Question

Which among the following statements about $[\text{Ni}(\text{CN})_4]^{2-}$ is NOT true?

• A. In this electrons are paired prior to hybridization.
• B. Ni undergoes $\text{dsp}^2$ hybridization.
• C. Oxidation state of Ni is +6 .
• D. It is a square planar complex.

Hint:

Whenever a question asks for a "NOT true" statement, calculate the oxidation states of the central transition metal first! Oxidation states like $+6$ are exceptionally rare for standard 3d metals like Nickel, making it an easy structural error to spot.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We must examine the coordinate complex ion $[\text{Ni}(\text{CN})_4]^{2-}$ using Valence Bond Theory (VBT) and crystal field parameters to locate the mathematically or conceptually false statement.

Step 2: Key Formula or Approach:
Let's first determine the basic chemical oxidation state of Nickel (Ni) in the complex by setting up a charge balance equation: $$\text{Oxidation State of Ni} + 4 \times (\text{Charge of CN}^-) = \text{Net Charge of Complex}$$

Step 3: Detailed Explanation:
Let the oxidation state of Ni be $x$. $$x + 4(-1) = -2$$ $$x - 4 = -2 \implies x = +2$$ Thus, the oxidation state of Ni is $+2$. This immediately reveals that statement (C) which claims the oxidation state is $+6$ is completely incorrect. Let's double-check the structural properties for conceptual validation: Ground state configuration of Ni ($Z=28$) is $[\text{Ar}]\,3\text{d}^8 4\text{s}^2$. For $\text{Ni}^{2+}$, the configuration is $[\text{Ar}]\,3\text{d}^8 4\text{s}^0$. Cyanide ($\text{CN}^-$) is a powerful

strong-field ligand. According to VBT, it forces the pairing of the two unpaired electrons residing in the $3\text{d}$ subshell prior to hybridization (Statement A is correct). This forced pairing leaves exactly one internal $3\text{d}$ orbital empty. The $\text{Ni}^{2+}$ ion blends this single $3\text{d}$ orbital with one $4\text{s}$ and two $4\text{p}$ orbitals to give a set of four

$\text{dsp}^2$ hybrid orbitals (Statement B is correct). Any coordination compound featuring $\text{dsp}^2$ hybrid orbitals displays a

square planar geometry (Statement D is correct).

Step 4: Final Answer:
The statement that is NOT true is option (C).