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?
Show 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 $0.2\text{ D}$.
Step 1 : Understanding the Question:
The question asks us to identify three $\text{AB}_3$-type molecules based on their measured dipole moments: $0.0\text{ D}$ (I), $0.2\text{ D}$ (II), and $1.5\text{ D}$ (III). Step 2 : Key Formulas and Approach:
The dipole moment ($\mu$) depends on molecular geometry (VSEPR theory) and the electronegativity differences between the constituent atoms.
Symmetric molecules with no lone pairs have a net dipole moment of zero.
For pyramidal molecules, we compare the direction of the bond dipoles with the lone-pair orbital dipole. Step 3 : Detailed Explanation:
• Molecule I ($\mu = 0.0\text{ D}$):
$\text{BF}_3$ has a symmetric, trigonal planar geometry ($sp^2$ hybridized boron with no lone pairs).
The three polar $\text{B-F}$ bond dipoles point toward the corners of an equilateral triangle and perfectly cancel each other out, resulting in a net dipole moment of zero.
Thus, Molecule I is $\text{BF}_3$.
• Comparing $\text{NH_3$ and $\text{NF}_3$ (Molecules II and III):}
Both molecules have a trigonal pyramidal geometry ($sp^3$ hybridized central nitrogen with one lone pair).
• In $\text{NH}_3$, nitrogen is more electronegative than hydrogen.
The three $\text{N-H}$ bond dipoles point toward the nitrogen atom (upward), acting in the same direction as the lone-pair orbital dipole. They reinforce each other, resulting in a high net dipole moment ($\mu \approx 1.47\text{ D} \approx 1.5\text{ D}$).
Thus, Molecule III is $\text{NH}_3$.
• In $\text{NF}_3$, fluorine is more electronegative than nitrogen.
The three $\text{N-F}$ bond dipoles point away from the nitrogen atom (downward), opposing the lone-pair orbital dipole. They partially cancel each other out, resulting in a very low net dipole moment ($\mu \approx 0.23\text{ D} \approx 0.2\text{ D}$).
Thus, Molecule II is $\text{NF}_3$. Step 4 : Final Answer:
The correct identities are: I: $\text{BF}_3$, II: $\text{NF}_3$, III: $\text{NH}_3$.
This matches Option (A).
