Import Question JSON

Current Question (ID: 8644)

Question:

$\text{Para hydrogen is-}$

Options:

1. $\text{Less stable than ortho hydrogen.}$
2. $\text{More stable than ortho hydrogen.}$
3. $\text{As stable as ortho hydrogen.}$
4. $\text{None of the above.}$

Solution:

$\text{Hint: Ordinary hydrogen consists of 75\% ortho and 25\% para}$ $\text{Explanation:}$ $\text{The hydrogen molecule is diatomic, i.e., it contains two hydrogen atoms. Each atom has one proton in the nucleus with an electron moving around it.}$ $\text{Like an electron, the proton is also spinning about an axis. Two protons in the hydrogen molecule may have either their spins in the same direction or in opposite directions giving rise to two forms ortho and para.}$ $\text{When the proton spins are in the same direction, the form is termed ortho hydrogen and when the proton spins are in the opposite directions, the form is known as para hydrogen. Ortho form is more stable than para form.}$ $\text{Hence, the answer is option 1.}$

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
}