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:

[C o (N H_{3})_{5} (N O_{2})] C l_{2}

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[C o (N H_{3})_{5} C l] C l (N O_{2})

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[C o (N H_{3})_{4} (N O_{2}) C l] (N H_{3}) C l

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[C o (N H_{3})_{5}] (N O_{2})_{2} C l_{2}

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Solution

The correct option is A $[C o (N H_{3})_{5} (N O_{2})] C l_{2}$
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 $[C o (N H_{3})_{5} (N O_{2})] C l_{2}$ .
$[C o (N H_{3})_{5} N O_{2}] C l_{2} \to {[C o (N H_{3})_{5} N O_{2}]}_{3}^{2 +} + 2 C l^{-}$
$one mole complex \to 3 mole ions$

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

N H_{3}

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A g N O_{3}

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