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:

A quick shortcut: If the central atom has no lone pairs and all the surrounding atoms are the same, the molecule will almost always have a zero dipole moment due to its perfect symmetry!

Answer 1

The Correct Option is D

Concept: The dipole moment ($ \mu $) of a molecule depends on both the polarity of the individual bonds and the geometric shape (symmetry) of the molecule. A molecule has a zero dipole moment if it is highly symmetrical, such that the individual bond dipoles cancel each other out as vectors.
• Polar Molecules: Contain lone pairs on the central atom or have asymmetrical surrounding atoms (e.g., $ \text{NH}_3, \text{H}_2\text{O}, \text{NF}_3 $).
• Non-Polar Molecules: Symmetrical shapes where bond dipoles cancel (e.g., Linear, Trigonal Planar, Tetrahedral with identical atoms).

Step 1: Analyzing molecules in the options. Let's examine the symmetry and dipole cancellation for each unique molecule mentioned:
• $ \text{CO_2 $: It has a linear geometry ($ \text{O=C=O} $). The two $ \text{C=O} $ bond dipoles are equal in magnitude but opposite in direction. They cancel out perfectly. $ \mu = 0 $.
• $ \text{BF_3 $: It has a trigonal planar geometry. The three $ \text{B-F} $ bond dipoles are oriented at 120$^\circ$ to each other. Their vector sum is zero. $ \mu = 0 $.
• $ \text{CH_4 $: It has a tetrahedral geometry. All four $ \text{C-H} $ bonds are identical and symmetrically arranged. The vector sum of these four dipoles is zero. $ \mu = 0 $.

Step 2: Identifying molecules with non-zero dipole moments. Now, let's see why the other sets are incorrect:
• $ \text{NH_3 $ and $ \text{NF}_3 $: These have pyramidal geometry due to the presence of a lone pair on Nitrogen. The lone pair contributes to the dipole and prevents the bond dipoles from canceling. $ \mu \neq 0 $.
• $ \text{H_2\text{O} $: It has a bent (V-shape) geometry because of two lone pairs on Oxygen. The dipoles do not cancel. $ \mu \neq 0 $.
• $ \text{PF_3 $: Similar to $ \text{NH}_3 $, it is pyramidal with a lone pair. $ \mu \neq 0 $.

Step 3: Final Selection.
Comparing our findings with the options:
• (a) Contains $ \text{NH}_3 $ and $ \text{H}_2\text{O} $ ($ \mu \neq 0 $).
• (b) Contains $ \text{NH}_3 $ and $ \text{NF}_3 $ ($ \mu \neq 0 $).
• (c) Contains $ \text{PF}_3 $ and $ \text{NH}_3 $ ($ \mu \neq 0 $).
• (d) Contains $ \text{CH}_4, \text{BF}_3, $ and $ \text{CO}_2 $. All three are symmetrical and have zero dipole moment.