For the square planar complex $[M{\rm{ }}(a)(b)(c)(d)]$ (where , $M{\rm{ }} = $ central metal and $a,{\rm{ }}b,{\rm{ }}c,$ and $d$ are monodentate ligands), the number of possible geometrical isomers are

JEE CHEMISTRY MCQ: CO-ORDINATION COMPOUNDS

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For the square planar complex

[M\mathrm{ }(a)(b)(c)(d)]

(where ,

M\mathrm{ } =

central metal and

a,\mathrm{ }b,\mathrm{ }c,

and

d

are monodentate ligands), the number of possible geometrical isomers are

View answer and explanation

Correct option: C

Option C (3) is correct because for a square planar complex with four different monodentate ligands, there are 3 possible geometrical isomers.

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\mathrm{K}\left[ {\mathrm{Ag}{{\left( {\mathrm{CN}} \right)}_2}} \right]

\left[ {\mathrm{PtC}{\mathrm{l}_2}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]\mathrm{B}{\mathrm{r}_2}\mathrm{\;and\;}\left[ {\mathrm{PtB}{\mathrm{r}_2}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]\mathrm{C}{\mathrm{l}_2}\mathrm{\;}

\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right]\left[\mathrm{PtCl}_4\right]

\left[{Co}\left({NH}_3\right)_5\left({NO}_2\right)\right] {Cl}_2

\left[{Co}\left({NH}_3\right)_5({ONO})\right] {Cl}_2 \ldots \ldots \ldots

\left[{Cu}\left({NH}_3\right)_4\right]\left[{PtCl}_4\right] \,\, \& \,\, \left[{Pt}\left({NH}_3\right)_4\right]\left[{CuCl}_4\right] \ldots \ldots \ldots

\left[{Pt}\left({NH}_3\right)_4 {Cl}_2\right] {Br}_2 \, \, \& \, \, \left[{Pt}\left({NH}_3\right)_4 {Br}_2\right]{Cl}_2 \ldots \ldots \ldots