A solution containing $2.675 \mathrm{~g}$ of $\mathrm{CoCl}_3 .6 \mathrm{NH}_3$ (molar mass $=267.5…

A solution containing $2.675 \mathrm{~g}$ of $\mathrm{CoCl}_3 .6 \mathrm{NH}_3$ (molar mass $=267.5 \mathrm{~g} \mathrm{~mol}^{-1}$ ) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of $\mathrm{AgNO}_3$ to give $4.78 \mathrm{~g}$ of $\mathrm{AgCl}$ (molar mass $=143.5 \mathrm{~g} \mathrm{~mol}^{-1}$ ). The formula of the complex is (At. Mass of $\mathrm{Ag}=108 \mathrm{u}$ )
  1. $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3$
  2. $\left[\mathrm{CoCl}_2\left(\mathrm{NH}_3\right)_4\right] \mathrm{Cl}$
  3. $\left[\mathrm{CoCl}_3\left(\mathrm{NH}_3\right)_3\right]$
  4. $\left[\mathrm{CoCl}\left(\mathrm{NH}_3\right)_5\right] \mathrm{Cl}_2$

Solution

$\mathrm{CoCl}_3 .6 \mathrm{NH}_3 \rightarrow \times \mathrm{Cl}^{-} \stackrel{\mathrm{AgNO}_3}{\longrightarrow} \times \mathrm{AgCl} \downarrow$ $\mathrm{n}(\mathrm{AgCl})=\mathrm{x}\left(\mathrm{CoCl}_3 .6 \mathrm{NH}_3\right)$ $\frac{4.78}{143.5}=\times \frac{2.675}{267.5} \quad \therefore \mathrm{x}=3$ $\therefore$ The complex is $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3$

Asked in: JEE Main 2010

Practice more Coordination Compounds questions on Aicharya