Question:
The hybridisation involved in nickel of the complex ion \([Ni(CN)_4]^{2-}\) is.
The hybridisation involved in nickel of the complex ion \([Ni(CN)_4]^{2-}\) is.
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$[Ni(Cl)_4]^{2-}$ is $sp^3$ (tetrahedral) because $Cl^-$ is a weak field ligand.
Updated On: Apr 27, 2026
- $sp^{3}$
- $d^{2}sp^{3}$
- $dsp^{2}$
- $sp^{3}d$
- $sp^{2}$
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The Correct Option is C
Solution and Explanation
Step 1: Concept
Nickel in $[Ni(CN)_{4}]^{2-}$ is in the +2 oxidation state ($3d^8$ configuration).
Step 2: Meaning
$CN^{-}$ is a strong field ligand, causing pairing of electrons in the $3d$ subshell.
Step 3: Analysis
Pairing leaves one $3d$ orbital vacant. Hybridization occurs using one $3d$, one $4s$, and two $4p$ orbitals.
Step 4: Conclusion
The resulting hybridization is $dsp^2$, leading to a square planar geometry.
Final Answer: (C)
Nickel in $[Ni(CN)_{4}]^{2-}$ is in the +2 oxidation state ($3d^8$ configuration).
Step 2: Meaning
$CN^{-}$ is a strong field ligand, causing pairing of electrons in the $3d$ subshell.
Step 3: Analysis
Pairing leaves one $3d$ orbital vacant. Hybridization occurs using one $3d$, one $4s$, and two $4p$ orbitals.
Step 4: Conclusion
The resulting hybridization is $dsp^2$, leading to a square planar geometry.
Final Answer: (C)
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