Step 1: Analyze Option (1) - BF₃
BF₃ is a planar molecule with B-F bonds arranged symmetrically. The dipole moments of the B-F bonds cancel each other, resulting in a zero dipole moment for the molecule. Thus, option (1) is false.
Step 2: Analyze Option (2) - Dipole Moment of NF₃ vs NH₃
In NF₃, fluorine is more electronegative than nitrogen, and the dipoles do not cancel, so NF₃ has a non-zero dipole moment.
NH₃ also has a non-zero dipole moment, but NF₃ has a greater dipole moment due to the greater electronegativity of fluorine. Thus, option (2) is true.
Step 3: Analyze Options (3) and (4)
Option (3): The Carbonate Ion (CO₃²⁻)
The carbonate ion (CO₃²⁻) has three resonance structures with equal contribution, so option (3) is true.
Option (4): The Ozone Molecule (O₃)
The ozone molecule (O₃) has two major resonance structures, so option (4) is true.
Step 4: Conclusion
The correct answer is: Option (4): Three canonical forms can be drawn for the CO₃²⁻ ion.
Complete the following reactions by writing the structure of the main products:
A bob of heavy mass \(m\) is suspended by a light string of length \(l\). The bob is given a horizontal velocity \(v_0\) as shown in figure. If the string gets slack at some point P making an angle \( \theta \) from the horizontal, the ratio of the speed \(v\) of the bob at point P to its initial speed \(v_0\) is :