Match the following hybridization in Column I | vedclass

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Question

(A) The hybridization of the given complexes is determined as follows:$A. sp^3$: $[Ni(CO)_4]$ is a tetrahedral complex with $sp^3$ hybridization. Thus

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$A-(ii), B-(iii), C-(iv), D-(i)$

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(A) The hybridization of the given complexes is determined as follows:$A. sp^3$: $[Ni(CO)_4]$ is a tetrahedral complex with $sp^3$ hybridization. Thus,$A-(ii)$.$B. dsp^2$: $[Pt(NH_3)_2Cl_2]$ is a square planar complex with $dsp^2$ hybridization. Thus,$B-(iii)$.$C. sp^3d^2$: $[CoF_6]^{3-}$ is an outer orbital octahedral complex with $sp^3d^2$ hybridization. Thus,$C-(iv)$.$D. d^2sp^3$: $[Co(NH_3)_6]^{3+}$ is an inner orbital octahedral complex with $d^2sp^3$ hybridization. Thus,$D-(i)$.Therefore,the correct matching is $A-(ii), B-(iii), C-(iv), D-(i)$.

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

List-$I$ (Complex/Species)	List-$II$ (Shape and magnetic moment)
$A$. $[Ni(CO)_4]$	$I$. Tetrahedral,$2.8 \ BM$
$B$. $[Ni(CN)_4]^{2-}$	$II$. Square planar,$0 \ BM$
$C$. $[NiCl_4]^{2-}$	$III$. Tetrahedral,$0 \ BM$
$D$. $[MnBr_4]^{2-}$	$IV$. Tetrahedral,$5.9 \ BM$

Match the following hybridization in Column $I$ with the corresponding coordination complexes in Column $II$.
$A. sp^3$ $(i). [Co(NH_3)_6]^{3+}$
$B. dsp^2$ $(ii). [Ni(CO)_4]$
$C. sp^3d^2$ $(iii). [Pt(NH_3)_2Cl_2]$
$D. d^2sp^3$ $(iv). [CoF_6]^{3-}$
$(v). [Fe(CO)_5]$

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Match the following hybridization in Column $I$ with the corresponding coordination complexes in 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]$