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Current Question (ID: 8610)

Question:

$\text{Identify the disproportionation reaction among the following:}$

Options:

1. $\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$
2. $\text{CH}_4 + 4\text{Cl}_2 \rightarrow \text{CCl}_4 + 4\text{HCl}$
3. $\text{2F}_2 + 2\text{OH}^{-} \rightarrow 2\text{F}^{-} + \text{OF}_2 + \text{H}_2\text{O}$
4. $\text{2NO}_2 + 2\text{OH}^{-} \rightarrow \text{NO}_2^{-} + \text{NO}_3^{-} + \text{H}_2\text{O}$

Solution:

$\text{Hint: A disproportionation reaction is a type of redox reaction in which the same substance is simultaneously oxidized and reduced.}$ $\text{Let's analyze the oxidation numbers of the elements in each reaction:}$ $1.\ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$ $\text{In } \text{CH}_4\text{, C is } -4\text{, H is } +1\text{.}$ $\text{In } \text{O}_2\text{, O is } 0\text{.}$ $\text{In } \text{CO}_2\text{, C is } +4\text{, O is } -2\text{.}$ $\text{In } \text{H}_2\text{O}\text{, H is } +1\text{, O is } -2\text{.}$ $\text{Carbon is oxidized (} -4 \rightarrow +4 \text{) and oxygen is reduced (} 0 \rightarrow -2 \text{). This is not a disproportionation reaction as two different elements undergo oxidation and reduction.}$ $2.\ \text{CH}_4 + 4\text{Cl}_2 \rightarrow \text{CCl}_4 + 4\text{HCl}$ $\text{In } \text{CH}_4\text{, C is } -4\text{, H is } +1\text{.}$ $\text{In } \text{Cl}_2\text{, Cl is } 0\text{.}$ $\text{In } \text{CCl}_4\text{, C is } +4\text{, Cl is } -1\text{.}$ $\text{In } \text{HCl}\text{, H is } +1\text{, Cl is } -1\text{.}$ $\text{Carbon is oxidized (} -4 \rightarrow +4 \text{) and chlorine is reduced (} 0 \rightarrow -1 \text{). This is not a disproportionation reaction.}$ $3.\ 2\text{F}_2 + 2\text{OH}^{-} \rightarrow 2\text{F}^{-} + \text{OF}_2 + \text{H}_2\text{O}$ $\text{In } \text{F}_2\text{, F is } 0\text{.}$ $\text{In } \text{OH}^{-}\text{, O is } -2\text{, H is } +1\text{.}$ $\text{In } \text{F}^{-}\text{, F is } -1\text{.}$ $\text{In } \text{OF}_2\text{, O is } +2\text{, F is } -1\text{.}$ $\text{In } \text{H}_2\text{O}\text{, H is } +1\text{, O is } -2\text{.}$ $\text{Fluorine is reduced (} 0 \rightarrow -1 \text{ in } \text{F}^{-} \text{) and oxygen is oxidized (} -2 \rightarrow +2 \text{ in } \text{OF}_2 \text{). This is not a disproportionation reaction for a single element.}$ $4.\ 2\text{NO}_2 + 2\text{OH}^{-} \rightarrow \text{NO}_2^{-} + \text{NO}_3^{-} + \text{H}_2\text{O}$ $\text{Let's find the oxidation state of Nitrogen in each compound:}$ $\text{In } \text{NO}_2\text{, let O.S. of N be } x\text{. } x + 2(-2) = 0 \Rightarrow x = +4\text{.}$ $\text{In } \text{NO}_2^{-}\text{, let O.S. of N be } y\text{. } y + 2(-2) = -1 \Rightarrow y = +3\text{.}$ $\text{In } \text{NO}_3^{-}\text{, let O.S. of N be } z\text{. } z + 3(-2) = -1 \Rightarrow z = +5\text{.}$ $\text{In this reaction, nitrogen in } \text{NO}_2 \text{ (oxidation state } +4 \text{) is converted to nitrogen in } \text{NO}_2^{-} \text{ (oxidation state } +3 \text{), which is a reduction.}$ $\text{Nitrogen in } \text{NO}_2 \text{ (oxidation state } +4 \text{) is also converted to nitrogen in } \text{NO}_3^{-} \text{ (oxidation state } +5 \text{), which is an oxidation.}$ $\text{Since the same element, Nitrogen, is both oxidized (from } +4 \text{ to } +5 \text{) and reduced (from } +4 \text{ to } +3 \text{) in this reaction, it is a disproportionation reaction.}$

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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
}