Question

Identify the correct set of molecules with zero dipole moment:

• A. $ \text{CO}_2, \text{NH}_3, \text{H}_2\text{O} $
• B. $ \text{NH}_3, \text{NF}_3, \text{BF}_3 $
• C. $ \text{PF}_3, \text{NH}_3, \text{CH}_4 $
• D. $ \text{CH}_4, \text{BF}_3, \text{CO}_2 $

Hint:

Symmetrical molecules with identical surrounding atoms and no lone pairs usually possess zero dipole moment. Examples:
• Linear ($AX_2$)
• Trigonal planar ($AX_3$)
• Tetrahedral ($AX_4$)

Answer 1

The Correct Option is D

Concept: A molecule possesses zero dipole moment when the vector sum of all bond dipoles becomes zero. This generally occurs in highly symmetrical molecules.

Step 1: Analyzing molecules in option (d).
$\text{CH_4$ (Methane):}
• Geometry: Tetrahedral
• Hybridization: $sp^3$
• All four hydrogen atoms are identical. The bond dipoles cancel due to perfect symmetry. \[ \mu = 0 \] $\text{BF_3$ (Boron trifluoride):}
• Geometry: Trigonal planar
• Hybridization: $sp^2$
• Three identical fluorine atoms arranged symmetrically. All bond dipoles cancel. \[ \mu = 0 \] $\text{CO_2$ (Carbon dioxide):}
• Geometry: Linear
• Hybridization: $sp$
• Two equal and opposite C=O dipoles. Therefore, \[ \mu = 0 \]

Step 2: Checking why the other molecules are polar.
$\text{NH_3$:} Trigonal pyramidal geometry due to one lone pair. \[ \mu \neq 0 \] $\text{H_2\text{O}$:} Bent shape due to two lone pairs. \[ \mu \neq 0 \] $\text{NF_3$ and $\text{PF}_3$:} Both possess lone pairs and pyramidal geometry. Hence they are polar molecules.

Step 3: Final conclusion.
Only: \[ \text{CH}_4,\ \text{BF}_3,\ \text{CO}_2 \] have symmetrical structures with complete dipole cancellation. Therefore, the correct answer is: \[ \boxed{(d)\ \text{CH}_4,\ \text{BF}_3,\ \text{CO}_2} \]