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Current Question (ID: 10643)

Question:

$\text{Which of the following reactions does not exhibit the oxidising behaviour of H}_2\text{SO}_4\text{?}$

Options:

1. $\text{Cu + 2H}_2\text{SO}_4 \rightarrow \text{CuSO}_4 + \text{SO}_2 + 2\text{H}_2\text{O}$
2. $3\text{S + 2H}_2\text{SO}_4 \rightarrow 3\text{SO}_2 + 2\text{H}_2\text{O}$
3. $\text{C + 2H}_2\text{SO}_4 \rightarrow \text{CO}_2 + 2\text{SO}_2 + 2\text{H}_2\text{O}$
4. $\text{CaF}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{CaSO}_4 + 2\text{HF}$ (Correct)

Solution:

$\text{Hint: Compound oxidizes others and itself gets reduced. The oxidation state must be decreased.}$ $\text{Explanation:}$ $\text{(a) An Oxidising agent oxidizes the other species and itself gets reduced.}$ $\text{Cu + 2H}_2\text{S}^{+6}\text{O}_4 \rightarrow \text{CuSO}_4 + \text{S}^{+4}\text{O}_2 + 2\text{H}_2\text{O}$ $3\text{S + 2H}_2\text{S}^{+6}\text{O}_4 \rightarrow 3\text{S}^{+4}\text{O}_2 + 2\text{H}_2\text{O}$ $\text{C + H}_2\text{S}^{+6}\text{O}_4 \rightarrow \text{CO}_2 + 2\text{S}^{+4}\text{O}_2 + 2\text{H}_2\text{O}$ $\text{CaF}_2 + \text{H}_2\text{S}^{+6}\text{O}_4 \rightarrow \text{CaS}^{+6}\text{O}_4 + 2\text{HF}$ $\text{(b) In reaction (iv), the oxidation number of elements remains unchanged. Thus, in this reaction, H}_2\text{SO}_4\text{ does NOT act as an oxidizing agent. While in other reactions the oxidation state of sulphur in H}_2\text{SO}_4\text{ decreases.}$

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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
}