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Current Question (ID: 8599)

Question:

$\text{The change in oxidation number of chlorine when } \text{Cl}_2 \text{ gas reacts with hot and concentrated sodium hydroxide solution is:}$

Options:

1. $\text{Zero to +1 and Zero to -5}$
2. $\text{Zero to -1 and Zero to +5}$
3. $\text{Zero to -1 and Zero to +3}$
4. $\text{Zero to +1 and Zero to -3}$

Solution:

$\text{In a disproportionation reaction, the same element is oxidized and reduced simultaneously.}$ $\text{Therefore, when the chlorine gas is reacted with hot and concentrated NaOH solution, it is seen to disproportionate into chloride (Cl}^-\text{) and chlorate (ClO}_3^-\text{) ions.}$ $\text{During this process, chlorine is oxidized to chlorate (ClO}_3^-\text{) ions (oxidation state = +5) and reduced to chloride ion (oxidation state = -1).}$ $3\text{Cl}_2 + 6\text{NaOH} \rightarrow 5\text{NaCl} + \text{NaClO}_3 + 3\text{H}_2\text{O}$ $\text{The oxidation state changes from 0 to -1 (reduction) and 0 to +5 (oxidation).}$

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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
}