Import Question JSON

Current Question (ID: 7655)

Question:

$\text{Consider the following curve showing atomic radius vs atomic number. X represents group with highest atomic radius. The element that belongs to Y group can be:}$

Options:

1. $\text{Na}$
2. $\text{Mg}$
3. $\text{Br}$ (Correct)
4. $\text{Si}$

Solution:

$\text{HINT: Left to right atomic radius decreases.}$\n\n$\text{Explanation:}$\n\n$\text{Atomic size gradually decreases from left to right across a period of elements. This is because, within a period or family of elements, all electrons are added to the same shell. The valence electrons are held closer towards the nucleus of the atom. As a result, the atomic radius decreases.}$\n\n$\text{The 1}^{\text{st}} \text{ group represents s block elements and 17}^{\text{th}} \text{ group represents halogen. Left to right atomic radius decreases.}$\n\n$\text{So, the 1}^{\text{st}} \text{ group element presents at the extreme left, and 17}^{\text{th}} \text{ group elements present at the extreme right.}$\n\n$\text{From the graph, X represents the group with highest atomic radius (extreme left - Group 1), and Y represents the group with lower atomic radius (extreme right - Group 17). Among the given options, Br (Bromine) belongs to Group 17 (halogens), which corresponds to curve Y.}$

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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
}