Skip to content
NCERT Exemplar · Q15

Q.Arrange the following complexes in the increasing order of conductivity of their solution: [Co(NH3)3Cl3][Co(NH_3)_3Cl_3], [Co(NH3)4Cl2]Cl[Co(NH_3)_4Cl_2]Cl, [Co(NH3)6]Cl3[Co(NH_3)_6]Cl_3, [Cr(NH3)5Cl]Cl2[Cr(NH_3)_5Cl]Cl_2

Odisha ChseShort· 2mImportance★★★★★
62% · 63/101 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Conductivity depends on the number of ions produced per formula unit in solution. Counting the ions from each complex gives the order: [Co(NH3)3Cl3]<[Co(NH3)4Cl2]Cl<[Cr(NH3)5Cl]Cl2<[Co(NH3)6]Cl3[Co(NH_3)_3Cl_3] < [Co(NH_3)_4Cl_2]Cl < [Cr(NH_3)_5Cl]Cl_2 < [Co(NH_3)_6]Cl_3.

Why conductivity tells us about the complex

When a coordination compound dissolves in water, the complex ion stays intact (usually), but the counter ions — the ones outside the coordination sphere — break off and become free ions in solution. The more free ions a compound releases, the higher its electrical conductivity.

So the trick is simple: look at the formula, identify which ligands are inside the coordination sphere (written inside the square brackets) and which are outside. The ones outside are the ones that will dissociate.

Watch out

A common mistake is to count all chlorine atoms as dissociating. But if chlorine is inside the coordination sphere (like in [Co(NH3)3Cl3][Co(NH_3)_3Cl_3]), it is covalently bonded to the metal and does not dissociate. Only the chlorines written outside the brackets become chloride ions.

Step-by-step breakdown

1. [Co(NH3)3Cl3][Co(NH_3)_3Cl_3] — This complex has no ions outside the brackets. Everything is inside the coordination sphere. When dissolved, it remains as a neutral molecule. It produces 0 ions in solution. Conductivity: negligible.

2. [Co(NH3)4Cl2]Cl[Co(NH_3)_4Cl_2]Cl — Here, one chlorine is outside the brackets. That chlorine dissociates as Cl−\text{Cl}^-. The complex ion [Co(NH3)4Cl2]+[Co(NH_3)_4Cl_2]^+ stays intact. So total ions per formula unit = 2 (one cation, one anion).

3. [Cr(NH3)5Cl]Cl2[Cr(NH_3)_5Cl]Cl_2 — Two chlorines are outside the brackets. They dissociate as 2Cl−2\text{Cl}^-. The complex ion [Cr(NH3)5Cl]2+[Cr(NH_3)_5Cl]^{2+} remains. Total ions = 3 (one cation, two anions). …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.