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(D) For $[Ni(CO)_4]$: The oxidation state of $Ni$ is $0$. The electronic configuration is $[Ar] 3d^8 4s^2$. Since $CO$ is a strong field ligand,it cau

Vedclass · Accepted Answer

$[Ni(CO)_4]$ diamagnetic,$[NiCl_4]^{2-}$ paramagnetic

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(D) For $[Ni(CO)_4]$: The oxidation state of $Ni$ is $0$. The electronic configuration is $[Ar] 3d^8 4s^2$. Since $CO$ is a strong field ligand,it causes pairing of electrons,resulting in $sp^3$ hybridization with $0$ unpaired electrons,making it diamagnetic.For $[NiCl_4]^{2-}$: The oxidation state of $Ni$ is $+2$. The electronic configuration is $[Ar] 3d^8$. Since $Cl^-$ is a weak field ligand,it does not cause pairing of electrons,resulting in $sp^3$ hybridization with $2$ unpaired electrons,making it paramagnetic.

List-$I$	List-$II$
$(1)$ $dsp^2$	$(A)$ $BrF_5$
$(2)$ $sp^3d^2$	$(B)$ $[Co(NH_3)_6]^{3+}$
$(3)$ $sp^3d$	$(C)$ $[PtCl_4]^{2-}$
$(4)$ $d^2sp^3$	$(D)$ $PCl_5$

Select the option with the correct property -

Similar Questions

Match the type of hybridization given in List-$I$ with the molecules/ions given in List-$II$.

List-$I$ List-$II$

$(1)$ $dsp^2$ $(A)$ $BrF_5$

$(2)$ $sp^3d^2$ $(B)$ $[Co(NH_3)_6]^{3+}$

$(3)$ $sp^3d$ $(C)$ $[PtCl_4]^{2-}$

$(4)$ $d^2sp^3$ $(D)$ $PCl_5$

Identify the number of unpaired electrons present and the geometry,respectively,of the complex $[CoF_6]^{3-}$ (given atomic number of $Co = 27$).

$A$ list of species having the formula $XZ_4$ is given below.
$XeF_4, SF_4, SiF_4, BF_4^-, BrF_4^-, [Cu(NH_3)_4]^{2+}, [FeCl_4]^{2-}, [CoCl_4]^{2-}$ and $[PtCl_4]^{2-}$.
Defining shape on the basis of the location of $X$ and $Z$ atoms,the total number of species having a square planar shape is

The number of species from the following that are involved in $sp^3 d^2$ hybridization is $[Co(NH_3)_6]^{3+}, SF_6, [CrF_6]^{3-}, [CoF_6]^{3-}, [Mn(CN)_6]^{3-}$ and $[MnCl_6]^{3-}$.

In which of the following configurations will there be the possibility of both paramagnetism and diamagnetism,depending on the nature of the ligands?

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