Import Question JSON

Current Question (ID: 21197)

Question:

$\text{Match the redox reactions in List-I with their corresponding type of redox reaction in List-II.}$ $\text{List-I (Redox Reaction)}$ \begin{array}{ll} \text{(A)} & \text{CH}_4 (g) + 2\text{O}_2 (g) \xrightarrow{\Delta} \text{CO}_2 (g) + 2\text{H}_2\text{O} (l) \\ \text{(B)} & 2\text{NaH} \xrightarrow{\Delta} 2\text{Na} (s) + \text{H}_2 (g) \\ \text{(C)} & \text{V}_2\text{O}_5 (s) + 5\text{Ca} \xrightarrow{} 2\text{V} (s) + 5\text{CaO} (s) \\ \text{(D)} & 2\text{H}_2\text{O}_2 (aq) \xrightarrow{} 2\text{H}_2\text{O} (l) + \text{O}_2 (g) \end{array}$ $\text{List-II (Type of Redox Reaction)}$ \begin{array}{ll} \text{(I)} & \text{Disproportionation reaction} \\ \text{(II)} & \text{Combination reaction} \\ \text{(III)} & \text{Decomposition reaction} \\ \text{(IV)} & \text{Displacement reaction} \end{array}$ $\text{Choose the correct answer from the options given below:}$

Options:

1. $\text{A-II, B-III, C-IV, D-I}$
2. $\text{A-II, B-I, C-I, D-IV}$
3. $\text{A-III, B-IV, C-I, D-II}$
4. $\text{A-IV, B-I, C-II, D-III}$

Solution:

$\text{Explanation:}$ $\text{(1) Disproportionation reactions are redox processes where a single species undergoes both oxidation and reduction, forming products with different oxidation states. In the decomposition of hydrogen peroxide (2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2), oxygen's oxidation state changes from -1 (in \text{H}_2\text{O}_2) to -2 (in \text{H}_2\text{O}) and 0 (in \text{O}_2).}$ $\text{(2) The combination reaction of methane (\text{CH}_4) and oxygen (\text{O}_2) is a combustion reaction, which is a type of chemical reaction where a hydrocarbon reacts with oxygen to produce carbon dioxide (\text{CO}_2), water (\text{H}_2\text{O}), and energy in the form of heat.}$ $\text{(3) The decomposition reaction of sodium hydride (\text{NaH}) involves its breakdown into sodium (\text{Na}) and hydrogen gas (\text{H}_2) when heated.}$ $\text{(4) The reaction between vanadium pentoxide (\text{V}_2\text{O}_5) and calcium (\text{Ca}) is a displacement reaction where calcium, being more reactive, displaces vanadium from its oxide.}$ $\text{In summary:}$ $\text{(A) Combustion of hydrocarbon or combination reaction.}$ $\text{(B) Decomposition into gaseous product.}$ $\text{(C) Displacement of 'V' by 'Ca' atom.}$ $\text{(D) Disproportionation of } \text{H}_2\text{O}_2^{-1} \text{ into } \text{O}^{-2} \text{ and } \text{O}^0 \text{ oxidation states.}$ $\text{Hence, option 1 is the correct answer.}$

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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
}