In [Pt(NH_3)_2Cl_2],the Pt-Cl bond length is | vedclass

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(C) The complex $[Pt(NH_3)_2Cl_2]$ exhibits square planar geometry.For a square planar complex with bond length $d = 2 \ \mathring{A}$,the distance be

Vedclass · Accepted Answer

cis-square planar

Vedclass · Answer

(C) The complex $[Pt(NH_3)_2Cl_2]$ exhibits square planar geometry.For a square planar complex with bond length $d = 2 \ \mathring{A}$,the distance between adjacent $Cl$ atoms (cis-position) is $d\sqrt{2} = 2 	imes 1.414 = 2.828 \ \mathring{A} \approx 2.88 \ \mathring{A}$.The distance between opposite $Cl$ atoms (trans-position) would be $2d = 2 	imes 2 = 4 \ \mathring{A}$.Since the given $Cl-Cl$ distance is $2.88 \ \mathring{A}$,which corresponds to the cis-configuration,the complex is cis-square planar.Hence,the correct option is $C$.

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

In $[Pt(NH_3)_2Cl_2]$,the $Pt-Cl$ bond length is $2 \ \mathring{A}$ and the $Cl-Cl$ distance is $2.88 \ \mathring{A}$. What is the geometry of the compound?

Similar Questions

The number of unpaired electrons responsible for the paramagnetic nature of the following complex species are respectively $:$
$[Fe(CN)_6]^{3-}, [FeF_6]^{3-}, [CoF_6]^{3-}, [Mn(CN)_6]^{3-}$

The complex ion $[Co(C_2O_4)_3]^{3-}$ has which of the following hybridizations?

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

List $-I$ List $-II$

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

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

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

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

The hybridization of the central atom in $NH_3, [PtCl_4]^{2-}, PCl_5$,and $BCl_3$ respectively is:

The hybridisation and unpaired electrons in $[Fe(H_2O)_6]^{2+}$ ion are:

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