Question:
The correct order of dipole moment of the molecules \(NH_3\) (I), \(BF_3\) (II), \(H_2O\) (III), \(NF_3\) (IV) is:
The correct order of dipole moment of the molecules \(NH_3\) (I), \(BF_3\) (II), \(H_2O\) (III), \(NF_3\) (IV) is:
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Symmetrical molecules like \(BF_3\) have zero dipole moment because bond dipoles cancel. In \(NF_3\), bond dipoles oppose the lone pair dipole, so its dipole moment is smaller than \(NH_3\).
Updated On: Jun 26, 2026
- III \(\gt \) I \(\gt \) IV \(\gt \) II
- IV \(\gt \) I \(\gt \) III \(\gt \) II
- I \(\gt \) IV \(\gt \) II \(\gt \) III
- III \(\gt \) II \(\gt \) I \(\gt \) IV
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The Correct Option is A
Solution and Explanation
Step 1: Understand dipole moment.
Dipole moment depends on bond polarity and molecular geometry.
If bond dipoles cancel each other due to symmetry, the net dipole moment becomes zero.
If bond dipoles do not cancel, the molecule has a non-zero dipole moment.
Step 2: Analyze \(BF_3\).
\(BF_3\) has trigonal planar geometry.
The three \(B-F\) bond dipoles are equal and symmetrically arranged at \(120^\circ\).
Therefore, they cancel each other.
Hence, \[ \mu(BF_3)=0 \] So, \(BF_3\) has the least dipole moment.
Step 3: Analyze \(H_2O\).
\(H_2O\) has bent shape due to two lone pairs on oxygen.
The \(O-H\) bond dipoles do not cancel each other.
Therefore, water has a high dipole moment: \[ \mu(H_2O)\approx1.85\ \text{D} \]
Step 4: Analyze \(NH_3\).
\(NH_3\) has trigonal pyramidal geometry due to one lone pair on nitrogen.
The \(N-H\) bond dipoles and lone pair contribution act in the same general direction.
Thus, \[ \mu(NH_3)\approx1.47\ \text{D} \]
Step 5: Analyze \(NF_3\).
\(NF_3\) also has trigonal pyramidal geometry.
However, fluorine is more electronegative than nitrogen, so the \(N-F\) bond dipoles are directed opposite to the lone pair dipole.
This causes partial cancellation.
Hence, \[ \mu(NF_3) \] is smaller than \[ \mu(NH_3) \]
Step 6: Write the correct order.
Thus, the correct order of dipole moment is \[ H_2O\gt NH_3\gt NF_3\gt BF_3 \] Using the given labels: \[ \text{III}\gt \text{I}\gt \text{IV}\gt \text{II} \]
Step 7: Final conclusion.
Therefore, the correct order is \[ \boxed{\text{III}\gt \text{I}\gt \text{IV}\gt \text{II}} \] Hence, the correct option is \[ \boxed{(1)} \]
Dipole moment depends on bond polarity and molecular geometry.
If bond dipoles cancel each other due to symmetry, the net dipole moment becomes zero.
If bond dipoles do not cancel, the molecule has a non-zero dipole moment.
Step 2: Analyze \(BF_3\).
\(BF_3\) has trigonal planar geometry.
The three \(B-F\) bond dipoles are equal and symmetrically arranged at \(120^\circ\).
Therefore, they cancel each other.
Hence, \[ \mu(BF_3)=0 \] So, \(BF_3\) has the least dipole moment.
Step 3: Analyze \(H_2O\).
\(H_2O\) has bent shape due to two lone pairs on oxygen.
The \(O-H\) bond dipoles do not cancel each other.
Therefore, water has a high dipole moment: \[ \mu(H_2O)\approx1.85\ \text{D} \]
Step 4: Analyze \(NH_3\).
\(NH_3\) has trigonal pyramidal geometry due to one lone pair on nitrogen.
The \(N-H\) bond dipoles and lone pair contribution act in the same general direction.
Thus, \[ \mu(NH_3)\approx1.47\ \text{D} \]
Step 5: Analyze \(NF_3\).
\(NF_3\) also has trigonal pyramidal geometry.
However, fluorine is more electronegative than nitrogen, so the \(N-F\) bond dipoles are directed opposite to the lone pair dipole.
This causes partial cancellation.
Hence, \[ \mu(NF_3) \] is smaller than \[ \mu(NH_3) \]
Step 6: Write the correct order.
Thus, the correct order of dipole moment is \[ H_2O\gt NH_3\gt NF_3\gt BF_3 \] Using the given labels: \[ \text{III}\gt \text{I}\gt \text{IV}\gt \text{II} \]
Step 7: Final conclusion.
Therefore, the correct order is \[ \boxed{\text{III}\gt \text{I}\gt \text{IV}\gt \text{II}} \] Hence, the correct option is \[ \boxed{(1)} \]
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