Consider the following complexes:(a) K_2PtCl_ | vedclass

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(A) The electrical conductance of a complex in aqueous solution depends on the number of ions produced upon dissociation.$(a) K_2[PtCl_6] \rightarrow

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$256, 0, 97, 404$

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(A) The electrical conductance of a complex in aqueous solution depends on the number of ions produced upon dissociation.$(a) K_2[PtCl_6] \rightarrow 2K^+ + [PtCl_6]^{2-}$. Total ions $= 3$. Conductance $\approx 256 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.$(b) [Pt(NH_3)_2Cl_4]$. This is a neutral complex,so it does not dissociate. Total ions $= 0$. Conductance $= 0 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.$(c) [Pt(NH_3)_3Cl_3]Cl \rightarrow [Pt(NH_3)_3Cl_3]^+ + Cl^-$. Total ions $= 2$. Conductance $\approx 97 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.$(d) [Pt(NH_3)_5Cl]Cl_3 \rightarrow [Pt(NH_3)_5Cl]^{3+} + 3Cl^-$. Total ions $= 4$. Conductance $\approx 404 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.Thus,the order is $256, 0, 97, 404$.

Consider the following complexes:
$(a) K_2PtCl_6$
$(b) PtCl_4 \cdot 2NH_3$
$(c) PtCl_4 \cdot 3NH_3$
$(d) PtCl_4 \cdot 5NH_3$
Their electrical conductances in aqueous solutions are:

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Consider the following complexes:$(a) K_2PtCl_6$$(b) PtCl_4 \cdot 2NH_3$$(c) PtCl_4 \cdot 3NH_3$$(d) PtCl_4 \cdot 5NH_3$Their electrical conductances in aqueous solutions are: