Identify the non-ionisable octahedral complex | vedclass

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Question

(A) For a complex to be non-ionisable,all ligands must be present within the coordination sphere,and there should be no counter-ions outside the squar

Vedclass · Accepted Answer

$[Pt(NH_3)_2Cl_4]$

Vedclass · Answer

(A) For a complex to be non-ionisable,all ligands must be present within the coordination sphere,and there should be no counter-ions outside the square brackets.$Pt$ typically exhibits a coordination number $(C.N.)$ of $6$ in its octahedral complexes.In option $A$,$[Pt(NH_3)_2Cl_4]$,the total number of ligands is $2 + 4 = 6$. This satisfies the $C.N.$ of $6$ and has no counter-ions,making it non-ionisable.In option $B$,$[Pt(NH_3)_4Cl_2]Cl_2$,the $Cl_2$ outside the bracket makes it ionisable.In option $C$,$[Pt(NH_3)_4Cl_4]$,the total number of ligands is $8$,which is not standard for $Pt(IV)$ octahedral geometry.In option $D$,$[Pt(NH_3)_3Cl_3]Cl$,the $Cl$ outside the bracket makes it ionisable.Therefore,the correct non-ionisable complex is $[Pt(NH_3)_2Cl_4]$.

Identify the non-ionisable octahedral complex among the following.

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