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Current Question (ID: 8565)

Question:

\text{Fluorine reacts with ice as per the following reaction:} \text{H}_2\text{O (s) + F}_2\text{ (g) → HF (g) + HOF (g)} \text{This reaction is a redox reaction because-}

Options:

1. $\text{F}_2 \text{ is getting oxidized.}$
2. $\text{F}_2 \text{ is getting reduced.}$
3. $\text{Both (1) and (2)}$
4. $\text{None of the above.}$

Solution:

$\text{Hint: F is most electronegative element.}$ $\text{Explanation:}$ $\text{Let us write the oxidation number of each atom involved in the given reaction above its symbol as:}$ $\overset{+1 \, -2}{\text{H}_2\text{O}} + \overset{0}{\text{F}_2} \rightarrow \overset{+1 \, -1}{\text{HF}} + \overset{+1 \, -2 \, +1}{\text{HOF}}$ $\text{Here, we have observed that the oxidation number of F increases from 0 in F}_2 \text{ to +1 in HOF.}$ $\text{Also, the oxidation number decreases from 0 in F}_2 \text{ to -1 in HF.}$ $\text{Thus, in the above reaction, F is reduced and O gets oxidised from -2 to 0.}$ $\text{Hence, the given reaction is a redox reaction.}$ $\text{Analysis:}$ $\text{F}_2 \text{ undergoes both oxidation (0 to +1 in HOF) and reduction (0 to -1 in HF).}$ $\text{This is a disproportionation reaction where the same element (F) simultaneously gets oxidized and reduced.}$ $\text{Generally F shows -1 oxidation state, but in HOF fluorine shows +1 oxidation state.}$ $\text{The electron charge density in fluorine is less as compared to oxygen hence, it is an exception case.}$

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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
}