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Current Question (ID: 8598)

Question:
$\text{The incorrect statement regarding the rule to find the oxidation number among the following is:}$
Options:
  • 1. $\text{The oxidation number of hydrogen is always } +1\text{.}$
  • 2. $\text{The algebraic sum of all the oxidation numbers carried by elements in a compound is zero.}$
  • 3. $\text{An element in its free or uncombined state has an oxidation number of zero.}$
  • 4. $\text{Generally, in all its compounds, the oxidation number of fluorine is } -1\text{.}$
Solution:
$\text{Let's analyze each statement:}$ $1.\ \text{The oxidation number of hydrogen is always } +1\text{.}$ $\text{This statement is incorrect.}$ $\text{While hydrogen's oxidation number is } +1 \text{ in most compounds (e.g., hydrogen halides, acids, and compounds with non-metals), it can also have an oxidation number of } -1 \text{ in metal hydrides (e.g., } \text{NaH}, \text{CaH}_2\text{).}$ $\text{In its elemental form } (\text{H}_2)\text{, the oxidation number of hydrogen is } 0\text{.}$ $2.\ \text{The algebraic sum of all the oxidation numbers carried by elements in a compound is zero.}$ $\text{This statement is correct.}$ $\text{For a neutral compound, the sum of the oxidation numbers of all atoms is zero.}$ $3.\ \text{An element in its free or uncombined state has an oxidation number of zero.}$ $\text{This statement is correct.}$ $\text{For example, the oxidation number of } \text{O} \text{ in } \text{O}_2 \text{, } \text{H} \text{ in } \text{H}_2 \text{, } \text{Cl} \text{ in } \text{Cl}_2 \text{, and } \text{Na} \text{ in solid sodium metal is } 0\text{.}$ $4.\ \text{Generally, in all its compounds, the oxidation number of fluorine is } -1\text{.}$ $\text{This statement is correct.}$ $\text{Fluorine is the most electronegative element in the periodic table. Due to its exceptionally high electronegativity, it always exhibits an oxidation state of } -1 \text{ in all its compounds.}$ $\text{Therefore, the incorrect statement is number 1.}$

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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
}