Question

Hybridisation of $[Ni(CN)_4]^{2-}$ is

• A. $dsp^3$
• B. $sp^3d$
• C. $dsp^2$
• D. $d^2sp^3$

Hint:

For coordination number 4:
- $Ni^{2+}$ with weak ligands (like $Cl^-$) $\rightarrow sp^3$ (Tetrahedral).
- $Ni^{2+}$ with strong ligands (like $CN^-$) $\rightarrow dsp^2$ (Square Planar).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks to identify the hybridization of the central metal atom (Nickel) in the coordination complex $[Ni(CN)_4]^{2-}$.

Step 3: Detailed Explanation:
1. Determine the oxidation state of Ni:
Let $x$ be the oxidation state.
$x + 4(-1) = -2$
$x - 4 = -2 \Rightarrow x = +2$.
The metal is in the $Ni^{2+}$ state.
2. Electronic configuration of $Ni^{2+$:}
Neutral $Ni$ ($Z=28$): $[Ar] 3d^8 4s^2$.
$Ni^{2+}$: $[Ar] 3d^8 4s^0$.
The $3d$ subshell contains 8 electrons.
3. Assess the nature of the ligand:
$CN^-$ is a strong field ligand. According to the Spectrochemical Series and Valance Bond Theory, strong field ligands cause the pairing of electrons in the $d$-orbitals if possible.
4. Orbital rearrangement:
In $Ni^{2+}$, there are two unpaired electrons in the $3d$ orbitals. The incoming strong field $CN^-$ ligands force these two electrons to pair up.
This results in one $3d$ orbital becoming empty.
5. Hybridization and Geometry:
With 4 $CN^-$ ligands, the metal needs 4 vacant hybrid orbitals. It uses:
- One $3d$ orbital
- One $4s$ orbital
- Two $4p$ orbitals
This results in $dsp^2$ hybridization.
The geometry associated with $dsp^2$ hybridization is square planar. Since all electrons are paired, the complex is diamagnetic.

Step 4: Final Answer:
The hybridization is $dsp^2$.