Import Question JSON

Current Question (ID: 7671)

Question:

$\text{The correct graph depicting ionization energies for } 2^{\text{nd}} \text{ period elements is:}$

Options:

1. $\text{Graph 1: Shows smooth increasing trend with slight dip at B, then continues increasing to Ne}$
2. $\text{Graph 2: Shows increasing trend with dip at B and another dip at O, then increases to Ne}$
3. $\text{Graph 3: Shows increasing trend with dip at B and peak at N higher than O, then increases to Ne}$ (Correct)
4. $\text{None of the above}$

Solution:

$\text{Generally left to right across the period ionization enthalpy increases. But not in a regular fashion. The ionization energy of beryllium is higher than boron because it has completely filled 1s and 2s orbitals.}$\n\n$\text{The ionization energy of oxygen is less than nitrogen because nitrogen has a half-filled 2p orbital, and it is more stable than a partially filled 2p orbital of oxygen. Hence, option third is the correct answer.}$\n\n$\text{Key anomalies in the 2nd period ionization trend:}$\n\n$\text{1. Be > B: Beryllium (} 1s^2 2s^2\text{) has a stable filled 2s subshell, while boron (} 1s^2 2s^2 2p^1\text{) starts filling the 2p orbital}$\n\n$\text{2. N > O: Nitrogen (} 1s^2 2s^2 2p^3\text{) has a stable half-filled 2p}^3 \text{ configuration, while oxygen (} 1s^2 2s^2 2p^4\text{) has electron pairing which reduces stability}$\n\n$\text{The correct graph (option 3) shows:}$\n\n$\text{- Overall increasing trend from Li to Ne}$\n\n$\text{- Dip at B (lower than Be)}$\n\n$\text{- Peak at N (higher than O)}$\n\n$\text{- Maximum at Ne (noble gas)}$

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
}