Match List-I (Species) with List-II (Hybrid o | vedclass

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(B) The hybridization of the central atom in the given species is as follows:$1$. $Ni(CN)_5^{3-}$: The central atom $Ni$ has a coordination number of

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$3-4-2-1$

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(B) The hybridization of the central atom in the given species is as follows:$1$. $Ni(CN)_5^{3-}$: The central atom $Ni$ has a coordination number of $5$. It exhibits $sp^3d_{z^2}$ hybridization (trigonal bipyramidal geometry).$2$. $CuCl_5^{3-}$: The central atom $Cu$ has a coordination number of $5$. It exhibits $d_{x^2-y^2}sp^3$ hybridization (square pyramidal geometry).$3$. $AuCl_4^-$: The central atom $Au$ has a coordination number of $4$. It exhibits $dsp^2$ hybridization (square planar geometry).$4$. $ClO_4^-$: The central atom $Cl$ has a coordination number of $4$. It exhibits $sp^3$ hybridization (tetrahedral geometry).Thus,the correct matching is $A-3, B-4, C-2, D-1$.

List-$I$	List-$II$
$A. \, Ni(CN)_5^{3-}$	$1. \, sp^3$
$B. \, CuCl_5^{3-}$	$2. \, dsp^2$
$C. \, AuCl_4^-$	$3. \, sp^3d_{z^2}$
$D. \, ClO_4^-$	$4. \, d_{x^2-y^2}sp^3$

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Match List-$I$ (Species) with List-$II$ (Hybrid orbitals used by the central atom in their formation) and select the correct answer. List-$I$ List-$II$ $A. \, Ni(CN)_5^{3-}$ $1. \, sp^3$ $B. \, CuCl_5^{3-}$ $2. \, dsp^2$ $C. \, AuCl_4^-$ $3. \, sp^3d_{z^2}$ $D. \, ClO_4^-$ $4. \, d_{x^2-y^2}sp^3$