Question

On treating 100 mL of 0.1 M aqueous solution of the complex CrCl₃.6H₂O with excess of AgNO₃, 2.86 g of AgCl was obtained. The complex is:

• A. [Cr(H₂O)₃Cl₃].3H₂O
• B. [Cr(H₂O)₄Cl₂]Cl.2H₂O
• C. [Cr(H₂O)₅Cl]Cl₂.H₂O
• D. [Cr(H₂O)₆]Cl₃

Hint:

The release of 2 moles of AgCl per mole of the complex confirms two chloride counter-ions outside the coordination sphere.

Answer 1

The Correct Option is C

To determine the correct formula of the complex based on the given conditions, let's analyze the given problem.

We have a 0.1 M aqueous solution of the complex CrCl₃.6H₂O, with a volume of 100 mL. This solution is treated with excess AgNO₃ and yields 2.86 g of AgCl. 

First, let's calculate the moles of AgCl formed:

The molar mass of AgCl is the sum of the molar masses of silver (Ag) and chlorine (Cl):

• Ag: 107.87 g/mol
• Cl: 35.45 g/mol

\(107.87 + 35.45 = 143.32 \,\text{g/mol}\)

Now, calculate the moles of AgCl:

\(\text{Moles of AgCl} = \frac{2.86 \, \text{g}}{143.32 \, \text{g/mol}} = 0.01995 \, \text{mol}\)

Since AgCl is formed from the chloride ions in the complex, and each chloride ion forms one AgCl, the moles of AgCl represent the moles of chloride ions.

The 0.1 M solution of the complex has a concentration of:

\(0.1 \, \text{mol/L} \times 0.1 \, \text{L} = 0.01 \, \text{mol of complex}\)

From the stoichiometry of the reaction, each complex releases two chloride ions because:

\(0.01 \, \text{mol of complex} \times 2 \, \text{Cl}^- = 0.02 \, \text{mol of Cl}^-\\)

This calculated amount aligns with the moles of AgCl formed, confirming our stoichiometric analysis.

Therefore, the correct formula of the complex that matches this stoichiometry is:

[Cr(H₂O)₅Cl]Cl₂.H₂O

This complex has two chloride ions outside the coordination sphere, which react with Ag⁺ to produce AgCl, consistent with the experiment.