Choose the correct statement for the [Ni(CN)_ | vedclass

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(C) $1$. The atomic number of $Ni$ is $28$. Its electronic configuration is $[Ar] 3d^8 4s^2$.$2$. In $[Ni(CN)_4]^{2-}$, the oxidation state of $Ni$ is

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The complex is square planar and diamagnetic

Vedclass · Answer

(C) $1$. The atomic number of $Ni$ is $28$. Its electronic configuration is $[Ar] 3d^8 4s^2$.$2$. In $[Ni(CN)_4]^{2-}$, the oxidation state of $Ni$ is $+2$, so the configuration of $Ni^{2+}$ is $[Ar] 3d^8$.$3$. $CN^-$ is a strong field ligand, which causes the pairing of electrons in the $3d$ orbitals.$4$. Due to this pairing, one $3d$ orbital becomes vacant, allowing $dsp^2$ hybridization.$5$. The $dsp^2$ hybridization results in a square planar geometry.$6$. Since all electrons are paired, the complex is diamagnetic.

Choose the correct statement for the $[Ni(CN)_4]^{2-}$ complex ion (Atomic number of $Ni=28$).

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