Question

A coordination complex compound of cobalt has 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 $AgN{O}_3$ solution two moles of $AgCl$ get precipitated. The ionic formula of this complex would be:

• A. $\left[Co{\left(N{H}_3\right)}_4\right(NO{)}_{2}Cl]\left[\left(N{H}_3\right)Cl\right]$
• B. $\left[Co{\left(N{H}_3\right)}_{5}Cl\right]\left[Cl\left(N{O}_2\right)\right]$
• C. $\left[Co{\left(N{H}_3\right)}_5\left(N{O}_2\right)\right]C{l}_2$
• D. $\left[Co{\left(N{H}_3\right)}_5\right]\left[{\left(N{O}_2\right)}_{2}C{l}_2\right]$

Answer 1

Answer: (C)

$\left[Co{\left(N{H}_3\right)}_5\left(N{O}_2\right)\right]C{l}_2$

The crux lies in the fact that on treating with $AgN{O}_3$ solution two moles of $AgCl$ precipitated indicating that both the chloride ions are present outside the coordination sphere hence answer is C.

The most probable complex which gives three moles ions in aqueous solution may be $\left[Co{\left(N{H}_3\right)}_5\left(N{O}_2\right)\right]C{l}_2$because it gives two chlorine atoms on ionisation.

$\left[Co{\left(N{H}_3\right)}_5\left(N{O}_2\right)\right]C{l}_2$ $\to$ ${\left[Co{\left(N{H}_3\right)}_5\left(N{O}_2\right)\right]}^{2+}+2C{l}^{-}$