Step 1: Understanding the Concept:
Dipole moment is a vector quantity that represents the separation of charge within a chemical bond or molecule.
A molecule has a zero dipole moment if it is electrically symmetrical.
Symmetry allows the individual polarities of each bond to counteract and perfectly cancel each other out.
In highly symmetrical geometries like Linear, Trigonal Planar, and Tetrahedral, if all surrounding atoms are identical, the net dipole moment is zero.
Lone pairs on the central atom usually break this symmetry, making the molecule polar.
Step 2: Key Formula or Approach:
Identify the VSEPR shape and check for lone pairs:
- \( CH_4 \): Tetrahedral \( (AX_4) \).
- \( BF_3 \): Trigonal Planar \( (AX_3) \).
- \( CO_2 \): Linear \( (AX_2) \).
Step 3: Detailed Explanation:
Let's analyze the properties of set (d):
1. \( CH_4 \) (Methane): Carbon is sp\(^3\) hybridized. The geometry is a perfect tetrahedron. The four C-H bond dipoles are oriented symmetrically in 3D space. Their vector sum is zero. Thus, methane is non-polar.
2. \( BF_3 \) (Boron Trifluoride): Boron is sp\(^2\) hybridized. It has a trigonal planar shape. The three polar B-F bond vectors are separated by \( 120^\circ \). The sum of three identical vectors at \( 120^\circ \) in a plane is zero. Thus, \( BF_3 \) is non-polar.
3. \( CO_2 \) (Carbon Dioxide): The molecule is linear (\( O=C=O \)). The two C=O bond dipoles point in exactly opposite directions. Since they are equal in magnitude, they cancel out. Thus, \( CO_2 \) is non-polar.
Comparison with other options:
- \( NH_3 \): Pyramidal shape due to a lone pair. Bond dipoles don't cancel. Polar.
- \( H_2O \): Bent shape due to two lone pairs. Highly polar.
- \( NF_3/PF_3 \): Pyramidal shape. Polar.
Since set (d) contains only non-polar molecules, it is the correct identification.
Step 4: Final Answer:
The set \( CH_4, BF_3, CO_2 \) has zero dipole moment.
This is Option (d).