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Karnataka Board PUCPUC Science 2nd PUC Class 12

When 1 mol CrClX3⋅6HX2O is treated with excess of AgNOX3, 3 mol of AgCl are obtained. The formula of the complex is ______.

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Question

When 1 mol \[\ce{CrCl3.6H2O}\] is treated with excess of \[\ce{AgNO3}\], 3 mol of \[\ce{AgCl}\] are obtained. The formula of the complex is ______.

Options

  • \[\ce{[CrCl3 (H2O)3].3H2O}\]

  • \[\ce{[CrCl2 (H2O)4]Cl.2H2O}\]

  • \[\ce{CrCl(H2O)5]Cl2.H2O}\]

  • \[\ce{[Cr(H2O)6]Cl3}\]

MCQ
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Solution

When 1 mol \[\ce{CrCl3.6H2O}\] is treated with excess of \[\ce{AgNO3}\], 3 mol of \[\ce{AgCl}\] are obtained. The formula of the complex is \[\ce{[Cr(H2O)6]Cl3}\].

Explanation:

One mol of \[\ce{AgCl}\] is precipitated by one mole of \[\ce{Cl^-}\], therefore three moles of \[\ce{AgCl}\] would get precipitared by three moles of chloride ions, \[\ce{Cl^-}\] and in this case \[\ce{3Cl^-}\] are present in ionization sphere (i.e. out side the coordination sphere) in complex at \[\ce{[Cr(H2O)6]Cl3}\], therefore when 1 mol of it is treated with excess of this complex 3 mols of \[\ce{AgCl}\] is precipitated.

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Chapter 9: Coordination Compounds - Exercises [Page 121]

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NCERT Exemplar Chemistry Exemplar [English] Class 12
Chapter 9 Coordination Compounds
Exercises | Q I. 4. | Page 121

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