Import Question JSON

To determine the correct order of increasing dipole moment, we need to analyze the molecular geometry and bond polarities of each molecule. 1. **$\text{CF}_4$ (Carbon Tetrafluoride):** * Structure: Tetrahedral. * Bond polarity: C-F bonds are polar due to the high electronegativity of fluorine. * Overall polarity: Despite polar bonds, the symmetrical tetrahedral geometry causes all bond dipoles to cancel each other out, resulting in a zero dipole moment ($\mu = 0$). 2. **$\text{NF}_3$ (Nitrogen Trifluoride):** * Structure: Trigonal pyramidal (due to one lone pair on nitrogen). * Bond polarity: N-F bonds are polar (F is more electronegative than N). The bond dipoles point towards the fluorine atoms. * Overall polarity: The lone pair on nitrogen contributes a dipole moment that opposes the resultant vector of the N-F bond dipoles. This leads to a relatively small net dipole moment ($\mu \approx 0.24 \text{ D}$). The bond moments are directed away from the lone pair, leading to partial cancellation. 3. **$\text{NH}_3$ (Ammonia):** * Structure: Trigonal pyramidal (due to one lone pair on nitrogen). * Bond polarity: N-H bonds are polar (N is more electronegative than H). The bond dipoles point towards the nitrogen atom. * Overall polarity: Both the lone pair dipole and the resultant vector of the N-H bond dipoles point in the same general direction (towards the nitrogen atom). This leads to a significant net dipole moment ($\mu \approx 1.46 \text{ D}$). The bond moments add up to the lone pair moment. 4. **$\text{H}_2\text{O}$ (Water):** * Structure: Bent (V-shaped) due to two lone pairs on oxygen. * Bond polarity: O-H bonds are highly polar (O is much more electronegative than H). The bond dipoles point towards the oxygen atom. * Overall polarity: The bent shape and the significant electronegativity difference, combined with the contribution from two lone pairs on oxygen, result in a large net dipole moment ($\mu \approx 1.85 \text{ D}$). The bond moments and lone pair moments reinforce each other. Comparing the dipole moments: * $\text{CF}_4 = 0 \text{ D}$ * $\text{NF}_3 \approx 0.24 \text{ D}$ * $\text{NH}_3 \approx 1.46 \text{ D}$ * $\text{H}_2\text{O} \approx 1.85 \text{ D}$ Therefore, the increasing order of dipole moment is: $\text{CF}_4 < \text{NF}_3 < \text{NH}_3 < \text{H}_2\text{O}$. The correct answer is Option 4.