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

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Q. When $1$ mol $CrCl_{3}\cdot 6H_{2}O$ is treated with excess of $AgNO_{3}$, $3$ mol of $AgCl$ are obtained. The formula of the complex is

Coordination Compounds

Solution:

As $3$ moles of $AgCl$ are obtained when $1$ mol of $CrCl_{3}\cdot 6H_{2}O$ is treated with excess of $AgNO_{3}$ which shows that one molecule of the complex gives three chloride ions in solution. Hence, formula of the complex is $[Cr(H_{2}O)_{6}]Cl_{3}.$