Question

On treatment of 100 mL of 0.1 M solution of $CoCl_2$. 6H₂O with excess $AgNO_3$; 1.2 x $10^{22}$ ions are precipitated. The complex is:

• A. $[Co(H_2O)_3Cl_3].3H_2O$
• B. $[Co(H_2O)_6]Cl_2$
• C. $[Co(H_2O)_5Cl]Cl_2.H_2O$
• D. $[Co(H_2O)_4Cl_2]Cl.2H_2O$

Answer 1

Answer: (B)

(2) $[Co(H_2O)_6]Cl_2$

Answer 2

To determine the complex structure, we first need to find out how many chloride ions are outside the coordination sphere (i.e., are ionizable) by reacting with $AgNO_3$.

1. Calculate the moles of the cobalt compound: Volume of solution = 100 mL = 0.1 L Molarity of solution = 0.1 M Moles of $CoCl_2 \cdot 6H_2O = \text{Molarity} \times \text{Volume} = 0.1 \text{ mol/L} \times 0.1 \text{ L} = 0.01 \text{ mol}$.

2. Calculate the moles of precipitated chloride ions: Number of ions precipitated = $1.2 \times 10^{22}$ ions. Using Avogadro's number ($N_A \approx 6.022 \times 10^{23} \text{ ions/mol}$ or approximately $6 \times 10^{23} \text{ ions/mol}$ for quick calculation, as often used in JEE problems): Moles of $Cl^- = \frac{\text{Number of ions}}{\text{Avogadro's number}} = \frac{1.2 \times 10^{22}}{6.022 \times 10^{23}} \approx 0.0199 \text{ mol} \approx 0.02 \text{ mol}$.

3. Determine the number of ionizable chloride ions per mole of the complex: Number of $Cl^-$ ions precipitated per mole of $CoCl_2 \cdot 6H_2O = \frac{\text{Moles of } Cl^-}{\text{Moles of } CoCl_2 \cdot 6H_2O} = \frac{0.02 \text{ mol}}{0.01 \text{ mol}} = 2$. This means that 2 chloride ions are outside the coordination sphere and will precipitate with $AgNO_3$.

4. Analyze the given options based on the formula $CoCl_2 \cdot 6H_2O$ and 2 ionizable chloride ions: The complex must contain 1 Co atom, 2 Cl atoms, and 6 H$_2$O molecules in total. It must also have 2 chloride ions outside the coordination sphere.

   • (1) $[Co(H_2O)_3Cl_3].3H_2O$: This complex has 3 Cl atoms and 6 H$_2$O molecules in total. It corresponds to $CoCl_3 \cdot 6H_2O$. Also, it has 0 ionizable $Cl^-$ ions. This option is incorrect as it does not match $CoCl_2 \cdot 6H_2O$.
   • (2) $[Co(H_2O)_6]Cl_2$:
     • This complex has 1 Co, 2 Cl, and 6 H$_2$O in total. This matches the formula $CoCl_2 \cdot 6H_2O$.
     • The coordination sphere is $[Co(H_2O)_6]^{2+}$, and the counter ions are $2Cl^-$.
     • It has 2 ionizable $Cl^-$ ions, which matches our calculation.
     • The coordination number of Co is 6 (octahedral), and Co is in the +2 oxidation state, which is common for cobalt(II). This option is consistent.
   • (3) $[Co(H_2O)_5Cl]Cl_2.H_2O$: This complex has 3 Cl atoms and 6 H$_2$O molecules in total. It corresponds to $CoCl_3 \cdot 6H_2O$. Also, it has 2 ionizable $Cl^-$ ions. This option is incorrect as it does not match $CoCl_2 \cdot 6H_2O$. (Note: This would be the correct answer if the question had stated $CoCl_3 \cdot 6H_2O$, as in the similar question).
   • (4) $[Co(H_2O)_4Cl_2]Cl.2H_2O$: This complex has 3 Cl atoms and 6 H$_2$O molecules in total. It corresponds to $CoCl_3 \cdot 6H_2O$. Also, it has 1 ionizable $Cl^-$ ion. This option is incorrect as it does not match $CoCl_2 \cdot 6H_2O$.

Based on the given information that the compound is $CoCl_2 \cdot 6H_2O$ and that 2 chloride ions are precipitated, option (2) is the only correct choice.