Considering the complexes (i) K_2PtCl_6,(ii) | vedclass

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Question

(A) The coordination number of $Pt$ is $6$. The complexes can be written as:$(i) K_2[PtCl_6] \rightarrow 2K^+ + [PtCl_6]^{2-}$ (Total $3$ ions)$(ii) [

Vedclass · Accepted Answer

$256, 0, 97, 404$

Vedclass · Answer

(A) The coordination number of $Pt$ is $6$. The complexes can be written as:$(i) K_2[PtCl_6] \rightarrow 2K^+ + [PtCl_6]^{2-}$ (Total $3$ ions)$(ii) [Pt(NH_3)_2Cl_4] \rightarrow$ Non-electrolyte (Total $0$ ions)$(iii) [Pt(NH_3)_3Cl_3]Cl \rightarrow [Pt(NH_3)_3Cl_3]^+ + Cl^-$ (Total $2$ ions)$(iv) [Pt(NH_3)_5Cl]Cl_3 \rightarrow [Pt(NH_3)_5Cl]^{3+} + 3Cl^-$ (Total $4$ ions)Since molar conductivity is directly proportional to the number of ions produced in solution,the order of conductivity corresponds to the number of ions: $4 > 3 > 2 > 0$.Matching the values: $(iv) = 404$,$(i) = 256$,$(iii) = 97$,$(ii) = 0$.Thus,the correct sequence is $256, 0, 97, 404$.

Considering the complexes $(i) K_2PtCl_6$,$(ii) PtCl_4 \cdot 2NH_3$,$(iii) PtCl_4 \cdot 3NH_3$,and $(iv) PtCl_4 \cdot 5NH_3$,the molar conductivities in aqueous solution follow the order (in $\Omega^{-1} \ cm^2 \ mol^{-1}$):

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