On treatment of $100\, mL$ of $0.1\, M$ solution of $CoCl _{3} \cdot 6 H _{2} O$ with excess $AgNO _{3} ; 1.2 \times 10^{22}$ ions are precipitated. The complex is

Question

Sum of oxidation state (magnitude) and coordination number of cobalt in $Na[Co(bpy)Cl_4]$ is____ .
(Given : bpy =

)

Answer 1

To solve this problem, we will determine the complex compound by analyzing the number of ions precipitated when a solution of the cobalt complex is treated with excess silver nitrate (AgNO_3). The relevant reaction is that silver nitrate precipitates chloride ions as silver chloride (AgCl). Let's solve this step-by-step:

First, determine the moles of chloride ions precipitated:

We are given that 1.2 \times 10^{22} ions are precipitated. Since each chloride ion is a monoanion, this value directly gives the number of chloride ions.

Convert the number of ions to moles:

The number of ions is converted to moles using Avogadro's number (6.022 \times 10^{23} \text{ ions/mol}):

\text{Moles of } Cl^- = \frac{1.2 \times 10^{22} \text{ ions}}{6.022 \times 10^{23} \text{ ions/mol}} \approx 0.02 \text{ mol}

Analyze the given complex compound:

We have a 100\, \text{mL} solution with a concentration of 0.1\, M, which equates to 0.01\, \text{mol} of the cobalt complex.

Correlate the moles of Cl^− precipitated with the complex compound options:

The complex \left[ Co \left( H _{2} O \right)_{5} Cl \right] Cl _{2} \cdot H _{2} O releases two chloride ions per molecule:

Since 0.01 \text{ mol} of the complex is in solution, it releases 0.01 \times 2 = 0.02 \text{ mol} of chloride ions.

Select the correct complex:

The precipitated chloride ions (0.02 mol) match the release from the complex with two chloride ions, confirming that the correct answer is:

\left[ Co \left( H _{2} O \right)_{5} Cl \right] Cl _{2} \cdot H _{2} O

Answer 2