When 1 mol CrCl3⋅ 6H2O is treated with excess of AgNO3, 3 mol of AgCl are obtained. The formula of the complex is

Question

When 1 mol CrCl3⋅ 6H2O is treated with excess of AgNO3, 3 mol of AgCl are obtained. The formula of the complex is (A) [CrCl3(H2O )3] ⋅ 3H2O (B) [CrCl2(H2O)4]Cl⋅ 2H2O (C) [CrCl(H2O )5]Cl2⋅ H2O (D) [Cr(H2O )6]Cl3

Tardigrade Admin · Accepted Answer

As 3 moles of AgCl are obtained when 1 mol of CrCl3⋅ 6H2O is treated with excess of AgNO3 which shows that one molecule of the complex gives three chloride ions in solution. Hence, formula of the complex is [Cr(H2O)6]Cl3.

Q. When $1$ mol $C r C l_{3} \cdot 6 H_{2} O$ is treated with excess of $A g N O_{3}$ , $3$ mol of $A g Cl$ are obtained. The formula of the complex is

$[C r C l_{3} (H_{2} O)_{3}] \cdot 3 H_{2} O$

$[C r C l_{2} (H_{2} O)_{4}] Cl \cdot 2 H_{2} O$

$[C r Cl (H_{2} O)_{5}] C l_{2} \cdot H_{2} O$

$[C r (H_{2} O)_{6}] C l_{3}$

As $3$ moles of $A g Cl$ are obtained when $1$ mol of $C r C l_{3} \cdot 6 H_{2} O$ is treated with excess of $A g N O_{3}$ which shows that one molecule of the complex gives three chloride ions in solution. Hence, formula of the complex is $[C r (H_{2} O)_{6}] C l_{3} .$

Q. When 1 mol CrCl3​⋅6H2​O is treated with excess of AgNO3​, 3 mol of AgCl are obtained. The formula of the complex is

[CrCl3​(H2​O)3​]⋅3H2​O

[CrCl2​(H2​O)4​]Cl⋅2H2​O

[CrCl(H2​O)5​]Cl2​⋅H2​O

[Cr(H2​O)6​]Cl3​