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Current Question (ID: 18523)

Question:

$\text{(i)} \text{Zr} + 2\text{I}_2 \xrightarrow{870\text{K}} \text{ZrI}_4 \xrightarrow{2075\text{K}} \text{Zr} + 2\text{I}_2$ $\text{(ii)} \text{Ni} + 4\text{CO} \xrightarrow{330\text{K}} \text{Ni(CO)}_4 \xrightarrow{450\text{K}} \text{Ni}_{(\text{pure})} + 4\text{CO}$ $\text{The processes (ii) and (i) above, respectively, are:}$

Options:

1. $\text{Mond's process; Van Arkel process}$
2. $\text{Van Arkel process; Mond's process}$
3. $\text{Goldschmidt process; Mond's process}$
4. $\text{Mond's process; Goldschmidt process}$

Solution:

$\text{STEP 1: Mond process:}$ $\text{In this process, nickel is heated in a stream of carbon monoxide forming a volatile complex named nickel tetracarbonyl. This complex is decomposed at a higher temperature to obtain pure metal.}$ $\text{The reaction is as follows:}$ $\text{Ni} + 4\text{CO} \rightarrow \text{Ni(CO)}_4$ $\text{Ni(CO)}_4 \rightarrow \text{Ni} + 4\text{CO}$ $\text{STEP 2: Van Arkel process:}$ $\text{This method is very useful for removing all the oxygen and nitrogen present in the form of impurity in certain metals like Zr and Ti. The crude metal is heated in an evacuated vessel with iodine.}$ $\text{The metal iodide is more covalent and volatilized.}$ $\text{Zr} + 2\text{I}_2 \rightarrow \text{ZrI}_4$ $\text{The metal iodide is decomposed on a tungsten filament, electrically heated to about 1800\text{K}. The pure metal deposits on the filament.}$ $\text{ZrI}_4 \rightarrow \text{Zr} + 2\text{I}_2$

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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
}