Question

Which of the following transition metal complexes is expected to be diamagnetic?

• A. [Ni(CN)$_4$]$^{2-}$
• B. [NiCl$_4$]$^{2-}$
• C. [Fe(H$_2$O)$_6$]$^{3+}$
• D. [CoF$_6$]$^{3-}$

Hint:

CN$^-$ is a strong field ligand that often produces low-spin, diamagnetic complexes especially for d$^8$ metals like Ni$^{2+}$.

Answer 1

The Correct Option is A

Step 1: Link magnetism to unpaired electrons.
A complex is diamagnetic only when every electron is paired. So we look for the complex whose central metal ion ends up with no unpaired electrons.

Step 2: Find the metal ion and d-count in each.
$[Ni(CN)_4]^{2-}$ and $[NiCl_4]^{2-}$ both have $Ni^{2+}$ which is $3d^8$. $[Fe(H_2O)_6]^{3+}$ has $Fe^{3+}$ which is $3d^5$. $[CoF_6]^{3-}$ has $Co^{3+}$ which is $3d^6$.

Step 3: Sort the ligands by field strength.
Cyanide is a strong field ligand, while chloride, water, and fluoride are weak field ligands. Strong field ligands force electrons to pair up.

Step 4: Analyse the two nickel complexes.
With strong cyanide, $[Ni(CN)_4]^{2-}$ is square planar, and the eight d-electrons all pair up, leaving none unpaired. With weak chloride, $[NiCl_4]^{2-}$ is tetrahedral and keeps two unpaired electrons.

Step 5: Analyse the iron and cobalt complexes.
$[Fe(H_2O)_6]^{3+}$ with weak water is high spin and has five unpaired electrons. $[CoF_6]^{3-}$ with weak fluoride is high spin and has four unpaired electrons. Both are paramagnetic.

Step 6: Choose the diamagnetic one.
Only $[Ni(CN)_4]^{2-}$ has all electrons paired, so it alone is diamagnetic.
\[ \boxed{[Ni(CN)_4]^{2-}} \]