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Question

(A) Electrical conductivity is directly proportional to the number of ions produced in the solution.$(I) [Pt(NH_3)_6]Cl_4 \rightarrow [Pt(NH_3)_6]^{4+

Vedclass · Accepted Answer

$III < IV < II < I$

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(A) Electrical conductivity is directly proportional to the number of ions produced in the solution.$(I) [Pt(NH_3)_6]Cl_4 \rightarrow [Pt(NH_3)_6]^{4+} + 4Cl^-$ (Total $5$ ions)$(II) [Cr(NH_3)_6]Cl_3 \rightarrow [Cr(NH_3)_6]^{3+} + 3Cl^-$ (Total $4$ ions)$(III) [Co(NH_3)_4Cl_2]Cl \rightarrow [Co(NH_3)_4Cl_2]^+ + Cl^-$ (Total $2$ ions)$(IV) K_2[PtCl_6] \rightarrow 2K^+ + [PtCl_6]^{2-}$ (Total $3$ ions)Comparing the number of ions: $III (2) Therefore,the correct increasing order of conductivity is $III < IV < II < I$.

Arrange the following complexes in the correct increasing order of their electrical conductivity in aqueous solution:
$(I) [Pt(NH_3)_6]Cl_4$
$(II) [Cr(NH_3)_6]Cl_3$
$(III) [Co(NH_3)_4Cl_2]Cl$
$(IV) K_2[PtCl_6]$

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Arrange the following complexes in the correct increasing order of their electrical conductivity in aqueous solution:$(I) [Pt(NH_3)_6]Cl_4$$(II) [Cr(NH_3)_6]Cl_3$$(III) [Co(NH_3)_4Cl_2]Cl$$(IV) K_2[PtCl_6]$