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.

Question

Accepted Answer

> [!TLDR]
> $CO_2$ is measured to have ZERO dipole moment — possible only if its two (polar) C=O bond dipoles cancel exactly, i.e. the molecule is linear. $H_2O$ has a large dipole moment (1.85 D) — its O–H bond dipoles clearly do not cancel, so the molecule must be bent, not linear.

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.

> [!FORMULA]
   > For $CO_2$: $\vec{\mu}_{	ext{net}} = \vec{\mu}_{C=O} + \vec{\mu}_{C=O} = 0$ 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

Steric Number	Electron-Pair Geometry	Lone Pairs	Molecular Geometry	Example	Bond Angle
2	Linear	0	Linear	$\text{CO}_2$	180°
3	Trigonal planar	0	Trigonal planar	$\text{BF}_3$	120°
3	Trigonal planar	1	Bent	$\text{SO}_2$	~119°
4	Tetrahedral	0	Tetrahedral	$\text{CH}_4$	109.5°
4	Tetrahedral	1	Trigonal pyramidal	$\text{NH}_3$	~107°
4	Tetrahedral	2	Bent	$\text{H}_2\text{O}$	~104.5°
5	Trigonal bipyramidal	0	Trigonal bipyramidal	$\text{PCl}_5$	90°, 120°
6	Octahedral	0	Octahedral	$\text{SF}_6$	90°

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 understanding — VSEPR Theory

VSEPR Theory: Why Molecules Have the Shapes They Do

The Core Intuition

The Precise Statement

How to Predict Shape in 3 Steps

The Common Geometries at a Glance

Why Lone Pairs Squeeze Bond Angles

1. ClF₃ (Chlorine trifluoride)

2. BrF₅ (Bromine pentafluoride)

3. SF₄ (Sulfur tetrafluoride)

Q.Although both CO2CO_2CO2​ and H2OH_2OH2​O are triatomic molecules, the shape of H2OH_2OH2​O molecule is bent while that of CO2CO_2CO2​ is linear. Explain this on the basis of dipole moment.

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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.

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