Three moles of AgCl get precipitated when one | vedclass

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(B) The number of moles of $AgCl$ precipitated corresponds to the number of $Cl^{-}$ ions present in the ionization sphere.Given the empirical formula

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(B) The number of moles of $AgCl$ precipitated corresponds to the number of $Cl^{-}$ ions present in the ionization sphere.Given the empirical formula $CrCl_{3} \cdot 3NH_{3} \cdot 3H_{2}O$,and the fact that $3$ moles of $AgCl$ are precipitated,all $3$ chloride ions must be outside the coordination sphere.Thus,the complex is $[Cr(H_{2}O)_{3}(NH_{3})_{3}]Cl_{3}$.In this complex,the coordination number of the central metal ion $Cr^{3+}$ is $6$,which is satisfied by $3$ $H_{2}O$ molecules and $3$ $NH_{3}$ molecules.Since all $3$ $Cl^{-}$ ions are in the ionization sphere,the number of chloride ions satisfying the secondary valency (coordination number) is $0$.

Three moles of $AgCl$ get precipitated when one mole of an octahedral coordination compound with empirical formula $CrCl_{3} \cdot 3NH_{3} \cdot 3H_{2}O$ reacts with excess of silver nitrate. The number of chloride ions satisfying the secondary valency of the metal ion is $......$

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