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Current Question (ID: 21557)

Question:

$\text{The most common oxidation state of Lanthanoid elements is } +3. \text{ Which of the following is likely to deviate easily from } +3 \text{ oxidation state?}$ $1. \text{Ce(At. No. 58)}$ $2. \text{La(At. No. 57)}$ $3. \text{Lu(At. No. 71)}$ $4. \text{Gd(At. No. 64)}$

Options:

1. $\text{Ce(At. No. 58)}$
2. $\text{La(At. No. 57)}$
3. $\text{Lu(At. No. 71)}$
4. $\text{Gd(At. No. 64)}$

Solution:

$\text{Hint: } \text{Ce}^{3+} = [\text{Xe}]4f^15 d^0$ $\text{Ce} \rightarrow [\text{Xe}]4f^15 d^16 s^2$ $\text{Ce}^{+4} \rightarrow [\text{Xe}]4f^05 d^06 s^0$ $\text{Cerium in } +4 \text{ oxidation state acquires inert gas configuration.}$

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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
}