Import Question JSON

Current Question (ID: 7675)

Question:
$\text{The element having very high electron affinity but zero ionisation enthalpy is:}$
Options:
  • 1. $\text{He (due to inert gas configuration)}$
  • 2. $\text{Be (due to fully filled subshell)}$
  • 3. $\text{H (due to presence of allotropes)}$
  • 4. $\text{None of the above}$ (Correct)
Solution:
$\text{HINT: Ionisation energy cannot be zero.}$ $\text{Explanation:}$ $\text{Ionisation energy is the minimum amount of energy required to remove the most loosely bound electron of an isolated neutral gaseous atom or molecule.}$ $\text{In the periodic table, no element has zero ionization energy. So, here option fourth is the correct answer.}$ $\text{Additional explanation:}$ $\text{The question asks for an element with very high electron affinity but zero ionisation enthalpy. However, this is physically impossible because:}$ $\text{1. Ionisation energy is always positive as energy is always required to remove an electron from an atom.}$ $\text{2. Even the most electropositive elements like cesium and francium have positive ionisation energies.}$ $\text{3. Zero ionisation energy would mean that electrons are not bound to the nucleus at all, which contradicts the basic atomic structure.}$ $\text{Therefore, no such element exists, making option 4 "None of the above" the correct answer.}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}