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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

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

Vedclass · Answer

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

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

List $-I$ List $-II$

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

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

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

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

Similar Questions

What is the hybridisation and shape of $Cr$ in chromyl chloride $(CrO_2Cl_2)$?

How many unpaired electrons are present in $Ni^{2+}$?

Select the incorrect match.

If an iron $(III)$ complex with the formula $[Fe(NH_3)_{x}(CN)_{y}]^{-}$ has no electron in its $e_g$ orbital, then the value of $x+y$ is

Identify from the following species in which $d^2 sp^3$ hybridization is shown by the central atom:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

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