The molar ionic conductances of the octahedra | vedclass

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(B) The molar ionic conductance depends on the number of ions produced in the solution.$(I)$ $[Pt(NH_3)_5Cl]Cl_3$ dissociates into $4$ ions ($1$ catio

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$IV < III < II < I$

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(B) The molar ionic conductance depends on the number of ions produced in the solution.$(I)$ $[Pt(NH_3)_5Cl]Cl_3$ dissociates into $4$ ions ($1$ cation + $3$ anions).$(II)$ $[Pt(NH_3)_4Cl_2]Cl_2$ dissociates into $3$ ions ($1$ cation + $2$ anions).$(III)$ $[Pt(NH_3)_3Cl_3]Cl$ dissociates into $2$ ions ($1$ cation + $1$ anion).$(IV)$ $[Pt(NH_3)_2Cl_4]$ is a neutral complex and does not dissociate into ions ($0$ ions).Since the number of ions follows the order $IV < III < II < I$,the molar ionic conductance follows the same order: $IV < III < II < I$.

The molar ionic conductances of the octahedral complexes:
$(I)$ $PtCl_4 \cdot 5NH_3$
$(II)$ $PtCl_4 \cdot 4NH_3$
$(III)$ $PtCl_4 \cdot 3NH_3$
$(IV)$ $PtCl_4 \cdot 2NH_3$
Follow the order $-$

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The molar ionic conductances of the octahedral complexes:$(I)$ $PtCl_4 \cdot 5NH_3$$(II)$ $PtCl_4 \cdot 4NH_3$$(III)$ $PtCl_4 \cdot 3NH_3$$(IV)$ $PtCl_4 \cdot 2NH_3$Follow the order $-$