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(C) The oxidation state of $Ni$ in $Ni(CO)_4$ is $0$.Thus,the electronic configuration of $Ni(0)$ is $[Ar] 3d^8 4s^2$.In the presence of the strong fi

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tetrahedral and diamagnetic

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(C) The oxidation state of $Ni$ in $Ni(CO)_4$ is $0$.Thus,the electronic configuration of $Ni(0)$ is $[Ar] 3d^8 4s^2$.In the presence of the strong field ligand $CO$,the $4s$ electrons pair up with the $3d$ electrons.This results in a fully filled $3d$ subshell and the use of one $4s$ and three $4p$ orbitals for $sp^3$ hybridisation.$Ni(CO)_4$ has tetrahedral geometry ($sp^3$ hybridisation) and is diamagnetic due to the absence of unpaired electrons.

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

$Ni(CO)_4$ is

Similar Questions

What is the coordination number of the central metal ion if it forms an octahedral complex?

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

List-$I$ List-$II$

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

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

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

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

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