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Question: When 1 mol \(CrCl_{3},6H_{2}O\)is treated with excess of \(AgNO_{3},\) 3 mol of AgCl are obtained. T...

When 1 mol CrCl3,6H2OCrCl_{3},6H_{2}Ois treated with excess of AgNO3,AgNO_{3}, 3 mol of AgCl are obtained. The formula of the complex is

A

[CrCl3(H2O)3].3H2O\lbrack CrCl_{3}(H_{2}O)_{3}\rbrack.3H_{2}O

B

[CrCl2(H2O)4]Cl.2H2O\lbrack CrCl_{2}(H_{2}O)_{4}\rbrack Cl.2H_{2}O

C

[CrCl(H2O)5]Cl2.H2O\lbrack CrCl(H_{2}O)_{5}\rbrack Cl_{2}.H_{2}O

D

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

Answer

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

Explanation

Solution

As 3 moles of AgCl are obtained when 1 mol of CrCl3,6H2OCrCl_{3},6H_{2}Ois treated with excess of AgNO3AgNO_{3}which shows that one molecules of the complex gives three chloride ions in solutions. Hence, formula of the complex is [Cr(H2O)6]Cl3.\lbrack Cr(H_{2}O)_{6}\rbrack Cl_{3}.