Import Question JSON

Current Question (ID: 10707)

Question:

$\text{Which of the following is incorrect regarding noble gases?}$

Options:

1. $\text{The atomic radii corresponds to van der Waal's radii.}$
2. $\text{They have the largest size in a period.}$
3. $\text{They are also called inert gases.}$
4. $\text{They combine to form molecules.}$ (Correct)

Solution:

$\text{HINT: Noble gases do not form molecules.}$ $\text{Explanation:}$ $\text{STEP 1: The general electronic configuration of noble gas elements is } \text{ns}^2\text{np}^6$ $\text{In the case of noble gases, the atomic radii correspond to van der Waal's radii. On the other hand, the atomic radii of other elements correspond to their covalent radii.}$ $\text{By definition, van der Waal's radii are larger than covalent radii. It is for this reason that noble gases are very large in size as compared to other atoms belonging to the same period.}$ $\text{STEP 2: The values of radius for second period element is as follows:}$ $\text{For Boron atomic radius is 85 pm, carbon radius is 77, nitrogen atomic radius is 70 pm, oxygen radius is 66 pm, fluorine radius is 64 pm and for He radius is 120 pm.}$ $\text{STEP 3: Due to high ionization enthalpy and stable electronic configuration noble gases are generally non reactive, that's why they are called inert gases and did not combine to form molecules.}$

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
}