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Current Question (ID: 18673)

Question:
$\text{Cobalt(III) chloride forms several octahedral complexes with ammonia.}$ $\text{A compound among the following that does not give a test for chloride ions}$ $\text{with silver nitrate at 25 } ^\circ \text{C is:}$
Options:
  • 1. $\text{CoCl}_3 \cdot 3\text{NH}_3$
  • 2. $\text{CoCl}_3 \cdot 4\text{NH}_3$
  • 3. $\text{CoCl}_3 \cdot 5\text{NH}_3$
  • 4. $\text{CoCl}_3 \cdot 6\text{NH}_3$
Solution:
$\text{HINT: If Cl is present outside the complex it will produce white precipitate}$ $\text{with silver nitrate}$ $\text{Explanation: Octahedral complexes will form. So the co-ordination number}$ $\text{of Co = 6}$ $\text{The reactions of complexes in the solution are as follows:}$ $\text{CoNH}_3{}_6\text{Cl}_3 \rightarrow \text{CoNH}_3{}_6^{3+} + \text{Cl}^- $ $\text{CoNH}_3{}_3\text{Cl}_3 \rightarrow \text{CoNH}_3{}_3\text{Cl}_3$ $\text{CoNH}_3{}_4\text{Cl}_2 \rightarrow \text{CoNH}_3{}_4\text{Cl}_2^+ + \text{Cl}^- $ $\text{CoNH}_3{}_5\text{ClCl}_2 \rightarrow \text{CoNH}_3{}_5\text{Cl}^{2+} + 2\text{Cl}^- $

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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
}