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Current Question (ID: 10676)

Question:

$\text{The electron gain enthalpy values for O} \rightarrow \text{O}^- \text{ and O} \rightarrow \text{O}^{2-} \text{ are -141 and 702 kJ mol}^{-1}\text{, respectively. The correct statement about the formation of oxides is-}$

Options:

1. $\text{Higher the electron gain enthalpy lower is the lattice energy.}$
2. $\text{Higher the electron gain enthalpy higher is the lattice energy.}$
3. $\text{Higher the electron gain enthalpy higher is the stability of oxide.}$ (Correct)
4. $\text{O}^- \text{ is more stable than O}_2^-$

Solution:

$\text{Hint: Consider the lattice energy factor in the formation of compounds.}$ $\text{Explanation:}$ $\text{Step 1:}$ $\text{The stability of an ionic compound depends on its lattice energy.}$ $\text{The more the lattice energy of a compound, the more stable it will be. Lattice energy is directly proportional to the charge carried by an ion.}$ $\text{When a metal combines with oxygen, the lattice energy of the oxide involving O}_2^- \text{ ion is much more than the oxide involving O}^- \text{ion.}$ $\text{Step 2:}$ $\text{Hence, the oxide having O}_2^- \text{ ions are more stable than oxides having O}^-.$ $\text{Hence, we can say that the formation of O}_2^- \text{ is energetically more favorable than the formation of O}^-.$ $\text{If EA of element is high than the difference of EA will be lesser and compound will have lesser covalent character,so stability will be lesser. since oxygen is having higher EA, so another element with higher EA will produce less stable compound.}$

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{
  "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
}