The molar ionic conductance of the octahedral | vedclass

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(D) The molar ionic conductance is directly proportional to the number of ions produced in the solution.$(i)$ $[Pt(NH_3)_5 Cl]Cl_3 \rightarrow [Pt(NH_

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$iv < iii < ii < i$

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(D) The molar ionic conductance is directly proportional to the number of ions produced in the solution.$(i)$ $[Pt(NH_3)_5 Cl]Cl_3 \rightarrow [Pt(NH_3)_5 Cl]^{3+} + 3Cl^-$ ($4$ ions)$(ii)$ $[Pt(NH_3)_4 Cl_2]Cl_2 \rightarrow [Pt(NH_3)_4 Cl_2]^{2+} + 2Cl^-$ ($3$ ions)$(iii)$ $[Pt(NH_3)_3 Cl_3]Cl \rightarrow [Pt(NH_3)_3 Cl_3]^+ + Cl^-$ ($2$ ions)$(iv)$ $[Pt(NH_3)_2 Cl_4]$ is a neutral complex and does not dissociate into ions ($0$ ions).Thus,the order of molar ionic conductance is $(iv) < (iii) < (ii) < (i)$.

The molar ionic conductance of the octahedral complexes:
$(i)$ $PtCl_4 \cdot 5 NH_3$ $(ii)$ $PtCl_4 \cdot 4 NH_3$ $(iii)$ $PtCl_4 \cdot 3 NH_3$ $(iv)$ $PtCl_4 \cdot 2 NH_3$
Follow the order:

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The molar ionic conductance of the octahedral complexes:$(i)$ $PtCl_4 \cdot 5 NH_3$ $(ii)$ $PtCl_4 \cdot 4 NH_3$ $(iii)$ $PtCl_4 \cdot 3 NH_3$ $(iv)$ $PtCl_4 \cdot 2 NH_3$Follow the order: