Question

The complex $[Ni(CN)_4]^{2-}$ is diamagnetic and the complex $[NiCl_4]^{2-}$ is paramagnetic. What can you conclude about their molecular geometries?

• A. Both complexes have square planar geometries
• B. Both complexes have tetrahedral geometries
• C. $[NiCl_4]^{2-}$ has a square planar geometry while $[Ni(CN)_4]^{2-}$ has a tetrahedral geometry.
• D. $[NiCl_4]^{2-}$ has a tetrahedral geometry while $[Ni(CN)_4]^{2-}$ has a square planar geometry

Answer 1

Answer: (D)

$[NiCl_4]^{2-}$ has a tetrahedral geometry while $[Ni(CN)_4]^{2-}$ has a square planar geometry

Answer 2

Here's the solution based on the magnetic properties of the complexes:

1. Determine the oxidation state of Ni: In both complexes, $[Ni(CN)_4]^{2-}$ and $[NiCl_4]^{2-}$, the oxidation state of Ni is +2.

   • For $[Ni(CN)_4]^{2-}$: $x + 4(-1) = -2 \implies x = +2$
   • For $[NiCl_4]^{2-}$: $x + 4(-1) = -2 \implies x = +2$

2. Determine the electronic configuration of $Ni^{2+}$: Nickel (atomic number 28) has the electronic configuration $[Ar] 3d^8 4s^2$. $Ni^{2+}$ has the configuration $[Ar] 3d^8$.

3. Consider the ligands: $CN^-$ is a strong field ligand (SFL), and $Cl^-$ is a weak field ligand (WFL).

4. Analyze $[Ni(CN)_4]^{2-}$:

   • $Ni^{2+}$ is $d^8$.
   • $CN^-$ is a SFL.
   • The complex is diamagnetic (no unpaired electrons).
   • For a $d^8$ ion, square planar geometry with SFLs leads to complete pairing of electrons, resulting in a diamagnetic complex. The hybridization is $dsp^2$. The crystal field splitting in a square planar field is large, pushing the $d_{x^2-y^2}$ orbital to a much higher energy. The 8 electrons fill the lower energy orbitals ($d_{xz}, d_{yz}, d_{z^2}, d_{xy}$) completely.
   • Tetrahedral geometry for a $d^8$ ion results in 2 unpaired electrons in the $t_{2g}$ orbitals $(e_g)^4 (t_{2g})^4$, making it paramagnetic.
   • Since $[Ni(CN)_4]^{2-}$ is diamagnetic, its geometry must be square planar.

5. Analyze $[NiCl_4]^{2-}$:

   • $Ni^{2+}$ is $d^8$.
   • $Cl^-$ is a WFL.
   • The complex is paramagnetic (has unpaired electrons).
   • For a $d^8$ ion, tetrahedral geometry with WFLs leads to 2 unpaired electrons $(e_g)^4 (t_{2g})^4$, making it paramagnetic. The hybridization is $sp^3$.
   • Square planar geometry for a $d^8$ ion with WFLs could theoretically be paramagnetic if the splitting was small enough to overcome pairing energy, but the typical behaviour for $d^8$ square planar complexes is diamagnetic. Given the observed paramagnetism with a WFL like $Cl^-$, the geometry is likely tetrahedral.
   • Since $[NiCl_4]^{2-}$ is paramagnetic, its geometry must be tetrahedral.

6. Conclusion: $[Ni(CN)_4]^{2-}$ has a square planar geometry, and $[NiCl_4]^{2-}$ has a tetrahedral geometry.