The geometry and magnetic property of [NiCl_4 | vedclass

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Question

(A) The oxidation state of $Ni$ in $[NiCl_4]^{2-}$ is $+2$. The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8 4s^0$. As $Cl^{-}$ is a weak field

Vedclass · Accepted Answer

tetrahedral,paramagnetic

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(A) The oxidation state of $Ni$ in $[NiCl_4]^{2-}$ is $+2$. The electronic configuration of $Ni^{2+}$ is $[Ar] 3d^8 4s^0$. As $Cl^{-}$ is a weak field ligand,pairing of electrons will not occur. The $Ni^{2+}$ ion in the complex undergoes $sp^3$ hybridisation using one $4s$ and three $4p$ orbitals. Thus,the geometry is tetrahedral. Since there are $2$ unpaired electrons in the $3d$ orbitals,the complex is paramagnetic.

The geometry and magnetic property of $[NiCl_4]^{2-}$,respectively are

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In the formation of $K_4[Fe(CN)_6]$,the hybridisation involved is

Which of the following statements are true about $[NiCl_4]^{2-}$?
$(a)$ The complex has tetrahedral geometry
$(b)$ Coordination number of $Ni$ is $2$ and oxidation state is $+4$
$(c)$ The complex is $sp^3$ hybridised
$(d)$ It is a high spin complex
$(e)$ The complex is paramagnetic

Which of the following statements is not correct?

Which of the following has a non-zero magnetic moment and $sp^3d^2$ hybridization?

Which of the following complexes is diamagnetic and square planar?

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