Import Question JSON

Current Question (ID: 21198)

Question:
$\text{Among the following statements, the correct set of statements are:}$ $\text{a. } \text{B}_2\text{H}_6 \text{ is Lewis acid.}$ $\text{b. } \text{B}_2\text{H}_6 \text{ has a planar structure.}$ $\text{c. All B-H Bond lengths are equal in } \text{B}_2\text{H}_6.$ $\text{d. In } \text{B}_2\text{H}_6, \text{ four } 3\text{C} - 2\text{e}^- \text{ bonds are present.}$ $\text{e. } \text{B}_2\text{H}_6 \text{ can be prepared by the reaction of } \text{BF}_3 \text{ and } \text{NaBH}_4.$
Options:
  • 1. $\text{a, b, c, d, e}$
  • 2. $\text{a, e}$
  • 3. $\text{a, b}$
  • 4. $\text{d, e}$
Solution:
$\text{Hint: Only banana bonds are } 3\text{C} - 2\text{e}^- \text{ bonds}$ $\text{B}_2\text{H}_6 \text{ have 4 2c-2e bonds and 2 3c-2e bonds. Bridging bonds have larger bond length than terminal bonds.}$ $\text{Angle between terminal bonds is more than angle between bridging bonds if all 4 terminal bonds are in one plane then bridging bonds are in perpendicular plane}$ $3\text{NaBH}_4 + 4\text{BF}_3 \xrightarrow{\text{ether} \ 450\text{K}} 3\text{NaBF}_4 + 2\text{B}_2\text{H}_6$

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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
}