The geometries of [Ni(CO)_4],[PtCl_4]^{2-} an | vedclass

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(D) $[Ni(CO)_4]$: The central metal $Ni$ is in $0$ oxidation state with configuration $3d^8 4s^2$. Due to the strong field ligand $CO$,the electrons p

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tetrahedral,square planar and octahedral

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(D) $[Ni(CO)_4]$: The central metal $Ni$ is in $0$ oxidation state with configuration $3d^8 4s^2$. Due to the strong field ligand $CO$,the electrons pair up to give $3d^{10} 4s^0$. It undergoes $sp^3$ hybridisation,resulting in a tetrahedral geometry.$[PtCl_4]^{2-}$: $Pt^{2+}$ is a $5d$ series metal ion. For $5d$ elements,the crystal field splitting energy is large,which favors square planar geometry for $d^8$ complexes,even with weak field ligands like $Cl^-$. Thus,it is square planar.$[Co(NH_3)_6]^{3+}$: $Co^{3+}$ has a $3d^6$ configuration. $NH_3$ acts as a strong field ligand,causing pairing of electrons to give $t_{2g}^6 e_g^0$. It undergoes $d^2sp^3$ hybridisation,resulting in an octahedral geometry.

Column $I$	Column $II$
$(A)$ $sp^3$	$(i)$ $[Co(NH_3)_6]^{3+}$
$(B)$ $dsp^2$	$(ii)$ $[Ni(CO)_4]$
$(C)$ $sp^3d^2$	$(iii)$ $[Pt(NH_3)_2Cl_2]$
$(D)$ $d^2sp^3$	$(iv)$ $[CoF_6]^{3-}$
	$(v)$ $[Fe(CO)_5]$

The geometries of $[Ni(CO)_4]$,$[PtCl_4]^{2-}$ and $[Co(NH_3)_6]^{3+}$ respectively are

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