Question

[NiCl₄]^2- is paramagnetic, while Ni(CO)₄ is diamagnetic, although both are tetrahedral. Why?

Hint:

Compare the ligand field strength: Cl^- is weak so Ni²⁺ (d⁸) keeps 2 unpaired electrons, while CO is strong and pairs all electrons to 3d¹⁰. Both remain sp³ tetrahedral.

Answer 1

Step 1: Find the oxidation state and d-electron count of nickel in each complex.
In [NiCl₄]^2-: each Cl carries -1 and the overall charge is -2, so Ni is +2. Ni (Z = 28) is [Ar]3d⁸4s²; removing the two 4s electrons gives Ni²⁺ = 3d⁸, which has 2 unpaired electrons.
In Ni(CO)₄: CO is neutral and the complex is neutral, so Ni is 0, i.e. 3d⁸4s².
Step 2: Compare the ligand field strengths.
Cl^- is a weak field ligand and cannot pair the d electrons, so the 3d⁸ set keeps its 2 unpaired electrons. Ni²⁺ uses sp³ hybridisation (one 4s + three 4p orbitals) giving a tetrahedral shape, and the 2 unpaired electrons make it paramagnetic.
CO is a strong field ligand; it forces the 4s² electrons of Ni(0) to pair and shift into 3d, giving 3d¹⁰4s⁰. The empty 4s and 4p orbitals form sp³ hybrids, again tetrahedral, but now every electron is paired, so it is diamagnetic.
Step 3: Conclusion. Both are sp³ tetrahedral, yet the weak Cl^- leaves 2 unpaired electrons (paramagnetic) while the strong CO pairs all electrons (diamagnetic).