Question

On treatment of $100mL$ of $0.1M$ solution of $CoC{l}_3.6H_{2}O$ with excess $AgN{O}_3;1.2\times {10}^{22}$ ions are precipitated. The complex is :

• A. $\left[Co{\left(H_{2}O\right)}_{3}C{l}_3\right].2H_{2}O$
• B. $\left[Co{\left(H_{2}O\right)}_6\right]C{l}_3$
• C. $\left[Co{\left(H_{2}O\right)}_{5}Cl\right]C{l}_2.H_{2}O$
• D. $\left[Co{\left(H_{2}O\right)}_{4}Cl\right]C{l}_2.H_{2}O$

Answer 1

The correct option is C $\left[Co{\left(H_{2}O\right)}_{5}Cl\right]C{l}_2.H_{2}O$

Avogadro's number is $6.023\times {10}^{23}$

$1.2\times {10}^{22}$ ions corresponds to $\frac{1.2\times {10}^{22}}{6.023\times {10}^{23}}=0.02$ moles.

$100$ mL of 0.1M solution of $CoC{l}_3.6H_{2}O$ corresponds to $\frac{100mL}{1000mL/L}\times 0.1M=0.01moles$.

$0.01$ moles of $CoC{l}_3.6H_{2}O$ reacts with excess $AgN{O}_3$ to form $0.02$ moles of $AgCl$ precipitate, which means $0.01$ moles of $A{g}^{+}$ and $0.01$ moles of $C{l}^{-}$ ions.

Thus, from one molecule of $CoC{l}_3.6H_{2}O$, one chlorine atom is displaced. Hence, the complex is

$\left[Co{\left(H_{2}O\right)}_{5}Cl\right]C{l}_2.H_{2}O$ as number of chloride ion outside co-ordination sphere should be one.

Hence, the option (C) is the correct answer.