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Current Question (ID: 10709)

Question:

$\text{Which of the following statements is correct?}$

Options:

1. $\text{Only type of interactions between particles of noble gases are due to weak dispersion forces.}$ (Correct)
2. $\text{Ionisation enthalpy of molecular oxygen is very high as compared to xenon.}$
3. $\text{Hydrolysis of } \text{XeF}_6 \text{ is a redox reaction.}$
4. $\text{Xenon fluorides are not reactive.}$

Solution:

$\text{HINT- Xenon fluorides are highly reactive in nature.}$ $\text{Explanation:}$ $\text{(1) Only one type of interactions between particles of noble gases are weak dispersion forces.}$ $\text{Noble gases have very little intermolecular forces acting between them since they are monoatomic and unpolarised. Thus, only London dispersion forces act and these directly depend on the number of electrons in a compound.}$ $\text{(2) Ionisation enthalpy of molecular oxygen is very close to that of xenon.}$ $\text{Enthalpy of molecular oxygen (1175 kJ mol}^{-1}\text{) is almost identical with that of xenon (1170 kJ mol}^{-1}\text{). This is the reason for the formation of xenon oxides.}$ $\text{(3) Hydrolysis of } \text{XeF}_6\text{:}$ $\text{Xe}^{-1} + 6\text{F}_6^{+1} + 3\text{H}_2\text{O}^{-2} \rightarrow \text{Xe}^{+6}\text{O}_3^{-2} + 3\text{H}^{+1}\text{F}^{-1}$ $\text{is not a redox reaction as the oxidation state of Xe remains same.}$ $\text{(4) Xenon fluorides are highly reactive hydrolysis readily even by traces of water.}$

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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
}