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Current Question (ID: 18882)

Question:

$\text{The unpaired electron in Eu}^{3+} \text{ will bear similarities to that found in:}$

Options:

1. $\text{Yb}^{2+}$
2. $\text{Ho}^{3+}$
3. $\text{Sm}^{2+}$
4. $\text{Gd}^{3+}$

Solution:

$\text{Eu}^{3+} \text{ has 6 unpaired electrons.}$ $\text{1. The atomic number of Eu is 63. The electronic configuration is as follows:}$ $[\text{Xe}]4f^76s^2$ $\text{The electronic configuration of Eu}^{3+} \text{ is as follows: } [\text{Xe}] 4f^6$ $\text{It has 6 unpaired electrons.}$ $\text{2. The atomic number of Sm is 62. The electronic configuration of Sm is as follows:}$ $[\text{Xe}] 4f^66s^2$ $\text{The electronic configuration of Sm}^{2+} \text{ is as follows: } [\text{Xe}] 4f^6$ $\text{So, Eu}^{3+} \text{ and Sm}^{2+} \text{ will have 6 unpaired electrons.}$ $\text{The atomic number of Ho is 67. The electronic configuration is as follows:}$ $[\text{Xe}]4f^{11}6s^2$ $\text{The electronic configuration of Ho}^{3+} \text{ is as follows: } [\text{Xe}] 4f^{10}$ $\text{It has 4 unpaired electrons.}$ $\text{The atomic number of Gd is 64. The electronic configuration is as follows:}$ $[\text{Xe}]4f^75d^16s^2$ $\text{The electronic configuration of Gd}^{3+} \text{ is as follows: } [\text{Xe}] 4f^7$ $\text{It has 7 unpaired electrons.}$ $\text{Hence, the correct answer is option third.}$

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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
}