Import Question JSON

Current Question (ID: 8015)

Question:

The least polar bond is present in which of the following?

Options:

1. $ \text{O}-\text{F} $ (Correct)
2. $ \text{P}-\text{F} $
3. $ \text{Si}-\text{N} $
4. $ \text{B}-\text{F} $

Solution:

HINT: Bond polarity $ \propto $ Electronegativity difference Explanation: Electronegativity order is $ \text{F}>\text{O}>\text{N}>\text{P}>\text{Si}>\text{B} $. Bond polarity is directly related to the difference in electronegativity of bonded atoms. Let's consider the electronegativity differences for each bond: 1. $ \text{O}-\text{F} $: Electronegativity of $ \text{F} \approx 3.98 $ Electronegativity of $ \text{O} \approx 3.44 $ Difference = $ 3.98 - 3.44 = 0.54 $ 2. $ \text{P}-\text{F} $: Electronegativity of $ \text{F} \approx 3.98 $ Electronegativity of $ \text{P} \approx 2.19 $ Difference = $ 3.98 - 2.19 = 1.79 $ 3. $ \text{Si}-\text{N} $: Electronegativity of $ \text{N} \approx 3.04 $ Electronegativity of $ \text{Si} \approx 1.90 $ Difference = $ 3.04 - 1.90 = 1.14 $ 4. $ \text{B}-\text{F} $: Electronegativity of $ \text{F} \approx 3.98 $ Electronegativity of $ \text{B} \approx 2.04 $ Difference = $ 3.98 - 2.04 = 1.94 $ Comparing the differences, Oxygen and Fluorine ($ \text{O}-\text{F} $) has the least electronegativity difference ($ 0.54 $). Thus, it is the least polar bond. The correct answer is option 1.

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
}