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?

Similar Questions

According to $V.B.T.$,in which of the following complexes does the transition of an electron occur from one shell to another shell of the central metal atom?

Predict the number of unpaired electrons in the square planar $[Pt(CN)_{4}]^{2-}$ ion.

Which of the following is true for the complex $[Fe(CN)_6]^{4-}$?

The geometries of some complex ions are given below. Select the correct answer.
$(1) [Ag(NH_3)_2]^+ - \text{Linear}$
$(2) [MnCl_4]^{2-} - \text{Tetrahedral}$
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$(4) [Ni(CN)_4]^{2-} - \text{Square planar}$

Which of the following compounds has a different geometry from the others?

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