Question

A complex compound of cobalt has molecular formula containing five $N{H}_3$ molecules, one $N{O}_2$ group and two $Cl$ atoms for one $Co$ atom. One mole of this compound produces three moles of ions in aqueous solution. On reacting with an excess of $AgN{O}_3$ solution, two moles of $AgCl$ get precipitated. The ionic formula of the compound is:

• A. $\left[Co\right(N{H}_3{)}_{4}N{O}_{2}Cl]$ $\left[\right(N{H}_3\left)Cl\right]$
• B. $\left[Co\right(N{H}_3{)}_{5}Cl]$ $\left[Cl\right(N{O}_2\left)\right]$
• C. $\left[Co\right(N{H}_3{)}_5\left(N{O}_2\right)]C{l}_2$
• D. $\left[Co\right(N{H}_3{)}_5\left]\right[\left(N{O}_2\right)C{l}_3]$

Answer 1

Answer: (C)

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

The ionic formula of the compound is $\left[Co\right(N{H}_3{)}_5\left(N{O}_2\right)]C{l}_2$.

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

Total 3 ions are produced in aqueous solution.

Hence, the correct option is $\text{C}$