A complex of platinum,ammonia,and chloride pr | vedclass

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(C) The number of ions produced in a solution depends on the dissociation of the coordination compound into a complex ion and counter ions.$1$. $[Pt(N

Vedclass · Accepted Answer

$[Pt(NH_3)_5Cl]Cl_3$

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(C) The number of ions produced in a solution depends on the dissociation of the coordination compound into a complex ion and counter ions.$1$. $[Pt(NH_3)_4]Cl_4 \rightarrow [Pt(NH_3)_4]^{4+} + 4Cl^-$ (Total $5$ ions)$2$. $[Pt(NH_3)_2Cl_4]$ is a neutral complex and does not dissociate into ions.$3$. $[Pt(NH_3)_5Cl]Cl_3 \rightarrow [Pt(NH_3)_5Cl]^{3+} + 3Cl^-$ (Total $4$ ions)$4$. $[Pt(NH_3)_4Cl_2]Cl_2 \rightarrow [Pt(NH_3)_4Cl_2]^{2+} + 2Cl^-$ (Total $3$ ions)Since the complex produces $4$ ions per molecule,the correct structure is $[Pt(NH_3)_5Cl]Cl_3$.

$A$ complex of platinum,ammonia,and chloride produces four ions per molecule in the solution. The structure consistent with this observation is:

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