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A coordination 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 moles of ions in an aqueous solution. The aqueous solution on treatment with an excess of ${AgNO}_{3}$ gives two equivalents of $AgCl$ as a precipitate. The formula of this complex would be :

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$[Co {({NH}_{3})}_{4} ({NO}_{2}) Cl] [({NH}_{3}) Cl]$

$[Co {({NH}_{3})}_{5} Cl] [Cl {NO}_{2}]$

$[Co {({NH}_{3})}_{5} ({NO}_{2})] {Cl}_{2}$

$[Co {({NH}_{3})}_{5}] [{({NO}_{2})}_{2} {Cl}_{2}]$

answer is C.

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Detailed Solution

Since, $[Co {({NH}_{3})}_{5} ({NO}_{2})] {Cl}_{2}$ produces three moles of ions in aqueous solution and two chlorine atoms upon ionization.

$[Co {({NH}_{3})}_{5} ({NO}_{2})] {Cl}_{2} \to {[Co {({NH}_{3})}_{5} ({NO}_{2})]}^{2 +} + 2 {Cl}^{-}$

The correct complex is $[Co {({NH}_{3})}_{5} ({NO}_{2})] {Cl}_{2}$

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