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Electronegativity is one of the most important concepts linking bonding to molecular properties. It determines whether a covalent bond is polar or non-polar, and whether a molecule as a whole has a dipole moment. Understanding polarity is essential for predicting solubility, intermolecular forces, and physical properties.
Electronegativity is the ability of an atom to attract the bonding pair of electrons in a covalent bond towards itself. It was first quantified by Linus Pauling, who developed the Pauling electronegativity scale.
Key trends in the periodic table:
| Element | H | C | N | O | F | P | S | Cl | Br | I |
|---|---|---|---|---|---|---|---|---|---|---|
| Value | 2.1 | 2.5 | 3.0 | 3.5 | 4.0 | 2.1 | 2.5 | 3.0 | 2.8 | 2.5 |
Noble gases are generally not assigned electronegativity values because they do not typically form covalent bonds.
When two atoms with different electronegativities form a covalent bond, the bonding electrons are not shared equally. The more electronegative atom pulls the electron density towards itself, creating an uneven distribution of charge.
This produces a polar covalent bond, which has a partial negative charge (δ−) on the more electronegative atom and a partial positive charge (δ+) on the less electronegative atom:
H^δ+ — Cl^δ−
The greater the difference in electronegativity, the more polar the bond. If the difference is very large (roughly > 1.7), the bond is considered ionic rather than covalent — the electron transfer is essentially complete.
If the two atoms have identical electronegativities (as in H₂, O₂, Cl₂), the bonding electrons are shared equally and the bond is non-polar.
A dipole exists wherever there is a separation of positive and negative charge. Each polar bond has a bond dipole, which can be represented as an arrow pointing from the positive end (δ+) to the negative end (δ−).
The dipole moment of an entire molecule depends on the vector sum of all the individual bond dipoles. This means we must consider both the magnitude and the direction of each bond dipole.
This is where molecular geometry becomes critically important.
A molecule can contain polar bonds yet still be non-polar overall if the bond dipoles cancel out due to symmetry. This concept of symmetry cancellation is essential.
Carbon dioxide (CO₂): Each C=O bond is polar (oxygen is more electronegative than carbon). However, the molecule is linear (180°), so the two bond dipoles point in exactly opposite directions and cancel completely. CO₂ is a non-polar molecule.
Carbon tetrachloride (CCl₄): Each C–Cl bond is polar, but the molecule is tetrahedral. The four bond dipoles are symmetrically arranged in three dimensions and cancel out exactly. CCl₄ is non-polar.
Boron trifluoride (BF₃): Three polar B–F bonds arranged in a trigonal planar shape (120°). The three dipoles cancel by symmetry. BF₃ is non-polar.
Water (H₂O): Each O–H bond is polar, and the molecule is bent (104.5°). The two bond dipoles point in roughly the same direction (both towards the oxygen). They do not cancel, so water has a net dipole moment. H₂O is a polar molecule.
Trichloromethane (CHCl₃): Three C–Cl dipoles pointing broadly the same way (towards the chlorines), and one C–H dipole pointing the other way. The three C–Cl dipoles are much larger than the single C–H dipole, so they do not cancel. CHCl₃ is polar.
Nitrogen trifluoride (NF₃): Three polar N–F bonds in a pyramidal arrangement, plus a lone pair. The three bond dipoles do not cancel because the pyramidal shape means they all have a component pointing in the same direction. NF₃ is polar.
Both molecules contain polar bonds to oxygen. CO₂ is linear, so the dipoles cancel → non-polar. H₂O is bent, so the dipoles do not cancel → polar. This is a classic comparison that illustrates why geometry matters as much as bond polarity.
CCl₄ is tetrahedral with four identical C–Cl bonds, so the dipoles cancel → non-polar. In CHCl₃, one chlorine is replaced by hydrogen. The different magnitude of C–H vs C–Cl dipoles means symmetry is broken → polar.
BF₃ is trigonal planar with no lone pair. The three B–F dipoles cancel by symmetry → non-polar. NF₃ is pyramidal because nitrogen has a lone pair. The asymmetric shape means the three N–F dipoles do not cancel → polar. This comparison shows the critical role of lone pairs in determining both shape and polarity.
In reality, bonding is not sharply divided into "ionic" and "covalent." Instead, there is a continuous spectrum:
| Electronegativity Difference | Bond Type | Example |
|---|---|---|
| 0 | Pure covalent (non-polar) | Cl–Cl, H–H |
| 0.1–0.4 | Weakly polar covalent | C–H (ΔEN = 0.4) |
| 0.5–1.7 | Polar covalent | H–Cl (ΔEN = 0.9), O–H (ΔEN = 1.4) |
| > 1.7 | Ionic (predominantly) | Na–Cl (ΔEN = 2.1), Mg–O (ΔEN = 2.3) |
Even in "ionic" compounds like NaCl, there is some sharing of electron density. And in "covalent" compounds like HF, there is significant charge separation. The boundary is not sharp but is a useful guide.
The polarity of molecules has profound consequences:
Intermolecular forces: Polar molecules can form dipole-dipole interactions, which are stronger than the London forces that non-polar molecules rely on. This affects boiling points and melting points.
Solubility: Polar molecules tend to dissolve in polar solvents (like water) and non-polar molecules tend to dissolve in non-polar solvents (like hexane). This is the "like dissolves like" principle.
Hydrogen bonding: The most important consequence of polarity. When hydrogen is bonded to a very electronegative atom (N, O, or F), the extreme polarity enables a special type of strong intermolecular interaction called a hydrogen bond. This will be explored in detail in the next lessons.
Common exam mistake: Students sometimes state that "CO₂ is non-polar because C–O bonds are non-polar." This is wrong — C–O bonds are very polar (ΔEN = 1.0). CO₂ is non-polar because the linear shape causes the two polar bond dipoles to cancel. Always distinguish between bond polarity and molecular polarity.
Edexcel 9CH0 Topic 2.5 — Electronegativity and bond polarity sits inside Topic 2 (Bonding and Structure) on Paper 1 (Advanced Inorganic and Physical Chemistry) but is one of the most genuinely synoptic ideas in the whole specification: virtually every later topic assumes you can predict where the δ+ and δ− centres of a bond lie and decide whether the molecule itself carries a net dipole. The sub-topic explicitly requires you to define electronegativity, recall trends across periods and down groups, predict whether a bond is non-polar, polar covalent, or predominantly ionic from the difference in electronegativity (ΔEN), and decide whether a molecule with polar bonds is itself polar by considering its shape (linking back to Topic 2.4).
The cross-paper reach is substantial. Paper 1 Topic 4 (Inorganic chemistry and the periodic table) uses the same logic to explain why the period 3 chlorides shift from predominantly ionic (NaCl, MgCl₂) toward predominantly covalent (AlCl₃, SiCl₄, PCl₅) — a textbook application of Fajans' rules where small, highly charged cations polarise the chloride anion until the bond looks more like a shared pair than a transferred electron. Paper 2 (Advanced Organic and Physical Chemistry) is built on bond polarity: the C–X bond in halogenoalkanes drives nucleophilic substitution in Topic 6, the C=O carbonyl polarity governs nucleophilic addition in Topic 16, and the C=C π-system polarisation underlies electrophilic addition to alkenes. Paper 3 (General and Practical Principles) uses the synoptic skill of taking an unfamiliar molecule, sketching its shape, and predicting overall polarity from first principles — typically as the second or third part of a multi-step practical interpretation. (refer to the official Pearson Edexcel specification document for exact wording).
Question (7 marks):
(a) Explain, with reference to Fajans' rules, why MgCl₂ shows significantly more covalent character than NaCl, even though both are usually classified as ionic compounds. (4)
(b) Predict the order of melting points of NaCl, MgCl₂ and AlCl₃, and account for your prediction in terms of bonding and structure. (3)
Solution with mark scheme:
(a) Step 1 — define the cationic property that drives polarisation.
Fajans' rules state that a cation will distort (polarise) the electron cloud of a neighbouring anion most strongly when the cation is small and carries a high charge, because both factors raise its charge density (charge per unit volume).
M1 — identifying small ionic radius and high charge as the two factors raising polarising power. Candidates who write only "Mg²⁺ is more charged" miss the size component and lose this mark in tougher mark schemes.
Step 2 — compare the cations.
Na⁺ has a charge of +1 and an ionic radius of about 102 pm; Mg²⁺ has a charge of +2 and a smaller ionic radius of about 72 pm. So Mg²⁺ has both a higher charge and a smaller radius, giving it a substantially higher charge density and therefore much greater polarising power than Na⁺.
A1 — explicit comparison of charge and size between Na⁺ and Mg²⁺.
Step 3 — describe the effect on the chloride anion.
The Mg²⁺ cation pulls electron density from the Cl⁻ anion back toward the internuclear region, distorting the spherical electron cloud of Cl⁻ and creating a degree of electron sharing along the Mg–Cl axis. This is the definition of covalent character superimposed on an ionic framework.
B1 — describing electron-cloud distortion of the anion (not just "the bond becomes more covalent" without mechanism).
Step 4 — link back to the question.
Because Mg²⁺ polarises Cl⁻ much more strongly than Na⁺ does, the Mg–Cl bond contains significantly more shared electron density than the Na–Cl bond. MgCl₂ therefore exhibits more covalent character even though ΔEN(Mg–Cl) ≈ 1.8 still places it nominally in the "ionic" band.
A1 — explicit comparative conclusion tied to Fajans' rules.
(b) Step 1 — predicted order.
Melting point: NaCl > MgCl₂ > AlCl₃ (in fact AlCl₃ sublimes at about 178 °C at atmospheric pressure, vastly lower than the ~801 °C of NaCl).
B1 — correct order.
Step 2 — bonding rationale.
NaCl is a giant ionic lattice with strong electrostatic attraction in three dimensions, requiring high energy to overcome. MgCl₂ is also predominantly ionic but with significant covalent character (Fajans), giving a layered lattice that melts at a lower temperature (~714 °C). AlCl₃ has Al³⁺ — small and highly charged — polarising Cl⁻ so strongly that the bonding becomes predominantly covalent; in the solid it forms Al₂Cl₆ dimers held together by weak intermolecular forces, hence the very low sublimation temperature.
M1 — linking ionic-vs-covalent character to lattice type.
A1 — correct connection between AlCl₃'s molecular (dimer) structure and its low melting/sublimation point.
Total: 7 marks (M2 A3 B2 across both parts).
Question (6 marks): For each of the following molecules, predict whether it is polar overall or non-polar overall. Justify each answer by reference to both the polarity of the individual bonds and the shape of the molecule.
(i) H₂O (ii) CO₂ (iii) NH₃ (iv) BF₃ (v) CHCl₃ (vi) CCl₄
Mark-scheme decomposition by AO:
Total: 6 marks (AO1 ≈ 2, AO2 ≈ 3, AO3 ≈ 1). A typical Edexcel-style question of this type would mix three "easy" (CO₂, NH₃, H₂O) with three "discriminator" molecules (BF₃, CHCl₃, CCl₄) to separate B/A grade candidates from A* candidates: A* answers must explicitly name the shape and the cancellation argument, not just state "yes/no" with a ΔEN value.
Connects to:
Topic 2.4 (shapes of simple molecules and ions): polarity of a molecule depends as much on shape as on bond polarity. The same C–F bond polarity gives a polar molecule (CH₃F, asymmetric tetrahedral about C) and a non-polar molecule (CF₄, fully symmetric tetrahedral). VSEPR is the prerequisite tool for every "is the molecule polar?" question.
Topic 2.6 (intermolecular forces): polar molecules form permanent dipole–dipole interactions in addition to London dispersion forces; non-polar molecules rely on dispersion alone. This explains why HCl (polar) boils at −85 °C while F₂ (non-polar, similar M_r) boils at −188 °C. Hydrogen bonding is the limiting case where extreme bond polarity (H bonded to N, O, or F) creates an exceptionally strong dipole-dipole interaction.
Topic 6 (organic chemistry — halogenoalkanes): the C–X bond polarity (X = halogen) creates a δ+ carbon that is attacked by nucleophiles. Counter-intuitively, C–F is the least reactive halogenoalkane in nucleophilic substitution despite being the most polar bond — because the C–F bond enthalpy (~485 kJ mol⁻¹) is too high to break. C–I is the most reactive despite the smallest dipole, because the long, weak C–I bond (~238 kJ mol⁻¹) breaks easily. Bond strength out-weighs bond polarity in determining reaction rate.
Topic 15 (transition metals — hydrolysis of aqua complexes): small, highly charged cations such as Al³⁺(aq) and Fe³⁺(aq) have such high polarising power that they distort the O–H bonds of coordinated water molecules, weakening them sufficiently for H⁺ to be released. This is exactly the same Fajans-style polarising-power argument that explains the covalent character of AlCl₃, now applied to give acidic solutions of [Al(H₂O)₆]³⁺ and [Fe(H₂O)₆]³⁺.
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