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-$I$ (Compound)	List-$II$ (Hybridization of Central atom)
$A. [Ni(NH_3)_6]^{2+}$	$1. sp^3$
$B. [PtCl_4]^{2-}$	$2. sp^3d^2$
$C. [Ni(CO)_4]$	$3. dsp^2$
$D. [Co(ox)_3]^{3-}$	$4. d^2sp^3$

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?

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