Import Question JSON

Current Question (ID: 7908)

Question:

$\text{A configuration with the lowest ionization enthalpy among the following is:}$

Options:

1. $1s^2 2s^2 2p^5$
2. $1s^2 2s^2 2p^3$
3. $1s^2 2s^2 2p^6 3s^1$ (Correct)
4. $1s^2 2s^2 2p^6$

Solution:

$\text{HINT: } 1s^2 2s^2 2p^6 3s^1 \text{ has lowest ionisation enthalpy among given configurations.}$\n\n$\text{Explanation:}$\n\n$\text{Ionisation enthalpy depends on the force of attraction with which the electrons are held by the nucleus. Due to exceptionally high stability of fully and exactly half-filled p orbitals like } p^3 \text{ and } p^6\text{, high IE is required.}$\n\n$2p^6 \text{ is also strongly held since it lies close to the nucleus(n=2) but } 3s^1 \text{ lies farther than 2p orbital and also contains only one electron. Hence it will have the lowest value of IE.}$\n\n$\text{Analysis of each configuration:}$\n\n$\text{1. } 1s^2 2s^2 2p^5 \text{ - Nearly filled p orbital, high stability}$\n\n$\text{2. } 1s^2 2s^2 2p^3 \text{ - Half-filled p orbital, exceptional stability}$\n\n$\text{3. } 1s^2 2s^2 2p^6 3s^1 \text{ - Single electron in 3s orbital, farthest from nucleus}$\n\n$\text{4. } 1s^2 2s^2 2p^6 \text{ - Completely filled p orbital, high stability}$\n\n$\text{The 3s}^1 \text{ electron is easiest to remove due to greater distance from nucleus and lower effective nuclear charge, making option 3 correct.}$

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
}