Bond polarity and dipole moment

A covalent bond is genuinely equal only between identical atoms, such as the two chlorines in $\text{Cl}_2$ — this is called a non-polar covalent bond, with zero net charge separation. The moment the two bonded atoms differ, the story changes.

Electronegativity difference creates polarity. Electronegativity is an atom's tendency to pull shared electrons toward itself. When two different atoms bond, the more electronegative one pulls the shared pair closer, acquiring a partial negative charge ($\delta^-$), while the other atom is left with an equal partial positive charge ($\delta^+$). This asymmetric bond is a polar covalent bond, and it behaves like a miniature electric dipole — two equal, opposite charges separated by a small distance across the bond.

Dipole moment quantifies the separation. For any such dipole,

where: $q$ is the magnitude of the partial charge at either end (in coulombs) and $d$ is the distance between the two charge centres, essentially the bond length (in metres). The product $\mu$ is measured in coulomb-metres in SI, but chemists almost always quote it in debye (D), where $1\ \text{D} = 3.33564\times10^{-30}\ \text{C m}$. Dipole moment is a vector: it has both size and a direction, conventionally drawn as an arrow from the positive end toward the negative end, with a small crossed tail marking the $\delta^+$ end.

A molecule's net dipole is a vector sum, not a bond count. A polyatomic molecule has several bonds, each with its own dipole vector pointing along that bond. The molecule's overall dipole moment is found by adding all the individual bond dipoles as vectors, taking the actual bond angles (the geometry, from VSEPR) into account. Two consequences follow directly: Symmetric shapes can cancel polar bonds entirely — a molecule with polar bonds arranged with perfect symmetry around a central atom (no lone pairs breaking that symmetry) can have equal and oppositely directed bond dipoles that sum to zero, making the whole molecule non-polar even though every bond in it is polar. Asymmetric shapes, often caused by lone pairs on the central atom, leave the bond dipoles unable to cancel, so the molecule keeps a net dipole moment and is genuinely polar overall.

Dipole moment also probes ionic character. Comparing a bond's experimentally measured dipole moment against the theoretical value expected for 100% ionic character (complete electron transfer) at the same bond length gives the percentage ionic character of that bond — a way to place real bonds somewhere on the spectrum between purely covalent and purely ionic.

Explain, quantitatively, why $\text{H}_2\text{O}$ (bond angle $104.5°$) has a much larger net dipole moment than $\text{CO}_2$ (bond angle $180°$) would if it were bent instead of linear, using the vector-addition result above.

Step 1. Confirm both bonds are polar. In $\text{H}_2\text{O}$, oxygen (electronegativity $3.44$) is more electronegative than hydrogen ($2.20$), so each O–H bond dipole points from H ($\delta^+$) toward O ($\delta^-$). In $\text{CO}_2$, oxygen ($3.44$) is more electronegative than carbon ($2.55$), so each C=O bond dipole points from C ($\delta^+$) toward O ($\delta^-$). Both molecules start from genuinely polar individual bonds.

Step 2. Apply the geometry from VSEPR. Oxygen in $\text{H}_2\text{O}$ carries two bond pairs and two lone pairs, which push the bonds together into a bent shape with $\theta \approx 104.5°$. Carbon in $\text{CO}_2$ carries two bond pairs and no lone pairs, giving a strictly linear shape with $\theta = 180°$.

Step 3. Use $\mu_{net} = 2\mu_1 \cos(\theta/2)$ from the derivation. For $\text{CO}_2$, $\cos(180°/2) = \cos(90°) = 0$, so $\mu_{net} = 0$ regardless of how polar each C=O bond is individually — this matches the measured $\text{CO}_2$ dipole moment of essentially zero. For $\text{H}_2\text{O}$, $\cos(104.5°/2) = \cos(52.25°) \approx 0.61$, a clearly positive number, so $\mu_{net} = 2\mu_1(0.61)$ is nonzero and adds constructively along the bisector toward oxygen.

Step 4. State the result. Because $\cos(90°) = 0$ but $\cos(52.25°) \approx 0.61 \neq 0$, the same basic ingredient — two polar bonds — gives a net dipole moment of zero for linear $\text{CO}_2$ and a substantial net dipole moment of about $1.85\ \text{D}$ for bent $\text{H}_2\text{O}$. The bond angle alone, through the factor $\cos(\theta/2)$, is what separates a non-polar molecule from a strongly polar one built from the same kind of polar bonds.

• Assuming any molecule with polar bonds must itself be polar — geometry, captured by the $\cos(\theta/2)$ factor, decides whether those bond dipoles survive vector addition or cancel.
• Treating dipole moment as a scalar and simply adding magnitudes — it is a vector, and bond angle changes both the size and direction of the resultant.
• Getting the arrow direction backwards — the dipole vector points from the $\delta^+$ end toward the $\delta^-$ end, that is, toward the more electronegative atom.
• Forgetting that lone pairs on the central atom often break symmetry and are the reason molecules like $\text{H}_2\text{O}$ and $\text{NH}_3$ end up polar rather than cancelling like their lone-pair-free, more symmetric relatives.

Ask the tutor: want to work through whether a specific molecule like $\text{CHCl}_3$, $\text{BF}_3$, or $\text{SO}_2$ ends up polar or non-polar using the same vector-addition approach?