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Current Question (ID: 8004)

Question:
A molecule among the following that has the maximum dipole moment is:
Options:
  • 1. $\text{CO}_2$
  • 2. $\text{CH}_4$
  • 3. $\text{NH}_3$ (Correct)
  • 4. $\text{NF}_3$
Solution:
$\text{HINT: Molecule dipole moment is zero for symmetrical molecule.} \text{Explanation:} \text{1. } \text{CO}_2 \text{ is a symmetrical linear molecule, so the dipole moment is zero.} \text{2. In } \text{CH}_4 \text{, which is a tetrahedral molecule, the individual bond dipoles cancel each other out due to its symmetrical structure, resulting in a net dipole moment of zero.} \text{3. For } \text{NH}_3 \text{ (Ammonia), it has a trigonal pyramidal shape with a lone pair on the nitrogen atom. The bond dipoles of the N-H bonds and the lone pair moment add up in the same direction, resulting in a significant net dipole moment (approximately 1.46 D).} \text{4. For } \text{NF}_3 \text{ (Nitrogen trifluoride), it also has a trigonal pyramidal shape and a lone pair on nitrogen. However, the direction of the N-F bond dipoles is opposite to that of the lone pair moment on the nitrogen atom. This leads to a partial cancellation of dipoles, resulting in a much smaller net dipole moment (approximately 0.24 D) compared to } \text{NH}_3\text{.} \text{Comparing the dipole moments:} \text{CO}_2 \text{ (0 D)} \text{CH}_4 \text{ (0 D)} \text{NF}_3 \text{ (}\approx\text{0.24 D)} \text{NH}_3 \text{ (}\approx\text{1.46 D)} \text{Therefore, } \text{NH}_3 \text{ has the maximum dipole moment.}$

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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
}