Import Question JSON

Current Question (ID: 10686)

Question:

$\text{What order does the stability of interhalogen compounds follow?}$

Options:

1. $\text{IF}_3 > \text{BrF}_3 > \text{ClF}_3$ (Correct)
2. $\text{BrF}_3 > \text{IF}_3 > \text{ClF}_3$
3. $\text{ClF}_3 > \text{BrF}_3 > \text{IF}_3$
4. $\text{ClF}_3 > \text{IF}_3 > \text{BrF}_3$

Solution:

$\text{Hint: Stability of interhalogen compound is directly proportional to the difference of electronegativity and inversely proportional to steric hindrance.}$ $\text{The interhalogen compounds of type AX and AX}_3 \text{ are formed between the halogen having very low electronegative differences}$ $\text{(e.g., ClF, ClF}_3\text{). The interhalogen compounds of type AX}_5 \text{ and AX}_7 \text{ are formed by larger atoms having low electronegativity}$ $\text{with the smaller atoms having high electronegativity. This is because it is possible to fit the greater number of smaller atoms}$ $\text{around a larger one (e.g. BrF}_5\text{, IF}_7\text{).}$ $\text{All Interhalogens are volatile at room temperature. All are polar due to differences in their electronegativity. These are usually}$ $\text{covalent liquids or gases due to the small electronegativity difference among them.}$ $\text{Hence, option one is the correct answer.}$

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
}