Import Question JSON

Current Question (ID: 10916)

Question:

$\text{Consider the following resonating structures of HCOOH}$ $\text{I. H-C-O-H (with double bond between C and first O, lone pairs shown)}$ $\text{II. H-C=O-H (with positive charge on C, negative charge on first O)}$ $\text{III. H-C-O-H (with negative charge on first O, positive charge on C)}$ $\text{IV. H-C-O-H (with positive charge on first O, negative charge on second O)}$ $\text{The order of stability is-}$

Options:

1. $\text{I>II>III>IV}$ (Correct)
2. $\text{IV>I>II>III}$
3. $\text{I>III>II>IV}$
4. $\text{II>I>III>IV}$

Solution:

$\text{Hint: Neutral species is more stable than charged species.}$ $\text{The features that decreases the predicted stability of a contributing resonance structure as follows:}$ $\text{1. An atom with an incomplete octet}$ $\text{2. A negative charge that is not on the most electronegative atom or a positive charge that is not on the least electronegative atom}$ $\text{3. Charge separation}$ $\text{Structure one is the most stable because it is neutral. After that structure II is the most stable because all atoms has complete octet.}$ $\text{The structure III is more stable than structure IV because in third structure negative charge is present at the more electronegative element. Hence, the order is as follows:}$ $\text{I>II>III>IV}$

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
}