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Current Question (ID: 7907)

Question:

$\text{The graph between ionization energy and atomic number for the first group elements is shown below:}$\n\n$\text{[Graph shows points: Li (~525 kJ/mol), X (~495 kJ/mol), Y (~420 kJ/mol), Z (~400 kJ/mol), and Cs (~375 kJ/mol) with decreasing ionization energy as atomic number increases]}$\n\n$\text{The element represented by Y in the graph above is -}$

Options:

1. $\text{Cs}$
2. $\text{Rb}$
3. $\text{Ca}$
4. $\text{K}$ (Correct)

Solution:

$\text{HINT: Down the group ionization energy decreases.}$\n\n$\text{Explanation:}$\n\n$\text{In the first group element, down the group ionization energy decreases because of the size of the atom increases.}$\n\n$\text{Thus, due to the increase in the size of the atom the effective nuclear charge decreases, and because of it ionization energy decreases. The correct order of ionization enthalpy is Li < Na < K < Rb < Cs.}$\n\n$\text{From the graph, we can see the decreasing trend of ionization energy with increasing atomic number for Group 1 elements:}$\n\n$\text{Li (highest ionization energy) → X (Na) → Y → Z (Rb) → Cs (lowest ionization energy)}$\n\n$\text{Based on the position of Y in the graph and the decreasing trend, Y corresponds to K (Potassium), which comes after Na and before Rb in Group 1. The ionization energy value of Y (~420 kJ/mol) matches the expected value for K.}$

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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
}