Question

On the treatment of $100\mathrm{ml}$ of $0.1\mathrm{M}$ solution of ${\mathrm{CoCl}}_3.6{\mathrm{H}}_2\mathrm{O}$ with excess ${\mathrm{AgNO}}_3$, $1.2\times {10}^{22}$ions are precipitated. The complex is

• A. $\left[\mathrm{Co}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]{\mathrm{Cl}}_3$
• B. $\left[\mathrm{Co}{\left({\mathrm{H}}_2\mathrm{O}\right)}_5\mathrm{Cl}\right]{\mathrm{Cl}}_2.{\mathrm{H}}_2\mathrm{O}$
• C. $\left[\mathrm{Co}{\left({\mathrm{H}}_2\mathrm{O}\right)}_4{\mathrm{Cl}}_2\right]\mathrm{Cl}.2{\mathrm{H}}_2\mathrm{O}$
• D. $\left[\mathrm{Co}{\left({\mathrm{H}}_2\mathrm{O}\right)}_3{\mathrm{Cl}}_3\right].3{\mathrm{H}}_2\mathrm{O}$

Answer 1

The correct option is B

$\left[\mathrm{Co}{\left({\mathrm{H}}_2\mathrm{O}\right)}_5\mathrm{Cl}\right]{\mathrm{Cl}}_2.{\mathrm{H}}_2\mathrm{O}$

Step 1: Given data

Molarity of complex$=$$0.1\mathrm{M}$

Volume of complex$=100\mathrm{ml}$

${\mathrm{AgNO}}_3$ ions precipitated$=1.2\times {10}^{22}$

Step 2: Determining the complex

Moles of complex$=\frac{\mathrm{Molarity}\times \mathrm{volume}}{1000}$

$=\frac{0.1\times 100}{1000}\Rightarrow 0.01\mathrm{mol}$

Moles of ions precipitated with an excess of ${\mathrm{AgNO}}_3$$=\frac{\mathrm{ions}\mathrm{precipitated}}{\mathrm{Avogadro}\mathrm{number}}$

$=\frac{1.2\times {10}^{22}}{6.02\times {10}^{23}}\Rightarrow 0.02\mathrm{moles}$

Number of the ${\mathrm{Cl}}^{-}$ present in the ionization sphere$=\frac{\mathrm{moles}\mathrm{of}\mathrm{ions}\mathrm{precipitated}\mathrm{with}\mathrm{excess}{\mathrm{AgNO}}_3}{\mathrm{Moles}\mathrm{of}\mathrm{complex}}$

$=\frac{0.02}{0.01}\Rightarrow 2$

It means two chlorine ions must be present in the in the solution.

Therefore, $\left[\mathrm{Co}{\left({\mathrm{H}}_2\mathrm{O}\right)}_5\mathrm{Cl}\right]{\mathrm{Cl}}_2.{\mathrm{H}}_2\mathrm{O}$ is the final complex.