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Current Question (ID: 8608)

Question:

$\text{The highest oxidation number of nitrogen among the following compounds is:}$

Options:

1. $\text{N}_2\text{H}_4$
2. $\text{NH}_3$
3. $\text{N}_3\text{H}$
4. $\text{NH}_2\text{OH}$

Solution:

$\text{Hint: Use the general rule to calculate the oxidation number.}$ $\text{Let the oxidation state of nitrogen in the given compound be } x.$ $(1)\ \text{N}_2\text{H}_4 \text{ (Hydrazine):}$ $\text{Hydrogen has an oxidation state of } +1\text{.}$ $2(x) + 4(+1) = 0$ $2x + 4 = 0$ $2x = -4$ $x = -2$ $(2)\ \text{NH}_3 \text{ (Ammonia):}$ $\text{Hydrogen has an oxidation state of } +1\text{.}$ $x + 3(+1) = 0$ $x + 3 = 0$ $x = -3$ $(3)\ \text{N}_3\text{H} \text{ (Hydrazoic acid):}$ $\text{Hydrogen has an oxidation state of } +1\text{.}$ $3x + (+1) = 0$ $3x = -1$ $x = -\frac{1}{3}$ $(4)\ \text{NH}_2\text{OH} \text{ (Hydroxylamine):}$ $\text{Hydrogen has an oxidation state of } +1\text{.}$ $\text{Oxygen has an oxidation state of } -2\text{.}$ $x + 2(+1) + (-2) + (+1) = 0$ $x + 2 - 2 + 1 = 0$ $x + 1 = 0$ $x = -1$ $\text{Comparing the oxidation states of nitrogen:}$ $\text{N}_2\text{H}_4: -2$ $\text{NH}_3: -3$ $\text{N}_3\text{H}: -\frac{1}{3} \approx -0.33$ $\text{NH}_2\text{OH}: -1$ $\text{The highest oxidation state among these values is } -\frac{1}{3}.$ $\text{Thus, the oxidation state of nitrogen is highest in } \text{N}_3\text{H}.$

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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
}