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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
\[\ce{[CrCl3 (H2O)3].3H2O}\]
\[\ce{[CrCl2 (H2O)4]Cl.2H2O}\]
\[\ce{CrCl(H2O)5]Cl2.H2O}\]
\[\ce{[Cr(H2O)6]Cl3}\]
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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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