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Current Question (ID: 7985)

Question:

Which of the following atomic orbitals would have the greatest overlap to form the strongest covalent bond?

Options:

1. $2s-2s (\sigma)$
2. $2s-2p (\sigma)$
3. $2p-2p (\pi)$
4. $2p-2p (\sigma)$ (Correct)

Solution:

$\text{HINT: For the same n value, the strength of overlapping is p-p > s-p > s-s}$ $\text{Explanation:}$ $\text{Bond strength depends upon two factors:}$ $\text{(i) Size of orbitals (ii) Extent of overlapping}$ $\text{Thereby, } \sigma \text{ bond is always stronger than } \pi \text{ bonds. The smaller the size of orbitals, stronger is the bond.}$ $\text{When the n value is the same then p-p overlapping is stronger than s-p and s-s.}$ $\text{The order of bond strength can be explained on the basis of the fact that the more the area of overlap, the higher will be the strength. Since the p orbital has directional property, p} - \text{ p overlap provides more area for overlap for the same internuclear distance.}$ $\text{Thus, option 4th is the correct answer.}$

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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
}