Import Question JSON

Current Question (ID: 21556)

Question:

$\text{Match the following ions from List-I with their corresponding electron}$ $\text{configuration from List-II and select the correct option.}$ $\begin{array}{|c|c|} \hline \text{List-I (Ion)} & \text{List-II (electron configuration)} \\ \hline \text{a. } \text{Mn}^{2+} & \text{(i) } 3d^3 4s^1 \\ \text{b. } \text{V}^{+} & \text{(ii) } 3d^5 4s^0 \\ \text{c. } \text{Cr}^{+} & \text{(iii) } 3d^6 4s^0 \\ \text{d. } \text{Fe}^{2+} & \\ \hline \end{array}$

Options:

1. $\text{a} \rightarrow \text{(i)}, \text{b} \rightarrow \text{(ii)}, \text{c} \rightarrow \text{(iii)}, \text{d} \rightarrow \text{(iii)}$
2. $\text{a} \rightarrow \text{(iv)}, \text{b} \rightarrow \text{(ii)}, \text{c} \rightarrow \text{(i)}, \text{d} \rightarrow \text{(i)}$
3. $\text{a} \rightarrow \text{(ii)}, \text{b} \rightarrow \text{(i)}, \text{c} \rightarrow \text{(iii)}, \text{d} \rightarrow \text{(iii)}$
4. $\text{a} \rightarrow \text{(ii)}, \text{b} \rightarrow \text{(i)}, \text{c} \rightarrow \text{(iii)}, \text{d} \rightarrow \text{(iii)}$

Solution:

$\text{Hint: Determine the electron configurations of the ions like}$ $\text{Mn}^{2+} : 3d^5 4s^0$ $\text{Mn}^{2+} : 3d^5 4s^0$ $\text{V}^{+} : 3d^3 4s^1$ $\text{Cr}^{+} : 3d^5 4s^0$ $\text{Fe}^{2+} : 3d^6 4s^0$

Import JSON File

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
}