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
A body of mass 1000 kg is moving horizontally with a velocity of 6 m/s. If 200 kg extra mass is added, the final velocity (in m/s) is: