Co-ordination number of Cr in CrCl3⋅5H2O is six. The maximum volume of 0.1 N AgNO₃ needed to precipitate the chlorine in outer sphere in 200 ml of 0.01 M solution of the complex is/are:

Given that the coordination number of chromium (Cr) is six, the complex can be represented as [Cr(H₂O)₅Cl]Cl₂·H₂O. In this structure:

• Five water molecules and one chloride ion are coordinated to the Cr³⁺ ion, completing its coordination sphere.
• Two chloride ions are present outside the coordination sphere (outer sphere) as counterions.

Explanation

1. Moles of the Complex in Solution:
   • Volume of solution = 200 ml = 0.2 L
   • Molarity (M) = 0.01 M
   • Moles of complex = Molarity × Volume = 0.01 M × 0.2 L = 0.002 moles
2. Moles of Outer Sphere Chloride Ions:
   • Each complex unit has two chloride ions in the outer sphere.
   • Total moles of outer sphere Cl⁻ = 0.002 moles complex × 2 Cl⁻/complex = 0.004 moles
3. Volume of 0.1 N AgNO₃ Required:
   • AgNO₃ reacts with Cl⁻ in a 1:1 molar ratio to form AgCl precipitate.
   • Normality (N) of AgNO₃ = 0.1 N, which means 0.1 equivalents/L.
   • Volume of AgNO₃ needed = Equivalents of Cl⁻ / Normality of AgNO₃
   • Equivalents of Cl⁻ = Moles of Cl⁻ (since valency of Cl⁻ is 1) = 0.004 equivalents
   • Volume = 0.004 equivalents / 0.1 N = 0.04 L = 40 ml

Final Answer

The maximum volume of 0.1 N AgNO₃ required to precipitate the outer sphere chloride ions from 200 ml of a 0.01 M solution of CrCl₃·5H₂O is 40 ml.