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Current Question (ID: 18805)

Question:
$\text{The number of unpaired electrons in the gaseous ions } \text{Mn}^{3+}, \text{Cr}^{3+}, \text{V}^{3+}, \text{ and } \text{Ti}^{3+} \text{ are, respectively:}$
Options:
  • 1. $3, 2, 3, \text{ and } 1$
  • 2. $2, 2, 3, \text{ and } 1$
  • 3. $4, 3, 2, \text{ and } 1$
  • 4. $1, 2, 3, \text{ and } 4$
Solution:
$\text{Hint: Follow the electronic configuration of given ions.}$ $\text{Explanation:}$ $\begin{array}{ll} \text{Gaseous ions} & \text{Number of unpaired electrons} \\ \text{(i) } \text{Mn}^{2+}, [\text{Ar}]3d^4 & 4 \\ \text{(ii) } \text{Cr}^{2+}, [\text{Ar}]3d^3 & 3 \\ \text{(iii) } \text{V}^{3+}, [\text{Ar}]3d^2 & 2 \\ \text{(iv) } \text{Ti}^{3+}, [\text{Ar}]3d^1 & 1 \end{array}$

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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
}