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Current Question (ID: 7969)

Question:

Compound that is isoelectronic with an adduct of "Diborane and Ammonia" is-

Options:

1. Hydrogen peroxide. (Correct)
2. Diborane (at absolute zero temp.).
3. First member of alkyne series.
4. Diborane (at $10 \, ^\circ \text{C}$ temp).

Solution:

HINT: Isoelectronic species contain the same number of electrons. Explanation: STEP 1: Diborane is highly reactive when exposed to air and catches fire spontaneously. When diborane reacts with ammonia, an addition product, that is, an adduct is formed which further decomposes on heating at $473\,\text{K}$ to form a volatile compound called borazine or borazole. STEP 2: $\text{H}_3\text{N} \to \text{BH}_3$ is the adduct formed on the reaction of ammonia and $\text{BH}_3$. $\text{H}_3\text{N} \to \text{BH}_3$ contains $18$ electrons ($8$ electrons of $\text{BH}_3$ and $10$ electrons in $\text{NH}_3$). $\text{H}_2\text{O}_2$ also has $18$ electrons. Thus, both are isoelectronic.

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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
}