Import Question JSON

Current Question (ID: 7860)

Question:

$\text{The formula of the stable binary compounds that would be formed by an}$ $\text{element having atomic number 71 and fluorine:}$

Options:

1. $\text{LuF}_6$
2. $\text{LuF}_2$
3. $\text{LuF}_3$ (Correct)
4. $\text{LuF}_4$

Solution:

$\text{HINT: The valency of Lu is 3.}$ $\text{Explanation:}$ $\text{The element with the atomic number 71 is Lutetium (Lu). Electronic}$ $\text{configuration of Lu is [Xe]4f}^{14}\text{5d}^1\text{6s}^2$ $\text{It has valency 3. Hence, the formula of the compound is LuF}_3\text{.}$ $\text{To determine the correct formula:}$ $\text{- Lutetium (Lu) has atomic number 71 and belongs to the lanthanide series}$ $\text{- Its electronic configuration shows it has 3 valence electrons}$ $\text{(one 5d electron and two 6s electrons)}$ $\text{- Fluorine (F) has a valency of -1 (needs one electron to complete its octet)}$ $\text{- To form a stable compound, Lu}^{3+} \text{ combines with three F}^- \text{ ions}$ $\text{- Therefore, the formula is LuF}_3$ $\text{This follows the general principle that for ionic compounds, the total}$ $\text{positive charge must equal the total negative charge to achieve electrical}$ $\text{neutrality.}$

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
}