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Q.One mole of CrCl3⋅6H2OCrCl_3 \cdot 6H_2O compound reacts with excess AgNO3AgNO_3 solution to yield two moles of AgClAgCl (s). The structural formula of the compound is (A) [Cr(H2O)5Cl]Cl2⋅H2O[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O (B) [Cr(H2O)3Cl3]⋅3H2O[Cr(H_2O)_3Cl_3] \cdot 3H_2O (C) [Cr(H2O)4Cl2]Cl⋅2H2O[Cr(H_2O)_4Cl_2]Cl \cdot 2H_2O (D) [Cr(H2O)6]Cl3[Cr(H_2O)_6]Cl_3

CBSECBSE Class XII Board 2020MCQ· 1mImportance★★★★★
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The key idea is that only chloride ions outside the coordination sphere (ionisable chlorides) precipitate with AgNO3AgNO_3. Since 1 mole of the complex gives 2 moles of AgClAgCl, there must be 2 ionisable Cl⁻ ions outside the coordination sphere. The correct option is (A) [Cr(H2O)5Cl]Cl2⋅H2O[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O.

Why Werner Coordination Theory is the key

Before we touch a single calculation, let's understand the principle. Alfred Werner showed that in coordination compounds, some groups are directly bonded to the central metal ion (inside the coordination sphere, written in square brackets), while others are outside the sphere as counter-ions.

Here's the critical exam point: Only the chloride ions outside the coordination sphere are free to precipitate with AgNO3AgNO_3 as AgClAgCl. Chloride ions inside the sphere are covalently bonded to chromium and do not dissociate in solution.

So the problem reduces to a simple count: how many chlorides are outside the brackets?

Step-by-step reasoning

1. Identify the total composition

The formula CrCl3⋅6H2OCrCl_3 \cdot 6H_2O tells us we have:

  • 1 Cr atom
  • 3 Cl atoms
  • 6 H₂O molecules

Every option must contain exactly these atoms — just arranged differently between inside and outside the coordination sphere.

2. Interpret the experimental result

1 mole of compound + excess AgNO3AgNO_3 → 2 moles of AgClAgCl (s)

This means exactly 2 chloride ions from each formula unit are free to react. These are the ionisable chlorides — the ones outside the square brackets.

3. Apply the logic to each option

Let’s count the outside chlorides for each choice:

  • (A) [Cr(H2O)5Cl]Cl2⋅H2O[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O

    Inside: 1 Cl. Outside: 2 Cl. → 2 ionisable Cl⁻ ✓

  • (B) [Cr(H2O)3Cl3]⋅3H2O[Cr(H_2O)_3Cl_3] \cdot 3H_2O

    Inside: 3 Cl. Outside: 0 Cl. → 0 ionisable Cl⁻ ✗

  • (C) [Cr(H2O)4Cl2]Cl⋅2H2O[Cr(H_2O)_4Cl_2]Cl \cdot 2H_2O

    Inside: 2 Cl. Outside: 1 Cl. → 1 ionisable Cl⁻ ✗

  • (D) [Cr(H2O)6]Cl3[Cr(H_2O)_6]Cl_3 …

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