Question:
Each of the following compounds has been dissolved in water to make its 0.001 M solution. Rank them in order of their increasing conductivity in solution (assume 100% ionisation in each case)
a) [Pt(NH\(_3\))\(_6\)]Cl\(_4\)
b) [Cr(NH\(_3\))\(_6\)]Cl\(_3\)
c) [Co(NH\(_3\))\(_4\)Cl\(_2\)]Cl
d) K\(_2\)[PtCl\(_6\)]
Each of the following compounds has been dissolved in water to make its 0.001 M solution. Rank them in order of their increasing conductivity in solution (assume 100% ionisation in each case)
a) [Pt(NH\(_3\))\(_6\)]Cl\(_4\)
b) [Cr(NH\(_3\))\(_6\)]Cl\(_3\)
c) [Co(NH\(_3\))\(_4\)Cl\(_2\)]Cl
d) K\(_2\)[PtCl\(_6\)]
a) [Pt(NH\(_3\))\(_6\)]Cl\(_4\)
b) [Cr(NH\(_3\))\(_6\)]Cl\(_3\)
c) [Co(NH\(_3\))\(_4\)Cl\(_2\)]Cl
d) K\(_2\)[PtCl\(_6\)]
Show Hint
Conductivity \(\propto\) number of ions \(\times\) mobility. Here assuming equal mobility, more ions = more conductivity.
Updated On: Apr 23, 2026
- $c<d<a<b$
- $c<d<b<a$
- $a<b<c<d$
- $d<c<a<b$
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
Conductivity depends on number of ions produced per formula unit. More ions \(\Rightarrow\) higher conductivity.
Step 2: Detailed Explanation:
a) [Pt(NH\(_3\))\(_6\)]Cl\(_4\) \(\rightarrow\) [Pt(NH\(_3\))\(_6\)]\(^{4+}\) + 4Cl\(^-\) \(\Rightarrow\) 5 ions
b) [Cr(NH\(_3\))\(_6\)]Cl\(_3\) \(\rightarrow\) [Cr(NH\(_3\))\(_6\)]\(^{3+}\) + 3Cl\(^-\) \(\Rightarrow\) 4 ions
c) [Co(NH\(_3\))\(_4\)Cl\(_2\)]Cl \(\rightarrow\) [Co(NH\(_3\))\(_4\)Cl\(_2\)]\(^+\) + Cl\(^-\) \(\Rightarrow\) 2 ions
d) K\(_2\)[PtCl\(_6\)] \(\rightarrow\) 2K\(^+\) + [PtCl\(_6\)]\(^{2-}\) \(\Rightarrow\) 3 ions
Increasing order (lowest to highest ions): $c (2)<d (3)<b (4)<a (5)$
Step 3: Final Answer:
Thus, order = $ c<d<b<a.$
Conductivity depends on number of ions produced per formula unit. More ions \(\Rightarrow\) higher conductivity.
Step 2: Detailed Explanation:
a) [Pt(NH\(_3\))\(_6\)]Cl\(_4\) \(\rightarrow\) [Pt(NH\(_3\))\(_6\)]\(^{4+}\) + 4Cl\(^-\) \(\Rightarrow\) 5 ions
b) [Cr(NH\(_3\))\(_6\)]Cl\(_3\) \(\rightarrow\) [Cr(NH\(_3\))\(_6\)]\(^{3+}\) + 3Cl\(^-\) \(\Rightarrow\) 4 ions
c) [Co(NH\(_3\))\(_4\)Cl\(_2\)]Cl \(\rightarrow\) [Co(NH\(_3\))\(_4\)Cl\(_2\)]\(^+\) + Cl\(^-\) \(\Rightarrow\) 2 ions
d) K\(_2\)[PtCl\(_6\)] \(\rightarrow\) 2K\(^+\) + [PtCl\(_6\)]\(^{2-}\) \(\Rightarrow\) 3 ions
Increasing order (lowest to highest ions): $c (2)<d (3)<b (4)<a (5)$
Step 3: Final Answer:
Thus, order = $ c<d<b<a.$
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