2A.13-2A.15 - Electronegativity And Polarity

Bond Polarity And Molecular Polarity

Suppose two molecules both contain polar bonds. One is bent, like $\ce{H2O}$. Another is linear, like $\ce{CO2}$. It is tempting to say both molecules must be polar because both contain bonds between atoms with different electronegativities, but that skips the spatial step.

In this lesson, keep two levels separate:

• Bond polarity is about one covalent bond and how its bonding electrons are shared.
• Molecular polarity is about the whole molecule and whether all the bond dipoles cancel.

The exam-relevant trap is simple: a molecule can contain polar bonds and still be non-polar overall. Questions often test this by giving a familiar shape, asking for a prediction, and expecting the answer to use both electronegativity and shape.

Quick contrast

$\ce{Cl2}$ has a non-polar covalent bond because the two atoms are identical, so they attract the bonding electrons equally. $\ce{HCl}$ has a polar covalent bond because chlorine attracts the bonding pair more strongly than hydrogen. $\ce{CO2}$ has two polar C=O bonds, but the molecule is linear, so the two equal bond dipoles act in opposite directions and cancel.

Electronegativity And Bond Dipoles

When two bonded atoms attract their shared electrons unequally, the bond develops partial positive and negative ends.

The definition is deliberately about a bonded atom, not an isolated atom. It asks: once a shared pair exists between two atoms, which atom pulls that pair more strongly?

The Data Booklet gives Pauling electronegativities. Useful values for this lesson include H 2.1, C 2.5, N 3.0, O 3.5, F 4.0, and Cl 3.0. Larger values mean a stronger attraction for bonding electrons.

When two bonded atoms have different electronegativities, the bonding electrons are not shared equally:

• the more electronegative atom becomes partially negative, written $\delta^{-}$;
• the less electronegative atom becomes partially positive, written $\delta^{+}$;
• the bond has a bond dipole, pointing towards the more electronegative atom.

For H-Cl, chlorine is more electronegative than hydrogen, so the bonding pair is pulled closer to chlorine. Hydrogen is the $\delta^{+}$ end and chlorine is the $\delta^{-}$ end. The dipole is commonly shown with the cross at the positive end and the arrow head towards the negative end.

Worked example: decide the direction of a bond dipole

Question: Use Pauling electronegativities to decide the polarity of the C-F bond.

Route:

1. Identify the bonded atoms: C and F.
2. Read or recall the electronegativities: C = 2.5 and F = 4.0.
3. Compare them: F has the larger value.
4. Assign partial charges: C is $\delta^{+}$ and F is $\delta^{-}$.
5. Draw or describe the dipole: C($\delta^{+}$) -> F($\delta^{-}$), with the arrow head towards F.

Chemical interpretation: the bond is polar covalent because the electrons are still shared, but the sharing is unequal.

Worked comparisons

N is $\delta^{-}$ in N-H; O is $\delta^{-}$ in C-O; C is slightly $\delta^{-}$ in C-H because C 2.5 is higher than H 2.1. The C-H bond is only weakly polar, so do not exaggerate its effect unless the molecular context requires it.

Common error: full charge versus partial charge

Saying "chlorine is negative in $\ce{HCl}$" is too loose. The better explanation is that chlorine is partially negative in the covalent H-Cl bond because it attracts the shared bonding pair more strongly. $\ce{HCl}$ is not made of $\ce{H+}$ and $\ce{Cl-}$ ions in the same way as an ionic lattice.

Ionic And Covalent Are Extremes Of A Continuum

Bonding type is not a set of perfectly separate boxes. Ionic bonding and covalent bonding are useful extremes, and many real bonds sit between them. The specification treats them as a continuum because electronegativity difference changes gradually, not by a magic step.

[DIAGRAM: asset_name: Bonding Continuum; asset_slug: edexcel_a_level_chemistry_l011_bonding_continuum; recommended_method: matplotlib; description: A monochrome continuum showing equal sharing, polar covalent unequal sharing, and an ionic extreme as electronegativity difference increases, with no fixed boundary.]

At one extreme, identical atoms such as Cl-Cl have no electronegativity difference, so the bonding electrons are shared equally. In the middle, bonds such as H-Cl or C-F have unequal sharing and are polar covalent. At the other extreme, a very large electronegativity difference favours electron transfer and ionic bonding, as in a metal-non-metal ionic compound.

Avoid learning a hard numeric cut-off for "ionic" versus "covalent" in this specification. The useful reasoning is comparative: as electronegativity difference increases, the bond becomes more polar and has more ionic character. The exact structure also depends on the elements and the chemical context.

Worked example: use the continuum without a cut-off

Question: Compare the bonding in $\ce{Cl2}$, $\ce{HCl}$, and $\ce{NaCl}$ using electronegativity.

Route:

1. $\ce{Cl2}$: both atoms are chlorine, so the electronegativity difference is zero. The bond is non-polar covalent.
2. $\ce{HCl}$: Cl is more electronegative than H, so the bonding pair is pulled towards Cl. The bond is polar covalent.
3. $\ce{NaCl}$: Na has a much lower electronegativity than Cl, so the bonding is best described as ionic, with strong electrostatic attraction between oppositely charged ions in a lattice.

Chemical interpretation: these examples do not create three unrelated definitions. They show movement along the same idea: increasing inequality in attraction for electrons.

From Bond Dipoles To Molecular Polarity

To predict whether a molecule is polar, use a two-stage test.

1. Are there polar bonds? Use electronegativity differences.
2. Do the bond dipoles cancel? Use the molecular shape from electron-pair repulsion theory.

The second stage is why molecular shape matters here. Molecules are three-dimensional, so bond dipoles are vectors: they have size and direction. Equal dipoles can cancel if the molecule is symmetrical, but they add to give a net dipole if the shape is unsymmetrical.

[DIAGRAM: asset_name: Bond Dipoles Molecular Shape; asset_slug: edexcel_a_level_chemistry_l011_bond_dipoles_molecular_shape; recommended_method: matplotlib; description: A monochrome comparison of a linear A-B-A molecule where equal bond dipoles cancel and a bent A-B-A molecule where bond dipoles add to give a net dipole.]

Worked example: $\ce{CO2}$ and $\ce{H2O}$

$\ce{CO2}$:

• Each C=O bond is polar because O is more electronegative than C.
• $\ce{CO2}$ is linear.
• The two C=O bond dipoles are equal and opposite.
• They cancel, so $\ce{CO2}$ is not a polar molecule.

$\ce{H2O}$:

• Each O-H bond is polar because O is more electronegative than H.
• $\ce{H2O}$ is bent because oxygen has two bonding pairs and two lone pairs.
• The two O-H bond dipoles do not point in exactly opposite directions.
• They add to give a net dipole, so $\ce{H2O}$ is a polar molecule.

Notice the key contrast: both molecules contain polar bonds, but only $\ce{H2O}$ has a permanent overall molecular dipole.

Common-error contrast

Weak answer: "$\ce{CO2}$ is non-polar because it has no polar bonds."

Better answer: "$\ce{CO2}$ has polar C=O bonds, but the molecule is linear and symmetrical, so the equal bond dipoles cancel."

The better answer earns the chemistry because it keeps bond polarity and molecular polarity separate.

Predicting Molecular Polarity

The reliable prediction sequence is:

1. Draw or recall the shape.
2. Mark the polar bonds using electronegativity.
3. Ask whether the dipoles cancel by symmetry.
4. State whether the molecule is polar and justify it.

Worked example: predict whether $\ce{BF3}$ is polar

1. Shape: $\ce{BF3}$ is trigonal planar, with three B-F bonds at about 120 degrees.
2. Bond polarity: F is more electronegative than B, so each B-F bond is polar towards F.
3. Symmetry: the three identical B-F bond dipoles are arranged evenly around the central B atom.
4. Conclusion: $\ce{BF3}$ is not a polar molecule because the three bond dipoles cancel.

Worked example: predict whether $\ce{NH3}$ is polar

1. Shape: $\ce{NH3}$ is pyramidal, with three bonding pairs and one lone pair around N.
2. Bond polarity: N is more electronegative than H, so each N-H bond is polar towards N.
3. Symmetry: the pyramidal shape is not symmetrical enough for the three bond dipoles to cancel.
4. Conclusion: $\ce{NH3}$ is a polar molecule because it has a net dipole.

Worked polarity comparisons

1. $\ce{CCl4}$ is not polar overall: each C-Cl bond is polar, but the tetrahedral molecule is symmetrical and the four equal dipoles cancel.
2. $\ce{CH3Cl}$ is polar: the tetrahedral molecule is not symmetrical because one substituent is Cl and three are H, so the bond dipoles do not cancel. The weak C-H dipoles point towards carbon, while the stronger C-Cl dipole points towards chlorine; these are not four identical dipoles.
3. $\ce{BeCl2(g)}$ is not polar overall: each Be-Cl bond is polar, but the molecule is linear and the two equal dipoles cancel.

Command-word habit

For predict, do not just name "polar" or "non-polar". Give the expected result and support it from the model. A concise answer usually needs both the bond evidence and the shape evidence.

Quick Recap

Electronegativity concerns attraction for bonding electrons. Differences create bond dipoles along a continuum of bond character. Molecular polarity depends on the net effect of all bond dipoles in the actual three-dimensional shape.