Consider the following complexes:(a) K_2PtCl_ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$256, 0, 97, 404$

Vedclass · Answer

(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:

Similar Questions

Which of the following is a strong field ligand?

Which of the following complexes has the highest $C-O$ bond length?

Silver chloride dissolves in excess of $NH_4OH$. The cation present in this solution is

Cobalt $(III)$ chloride forms several octahedral complexes with ammonia. Which of the following will not give a test for chloride ions with silver nitrate at $25 \ ^oC$?

Which one will give $Fe^{3+}$ ions in solution?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

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: