Match List -I with List -II: List -I L | vedclass

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(B) The hybridization of the given species is as follows:$1$. $[PtCl_4]^{2-}$: $Pt^{2+}$ has a $d^8$ configuration. It forms a square planar complex w

Vedclass · Accepted Answer

$A-III, B-IV, C-I, D-II$

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(B) The hybridization of the given species is as follows:$1$. $[PtCl_4]^{2-}$: $Pt^{2+}$ has a $d^8$ configuration. It forms a square planar complex with $dsp^2$ hybridization. $(A-III)$$2$. $BrF_5$: The central atom $Br$ has $7$ valence electrons. It forms $5$ bonds and has $1$ lone pair,resulting in $sp^3d^2$ hybridization. $(B-IV)$$3$. $PCl_5$: The central atom $P$ has $5$ valence electrons. It forms $5$ bonds with no lone pairs,resulting in $sp^3d$ hybridization. $(C-I)$$4$. $[Co(NH_3)_6]^{3+}$: $Co^{3+}$ has a $d^6$ configuration. $NH_3$ is a strong field ligand,forcing pairing of electrons,resulting in $d^2sp^3$ hybridization. $(D-II)$Thus,the correct match is $A-III, B-IV, C-I, D-II$.

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

List-$I$ (Complexes)	List-$II$ (Hybridisation)
$A$. $[Ni(CO)_4]$	$I$. $sp^3$
$B$. $[Cu(NH_3)_4]^{2+}$	$II$. $dsp^2$
$C$. $[Fe(NH_3)_6]^{2+}$	$III$. $sp^3d^2$
$D$. $[Fe(H_2O)_6]^{2+}$	$IV$. $d^2sp^3$

Match List $-I$ with List $-II$:

List $-I$ List $-II$

$A$. $[PtCl_4]^{2-}$ $I$. $sp^3d$

$B$. $BrF_5$ $II$. $d^2sp^3$

$C$. $PCl_5$ $III$. $dsp^2$

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

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Match List-$I$ with List-$II$:

List-$I$ (Complexes) List-$II$ (Hybridisation)

$A$. $[Ni(CO)_4]$ $I$. $sp^3$

$B$. $[Cu(NH_3)_4]^{2+}$ $II$. $dsp^2$

$C$. $[Fe(NH_3)_6]^{2+}$ $III$. $sp^3d^2$

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$[FeF_6]^{3-}$ has ........ hybridisation and ....... geometry. Select the best option to fill the blank space respectively:

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Match List $-I$ with List $-II$: List $-I$ List $-II$ $A$. $[PtCl_4]^{2-}$ $I$. $sp^3d$ $B$. $BrF_5$ $II$. $d^2sp^3$ $C$. $PCl_5$ $III$. $dsp^2$ $D$. $[Co(NH_3)_6]^{3+}$ $IV$. $sp^3d^2$