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Current Question (ID: 18887)

Question:
$\text{A pair among the following that have the same size is:}$
Options:
  • 1. $\text{Fe}^{2+}, \text{Ni}^{2+}$
  • 2. $\text{Zr}^{4+}, \text{Ti}^{4+}$
  • 3. $\text{Zr}^{4+}, \text{Hf}^{4+}$
  • 4. $\text{Zn}^{2+}, \text{Hf}^{4+}$
Solution:
$\text{Zr}^{4+}, \text{Hf}^{4+} \text{ have almost same size.}$ $\text{In general, the atomic and ionic radii increases on moving down a group.}$ $\text{But the elements of second transition series (eg, Zr, Nb, Mo etc) have the almost same radii as the elements of third transition series (eg, Hf, Ta, W etc).}$ $\text{This is because of lanthanide contraction i.e., imperfect shielding of one 4f-electron by another.}$ $\text{Size of : Zr}^{4+} = 0.74 \text{ Å and Hf}^{4+} = 0.75 \text{ Å.}$ $\text{Hence, option third is the correct answer.}$

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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
}