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Chemistry Question on coordination compounds

When 11 mol CrCl3⋅6H2OCrCl_{3}\cdot 6H_{2}O is treated with excess of AgNO3AgNO_{3}, 33 mol of AgClAgCl are obtained. The formula of the complex is

A

[CrCl3(H2O)3]⋅3H2O[CrCl_{3}(H_{2}O )_{3}] \cdot 3H_{2}O

B

[CrCl2(H2O)4]Cl⋅2H2O[CrCl_{2}(H_{2}O)_{4}]Cl\cdot 2H_{2}O

C

[CrCl(H2O)5]Cl2⋅H2O[CrCl(H_{2}O )_{5}]Cl_{2}\cdot H_{2}O

D

[Cr(H2O)6]Cl3[Cr(H_{2}O )_{6}]Cl_{3}

Answer

[Cr(H2O)6]Cl3[Cr(H_{2}O )_{6}]Cl_{3}

Explanation

Solution

As 33 moles of AgClAgCl are obtained when 11 mol of CrCl3⋅6H2OCrCl_{3}\cdot 6H_{2}O is treated with excess of AgNO3AgNO_{3} which shows that one molecule of the complex gives three chloride ions in solution. Hence, formula of the complex is [Cr(H2O)6]Cl3.[Cr(H_{2}O)_{6}]Cl_{3}.