Question

Which of the following complexes is diamagnetic and square planar?

• A. $[\text{CoF}_6]^{3-}$
• B. $[\text{Co}(\text{NH}_3)_6]^{3+}$
• C. $[\text{NiCl}_4]^{2-}$
• D. $[\text{Ni}(\text{CN})_4]^{2-}$

Hint:

$[\text{Ni}(\text{CN})_4]^{2-}$ and $[\text{NiCl}_4]^{2-}$ are the classic textbook comparisons for coordination chemistry. Always remember: Nickel(II) with a strong ligand ($\text{CN}^-$) goes square planar and diamagnetic, while Nickel(II) with a weak ligand ($\text{Cl}^-$) stays tetrahedral and paramagnetic.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are looking for a coordination complex that simultaneously possesses a square planar geometry and is diamagnetic (has zero unpaired electrons).

Step 2: Key Formula or Approach:
According to Valence Bond Theory (VBT) and Crystal Field Theory (CFT): 1. A coordination number of 4 can yield either tetrahedral ($\text{sp}^3$) or square planar ($\text{dsp}^2$) geometry.
2. Strong field ligands (like $\text{CN}^-$) cause maximum pairing of electrons. For a $\text{d}^8$ configuration metal ion like $\text{Ni}^{2+}$, strong ligands force pairing, freeing up an inner d-orbital to form a square planar ($\text{dsp}^2$) arrangement with zero unpaired electrons (diamagnetic).

Step 3: Detailed Explanation:
Let's analyze each choice: 3.

$[\text{CoF}_6]^{3-}$: Here, $\text{Co}^{3+}$ has a $\text{d}^6$ configuration. Since $\text{F}^-$ is a weak-field ligand, it forms an outer-orbital high-spin complex ($\text{sp}^3\text{d}^2$) which is octahedral and strongly paramagnetic (4 unpaired electrons).
4.

$[\text{Co}(\text{NH}_3)_6]^{3+}$: $\text{Co}^{3+}$ is $\text{d}^6$. $\text{NH}_3$ behaves as a strong field ligand here, causing pairing to form an inner-orbital low-spin complex ($\text{d}^2\text{sp}^3$). It is diamagnetic, but its structural geometry is octahedral (coordination number = 6).
5.

$[\text{NiCl}_4]^{2-}$: $\text{Ni}^{2+}$ has a $\text{d}^8$ configuration. $\text{Cl}^-$ is a weak-field ligand, meaning no electron pairing takes place. It utilizes $\text{sp}^3$ hybridization, creating a tetrahedral geometry with 2 unpaired electrons (paramagnetic).
6.

$[\text{Ni}(\text{CN})_4]^{2-}$: $\text{Ni}^{2+}$ has a $\text{d}^8$ configuration: $3\text{d}^8 \, 4\text{s}^0 \, 4\text{p}^0$. $\text{CN}^-$ is an exceptionally strong field ligand. It forces the two unpaired electrons in the $3\text{d}$ subshell to pair up together. This completely clears out one single $3\text{d}$ orbital. The metal ion then uses this inner orbital for hybridization: $\text{dsp}^2$, giving it a perfectly square planar geometry. Because all electrons are fully paired, it is diamagnetic.

Step 4: Final Answer:
The complex that is both diamagnetic and square planar is $[\text{Ni}(\text{CN})_4]^{2-}$, matching option (D).