A coordination complex compound of cobalt has the molecular formula containing five ammonia molecules, one nitro group and two chlorine atoms for one cobalt atom. One mole of this compound produces three mole ions in an aqueous solution. On reacting this solution with excess of $A g N O_{3}$ solution, we get two moles of $A g C l$ precipitate. The ionic formula for this complex would be:

Open in App

Solution

As the complex gives two moles of $A g C l$ precipitate with $A g N O_{3}$ solution, so the complex must have two ionisable $C l$ atoms. Hence, the probable complex, which gives three mole ions may be $[C o (N H_{3})_{5} (N O_{2})] C l_{2}$ .
$[C o (N H_{3})_{5} N O_{2}] C l_{2} \to {[C o (N H_{3})_{5} N O_{2}]}_{3}^{2 +} + 2 C l^{-}$
$one mole complex \to 3 mole ions$

Suggest Corrections

Similar questions

Q. A complex compound of cobalt has molecular formula containing five

N H_{3}

molecules, one nitro group and two chlorine atom for one cobalt atom. One mole of this compound produces three moles of ion in aqueous solution. On reaction with excess of

A g N O_{3}

solution two moles of

A g C l

get precipitated. The ionic formula of the compound is :

One mole of the complex compound $C o (N H_{3})_{5} C l_{3}$ gives three moles of ions on dissolution in water. One mole of the same complex reacts with two moles of $A g N O_{3}$ solution to yield two moles of $A g C l (s)$ . The structure of complex is: