Match the following:List-I (Hybridisation)Lis | vedclass

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(A-IV, B-II, C-I, D-V) To find the correct match,we determine the hybridisation of each compound/ion:$A. sp^3d$: $PCl_5$ $(IV)$ has $5$ bond pairs and

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(A-IV, B-II, C-I, D-V) To find the correct match,we determine the hybridisation of each compound/ion:$A. sp^3d$: $PCl_5$ $(IV)$ has $5$ bond pairs and $0$ lone pairs,total $5$ hybrid orbitals,corresponding to $sp^3d$ hybridisation.$B. sp^3d^2$: $SF_6$ $(II)$ has $6$ bond pairs and $0$ lone pairs,total $6$ hybrid orbitals,corresponding to $sp^3d^2$ hybridisation.$C. dsp^2$: $[PtCl_4]^{2-}$ $(I)$ has $4$ bond pairs and $0$ lone pairs,total $4$ hybrid orbitals,corresponding to $dsp^2$ hybridisation.$D. dsp^3$: $ClF_3$ $(V)$ has $3$ bond pairs and $2$ lone pairs,total $5$ hybrid orbitals,which can be $sp^3d$ or $dsp^3$ (depending on the model,but $ClF_3$ is commonly associated with $sp^3d$ geometry). However,based on standard matching options provided in such questions,$D$ matches $V$ $(ClF_3)$.Thus,the correct matching is: $A-IV, B-II, C-I, D-V$.

List-$I$ (Hybridisation)	List-$II$ (Compound/ion)
$A. sp^3d$	$I. [PtCl_4]^{2-}$
$B. sp^3d^2$	$II. SF_6$
$C. dsp^2$	$III. BCl_3$
$D. dsp^3$	$IV. PCl_5$
	$V. ClF_3$

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

Match the following:
List-$I$ (Hybridisation) List-$II$ (Compound/ion)
$A. sp^3d$ $I. [PtCl_4]^{2-}$
$B. sp^3d^2$ $II. SF_6$
$C. dsp^2$ $III. BCl_3$
$D. dsp^3$ $IV. PCl_5$
$V. ClF_3$

The correct match is:

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Match the following:List-$I$ (Hybridisation)List-$II$ (Compound/ion)$A. sp^3d$$I. [PtCl_4]^{2-}$$B. sp^3d^2$$II. SF_6$$C. dsp^2$$III. BCl_3$$D. dsp^3$$IV. PCl_5$$V. ClF_3$The correct match is: