A six-coordinate complex of formula $C r C l_{6} .6 H_{2}$ O is green in colour. A $0.1 M$ solution of the complex when treated with excess of $A g N O_{3}$ gave $28.7 g$ of white precipitate. The formula of the complex would be:

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

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[C r {(H_{2} O)}_{5} C l] C l_{2} . H_{2} O

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[C r (H_{2} O)_{4} C l_{2}] C l .2 H_{2} O

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

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Solution

The correct option is B $[C r {(H_{2} O)}_{5} C l] C l_{2} . H_{2} O$
Let us call the six-coordinate compound of Cr 'x'. When a 0.1 M solution of x is treated with excess $A g N O_{3}$ , we get white precipitate of $A g C l$ . The number of moles of AgCl found will give us an estimate of the number of moles of Cl outside the coordination sphere in x.
We know that the molecular weight of $A g C l$ is 143 g.
Obtained weight of precipitate is 28.7 g.
Hence, the number of moles of $A g C l$ will be $\frac{28.7 / 143}{0.1} =$ 2 mol
Hence, we know that 2 moles of Cl atoms are present outside the coordination sphere in the solution. This tells us that the formula of the six coordination compound has to be $[C r {(H_{2} O)}_{5} C l] C l_{2} . H_{2} O$

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Similar questions

C r C l_{6} .6 H_{2}

0.1 M

A g N O_{3}

28.7 g

When one mole of a complex is treated with excess of $A g N O_{3}$ solution, 143.5 g of white precipitate is formed, the complex is :

M C l_{3} .4 H_{2} O

A g N O_{3}

F e {C l}_{3} \cdot 4 N H_{3}

A g N O_{3}

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