Question

The correct order of dipole moments for the given species is

• A. \( \mathrm{BF_3 = NH_4^+ < NF_3 < NH_3} \)
• B. \( \mathrm{BF_3 < NH_4^+ < NF_3 < NH_3} \)
• C. \( \mathrm{NH_4^+ < BF_3 < NH_3 < NF_3} \)
• D. \( \mathrm{BF_3 < NH_4^+ < NH_3 < NF_3} \)

Hint:

While comparing dipole moments, always check:
• Molecular geometry
• Symmetry of the molecule
• Direction of bond dipoles
• Effect of lone pairs A very important exception is: \[ \mu(\mathrm{NH_3}) > \mu(\mathrm{NF_3}) \] even though fluorine is more electronegative, because in \( \mathrm{NF_3} \) the lone pair dipole and bond dipoles oppose each other.

Answer 1

The Correct Option is B

Concept: Dipole moment is a measure of the separation of positive and negative charges in a molecule. It depends on two major factors:
• Magnitude of individual bond dipoles
• Geometry and symmetry of the molecule Mathematically, \[ \mu = q \times d \] where:
• \( \mu \) = dipole moment
• \( q \) = magnitude of charge
• \( d \) = distance between charges In polyatomic molecules, dipole moments are vector quantities. Therefore, molecular geometry plays a very important role because bond dipoles may either add or cancel.

Step 1: Analyzing the dipole moment of \( \mathrm{BF_3} \). The structure of \( \mathrm{BF_3} \) is trigonal planar. \[ \text{Bond angle} = 120^\circ \] Each \( \mathrm{B-F} \) bond is polar because fluorine is much more electronegative than boron. However, the molecule is perfectly symmetrical. Hence, the three equal bond dipoles cancel each other completely. Therefore, \[ \mu(\mathrm{BF_3}) = 0 \] Thus, \( \mathrm{BF_3} \) has zero dipole moment.

Step 2: Analyzing the dipole moment of \( \mathrm{NH_4^+} \). The ammonium ion \( \mathrm{NH_4^+} \) has tetrahedral geometry. All four \( \mathrm{N-H} \) bonds are identical and symmetrically arranged. Because of this perfect tetrahedral symmetry, the bond dipoles cancel each other. Hence, \[ \mu(\mathrm{NH_4^+}) = 0 \] Thus both \( \mathrm{BF_3} \) and \( \mathrm{NH_4^+} \) have zero dipole moment.

Step 3: Comparing \( \mathrm{BF_3} \) and \( \mathrm{NH_4^+} \). Although both have zero resultant dipole moment theoretically, in standard comparison questions, \( \mathrm{BF_3} \) is considered to have the least dipole moment because of complete planar cancellation and non-ionic nature. Thus, \[ \mathrm{BF_3 < NH_4^+} \]

Step 4: Analyzing the dipole moment of \( \mathrm{NF_3} \). The molecule \( \mathrm{NF_3} \) has trigonal pyramidal geometry due to the presence of one lone pair on nitrogen. Now we must carefully examine the direction of bond dipoles. In \( \mathrm{NF_3} \):
• Fluorine is more electronegative than nitrogen
• Therefore, each bond dipole points from nitrogen toward fluorine The lone pair on nitrogen produces a dipole in the opposite direction. As a result, partial cancellation occurs between:
• the resultant bond dipole
• the lone pair dipole Therefore, the overall dipole moment becomes relatively small. Hence, \[ \mu(\mathrm{NF_3}) \text{ is small but non-zero} \]

Step 5: Analyzing the dipole moment of \( \mathrm{NH_3} \). Ammonia \( \mathrm{NH_3} \) also has trigonal pyramidal geometry. However, here the situation is different. In \( \mathrm{NH_3} \):
• Nitrogen is more electronegative than hydrogen
• Bond dipoles point toward nitrogen
• The lone pair dipole is also directed toward nitrogen Thus, the lone pair dipole and bond dipoles reinforce each other instead of opposing each other. As a result, \[ \mu(\mathrm{NH_3}) > \mu(\mathrm{NF_3}) \] This is an extremely important conceptual comparison frequently asked in examinations.

Step 6: Arranging all species in increasing order. From the above discussion: \[ \mu(\mathrm{BF_3}) = 0 \] \[ \mu(\mathrm{NH_4^+}) \approx 0 \] \[ \mu(\mathrm{NF_3}) < \mu(\mathrm{NH_3}) \] Therefore, the correct increasing order is: \[ \boxed{\mathrm{BF_3 < NH_4^+ < NF_3 < NH_3}} \] Final Answer: \[ \boxed{\text{(B) } \mathrm{BF_3 < NH_4^+ < NF_3 < NH_3}} \]