Question

The dipole moments of three $\text{AB}_3$-type molecules I, II, and III are measured to be 0.0 D, 0.2 D, and 1.5 D, respectively. Which one of the following options is correct regarding the identity of I, II, and III?

• A. I: $\text{BF}_3$, II: $\text{NF}_3$, III: $\text{NH}_3$
• B. I: $\text{BF}_3$, II: $\text{NH}_3$, III: $\text{NF}_3$
• C. I: $\text{ClF}_3$, II: $\text{NF}_3$, III: $\text{NH}_3$
• D. I: $\text{BCl}_3$, II: $\text{NH}_3$, III: $\text{NF}_3$

Hint:

The comparison between $\text{NH}_3$ and $\text{NF}_3$ is a classic concept.
In $\text{NH}_3$, bond dipoles and lone-pair dipoles assist each other, whereas in $\text{NF}_3$, they oppose each other, which reduces its dipole moment significantly down to $0.2\text{ D}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The net dipole moment (\( \mu \)) of a multi-atomic molecule depends on both individual bond polarities and the overall geometric arrangement of the bonds (molecular symmetry).
Step 2: Detailed Explanation:
Let us evaluate each compound to match their molecular structures with the given dipole moments:
Molecule I (\( \mu = 0.0 \text{ D} \)):
- \( \text{BF}_3 \) has a central Boron atom with three valence electrons forming three \( \sigma \) bonds with Fluorine. Its hybridization is \( \text{sp}^2 \), giving it a perfectly symmetric trigonal planar geometry.
- Because of this symmetric shape, the individual \( \text{B}-\text{F} \) bond dipoles cancel each other out completely, resulting in a net dipole moment of exactly \( 0.0 \text{ D} \). This matches Molecule I.
Molecules II (\( \mu = 0.2 \text{ D} \)) and III (\( \mu = 1.5 \text{ D} \)):
- Both \( \text{NF}_3 \) and \( \text{NH}_3 \) have \( \text{sp}^3 \) hybridization with three bond pairs and one lone pair, yielding a trigonal pyramidal geometry. The presence of the lone pair ensures neither molecule has a zero dipole moment.
- In \( \text{NH}_3 \), Nitrogen is more electronegative than Hydrogen. The three individual \( \text{N}-\text{H} \) bond dipoles point toward Nitrogen and act in the same direction as the lone pair dipole. This reinforcement creates a large net dipole moment (\( 1.47 \text{ D} \approx 1.5 \text{ D} \)). Thus, III is \( \text{NH}_3 \).
- In \( \text{NF}_3 \), Fluorine is far more electronegative than Nitrogen. The individual \( \text{N}-\text{F} \) bond dipoles point away from Nitrogen, directly opposing the upward vector of the lone pair dipole. This structural opposition largely cancels the net dipole moment, leaving a very small residual value (\( 0.23 \text{ D} \approx 0.2 \text{ D} \)). Thus, II is \( \text{NF}_3 \).
Step 3: Final Answer:
The correct order matching the given values is I: \( \text{BF}_3 \), II: \( \text{NF}_3 \), III: \( \text{NH}_3 \).