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Current Question (ID: 18703)

Question:

$\text{The correct order of the stoichiometrics of AgCl formed when AgNO}_3 \text{ in excess is treated with the complexes:}$ $\text{CoCl}_3 \cdot 6\text{NH}_3, \text{CoCl}_3 \cdot 5\text{NH}_3, \text{CoCl}_3 \cdot 4\text{NH}_3,$ $\text{respectively, is:}$

Options:

1. $1 \text{ AgCl}, 3 \text{ AgCl}, 2 \text{ AgCl}$
2. $3 \text{ AgCl}, 1 \text{ AgCl}, 2 \text{ AgCl}$
3. $3 \text{ AgCl}, 2 \text{ AgCl}, 1 \text{ AgCl}$
4. $2 \text{ AgCl}, 3 \text{ AgCl}, 1 \text{ AgCl}$

Solution:

$\text{STEP 1: According to Werner's theory,}$ $\text{CoCl}_3 \cdot 6\text{NH}_3 \rightarrow [\text{Co(NH}_3)_6]^{3+} + 3 \text{ Cl}^- $ $\text{CoCl}_3 \cdot 5\text{NH}_3 \rightarrow [\text{Co(NH}_3)_5\text{Cl}]^{2+} + 2 \text{ Cl}^- $ $\text{CoCl}_3 \cdot 4\text{NH}_3 \rightarrow [\text{Co(NH}_3)_4\text{Cl}_2]^+ + \text{ Cl}^- $ $\text{STEP 2: When AgNO}_3 \text{ in excess is treated with these complexes then the following reactions take place:}$ $1. [\text{Co(NH}_3)_6]^{3+} + 3\text{Cl}^- + \text{AgNO}_3(\text{Excess}) \rightarrow 3\text{AgCl} + [\text{Co(NH}_3)_6]^{3+}$ $2. [\text{Co(NH}_3)_5]^{2+} 2\text{Cl}^- + \text{AgNO}_3(\text{Excess}) \rightarrow 2\text{AgCl} + [\text{Co(NH}_3)_5]^{2+}$ $3. [\text{Co(NH}_3)_4\text{Cl}_2]^+ \text{Cl}^- + \text{AgNO}_3(\text{Excess}) \rightarrow \text{AgCl} + [\text{Co(NH}_3)_4 \text{Cl}_2]^+$

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}