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 way to identify non-polar molecules is to check:
• Is the geometry symmetrical?
• Are all surrounding atoms identical?
• Does the central atom contain no lone pair? If the answer is yes, the molecule usually has: \[ \mu = 0 \] Common non-polar shapes are:
• Linear ($AX_2$)
• Trigonal planar ($AX_3$)
• Tetrahedral ($AX_4$)

Answer 1

The Correct Option is D

Concept: Dipole moment ($\mu$) is a measure of the separation of positive and negative charges in a molecule. It depends on two important factors:
• The polarity of individual bonds.
• The geometry (shape) of the molecule. Even if a molecule contains polar bonds, the overall dipole moment can become zero if the molecule possesses a highly symmetrical geometry such that all bond dipoles cancel one another vectorially. Therefore, molecules having:
• identical surrounding atoms,
• symmetrical geometry,
• and no lone pair distortion generally show zero dipole moment.

Step 1: Analyzing the molecules in Option (d).
We examine each molecule one by one. (i) $\text{CH_4$ (Methane)}
In methane:
• Carbon is the central atom.
• Carbon undergoes $sp^3$ hybridization.
• The geometry is tetrahedral.
• All four hydrogen atoms are identical. Although each C--H bond has a small bond dipole, the tetrahedral symmetry ensures complete cancellation of all dipole vectors. Hence, \[ \mu = 0 \] Thus, methane is a non-polar molecule. (ii) $\text{BF_3$ (Boron trifluoride)}
In boron trifluoride:
• Boron is $sp^2$ hybridized.
• The geometry is trigonal planar.
• The three fluorine atoms are arranged symmetrically at $120^\circ$. Each B--F bond is highly polar because fluorine is very electronegative. However, due to perfect trigonal planar symmetry, the bond dipoles cancel completely. Therefore, \[ \mu = 0 \] Hence, $\text{BF}_3$ is non-polar. (iii) $\text{CO_2$ (Carbon dioxide)}
In carbon dioxide:
• Carbon is $sp$ hybridized.
• The molecule is linear.
• The two oxygen atoms lie opposite to each other at $180^\circ$. The two C=O bonds are polar, but the dipoles act in opposite directions with equal magnitude. Therefore, they cancel each other completely. Hence, \[ \mu = 0 \] Thus, carbon dioxide is also non-polar.

Step 2: Checking why the other options are incorrect.
Now we analyze the molecules present in the remaining options. $\text{NH_3$ (Ammonia)}

• Nitrogen possesses one lone pair.
• Geometry becomes trigonal pyramidal.
• Bond dipoles do not cancel. Hence ammonia is polar. \[ \mu \neq 0 \] $\text{H_2\text{O}$ (Water)}

• Oxygen contains two lone pairs.
• Geometry becomes bent or V-shaped.
• O--H dipoles cannot cancel. Therefore water is highly polar. \[ \mu \neq 0 \] $\text{NF_3$ and $\text{PF}_3$}
Both molecules:
• contain one lone pair on the central atom,
• possess trigonal pyramidal geometry,
• and therefore have non-zero resultant dipole moments. Thus they are polar molecules.

Step 3: Final conclusion.
Among all the given options, only: \[ \text{CH}_4,\ \text{BF}_3,\ \text{CO}_2 \] have perfectly symmetrical geometries causing complete cancellation of bond dipoles. Hence all these molecules possess zero dipole moment. Therefore, the correct answer is: \[ \boxed{(d)\ \text{CH}_4,\ \text{BF}_3,\ \text{CO}_2} \]