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Current Question (ID: 18869)

Question:

$\text{The most common oxidation state for lanthanoids and actinoids is:}$

Options:

1. $+5$
2. $-3$
3. $+1$
4. $+3$

Solution:

$\text{HINT: Lanthanoids and actinoids shows various oxidation states.}$ $\text{Explanation:}$ $\text{STEP 1: Lanthanoids primarily show three oxidation states (+2, +3, +4).}$ $\text{Among these oxidation states, +3 state is the most common.}$ $\text{Lanthanoids display a limited number of oxidation states because the}$ $\text{energy difference between 4f, 5d, and 6s orbitals is quite large.}$ $\text{STEP 2: On the other hand, the energy difference between 5f, 6d, and 7s}$ $\text{orbitals is very less. Hence, actinoids display a large number of oxidation}$ $\text{states.}$ $\text{For example, uranium and plutonium display +3, +4, +5, and +6 oxidation}$ $\text{states while neptunium displays +3, +4, +5, and +7. The most common}$ $\text{oxidation state in case of actinoids is also +3.}$

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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
}