Import Question JSON

Current Question (ID: 7918)

Question:

$\text{In the following graph of variation of ionization energy with atomic number, X, Y, and Z represent elements.}$\n\n$\text{[Graph shows ionization energy vs atomic number with peaks labeled X (~2300 kJ/mol), Y (~1300 kJ/mol), and Z (~1000 kJ/mol) representing the highest points in their respective periods]}$\n\n$\text{The group number of X, Y, and Z elements is:}$

Options:

1. $1^{\text{st}} \text{ group}$
2. $15^{\text{th}} \text{ group}$
3. $17^{\text{th}} \text{ group}$
4. $18^{\text{th}} \text{ group}$ (Correct)

Solution:

$\text{HINT: Along the period noble gas has the highest ionization energy.}$\n\n$\text{Explanation:}$\n\n$\text{The graph shows the relation between ionization enthalpy and atomic number.}$\n\n$\text{It represents the ionization enthalpy of 60 elements. The ionization energy of noble gas is highest because of completely filled valence shells in the periodic table.}$\n\n$\text{So, more energy will be required to remove even a single electron from the noble gas elements.}$\n\n$\text{Hence, X, Y, and Z belongs to } 18^{\text{th}} \text{ group.}$\n\n$\text{Analysis of the graph:}$\n\n$\text{1. The graph shows periodic peaks corresponding to noble gases}$\n\n$\text{2. X, Y, and Z represent the highest ionization energy values in their respective periods}$\n\n$\text{3. These peaks correspond to elements with completely filled valence shells}$\n\n$\text{4. Group 18 elements (noble gases) have the highest ionization energies in each period due to their stable electronic configurations}$\n\n$\text{The pattern clearly shows that X, Y, and Z are noble gas elements belonging to Group 18.}$

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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
}