Question

An aqueous solution of \(CrCl_3 \cdot 6H_2O\) having molar mass \(266.5 \, g \, mol^{-1}\), containing \(2.665 \, g\) of the solute after processing, when treated with excess of \(AgNO_3\), gives \(2.87 \, g\) of \(AgCl\). Given molar mass of \(AgCl = 143.5 \, g \, mol^{-1}\). Choose the correct formula of the compound which gives these results.

• A. \([Cr(H_2O)_4Cl_2]Cl \cdot 2H_2O\)
• B. \([Cr(H_2O)_3Cl_3] \cdot 3H_2O\)
• C. \([Cr(H_2O)_5Cl]Cl_2 \cdot H_2O\)
• D. \([Cr(H_2O)_6]Cl_3\)

Hint:

Only chloride ions present outside the coordination sphere react with \(AgNO_3\) to form \(AgCl\). Chloride ions inside the coordination sphere are non-ionisable and do not immediately precipitate with \(Ag^+\).

Answer 1

The Correct Option is C

Step 1: Calculate moles of \(CrCl_3 \cdot 6H_2O\).
Given mass of solute is:
\[ 2.665 \, g \] Molar mass of \(CrCl_3 \cdot 6H_2O\) is:
\[ 266.5 \, g \, mol^{-1} \] So, moles of solute are:
\[ \text{Moles of solute} = \frac{2.665}{266.5} \] \[ \text{Moles of solute} = 0.01 \, mol \]

Step 2: Calculate moles of \(AgCl\) formed.
Given mass of \(AgCl\) is:
\[ 2.87 \, g \] Molar mass of \(AgCl\) is:
\[ 143.5 \, g \, mol^{-1} \] So, moles of \(AgCl\) formed are:
\[ \text{Moles of } AgCl = \frac{2.87}{143.5} \] \[ \text{Moles of } AgCl = 0.02 \, mol \]

Step 3: Relate \(AgCl\) formation with ionisable chloride ions.
When the complex is treated with excess \(AgNO_3\), only ionisable chloride ions present outside the coordination sphere react with \(Ag^+\).
Each ionisable \(Cl^-\) ion gives one mole of \(AgCl\).
\[ Ag^+ + Cl^- \rightarrow AgCl \downarrow \] Therefore, moles of ionisable chloride ions are equal to moles of \(AgCl\) formed.
\[ \text{Moles of ionisable } Cl^- = 0.02 \, mol \]

Step 4: Find the number of ionisable chloride ions per mole of complex.
We have:
\[ 0.01 \, mol \text{ complex gives } 0.02 \, mol \, AgCl \] So, \(1 \, mol\) complex will give:
\[ \frac{0.02}{0.01} = 2 \, mol \, AgCl \] This means each formula unit of the complex has \(2\) ionisable chloride ions outside the coordination sphere.

Step 5: Identify the correct coordination formula.
The compound must have \(2\) chloride ions outside the coordination sphere.
Now checking the options:
\[ [Cr(H_2O)_4Cl_2]Cl \cdot 2H_2O \] has only \(1\) ionisable chloride ion.
\[ [Cr(H_2O)_3Cl_3] \cdot 3H_2O \] has no ionisable chloride ion outside the coordination sphere.
\[ [Cr(H_2O)_5Cl]Cl_2 \cdot H_2O \] has \(2\) ionisable chloride ions outside the coordination sphere.
\[ [Cr(H_2O)_6]Cl_3 \] has \(3\) ionisable chloride ions outside the coordination sphere.

Step 6: Choose the correct option.
Since the experiment gives \(2\) moles of \(AgCl\) per mole of complex, the complex must contain \(2\) ionisable chloride ions.
Hence, the correct formula is:
\[ [Cr(H_2O)_5Cl]Cl_2 \cdot H_2O \] Therefore, the correct option is (C).
Final Answer:
The correct formula of the compound is:
\[ \boxed{[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O} \]