A lithium battery in your phone relies on lithium being an extremely reactive metal — it gives up electrons readily. The fluorine in non-stick Teflon is there because fluorine is the most electronegative element — it holds onto electrons so tightly that almost nothing can react with it. Reactivity and electronegativity are the chemical "personality" of each element, and they follow perfect periodic trends. Once you understand these trends, you can predict how any element will behave in a chemical reaction — without ever running the experiment.
Wrong: Electronegativity and electron affinity are the same thing.
Right: Electronegativity is an atom's ability to attract bonding electrons in a covalent bond (a relative scale). Electron affinity is the energy change when an atom gains an electron (a measurable thermodynamic quantity). They are related but distinct concepts.
| Selected Pauling electronegativity values | ||||
|---|---|---|---|---|
| H: 2.2 | Li: 1.0 | Be: 1.6 | B: 2.0 | C: 2.6 |
| N: 3.0 | O: 3.4 | F: 4.0 | Na: 0.9 | Mg: 1.3 |
| Al: 1.6 | Si: 1.9 | P: 2.2 | S: 2.6 | Cl: 3.2 |
The difference in electronegativity between two bonded atoms (|Δχ|) determines whether the bond is non-polar covalent, polar covalent, or ionic. These are approximate thresholds — bond character is actually a continuum from fully covalent to fully ionic.
| |Δχ| range | Bond type | Electron distribution | Example |
|---|---|---|---|
| < 0.4 | Non-polar covalent | Equally (or nearly equally) shared | H₂ (Δχ=0), Cl₂ (Δχ=0), CH₄ (Δχ=0.4) |
| 0.4–1.7 | Polar covalent | Unequally shared; δ+ on less electronegative, δ− on more electronegative | HCl (Δχ=1.0), H₂O (O–H: Δχ=1.2), NH₃ (N–H: Δχ=0.8) |
| ≥ 1.7 | Ionic | Electron effectively transferred; full charges form (M⁺ and X⁻) | NaCl (Δχ=2.3), MgO (Δχ=2.1), KF (Δχ=3.3) |
A more reactive metal will displace a less reactive metal from its salt solution. A more reactive halogen will displace a less reactive halide from its solution. These reactions are direct evidence of relative reactivity (covered in more depth in Module 2).
Worked Examples
Method A — Direct Δχ calculation
(a) Na–Cl
χ(Na) = 0.9, χ(Cl) = 3.2 → |Δχ| = 3.2 − 0.9 = 2.3 ≥ 1.7 → Ionic
Na effectively transfers its 3s¹ electron to Cl → Na⁺ and Cl⁻ form ionic bond.
(b) N–H in NH₃
χ(N) = 3.0, χ(H) = 2.2 → |Δχ| = 3.0 − 2.2 = 0.8 (0.4–1.7) → Polar covalent
N is more electronegative → δ− on N, δ+ on H. This is why NH₃ is a hydrogen bond acceptor and has significant dipole moment.
(c) C–H in CH₄
χ(C) = 2.6, χ(H) = 2.2 → |Δχ| = 2.6 − 2.2 = 0.4 → borderline non-polar covalent
Some sources classify this as just inside non-polar; others as very weakly polar. Methane is generally treated as non-polar overall.
(d) O–F in OF₂
χ(O) = 3.4, χ(F) = 4.0 → |Δχ| = 4.0 − 3.4 = 0.6 (0.4–1.7) → Polar covalent
F is more electronegative → δ− on F, δ+ on O. Unusual case: O is δ+ here, not δ−.
Method B — Position-based reasoning
(a) Na vs K — Group 1 metals
Na is Period 3; K is Period 4 → K is further down Group 1
Going down Group 1: atomic radius increases (more shells) → valence electron further from nucleus and more shielded → weaker nuclear pull on valence electron → lower IE → easier to lose the valence electron → MORE reactive metal.
K (Period 4) > Na (Period 3) in metallic reactivity.
(b) Cl vs Br — Group 17 halogens
Cl is Period 3; Br is Period 4 → Br is further down Group 17
Going down Group 17: atomic radius increases (more shells) → valence shell further from nucleus → more shielded → weaker nuclear attraction for an incoming electron → HARDER to gain that electron → LESS reactive non-metal.
Cl (Period 3) > Br (Period 4) in non-metallic reactivity.
Key contrast
Metals MORE reactive going down; non-metals LESS reactive going down.
Same underlying reason (larger atom, more shielding) but opposite outcomes because metals LOSE electrons and non-metals GAIN electrons.
Activities
1 Using the Pauling electronegativity values given in the lesson, classify each bond as non-polar covalent, polar covalent, or ionic. State the direction of polarity (δ+/δ−) where relevant: (a) H–F, (b) Mg–O, (c) P–Cl, (d) S–O.
2 Arrange the following in order of decreasing electronegativity: F, Na, O, Al, Cl. Justify by reference to period and group position.
A Predict which element in each pair is more reactive, and give the mechanistic reason: (a) Li vs Cs (both Group 1), (b) F₂ vs I₂ (both Group 17), (c) Mg vs Ba (both Group 2).
B Bromine water (Br₂ dissolved in water, orange-brown) is added to a solution of potassium iodide (KI, colourless). A purple colour develops, indicating I₂ has been produced. Write a word equation and explain which halogen is more reactive. Relate this to the periodic trend.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
Multiple Choice
5 random questions from a replayable lesson bank — feedback shown immediately
Short Answer
6. The following observations are recorded in a halogen displacement experiment: Chlorine water added to KBr solution → orange-brown colour develops (Br₂ formed). Bromine water added to KI solution → purple colour develops (I₂ formed). Iodine solution added to KCl solution → no colour change. Arrange Cl₂, Br₂, and I₂ in order of decreasing reactivity as oxidising agents, and explain the trend using the periodic table. 4 MARKS
7. Explain, using electronegativity and bond polarity, why HF is a polar molecule but F₂ is non-polar. In your answer, include a Δχ calculation for each bond and describe the charge distribution. 4 MARKS
1. (a) H–F: |Δχ| = 4.0−2.2 = 1.8 ≥ 1.7 → ionic (or very strongly polar covalent — many sources classify H–F as polar covalent with very high polarity due to the discrete H–F bond; |Δχ|=1.8 sits right at the boundary. Accept "polar covalent" with δ− on F and δ+ on H). (b) Mg–O: |Δχ| = 3.4−1.3 = 2.1 ≥ 1.7 → ionic; Mg gives electrons to O → Mg²⁺ and O²⁻. (c) P–Cl: |Δχ| = 3.2−2.2 = 1.0 (0.4–1.7) → polar covalent; δ− on Cl, δ+ on P. (d) S–O: |Δχ| = 3.4−2.6 = 0.8 (0.4–1.7) → polar covalent; δ− on O, δ+ on S.
2. Decreasing electronegativity: F > Cl > O > Al > Na. F (Group 17, Period 2): highest χ=4.0. Cl (Group 17, Period 3): χ=3.2 — same group as F but lower period → lower than F. O (Group 16, Period 2): χ=3.4 — same period as F but one group to the left → lower than F; higher than Cl because Period 2 vs Period 3. Al (Group 13, Period 3): χ=1.6 — Period 3 but far left. Na (Group 1, Period 3): χ=0.9 — far left of Period 3.
A: (a) Cs more reactive than Li — both Group 1 metals; Cs is in Period 6, Li in Period 2. Going down Group 1: atomic radius increases, valence electron is further from the nucleus and in a higher shell with more shielding → lower IE → much easier to lose the electron → Cs is far more reactive (reacts explosively with water). (b) F₂ more reactive than I₂ — both Group 17; F is Period 2, I is Period 5. Going down Group 17: atomic radius increases, the valence shell is further from nucleus → weaker pull on incoming electron → harder to gain that electron → less reactive. F₂ is the most reactive non-metal known. (c) Ba more reactive than Mg — both Group 2; Ba is Period 6, Mg is Period 3. Larger radius, lower IE → easier to lose 2 valence electrons → Ba is more reactive.
B: Word equation: bromine + potassium iodide → iodine + potassium bromide (Br₂ + 2KI → I₂ + 2KBr). This shows Br₂ is more reactive than I₂ because Br₂ can displace I⁻ from solution (more reactive halogen displaces less reactive halide). Periodic trend: Br is in Period 4, I is in Period 5 (same Group 17). Going down the group, atomic radius increases → the valence shell is further from the nucleus → nuclear attraction on an incoming electron weakens → harder to gain an electron → I₂ is less reactive than Br₂ as an oxidising agent (electron acceptor).
1. C — F has χ=4.0, the highest of all elements. O is 3.4, N is 3.0, Cl is 3.2.
2. B — |Δχ| = 4.0−0.8 = 3.2 ≥ 1.7 → ionic. K effectively transfers its 4s¹ electron to F.
3. D — Cs is below Li in Group 1, larger atomic radius, more shielded valence electron, lower IE₁ → easier to lose the valence electron → more reactive metal. A is wrong: Cs has lower electronegativity (not higher). B is wrong: all Group 1 elements have 1 valence electron. C is wrong: Cs has a LARGER radius, not smaller.
4. A — K (Period 4) > Na (Period 3) for metallic reactivity. B incorrectly uses mass. C is wrong: Mg reacts less vigorously than Na with water. D is wrong: F₂ is more reactive than Cl₂ (halogen reactivity decreases down the group).
5. C — Si and P are in the same period (Period 3), same shell shielding. P has Z=15, Si has Z=14 → P has one more proton → higher Z_eff → stronger attraction for bonding electrons → higher electronegativity (Si: 1.9, P: 2.2).
Q6 (4 marks): Order (most to least reactive as oxidising agents): Cl₂ > Br₂ > I₂ (1 mark). Halogens react by gaining one electron to form a halide ion (X⁻); the more reactive halogen can displace a less reactive halide from solution (1 mark). Going down Group 17 from Cl (Period 3) to Br (Period 4) to I (Period 5), the atomic radius increases — the outer valence shell is further from the nucleus with more inner electron shielding (1 mark). This weakens the nuclear attraction on an incoming electron, making it progressively harder to gain that electron → reactivity as an oxidising agent decreases down the group. This explains why Cl₂ can displace Br⁻ and I⁻, Br₂ can displace I⁻ but not Cl⁻, and I₂ cannot displace either Br⁻ or Cl⁻ (1 mark).
Q7 (4 marks): HF bond: χ(H) = 2.2, χ(F) = 4.0. |Δχ| = 4.0 − 2.2 = 1.8 ≥ 1.7 (1 mark). The bond is ionic/highly polar (at boundary). Bonding electrons are strongly attracted toward F (more electronegative) → F end carries δ− (partial negative charge) and H end carries δ+ (partial positive charge) → the molecule has a net dipole → HF is polar (1 mark). F₂ bond: χ(F) = 4.0 for both F atoms. |Δχ| = 4.0 − 4.0 = 0 (1 mark). There is zero difference in electronegativity → bonding electrons are shared exactly equally → no partial charges → no dipole → F₂ is a non-polar molecule (1 mark). Both are diatomic molecules, but their polarity differs entirely because one has identical atoms (F₂) and one has different atoms with different electronegativities (HF).
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
Fluorine is the most electronegative element and reacts violently with almost everything. Cesium is one of the least electronegative elements and also reacts violently with water. Yet their reactivity comes from opposite tendencies — fluorine wants to gain electrons, while cesium wants to lose them. How can two elements at opposite ends of the periodic table both be considered "highly reactive"?
Before reading on, write your best answer. What does electronegativity tell you about how an element will behave in a chemical reaction?
Core Content
Return to your Think First response. You should now be able to explain why both extremes are highly reactive:
Defend your ship by blasting the correct answers for Periodic Trends: Electronegativity and Reactivity. Scores count toward the Asteroid Blaster leaderboard.
Play Asteroid Blaster →Periodic Trends: Electronegativity and Reactivity
Tick when you've finished all activities and checked your answers.