Solved: CHEMISTRY Question for NEET

NEET CHEMISTRYMedium

The hybridization involved in the complex $[Ni(CN)_4]^{2-}$ is: (Atomic number of Ni = 28)

dsp²

sp³

d²sp²

d²sp³

Oxidation State: In the complex ion $[Ni(CN)_4]^{2-}$ , the cyano group ( $CN^-$ ) is a monodentate negative ligand. Let $x$ be the oxidation state of Nickel. $x + 4(-1) = -2 \Rightarrow x = +2$ . Thus, Nickel is in the $+2$ oxidation state ( $Ni^{2+}$ ).
Electronic Configuration: The ground state configuration of Ni (Z=28) is $[Ar]3d^8 4s^2$ . For $Ni^{2+}$ , the configuration is $[Ar]3d^8$ (electrons are removed from $4s$ first).
Ligand Field Effect: The cyanide ion ( $CN^-$ ) is a strong field ligand according to the spectrochemical series. It causes the pairing of the two unpaired electrons present in the $3d$ orbitals against Hund's rule.
Hybridisation: After pairing, one $3d$ orbital becomes empty. This empty inner $3d$ orbital mixes with the empty $4s$ and two empty $4p$ orbitals to form four equivalent $dsp^2$ hybrid orbitals.
Geometry: The $dsp^2$ hybridisation corresponds to a square planar geometry. The complex is diamagnetic (no unpaired electrons) .

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