Xenon (VI) fluoride on complete hydrolysis gives an oxide of xenon 'O'. The total number of \( \sigma \) and \( \pi \) bonds in 'O' is:
Step 1: Understanding the Hydrolysis of Xenon (VI) Fluoride
- \( XeF_6 \) undergoes complete hydrolysis to form Xenon trioxide (\( XeO_3 \)): \[ XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF \] - The molecular structure of \( XeO_3 \) is trigonal pyramidal.
Step 2: Counting \( \sigma \) and \( \pi \) Bonds
- Each Xe–O bond contains one \( \sigma \)-bond.
- Each O has one \( \pi \)-bond with Xe.
- Total bonds: \( 3\sigma + 3\pi = 6 \) bonds.
Identify the correct orders against the property mentioned:
A. H$_2$O $>$ NH$_3$ $>$ CHCl$_3$ - dipole moment
B. XeF$_4$ $>$ XeO$_3$ $>$ XeF$_2$ - number of lone pairs on central atom
C. O–H $>$ C–H $>$ N–O - bond length
D. N$_2$>O$_2$>H$_2$ - bond enthalpy
Choose the correct answer from the options given below:
Observe the following data given in the table. (\(K_H\) = Henry's law constant)
Gas | CO₂ | Ar | HCHO | CH₄ |
---|---|---|---|---|
\(K_H\) (k bar at 298 K) | 1.67 | 40.3 | \(1.83 \times 10^{-5}\) | 0.413 |
The correct order of their solubility in water is
For a first order decomposition of a certain reaction, rate constant is given by the equation
\(\log k(s⁻¹) = 7.14 - \frac{1 \times 10^4 K}{T}\). The activation energy of the reaction (in kJ mol⁻¹) is (\(R = 8.3 J K⁻¹ mol⁻¹\))
Note: The provided value for R is 8.3. We will use the more precise value R=8.314 J K⁻¹ mol⁻¹ for accuracy, as is standard.