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Current Question (ID: 8607)

Question:

$\text{The oxidation number of phosphorous in ATP (adenosine triphosphate) is -}$

Options:

1. $2$
2. $3$
3. $4$
4. $5$

Solution:

$\text{Hint: Oxidation number is the same as oxidation state.}$ $\text{The total number of electrons that an atom either gains or loses in order to form a chemical bond with another atom determines its oxidation state.}$ $\text{ATP (adenosine triphosphate) contains three phosphate groups. The general formula for a phosphate group is } \text{PO}_4^{3-}\text{.}$ $\text{In the phosphate group } (\text{PO}_4^{3-})\text{, phosphorus is bonded to four oxygen atoms.}$ $\text{The oxidation state of oxygen is typically } -2\text{.}$ $\text{Let the oxidation number of phosphorus be } x\text{.}$ $\text{For the phosphate ion } \text{PO}_4^{3-}: \text{}$ $x + 4 \times (-2) = -3$ $x - 8 = -3$ $x = -3 + 8$ $x = +5$ $\text{Thus, the oxidation state of phosphorus in ATP is } +5\text{.}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}