The number of precipitable halide ions in the | vedclass

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

Question

(D) The given coordination compound is $[Pt(NH_3)Cl_2Br]Cl$. In an aqueous solution,the complex dissociates as follows: $[Pt(NH_3)Cl_2Br]Cl \rightleft

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) The given coordination compound is $[Pt(NH_3)Cl_2Br]Cl$. In an aqueous solution,the complex dissociates as follows: $[Pt(NH_3)Cl_2Br]Cl \rightleftharpoons [Pt(NH_3)Cl_2Br]^+ + Cl^-$. Only the $Cl^-$ ion present outside the coordination sphere is ionizable and can be precipitated by adding silver nitrate $(AgNO_3)$. Therefore,the number of precipitable halide ions is $1$.

The number of precipitable halide ions in the sample $[Pt(NH_3)Cl_2Br]Cl$ will be

Similar Questions

What is the coordination number of $Co$ in $[CoCl_2(NH_3)_4]^+$?

An aqueous solution of the complex $[Cu(NH_3)_4]SO_4$ gives the test for:

The coordination number of $Fe$ in the complexes $[Fe(CN)_6]^{4-}$,$[Fe(CN)_6]^{3-}$ and $[FeCl_4]^-$ would be respectively:

In $EDTA^{4-}$,the donor atoms are:

The bonds present in $K_4[Fe(CN)_6]$ are ........

Vedclass Test Series

Exam Paper Generator

Online Exam Module