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Current Question (ID: 7635)

Question:
$\text{The most common oxidation state of cerium (Ce) is:}$
Options:
  • 1. $+5, +3$
  • 2. $+5, +4$
  • 3. $+3, +4$ (Correct)
  • 4. $+3, +5$
Solution:
$\text{HINT: }\text{Ce}^{4+} \text{ is stable due to } 4f^0 \text{ configuration.}$ $\text{Explanation:}$ $\text{The electronic configuration of Ce is [Xe]}4f^1 5d^1 6s^2.$ $\text{The common stable oxidation state of all the lanthanides is +3. The oxidation states of +2 and +4 are also exhibited and these oxidation states are only stable in those cases where stable } 4f^0, 4f^7 \text{ or } 4f^{14} \text{ configurations are achieved. } \text{Ce}^{4+} \text{ is stable due to } 4f^0 \text{ configuration.}$

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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
}