Question

Answer the following questions about the complexes \[ [NiCl_4]^{2-} \quad \text{and} \quad [Ni(CN)_4]^{2-} \] (i) Write the hybridization involved in each case. (ii) Which of them is the inner orbital complex and which one is the outer orbital complex ? (iii) Compare their magnetic behaviour. (Atomic Number of Ni = 28)

Hint:

CN$^-$ is a strong field ligand and usually causes electron pairing, whereas Cl$^-$ is a weak field ligand and generally does not cause pairing.

Answer 1

Step 1: Electronic configuration of $Ni^{2+}$.
Ni (Z = 28): $[Ar]\,3d^8\,4s^2$. After losing 2 electrons: $Ni^{2+}: [Ar]\,3d^8$. Both complexes have the same $Ni^{2+}$ ion; the difference arises from the ligand field.
Step 2: Effect of ligand field strength on d-electron pairing.
$Cl^-$ is a weak field ligand and cannot pair the $3d$ electrons, leaving 2 unpaired electrons and no vacancy in the inner $3d$ subshell. $CN^-$ is a strong field ligand that forces pairing of all $3d$ electrons, creating one vacant $3d$ orbital available for hybridization.
Step 3: Hybridization, geometry, and magnetic properties.
$[NiCl_4]^{2-}$: $sp^3$ hybridization, tetrahedral, outer orbital complex, 2 unpaired electrons, paramagnetic. $[Ni(CN)_4]^{2-}$: $dsp^2$ hybridization, square planar, inner orbital complex, no unpaired electrons, diamagnetic. \[ \boxed{[NiCl_4]^{2-}:\ sp^3,\ \text{paramagnetic};\quad [Ni(CN)_4]^{2-}:\ dsp^2,\ \text{diamagnetic}} \]