Import Question JSON

Current Question (ID: 20954)

Question:

$\text{Match List-I (Molecules) with List-II (Bond order), and mark the appropriate choice}$ $\begin{array}{|c|c|} \hline \text{List-I (Molecule)} & \text{List-II (Bond order)} \\ \hline (a) \ \text{Ne}_2 & (i) \ 1 \\ (b) \ \text{N}_2 & (ii) \ 2 \\ (c) \ \text{F}_2 & (iii) \ 0 \\ (d) \ \text{O}_2 & (iv) \ 3 \\ \hline \end{array}$ $\text{Choose the correct match from the options given below:}$

Options:

1. $(a) \rightarrow (iii), (b) \rightarrow (iv), (c) \rightarrow (i), (d) \rightarrow (ii)$
2. $(a) \rightarrow (ii), (b) \rightarrow (iv), (c) \rightarrow (iii), (d) \rightarrow (i)$
3. $(a) \rightarrow (ii), (b) \rightarrow (i), (c) \rightarrow (iv), (d) \rightarrow (iii)$
4. $(a) \rightarrow (iv), (b) \rightarrow (iii), (c) \rightarrow (ii), (d) \rightarrow (i)$

Solution:

$\text{Hint:}$ $\text{The electronic configuration of Ne}_2 \text{ is as follows:}$ $\sigma 1s^2 \ \sigma^* 1s^2, \ \sigma 2s^2 \ \sigma^* 2s^2, \ \sigma px^2 \ \pi 2py^2 \ \pi^* 2py^2 \pi^* 2pz^2 \ \sigma^* 2px^2$ $(a) \ \text{Ne}_2 = \text{Total} e^- = 20$ $\text{B.O.} = \frac{10-10}{2} = 0$ $(b) \ \text{N}_2 = \text{Total} e^- = 14$ $\text{B.O.} = \frac{10-4}{2} = 3$ $(c) \ \text{F}_2 = \text{Total} e^- = 18$ $\text{B.O.} = \frac{10-8}{2} = 1$ $(d) \ \text{O}_2 = \text{Total} e^- = 16$ $\text{B.O.} = \frac{10-6}{2} = 2$

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
}