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Current Question (ID: 7974)

Question:

The bond length can be defined as -

Options:

1. The equilibrium distance between the nuclei of two bonded atoms in a molecule. (Correct)
2. The farthest distance between the nuclei of two bonded atoms in a molecule.
3. The shortest distance between the nuclei of two bonded atoms in a molecule.
4. None of the above.

Solution:

HINT: The average distance between nuclei of two bonded atoms in a molecule is bond length. Explanation: Bond length is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule. Bond lengths are expressed in terms of Angstrom ($10^{-10} \text{ m}$) or picometer ($10^{-12} \text{ m}$) and are measured by spectroscopic X-ray diffractions and electron-diffraction techniques. In an ionic compound, the bond length is the sum of the ionic radii of the constituting atoms ($d = r_{+} + r_{-}$.). In a covalent compound, it is the sum of their covalent radii ($d = r_{\text{A}} + r_{\text{B}}$)

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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
}