Import Question JSON

Current Question (ID: 18786)

Question:

$\text{Transition metals show paramagnetic behaviour due to:}$ $1. \text{Paired electrons in the (n-1) d-orbitals}$ $2. \text{Unpaired electrons in the (n-1) d-orbitals}$ $3. \text{Due to } d^{10}, \text{ and } d^0 \text{ configuration}$ $4. \text{None of the above}$

Options:

1. $\text{Paired electrons in the (n-1) d-orbitals}$
2. $\text{Unpaired electrons in the (n-1) d-orbitals}$
3. $\text{Due to } d^{10}, \text{ and } d^0 \text{ configuration}$
4. $\text{None of the above}$

Solution:

$\text{HINT: If unpaired electrons are present in the compound then it shows paramagnetic behaviour.}$ $\text{Step 1:}$ $\text{Transition metals show paramagnetic behaviour.}$ $\text{Paramagnetism arises due to the presence of unpaired electrons with each electron having a magnetic moment associated with its spin angular momentum and orbital angular momentum.}$ $\text{However, in the first transition series, the orbital angular momentum is quenched. Therefore, the resulting paramagnetism is only because of the unpaired electron.}$ $\text{Step 2:}$ $\text{Transition metals and many of their compounds show paramagnetic behaviour. Paramagnetism is a property that is shown by the elements having unpaired electrons.}$ $\text{In the case of transition metals, as they contain unpaired electrons in the (n-1) d-orbitals, most of the transition metal ions and their compounds are paramagnetic.}$ $\text{As the number of unpaired electrons increases from one to five, the paramagnetic character increases.}$

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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
}