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Current Question (ID: 10932)

Question:

$\text{The decreasing order of the stability of the given ions is-}$

Options:

1. $\text{I}>\text{II}>\text{III}$
2. $\text{II}>\text{III}>\text{I}$
3. $\text{III}>\text{I}>\text{II}$
4. $\text{II}>\text{I}>\text{III}$ (Correct)

Solution:

$\text{Hint: -I effect decreases the stability of carbocation}$ $\text{The second carbocation is the most stable structure because oxygen atom stabilize the positive charge on carbon atom. The structures are as follows:}$ $\text{H}_3\text{C}-\overset{\delta+}{\text{C}}-\text{O}-\text{CH}_3 \leftrightarrow \text{H}_3\text{C}-\overset{+}{\text{C}}=\overset{-}{\text{O}}-\text{CH}_3$ $\text{The third carbocation is the least stable carbocation because carbonyl group destabilize the positive charge via -I effect.}$ $\text{Hence, the order of the stability of the given ions is as follows:}$ $\text{II}>\text{I}>\text{III}$

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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
}