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Current Question (ID: 18785)

Question:

$\text{Mn}^{2+} \text{ compounds are more stable than } \text{Fe}^{2+} \text{ compounds towards oxidation in their } +3 \text{ state because:}$

Options:

1. $3d^5 \text{ configuration is more stable than } 3d^6 \text{ configuration.}$
2. $3d^6 \text{ configuration is more stable than } 3d^5 \text{ configuration.}$
3. $3p^5 \text{ configuration is less stable than } 3p^6 \text{ configuration.}$
4. $3p^6 \text{ configuration is less stable than } 3p^5 \text{ configuration.}$

Solution:

$\text{HINT: Half-filled and fully-filled orbitals have more stability.}$ $\text{Explanation:}$ $\text{Electronic configuration of } \text{Mn}^{2+} \text{ is } [\text{Ar}]^{18} 3d^5.$ $\text{Electronic configuration of } \text{Fe}^{2+} \text{ is } [\text{Ar}]^{18} 3d^6.$ $\text{It is known that half-filled and fully-filled orbitals are more stable.}$ $\text{Therefore, Mn in (+2) state has a stable } d^5 \text{ configuration. This is the reason } \text{Mn}^{2+} \text{ shows resistance to oxidation to } \text{Mn}^{3+}.$ $\text{Also, } \text{Fe}^{2+} \text{ has } 3d^6 \text{ configuration and by losing one electron, its configuration changes to a more stable } 3d^5 \text{ configuration. Therefore, } \text{Fe}^{2+} \text{ easily gets oxidized to } \text{Fe}^{+3} \text{ oxidation state.}$

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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
}