Question

On treatment of 100 mL 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. $\left[Co\right(H_{2}O{)}_6]C{l}_3$
  
• B. $\left[Co\right(H_{2}O{)}_{5}Cl]C{l}_2.H_{2}O$
• C. $\left[Co\right(H_{2}O{)}_{4}C{l}_2]Cl.2H_{2}O$
• D. $\left[Co\right(H_{2}O{)}_{3}C{l}_3].3H_{2}O$

Answer 1

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

Millimoles of $AgN{O}_3=\frac{1.2\times {10}^{22}}{6\times {10}^{23}}\times 1000=20$
Millimoles of $CoC{l}_3.6H_{2}O=0.1\times 100=10$
$\therefore$ Each mole of $CoC{l}_3.6H_{2}O$ gives two chloride ions.
$\therefore \left[Co\right(H_{2}O{)}_{5}Cl]C{l}_2.H_{2}O$