UNIT 2 CHEMISTRY · PEARSON AAQ · B1.6
Electronegativity and Polarity
FREE CHEMISTRY REVISION · TOPIC 15
From unequal sharing to the polarity of a whole molecule.
Connect periodic trends, partial charges and three-dimensional shape to explain when bond dipoles cancel.
Pearson BTEC Level 3 AAQ Applied Science
Unit 2: Principles and Applications of Chemistry · B1.6
What you will learn
Define electronegativity; explain its periodic trends; assign partial charges and bond-dipole directions; combine bond polarity with three-dimensional molecular shape to decide whether a molecule is polar. This topic builds directly on B1.5 molecular shapes.
1. Electronegativity: attraction within a bond
Electronegativity is the ability of an atom in a covalent bond to attract the shared pair of electrons towards itself. It is a relative measure, usually quoted on a scale such as the Pauling scale, with no unit. It describes a bonded atom, rather than the removal of an electron from an isolated gaseous atom (ionisation energy).
The shared pair is attracted by both nuclei. The more electronegative atom draws more of the bonding electron density towards itself. Equal atoms, such as Cl–Cl, have equal electronegativity and share symmetrically. Different atoms can share unequally while still forming a covalent bond. Electronegativity is a model for predicting this distribution; it is not an electron count or an ionic charge.
2. Trends across a period and down a group
Across a period, electronegativity generally increases. Nuclear charge rises as proton number increases. Added electrons enter the same principal shell, so shielding does not rise enough to cancel the stronger nuclear attraction. Atomic radius generally decreases and the bonded atom attracts the shared pair more strongly.
Down a group, electronegativity generally decreases. The bonding region is farther from the nucleus because more shells are occupied, and inner electrons provide more shielding. These effects outweigh the increased nuclear charge. Fluorine is the most electronegative element on the Pauling scale; it is small and attracts shared electrons strongly.
Use the general trends for the main-group comparisons normally set at this level. Values depend on the scale and chemical environment, and the pattern is not a rigid rule for every transition-metal comparison. Many introductory tables leave noble gases blank; do not assume that a blank entry means zero.
3. From electronegativity to bond polarity
In H–Cl, chlorine is more electronegative: H is δ+ and Cl is δ−. The symbol δ means partial. The molecule remains electrically neutral; these are not H⁺ and Cl⁻ ions. A bond dipole is a separation of charge within a bond.
A chemical dipole arrow has a crossed tail at the positive end and an arrowhead pointing towards the more electronegative, partially negative atom. Label the atoms as well as the arrow. For O–H the arrow points towards O; for C–Cl towards Cl. For a bond between identical atoms, there is no electronegativity difference and no permanent bond dipole.
If values are supplied, calculate Δχ = |χ(A) − χ(B)|. A larger difference generally indicates a more polar comparable bond. Bonding lies on a continuum: numerical cut-offs between non-polar covalent, polar covalent and ionic are useful conventions, not universal physical boundaries. Do not replace a structure explanation with a memorised cut-off.
4. Molecular polarity is a vector result
A polar molecule has a non-zero overall permanent dipole moment: the centres of positive and negative charge do not coincide. First identify the polar bonds, then establish the shape, then ask whether their dipoles cancel in three dimensions. A neutral molecule may still be polar.
Equal bond dipoles arranged symmetrically can cancel. CO₂ is linear with identical terminal O atoms, so its two C=O dipoles are equal and opposite. BF₃ is trigonal planar: three equal B–F dipoles cancel at 120°. CCl₄ is tetrahedral: four identical C–Cl dipoles cancel in 3D. SF₆ is octahedral and also has zero resultant in its symmetric structure.
H₂O is bent, so its two O–H dipoles have components in the same direction towards O; their resultant is non-zero. NH₃ is trigonal pyramidal, not planar, and is polar. CH₃Cl is tetrahedral around C but has different surrounding atoms: its bond dipoles do not cancel. A shape name alone is insufficient: compare the surrounding atoms and dipole magnitudes too.
| Molecule | Shape | Reason | Overall |
|---|---|---|---|
| CO₂ | Linear | Equal opposite C=O dipoles | Non-polar |
| H₂O | Bent | O–H dipoles do not cancel | Polar |
| BF₃ | Trigonal planar | Three equal dipoles cancel | Non-polar |
| CCl₄ | Tetrahedral | Four identical surroundings; 3D cancellation | Non-polar |
| NH₃ | Trigonal pyramidal | Asymmetric charge distribution | Polar |
| CH₃Cl | Tetrahedral around C | Different substituents; no cancellation | Polar |
5. A reliable method for unfamiliar molecules
1. Draw the bonding arrangement and identify lone pairs on the central atom.
2. Use electron-pair repulsion to establish the actual molecular shape, not just its flat structural formula.
3. Compare electronegativities and label δ+ and δ− on each relevant bond.
4. Compare dipoles as vectors. State explicitly whether they cancel.
5. Give the overall classification and a reason that links both bonding and shape.
Lone pairs can change shape and the total electron distribution, but “contains lone pairs” is not by itself a proof of polarity. For example, a symmetric square-planar molecule with identical surrounding atoms can have cancelling dipoles despite central lone pairs. Conversely CH₃Cl is polar without a lone pair on carbon. Use the full arrangement.
Worked examples
Example 1 · Explain the F–Cl comparison
F is above Cl in Group 17. Its bonding electrons are closer to the nucleus and less shielded. F therefore attracts a shared pair more strongly, despite having fewer protons. Explain both distance and shielding; proton number alone gives the wrong prediction.
Example 2 · Calculate and label a bond dipole
Use supplied illustrative values χ(H) = 2.2 and χ(Cl) = 3.2. Δχ = |3.2 − 2.2| = 1.0. Draw Hδ+—Clδ− and the crossed-tail dipole arrow from H towards Cl. This indicates polar covalent bonding, not complete transfer.
Example 3 · Decide whether CO₂ is polar
Each C=O bond is polar towards O. Electron regions give a linear O=C=O molecule, 180°. The identical dipoles are equal and opposite, so the vector sum is zero: non-polar. Do not conclude that the individual bonds are non-polar.
Example 4 · Decide whether H₂O is polar
O is more electronegative than H. The two bonding pairs and two lone pairs give a bent molecule, about 104.5°. The O–H dipoles do not cancel; the resultant points towards the oxygen end. Water is polar.
Example 5 · Compare CCl₄ and CH₃Cl
Both have tetrahedral arrangements around C. CCl₄ has four identical C–Cl bonds arranged symmetrically, so the resultant is zero. Replacing three Cl atoms with H removes that equality: CH₃Cl has a non-zero resultant and is polar. Tetrahedral does not automatically mean non-polar.
Common misconceptions and exam tips
“Electronegativity is energy to remove an electron.” That is ionisation energy; electronegativity concerns attraction for shared electrons.
“δ− means a full negative ion.” It is a partial charge.
“A molecule with polar bonds must be polar.” Symmetric bond dipoles may cancel.
“All tetrahedral molecules are non-polar.” The surrounding atoms must also give the necessary symmetry.
“Lone pairs always make a molecule polar.” Consider the complete charge distribution.
For a polarity explanation use the sequence electronegativity → bond dipoles → shape → cancellation or resultant. Draw a shape diagram when it clarifies your argument. Distinguish “no polar bonds” from “polar bonds whose dipoles cancel”.
Your revision summary
Electronegativity is attraction for a shared electron pair. It generally increases across a period and decreases down a group. Unequal sharing gives partial charges and a bond dipole. Overall molecular polarity depends on the three-dimensional vector sum of the dipoles; symmetry can cancel polar bonds.
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