Which of the following statements is incorrec | vedclass

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(A) $1$. In $K_4[Ni(CN)_4]$,the oxidation state of $Ni$ is $0$. The configuration is $[Ar] 3d^8 4s^2$. Due to the strong field ligand $CN^-$,the $4s$

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$K_4[Ni(CN)_4]$ is square planar while $K_2[Ni(CN)_4]$ is paramagnetic.

Vedclass · Answer

(A) $1$. In $K_4[Ni(CN)_4]$,the oxidation state of $Ni$ is $0$. The configuration is $[Ar] 3d^8 4s^2$. Due to the strong field ligand $CN^-$,the $4s$ electrons pair up in the $3d$ orbital,resulting in a $d^{10}$ configuration. It is tetrahedral and diamagnetic.$2$. In $K_2[Ni(CN)_4]$,the oxidation state of $Ni$ is $+2$. The configuration is $[Ar] 3d^8$. $CN^-$ is a strong field ligand,causing pairing of electrons,resulting in $dsp^2$ hybridization (square planar) and it is diamagnetic.$3$. In $K_2[NiCl_4]$,the oxidation state of $Ni$ is $+2$. The configuration is $[Ar] 3d^8$. $Cl^-$ is a weak field ligand,resulting in $sp^3$ hybridization (tetrahedral) and it is paramagnetic.$4$. Evaluating the options: Option $A$ is incorrect because $K_4[Ni(CN)_4]$ is tetrahedral,not square planar,and $K_2[Ni(CN)_4]$ is diamagnetic,not paramagnetic. Option $C$ is also incorrect as one is diamagnetic and the other is diamagnetic (both have $\mu = 0$),but the statement implies they are the same,which is true,yet $A$ is fundamentally incorrect regarding geometry and magnetism.

Which of the following statements is incorrect?

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