Q.Although both $CO_2$ and $H_2O$ are triatomic molecules, the shape of $H_2O$ molecule is bent while that of $CO_2$ is linear. Explain this on the basis of dipole moment.

Concept: VSEPR Theory — the shape of a molecule is determined by the repulsion between electron pairs (bonding and lone pairs) around the central atom.

Reasoning:

1. In $CO_2$, the central carbon has no lone pairs and two double bonds. The two regions of electron density repel equally, giving a linear geometry ($180^\circ$ bond angle). The individual C=O bond dipoles are equal and opposite, so the net dipole moment is zero. …

The question asks you to explain the shapes of $CO_2$ and $H_2O$ on the basis of dipole moment. This is a classic application of VSEPR theory combined with the idea that a molecule’s dipole moment is a direct experimental clue to its geometry.

The core idea: If a molecule has polar bonds, the overall dipole moment (the vector sum of all bond dipoles) depends on the molecular shape. If the bond dipoles cancel perfectly, the molecule is nonpolar and must be symmetric. If they don’t cancel, the molecule is polar and must be asymmetric.

Let’s apply this to the two molecules.

1. Start with $CO_2$.

   Carbon dioxide has two $C=O$ bonds. Oxygen is more electronegative than carbon, so each $C=O$ bond is polar — the oxygen end carries a partial negative charge ($\delta^-$) and the carbon end a partial positive charge ($\delta^+$). Each bond has a dipole moment vector pointing from carbon toward oxygen.

   Now, if the molecule were bent, these two vectors would add up to a net dipole pointing somewhere between them. But experimentally, $CO_2$ has zero dipole moment. The only way two equal bond dipoles can cancel is if they point in exactly opposite directions. That forces the molecule to be linear, with the oxygen atoms on opposite sides of carbon: $O=C=O$. The two dipoles are equal in magnitude and opposite in direction, so their vector sum is zero.

   For $CO_2$: $\vec{\mu}_{\text{net}} = \vec{\mu}_{C=O} + \vec{\mu}_{C=O} = 0$ only if the bond angle is $180^\circ$.

2. Now consider $H_2O$.

   Water also has two polar bonds: each $O-H$ bond is polar because oxygen is more electronegative than hydrogen. The bond dipole vectors point from hydrogen toward oxygen. If water were linear ($H-O-H$ at $180^\circ$), these two equal dipoles would point in opposite directions and cancel — giving zero net dipole. But experimentally, water does have a nonzero dipole moment ($\mu \approx 1.85 \, \text{D}$). So the molecule cannot be linear. …