For \(\text{OF}_2\) molecule consider the following:
[(A)] Number of lone pairs on oxygen is 2.
[(B)] F--O--F angle is less than \(104.5^\circ\).
[(C)] Oxidation state of O is \(-2\).
[(D)] Molecule is bent ‘V’-shaped.
[(E)] Molecular geometry is linear.
Correct options are:
The shape and bond angle of a molecule can be determined using VSEPR theory. Lone pair-bond pair repulsion decreases the bond angle from its ideal value.
A, B, D only
A, C, D only
C, D, E only
B, E, A only
The Lewis structure of \(\text{OF}_2\) shows that the central oxygen atom is bonded to two fluorine atoms and has two lone pairs of electrons. This gives the molecule a bent or ‘V’ shape due to lone pair-bond pair repulsion.
Step 2: Analysis of the Statements
Statement (A): The oxygen atom in \(\text{OF}_2\) has 6 valence electrons. After forming two single bonds with fluorine atoms, it retains two lone pairs. Hence, this statement is true
Statement (B): The F--O--F bond angle is slightly less than \(104.5^\circ\) (around \(102^\circ\)) due to increased lone pair-bond pair repulsion. Hence, this statement is true
Statement (C): The oxidation state of oxygen in \(\text{OF}_2\) is \(+2\), as fluorine is more electronegative. Hence, this statement is false
Statement (D): Due to the presence of two lone pairs, the molecular geometry is bent or ‘V’ shaped. Hence, this statement is true
Statement (E) The molecular geometry is not linear; it is bent. Hence, this statement is false}
Conclusion: The correct statements are A, B, D. Therefore, the correct answer is (4) A, B, D only
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:
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The equilibrium constant for decomposition of $ H_2O $ (g) $ H_2O(g) \rightleftharpoons H_2(g) + \frac{1}{2} O_2(g) \quad (\Delta G^\circ = 92.34 \, \text{kJ mol}^{-1}) $ is $ 8.0 \times 10^{-3} $ at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation ($ \alpha $) of water is _____ $\times 10^{-2}$ (nearest integer value). [Assume $ \alpha $ is negligible with respect to 1]