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Current Question (ID: 8564)

Question:

$\text{Which of the following reactions does not represent a redox change?}$

Options:

1. $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$ (Correct)
2. $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$
3. $\text{Na} + \text{H}_2\text{O} \rightarrow \text{NaOH} + \frac{1}{2}\text{H}_2$
4. $\text{MnCl}_3 \rightarrow \text{MnCl}_2 + \frac{1}{2}\text{Cl}_2$

Solution:

\text{A redox reaction involves a change in oxidation numbers of elements.} \text{Let's analyze each reaction:} \text{Option 1: } \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \text{Ca: +2} \rightarrow \text{+2 (no change), C: +4} \rightarrow \text{+4 (no change), O: -2} \rightarrow \text{-2 (no change)} \text{No oxidation number changes, so this is NOT a redox reaction.} \text{Option 2: } 2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} \text{H: 0} \rightarrow \text{+1, O: 0} \rightarrow \text{-2 (oxidation numbers change, this IS redox)} \text{Option 3: } \text{Na} + \text{H}_2\text{O} \rightarrow \text{NaOH} + \frac{1}{2}\text{H}_2 \text{Na: 0} \rightarrow \text{+1, H: +1} \rightarrow \text{0 (oxidation numbers change, this IS redox)} \text{Option 4: } \text{MnCl}_3 \rightarrow \text{MnCl}_2 + \frac{1}{2}\text{Cl}_2 \text{Mn: +3} \rightarrow \text{+2, Cl: -1} \rightarrow \text{0 (oxidation numbers change, this IS redox)}

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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
}