Which of the following ions forms complexes w | vedclass

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(A) The coordination number of a metal ion in a complex depends on its electronic configuration and the nature of the ligands. $Pt^{2+}$ (a $d^8$ meta

Vedclass · Accepted Answer

$Pt^{2+}$

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(A) The coordination number of a metal ion in a complex depends on its electronic configuration and the nature of the ligands. $Pt^{2+}$ (a $d^8$ metal ion) typically forms square planar complexes with a coordination number of $4$. $Cr^{3+}$ and $Fe^{3+}$ are $d^3$ and $d^5$ ions respectively,which typically form octahedral complexes with a coordination number of $6$. $Pt^{4+}$ is a $d^6$ ion which also typically forms octahedral complexes with a coordination number of $6$. Therefore,$Pt^{2+}$ is the correct answer.

List-$I$ (Complex/ion)	List-$II$ (Shape/geometry)
$A$. $[Pt(Cl)_2(NH_3)_2]$	$I$. Octahedral
$B$. $[Co(NH_3)_6]Cl_3$	$II$. Trigonal bipyramidal
$C$. $[NiCl_4]^{2-}$	$III$. Square planar
$D$. $[Fe(CO)_5]$	$IV$. Tetrahedral

Which of the following ions forms complexes with a coordination number of only $4$?

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Match List-$I$ with List-$II$:
List-$I$ (Complex/ion) List-$II$ (Shape/geometry)
$A$. $[Pt(Cl)_2(NH_3)_2]$ $I$. Octahedral
$B$. $[Co(NH_3)_6]Cl_3$ $II$. Trigonal bipyramidal
$C$. $[NiCl_4]^{2-}$ $III$. Square planar
$D$. $[Fe(CO)_5]$ $IV$. Tetrahedral

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