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Question:
$\text{Arrange the species } \text{Li} , \text{Li}_2 , \text{ and } \text{Li}_2^{+} \text{ in order of increasing stability:}$ $1. \text{Li} < \text{Li}_2^{-} < \text{Li}_2^{+}$ $2. \text{Li}_2 = \text{Li}_2^{+} = \text{Li}_2^{-}$ $3. \text{Li}_2 < \text{Li}_2^{+} < \text{Li}_2^{-}$ $4. \text{Li}_2^{-} < \text{Li}_2^{+} < \text{Li}_2$
Options:
  • 1. $\text{Li} < \text{Li}_2^{-} < \text{Li}_2^{+}$
  • 2. $\text{Li}_2 = \text{Li}_2^{+} = \text{Li}_2^{-}$
  • 3. $\text{Li}_2 < \text{Li}_2^{+} < \text{Li}_2^{-}$
  • 4. $\text{Li}_2^{-} < \text{Li}_2^{+} < \text{Li}_2$
Solution:
$\text{Hint: Stability depends on the number of antibonding electrons}$ $\text{Explanation:}$ $(i) \text{The stability of the molecule depends on the bond order of the species. Stability is directly proportional to the bond order.}$ $\text{The configuration of the given species according to MOT is as follows:}$ $\text{Li}_2 = \sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \text{; B.O.} = 1$ $\text{Li}_2^{+} = \sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^1 \text{; B.O.} = 0.5$ $\text{Li}_2^{-} = \sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*1} \text{; B.O.} = 0.5$ $\text{The } \text{Li}_2 \text{ is the most stable because it has the highest bond order.}$ $(ii) \text{Stability also depends on the number of antibonding electrons, as the number of antibonding electrons increases, the stability decreases.}$ $\text{Li}_2^{+} \text{ has 2 antibonding electrons but } \text{Li}_2^{-} \text{ has three electrons in the antibonding molecular orbital. Thus, } \text{Li}_2^{+} \text{ is most stable than } \text{Li}_2^{-}$ $\text{The order is as follows:}$ $\text{Li}_2^{-} < \text{Li}_2^{+} < \text{Li}_2$ $\text{So, option 4 is the correct answer.}$

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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
}