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Current Question (ID: 21536)

Question:

$\text{The pair that has similar atomic radii is:}$ $\text{1. Mo, and W}$ $\text{2. Mn, and Re}$ $\text{3. Ti, and Hf}$ $\text{4. Sc, and Ni}$

Options:

1. $\text{Mo, and W}$
2. $\text{Mn, and Re}$
3. $\text{Ti, and Hf}$
4. $\text{Sc, and Ni}$

Solution:

$\text{Hint: Lanthanoid contraction}$ $\text{In transition element, 4th period to 5th-period atomic size increases}$ $\text{considerably but hardly any increase between the 5th and 6th-period}$ $\text{elements.}$ $\text{The atomic radii of given elements are as follows:}$ $\text{1. Mo = 209 pm; W = 210 pm}$ $\text{2. Mn = 197 pm; Re = 217 pm}$ $\text{3. Ti = 187 pm; Hf = 212 pm}$ $\text{4. Sc = 211 pm; Ni = 163 pm}$ $\text{Mo belongs to the 5th period and belongs to the 6th period. It is due}$ $\text{to lanthanoid contraction.}$ $\text{Hence, option 1 is the correct answer}$

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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
}