The geometry of [Pt(NH_3)_4]Cl_2 is: | vedclass

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Question

(A) In the complex $[Pt(NH_3)_4]Cl_2$,the central metal ion is $Pt^{2+}$.$Pt$ $(Z=78)$ has an electronic configuration of $[Xe] 4f^{14} 5d^9 6s^1$. Fo

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Square planar

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(A) In the complex $[Pt(NH_3)_4]Cl_2$,the central metal ion is $Pt^{2+}$.$Pt$ $(Z=78)$ has an electronic configuration of $[Xe] 4f^{14} 5d^9 6s^1$. For $Pt^{2+}$,the configuration is $[Xe] 4f^{14} 5d^8$.Since $NH_3$ is a strong field ligand,it causes pairing of electrons in the $5d$ orbitals.The four $NH_3$ ligands donate electron pairs to the empty $5d$,$6s$,and $6p$ orbitals,resulting in $dsp^2$ hybridisation.$A$ $dsp^2$ hybridised complex always exhibits a square planar geometry.

List-$I$	List-$II$
$A$. $[MnBr_4]^{2-}$	$I$. $d^2sp^3$ and diamagnetic
$B$. $[FeF_6]^{3-}$	$II$. $sp^3d^2$ and paramagnetic
$C$. $[Co(C_2O_4)_3]^{3-}$	$III$. $sp^3$ and diamagnetic
$D$. $[Ni(CO)_4]$	$IV$. $sp^3$ and paramagnetic

List-$I$ (Hybridisation)	List-$II$ (Compound/ion)
$A. sp^3d$	$I. [PtCl_4]^{2-}$
$B. sp^3d^2$	$II. SF_6$
$C. dsp^2$	$III. BCl_3$
$D. dsp^3$	$IV. PCl_5$
	$V. ClF_3$

The geometry of $[Pt(NH_3)_4]Cl_2$ is:

Similar Questions

Match the following:
List-$I$ (Hybridisation) List-$II$ (Compound/ion)
$A. sp^3d$ $I. [PtCl_4]^{2-}$
$B. sp^3d^2$ $II. SF_6$
$C. dsp^2$ $III. BCl_3$
$D. dsp^3$ $IV. PCl_5$
$V. ClF_3$

The correct match is:

In the complex $[SbF_5]^{2-}$,$sp^3d$ hybridisation is present. The geometry of the complex is:

Which of the following pairs of structure and magnetic properties is correct for the tetracyanonickelate$(II)$ ion?

The geometry,hybridization,and magnetic moment of the ions $[Ni(CN)_4]^{2-}$,$[MnBr_4]^{2-}$,and $[FeF_6]^{4-}$ respectively are:

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