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Current Question (ID: 21535)

Question:

$\text{The lanthanide ion among the following that would show colour is:}$ $1.\ \text{La}^{3+}$ $2.\ \text{Gd}^{3+}$ $3.\ \text{Sm}^{3+}$ $4.\ \text{Lu}^{3+}$

Options:

1. $\text{La}^{3+}$
2. $\text{Gd}^{3+}$
3. $\text{Sm}^{3+}$
4. $\text{Lu}^{3+}$

Solution:

$\text{Hint: Generally, Ion contains unpaired electron will show colour}$ $\text{La} = [\text{Xe}]\ 5d^1 6s^2$ $\text{La}^{3+} = [\text{Xe}]$ $\text{Zero unpaired electron}$ $\text{Gd} = [\text{Xe}]\ 4f^7 5d^1 6s^2$ $\text{Gd}^{3+} = [\text{Xe}]\ 4f^7$ $\text{Gd}^{3+} \text{ is colorless even if it contains 7 unpaired electrons in 4f orbital. This is because, here 4f orbitals are half-filled, hence stable. As such f-f transitions are not possible because they will destroy the symmetry of 4f and make them unstable. Due to this, Gd}^{3+} \text{ ion does not absorb any wavelength from white light and hence it is colourless.}$ $\text{Sm} = [\text{Xe}]\ 4f^6 6s^2$ $\text{Sm}^{3+} = [\text{Xe}]\ 4f^5$ $\text{Sm}^{3+} \text{ contains 5 unpaired electrons. Hence, it will show colour.}$ $\text{Lu} = [\text{Xe}]\ 4f^{14} 5d^1 6s^2$ $\text{Lu}^{3+} = [\text{Xe}]\ 4f^{14}$ $\text{Lu}^{3+} \text{ does not contain any unpaired electrons hence it will not show any colour}$

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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
}