A solution containing 2.615 g of CoCl₃6NH₃ (molar mass = 267 g mol^-1) is passed through a cation exchanger' The chloride ions obtained in solution were treated with excess of AgNO₃ to give 4.78 g of AgCl. The formula of the complex is:

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${[Co {({NH}_{3})}_{3} {Cl}_{3}]}_{3} {NH}_{3}$

$[Co {({NH}_{3})}_{5} Cl] {Cl}_{2} {NH}_{3}$

$[Co {({NH}_{3})}_{6}] {Cl}_{3}$

$[Co {({NH}_{3})}_{4} {Cl}_{2}] {Cl}_{2} {NH}_{3}$

answer is A.

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Detailed Solution

No. of moles of complex $= \frac{2 .675}{267 .5} = 0 .01$
No. of moles of AgCI $= \frac{4 .78}{143 .5} = 0 .03$
This shows that three Cl^- ions are precipitated.
Thus, formula of the complex is $[Co {({NH}_{3})}_{6}] {Cl}_{3}$ .

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