Theory (CFT), which looks at how ligands split the d-orbitals of the central metal ion.
Strong field ligands (like \(CN^-\)) cause large splitting and favor pairing (low spin). Weak field ligands (like \(H_2O, F^-, Cl^-\)) cause small splitting and favor unpaired electrons (high spin).
Step 2: Detailed Explanation:
Let's analyze each complex:
(A) \([Fe(H_2O)_6]^{3+\):} Fe is in the \(+3\) state (\(3d^5\)). \(H_2O\) is a weak field ligand. Splitting is small, so electrons stay unpaired in all five d-orbitals. It is highly paramagnetic.
(B) \([CoF_6]^{3-}\): Co is in the \(+3\) state (\(3d^6\)). \(F^-\) is a weak field ligand. It forms a high-spin octahedral complex with 4 unpaired electrons. It is paramagnetic.
(C) \([Ni(CN)_4]^{2-\):} Ni is in the \(+2\) state (\(3d^8\)). \(CN^-\) is a very strong field ligand. The complex is 4-coordinate. For \(d^8\) with a strong field, the splitting is so large that the electrons pair up in four orbitals, leaving the highest energy orbital vacant. This results in square planar geometry and zero unpaired electrons. Hence, it is diamagnetic.
(D) \([MnCl_4]^{2-}\): Mn is in the \(+2\) state (\(3d^5\)). \(Cl^-\) is a weak field ligand. This is a tetrahedral complex with 5 unpaired electrons. It is paramagnetic.
Step 3: Final Answer:
Only \([Ni(CN)_4]^{2-}\) has all its electrons paired, making it diamagnetic.