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Current Question (ID: 8612)

Question:

$\text{The compound having oxygen in } -1 \text{ oxidation state is:-}$

Options:

1. $\text{H}_2\text{O}$
2. $\text{O}_2\text{F}_2$
3. $\text{Na}_2\text{O}$
4. $\text{BaO}_2$

Solution:

$\text{Let's determine the oxidation state (O.S.) of oxygen in each compound:}$ $1.\ \text{H}_2\text{O}:$ $\text{Hydrogen has an oxidation state of } +1 \text{.}$ $2 \times (+1) + \text{O.S. of O} = 0$ $2 + \text{O.S. of O} = 0$ $\text{O.S. of O} = -2$ $2.\ \text{O}_2\text{F}_2:$ $\text{Fluorine is the most electronegative element, so its oxidation state is always } -1 \text{.}$ $2 \times (\text{O.S. of O}) + 2 \times (-1) = 0$ $2 \times (\text{O.S. of O}) - 2 = 0$ $2 \times (\text{O.S. of O}) = 2$ $\text{O.S. of O} = +1$ $3.\ \text{Na}_2\text{O}:$ $\text{Sodium (an alkali metal) has an oxidation state of } +1 \text{.}$ $2 \times (+1) + \text{O.S. of O} = 0$ $2 + \text{O.S. of O} = 0$ $\text{O.S. of O} = -2$ $4.\ \text{BaO}_2:$ $\text{Barium (an alkaline earth metal) has an oxidation state of } +2 \text{.}$ $+2 + 2 \times (\text{O.S. of O}) = 0$ $2 \times (\text{O.S. of O}) = -2$ $\text{O.S. of O} = -1$ $\text{Therefore, } \text{BaO}_2 \text{ is the compound having oxygen in a } -1 \text{ oxidation state, as it is a peroxide.}$

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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
}