Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: The dipole moments of ${\text{CCl}}_4$, ${\text{CHCl}}_3$ and ${\text{CH}}_4$ are in the order

Imagine a microscopic game of tug-of-war happening inside every molecule. The players are the atoms, and the rope is the shared electron density in their chemical bonds. Some atoms, like Chlorine, are incredibly strong pullers (highly electronegative), while others, like Hydrogen, are relatively weak. This pulling creates a bond dipole, a vector pointing towards the stronger atom.

But here is the beautiful part: the overall polarity of a molecule—its net dipole moment ($\mu$)—isn't just about how strong the individual pulls are. It is entirely dependent on the 3D geometry and symmetry of the molecule. If the pulls are perfectly balanced in all directions, the molecule remains non-polar, no matter how fierce the internal tug-of-war is.

Let's look at our first two contenders: Methane (${\text{CH}}_4$) and Carbon Tetrachloride (${\text{CCl}}_4$). Both of these molecules feature a central carbon atom that is $s{p}^3$ hybridized. This hybridization forces the four attached atoms into a perfect tetrahedral geometry, with bond angles of exactly $109.5^{\circ }$.

In ${\text{CH}}_4$, the central Carbon is slightly more electronegative than the surrounding Hydrogens. Therefore, four bond dipole vectors point inward toward the Carbon. However, because the molecule is perfectly symmetrical, these four vectors perfectly cancel each other out mathematically. The net result is $\mu =0$.

In ${\text{CCl}}_4$, the situation is reversed. The highly electronegative Chlorine atoms pull electron density away from the central Carbon, creating four strong outward-pointing vectors. Yet, the magic of symmetry strikes again! The four identical outward pulls perfectly balance each other out. Despite having highly polar individual bonds, the molecule as a whole is non-polar, yielding $\mu =0$.

Now, let's examine Chloroform (${\text{CHCl}}_3$). This molecule breaks the perfect symmetry. We have one $\text{C}-\text{H}$ bond and three $\text{C}-\text{Cl}$ bonds arranged around the central Carbon.

The $\text{C}-\text{H}$ bond dipole points inward toward the Carbon, while the three $\text{C}-\text{Cl}$ bond dipoles point outward toward the Chlorines. When we add these vectors together, they no longer cancel. In fact, the inward push from the Hydrogen perfectly aligns with the downward pull of the three Chlorines, reinforcing the overall vector.

Because the symmetry is broken, the vector sum is non-zero, giving Chloroform a significant net dipole moment ($\mu eq 0$).

By comparing our findings, the conclusion is crystal clear. Both perfectly symmetrical molecules have a dipole moment of zero, while the asymmetrical molecule has a positive dipole moment.

Therefore, the correct order is: ${\text{CH}}_4={\text{CCl}}_4<{\text{CHCl}}_3$