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Current Question (ID: 8668)

Question:

$\text{The hydrides of carbon } (\text{C}_n\text{H}_{2n+2}) \text{ can act as -}$

Options:

1. $\text{Lewis acid.}$
2. $\text{Lewis base.}$
3. $\text{Both 1 and 2}$
4. $\text{None of the above}$

Solution:

$\text{Hint: Alkanes are neither Lewis acid nor Lewis base.}$ $\text{For carbon hydrides of type } \text{C}_n\text{H}_{2n+2}\text{, the following hydrides are possible}$ $\text{The molecular formulas of carbon hydride of } \text{C}_n\text{H}_{2n+2} \text{ type are as follows:}$ $n = 1 \Rightarrow \text{CH}_4$ $n = 2 \Rightarrow \text{C}_2\text{H}_6$ $n = 3 \Rightarrow \text{C}_3\text{H}_8$ $\text{For a hydride to act as a Lewis acid i.e., electron-accepting, it should be electron-deficient. Also, for it to act as a Lewis base i.e., electron-donating, it should be electron-rich. Taking } \text{C}_2\text{H}_6 \text{ as an example, the total number of electrons is 14 and the total covalent bonds are seven. Hence, the bonds are regular } 2e^- \text{-2 centered bonds.}$ $\text{Hence, hydride } \text{C}_2\text{H}_6 \text{ has sufficient electrons to be represented by a conventional Lewis structure. Therefore, it is an electron-precise hydride, having all atoms with complete octets.}$ $\text{Thus, it can neither donate nor accept electrons to act as a Lewis acid or Lewis base.}$

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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
}