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Current Question (ID: 8097)

Question:

$\text{Assuming that Hund's rule is violated, the bond order and magnetic nature of the diatomic molecule } \text{B}_2 \text{ is:}$

Options:

1. $1 \text{ and diamagnetic}$
2. $0 \text{ and diamagnetic}$
3. $1 \text{ and paramagnetic}$
4. $0 \text{ and paramagnetic}$

Solution:

$\text{HINT: } \text{B}_2 \text{ becomes diamagnetic on violation of Hund's Rule.}$ $\text{Explanation:}$ $\text{STEP 1:}$ $\text{Hund's rule: Every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied, and all electrons in singly occupied orbitals have the same spin.}$ $\text{STEP 2:}$ $\text{M.O. configuration of } \text{B}_2 \text{ in presence of Hund's rule } := \sigma_{1s^2}, \sigma_{1s^2}^*, \sigma_{2s^2}, \sigma_{2s^2}^*, \pi_{2p_x^1} \approx \pi_{2p_y^1}$ $\text{M.O. configuration of } \text{B}_2 \text{ in absence of Hund's rule } := \sigma_{1s^2}, \sigma_{1s^2}^*, \sigma_{2s^2}, \sigma_{2s^2}^*, \pi_{2p_x^2} \approx \pi_{2p_y}$ $\text{Thus, on violation of Hund's rule, Bond order of } \text{B}_2 = \frac{1}{2}[6 - 4] = 1 \text{ and it is diamagnetic.}$

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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
}