Question:
In which of the following, the molecules are arranged in correct order of their dipole moments?
In which of the following, the molecules are arranged in correct order of their dipole moments?
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Symmetry plays a key role: highly symmetrical molecules like BF\(_3\) always have zero dipole moment.
Updated On: Jul 18, 2026
- BF\(_3\) < NH\(_3\) < NF\(_3\)
- NF\(_3\) < NH\(_3\) < BF\(_3\)
- BF\(_3\) < NF\(_3\) < NH\(_3\)
- NF\(_3\) < BF\(_3\) < NH\(_3\)
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The Correct Option is C
Solution and Explanation
Step 1: Understand dipole moment concept.
Dipole moment depends on molecular geometry and polarity of bonds. It is a vector quantity, so resultant dipole depends on both bond polarity and symmetry of the molecule. Symmetrical molecules have zero dipole moment, while asymmetrical molecules have non-zero values.
Step 2: Analyze BF\(_3\).
BF\(_3\) has trigonal planar geometry with three identical B–F bonds arranged symmetrically at 120°. The individual bond dipoles cancel each other completely due to symmetry, giving: \[ \mu = 0 \] So BF\(_3\) has the smallest dipole moment.
Step 3: Analyze NF\(_3\).
NF\(_3\) has trigonal pyramidal geometry due to lone pair on nitrogen. Although N–F bonds are polar, fluorine is highly electronegative, and bond dipoles partially oppose the lone pair effect, resulting in a small net dipole moment. Hence NF\(_3\) has a low but non-zero dipole moment.
Step 4: Analyze NH\(_3\).
NH\(_3\) also has trigonal pyramidal geometry with a lone pair on nitrogen. N–H bonds are polar and their dipoles reinforce the lone pair effect rather than cancel it. Therefore, NH\(_3\) has a comparatively larger dipole moment than NF\(_3\).
Step 5: Compare all molecules.
From analysis: \[ BF_3 = 0 \lt NF_3 \lt NH_3 \] Thus, dipole moment increases in the order: \[ BF_3 \lt NF_3 \lt NH_3 \]
Step 6: Final conclusion.
Therefore, the correct order of dipole moments is: \[ \boxed{BF_3 \lt NF_3 \lt NH_3} \]
Dipole moment depends on molecular geometry and polarity of bonds. It is a vector quantity, so resultant dipole depends on both bond polarity and symmetry of the molecule. Symmetrical molecules have zero dipole moment, while asymmetrical molecules have non-zero values.
Step 2: Analyze BF\(_3\).
BF\(_3\) has trigonal planar geometry with three identical B–F bonds arranged symmetrically at 120°. The individual bond dipoles cancel each other completely due to symmetry, giving: \[ \mu = 0 \] So BF\(_3\) has the smallest dipole moment.
Step 3: Analyze NF\(_3\).
NF\(_3\) has trigonal pyramidal geometry due to lone pair on nitrogen. Although N–F bonds are polar, fluorine is highly electronegative, and bond dipoles partially oppose the lone pair effect, resulting in a small net dipole moment. Hence NF\(_3\) has a low but non-zero dipole moment.
Step 4: Analyze NH\(_3\).
NH\(_3\) also has trigonal pyramidal geometry with a lone pair on nitrogen. N–H bonds are polar and their dipoles reinforce the lone pair effect rather than cancel it. Therefore, NH\(_3\) has a comparatively larger dipole moment than NF\(_3\).
Step 5: Compare all molecules.
From analysis: \[ BF_3 = 0 \lt NF_3 \lt NH_3 \] Thus, dipole moment increases in the order: \[ BF_3 \lt NF_3 \lt NH_3 \]
Step 6: Final conclusion.
Therefore, the correct order of dipole moments is: \[ \boxed{BF_3 \lt NF_3 \lt NH_3} \]
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