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Current Question (ID: 7999)

Question:

$H_3PO_3$ can be represented by structures 1 and 2 shown below. These two structures cannot be taken as the canonical forms of the resonance hybrid, because :

Options:

1. The positions of the atoms have changed. (Correct)
2. The positions of the atoms are constant.
3. $H_3PO_3$ does not show resonance.
4. Two hydrogen atoms are missing.

Solution:

HINT: Resonance hybrid takes it's character from the average of all the individual resonance contributors. Explanation: The important point regarding resonance structure is as follows: i. All resonance contributors must have the same molecular formula, the same number of electrons, and the same net charge. ii. Resonance contributors only differ by the positions of pi bond and lone pair electrons. Sigma bonds are never broken or made, because of this atoms must maintain their same position. The given structures cannot be taken as the canonical forms of the resonance hybrid of $H_3PO_3$ because the positions of the atoms have changed. H is directly attached to P in 1, wherein 2nd it is attached with O attached with P directly.

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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
}