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Current Question (ID: 8585)

Question:

$\text{The oxidation number of carbon in } \text{C}_3\text{O}_2 \text{ and } \text{Mg}_2\text{C}_3 \text{ are respectively:}$

Options:

1. $-\frac{4}{3}, +\frac{4}{3}$
2. $+\frac{4}{3}, -\frac{4}{3}$
3. $-\frac{2}{3}, +\frac{2}{3}$
4. $-\frac{2}{3}, +\frac{4}{3}$

Solution:

$\text{Carbon has negative oxidation no. in } \text{Mg}_2\text{C}_3 \text{ and positive oxidation number in } \text{C}_3\text{O}_2. \text{ O is more electronegative than C.}$ $\text{Mg is more electropositive than C.}$ $\text{Let the oxidation state of C in } \text{C}_3\text{O}_2 \text{ be x.}$ $\text{Since, the overall charge on the complex is 0, the sum of oxidation states of all elements in it should be equal to 0.}$ $\text{Therefore, } 3x + 2(-2) = 0$ $\text{or, } x = \frac{4}{3}$ $\text{Let the oxidation state of C in } \text{Mg}_2\text{C}_3 \text{ be y.}$ $\text{Therefore, } 2 \times 2 + 3xy = 0$ $\text{or, } y = -\frac{4}{3}$

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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
}