When 1 mole of $C r C l_{3} .6 H_{2} O$ is treated with excess of $A g N O_{3},$ 3 moles of AgCl are obtained. The formula of the complex is
(a) $[C r C l_{3} (H_{2} O)_{3}] 3 H_{2} O$
(b) $[C r C l_{3} (H_{2} O)_{4}] C l_{2} .2 H_{2} O$
(c) $[C r C l_{3} (H_{2} O)_{5}] C l_{2} . H_{2} O$
(d) $[C r (H_{2} O)_{6}] C l_{3}$

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Solution

1 mole of $A g N O_{3}$ precipiates one free chloride ion $(C l^{-} .)$
Here, 3 moles of AgCl are precipitated by excess of $A g N O_{3} .$ Hence, there must be three free $C l^{-}$ ions.
So, the formula of the complex can be $[C r (H_{2} O)_{6}] C l_{3}$ and the correct choice is (d).

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solution -

When 1 mole of $C r C l_{3} .6 H_{2} O$ is treated with $A g N O_{3}$ , $x$ moles of $A g C l$ are obtained. The formula of the compound is $[C r (H_{2} O)_{6}] C l_{3}$ , $y$ is the oxidation number of $C r$ and $z$ is the coordination number of the compound. The value of $2 x + y - z$ is ___.

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When 1 mole of CrCl3.6H2O is treated with excess of AgNO3, 3 moles of AgCl are obtained. The formula of the complex is (a) [CrCl3(H2O)3]3H2O (b) [CrCl3(H2O)4]Cl2.2H2O (c) [CrCl3(H2O)5]Cl2.H2O (d) [Cr(H2O)6]Cl3

1 mole of AgNO3 precipiates one free chloride ion (Cl−.) Here, 3 moles of AgCl are precipitated by excess of AgNO3. Hence, there must be three free Cl− ions. So, the formula of the complex can be [Cr(H2O)6]Cl3 and the correct choice is (d).

1 mole of $A g N O_{3}$ precipiates one free chloride ion $(C l^{-} .)$
Here, 3 moles of AgCl are precipitated by excess of $A g N O_{3} .$ Hence, there must be three free $C l^{-}$ ions.
So, the formula of the complex can be $[C r (H_{2} O)_{6}] C l_{3}$ and the correct choice is (d).