How many geometrical isomers does [Pt(NH_3)(B | vedclass

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

Question

(A) The complex $[Pt(NH_3)(Br)(Cl)(Py)]$ is of the type $[Mabcd]$,where $M = Pt$,$a = NH_3$,$b = Br$,$c = Cl$,and $d = Py$.For a square planar complex

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) The complex $[Pt(NH_3)(Br)(Cl)(Py)]$ is of the type $[Mabcd]$,where $M = Pt$,$a = NH_3$,$b = Br$,$c = Cl$,and $d = Py$.For a square planar complex of the type $[Mabcd]$,the number of geometrical isomers is $3$.These isomers are formed by fixing one ligand (e.g.,$NH_3$) and arranging the other three ligands ($Br$,$Cl$,$Py$) in the $trans$ positions relative to the fixed ligand.

How many geometrical isomers does $[Pt(NH_3)(Br)(Cl)(Py)]$ have?

Similar Questions

The $[Cr(NH_3)_6]^{3+}$ ion is ...

The total number of unpaired electrons present in the complex $K_3[Cr(C_2O_4)_3]$ is ..........

Explain optical isomerism.

The increasing order of magnetic moments for the following complexes is:
$I$. $[Fe(CN)_6]^{4-}$
$II$. $[Fe(CN)_6]^{3-}$
$III$. $[Cr(NH_3)_6]^{3+}$
$IV$. $[Ni(H_2O)_4]^{2+}$

Which of the following complex ions is expected to absorb visible light?
[At. nos. $Zn = 30$,$Sc = 21$,$Ti = 22$,$Cr = 24$]

Vedclass Test Series

Exam Paper Generator

Online Exam Module