Import Question JSON

Current Question (ID: 7611)

Question:

$\text{The group number that has non-metal, liquid as well as gas at room temperature is:}$

Options:

1. $15$
2. $14$
3. $18$
4. $17$ (Correct)

Solution:

$\text{HINT: Group 17 has metal, non-metal, liquid as well as gas at room}$ $\text{temperature and group 17 elements are known as halogens.}$ $\text{Explanation:}$ $\text{Group 17 elements (halogens) exhibit various physical states and properties:}$ $\text{Non-metal - Fluorine is the most electronegative non-metal.}$ $\text{Metalloid - Tennessine of atomic no. 117 belong to halogen.}$ $\text{Liquid - Bromine is liquid at room temperature.}$ $\text{Chlorine - Behaves as gas also belongs to group 17.}$ $\text{The halogens (Group 17) are unique in that they contain elements in all three}$ $\text{physical states at room temperature:}$ $\text{- Fluorine (F}_2\text{) and Chlorine (Cl}_2\text{) are gases}$ $\text{- Bromine (Br}_2\text{) is a liquid}$ $\text{- Iodine (I}_2\text{) is a solid}$ $\text{- Astatine (At) is a radioactive solid}$ $\text{- Tennessine (Ts) is a synthetic superheavy element}$ $\text{Additionally, all halogens are non-metals (except for the heaviest ones which may}$ $\text{show some metallic character), making Group 17 the only group that satisfies}$ $\text{all the given criteria.}$

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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
}