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Current Question (ID: 8118)

Question:

The interaction energy of London force is inversely proportional to sixth power of the distance between two interacting particles. Their magnitude depends upon-

Options:

1. Charge of interacting particles.
2. Mass of interacting particles.
3. Polarizability of interacting particles.
4. Strength of permanent dipoles in the particles.

Solution:

Hint: $\text{London dispersion forces operate only over very short distances.}$ Explanation: $\text{The energy of interaction varies as } \frac{1}{(\text{distance})^6} \text{ between two interacting particles.}$ $\text{Large or more complex are the molecules, greater is the magnitude of London forces. This is obviously due to the fact that the large electron clouds are easily distorted or polarised.}$ $\text{Hence, the greater the polarizability of the interacting particles, the greater is the magnitude of the interaction energy.}$

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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
}