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(B) The electrical conductivity of an aqueous solution of coordination complexes depends on the number of ions produced upon dissociation.$1$. $[Pt(NH

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$A > B > D > C$

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(B) The electrical conductivity of an aqueous solution of coordination complexes depends on the number of ions produced upon dissociation.$1$. $[Pt(NH_3)_6]Cl_4 \rightarrow [Pt(NH_3)_6]^{4+} + 4Cl^-$. Total ions = $5$.$2$. $[Cr(NH_3)_6]Cl_3 \rightarrow [Cr(NH_3)_6]^{3+} + 3Cl^-$. Total ions = $4$.$3$. $[Co(NH_3)_4Cl_2]Cl \rightarrow [Co(NH_3)_4Cl_2]^+ + Cl^-$. Total ions = $2$.$4$. $K_2[PtCl_6] \rightarrow 2K^+ + [PtCl_6]^{2-}$. Total ions = $3$.Comparing the number of ions: $A (5) > B (4) > D (3) > C (2)$.Therefore,the order of electrical conductivity is $A > B > D > C$.

Give the correct order of electrical conductivity of aqueous solution of the following complexes:
$A. [Pt(NH_3)_6]Cl_4$
$B. [Cr(NH_3)_6]Cl_3$
$C. [Co(NH_3)_4Cl_2]Cl$
$D. K_2[PtCl_6]$

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Give the correct order of electrical conductivity of aqueous solution of the following complexes:$A. [Pt(NH_3)_6]Cl_4$$B. [Cr(NH_3)_6]Cl_3$$C. [Co(NH_3)_4Cl_2]Cl$$D. K_2[PtCl_6]$