Import Question JSON

Current Question (ID: 8239)

Question:

$\text{Dipole-dipole forces act between the molecules possessing permanent dipole. Ends of dipoles possess 'partial charges'. The partial charge will be-}$

Options:

1. $\text{More than unit electronic charge.}$
2. $\text{Equal to unit electronic charge.}$
3. $\text{Less than unit electronic charge.}$
4. $\text{Double the unit electronic charge.}$

Solution:

$\text{Hint: Permanent dipole moment should be there.}$ $\text{Explanation:}$ $\text{Dipole-dipole forces act between the molecules possessing permanent dipole and ends of dipoles possess 'partial charges'. Partial charges present on the ends of a dipole are always less than that unit electronic charge.}$ $\text{For example, in water (H}_2\text{O), the oxygen atom bears a partial negative charge (}\delta-\text{) while the hydrogen atoms bear partial positive charges (}\delta+\text{). The dipole moment of water is }\mu = 1.84\text{ D (Debye units).}$ $\text{These partial charges arise from the unequal sharing of electrons in covalent bonds due to differences in electronegativity between atoms. The magnitude of these partial charges is always less than a full unit charge (less than the charge of an electron or proton), which distinguishes them from ionic charges.}$ $\text{Therefore, the partial charge at the ends of a dipole will be less than unit electronic charge.}$

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
}