Import Question JSON

Current Question (ID: 10553)

Question:

$\text{NH}_3 \text{ forms a hydrogen bond but PH}_3 \text{ does not because:}$

Options:

1. $\text{Phosphorus is more electronegative as compared to nitrogen.}$
2. $\text{Nitrogen is more electronegative as compared to phosphorus.}$ (Correct)
3. $\text{Nitrogen and phosphorous have equal electronegativity.}$
4. $\text{Nitrogen is more stable as compared to phosphorus.}$

Solution:

$\text{HINT: N is more electronegative than P}$ $\text{Explanation:}$ $\text{STEP 1:}$ $\text{Hydrogen bonding occurs when hydrogen is attached to electronegative elements like F, N and O.}$ $\text{Nitrogen is highly electronegative as compared to phosphorus. The decrease in electronegativity on going down the group is due to increase in size of the atoms and shielding effect of inner electron shells on going down the group.}$ $\text{This causes a greater attraction of electrons towards nitrogen in NH}_3 \text{ than towards phosphorus in PH}_3\text{. Hence, the extent of hydrogen bonding in PH}_3 \text{ is very less as compared to NH}_3\text{.}$ $\text{STEP 2: H-Bonding in NH}_3 \text{ molecules can be represented as:}$ $\text{[The solution includes molecular diagrams showing hydrogen bonding between NH}_3 \text{ molecules with dotted lines representing hydrogen bonds]}$

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
}