Question

A solution containing 2.675 g of $CoC{l}_3.6N{H}_3$ (Molar mass = 267.5 g/mol) is passed through a cation exchanger. The chloride ions obtained in the solution is treated with an excess of $AgN{O}_3$ to give 4.78 g of $AgCl$ (Molar mass = 143.5 g/mol). The formula of the complex is:
(Atomic mass of Ag = 108 u)

• A. $\left[CoC{l}_3\right(N{H}_3{)}_3]$
• B. $\left[CoCl\right(N{H}_3{)}_5]C{l}_2$
• C. $\left[Co\right(N{H}_3{)}_6]C{l}_3$
• D. $\left[CoC{l}_2\right(N{H}_3{)}_4]Cl$

Answer 1

The correct option is C $\left[Co\right(N{H}_3{)}_6]C{l}_3$

Moles of complex $=\frac{2.675}{267.5}=0.01$
Moles of $AgCl$ precipitated $=\frac{4.78}{143.5}=0.03$
$\therefore C{l}^{-}$ ions produced = 0.03 moles.
It shows that 3 moles of $C{l}^{-}$ ions are produced per mole of complex i.e., 3 $C{l}^{-}$ ions are present outside the sphere.
$\therefore$ The possible structure of complex is $\left[Co\right(N{H}_3{)}_6]C{l}_3$.

$\left[Co\right(N{H}_3{)}_6]C{l}_3\to [Co(N{H}_3{)}_6{]}^{3+}+3C{l}^{-}$