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.
Core Content
| Direction | Trend | Explanation | Values |
|---|---|---|---|
| Across period (L→R) | Electronegativity increases | Z_eff increases (same shielding, more protons) → stronger nuclear pull on bonding electrons → greater tendency to attract shared electrons toward the atom | Period 2: Li(1.0) → F(4.0). Period 3: Na(0.9) → Cl(3.2) |
| Down a group | Electronegativity decreases | Atomic radius increases (new shells) → bonding electrons further from nucleus → more shielded → weaker ability to attract bonding electrons | Group 17: F(4.0) > Cl(3.2) > Br(3.0) > I(2.7) |
| 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) |
| Element type | Across a period (L→R) | Down a group | Explanation |
|---|---|---|---|
| Metals (left side) | Metallic reactivity decreases L→R (metals become less reactive; left side metals more reactive) | Metallic reactivity increases going down | Metals react by LOSING electrons. Down a group: atomic radius ↑, IE ↓ → easier to lose electrons → more reactive. Across period: IE increases → harder to lose electrons → less reactive. |
| Non-metals (right side) | Non-metallic reactivity increases L→R (right side non-metals more reactive) | Non-metallic reactivity decreases going down | Non-metals react by GAINING electrons. Down a group: atomic radius ↑ → harder to attract new electrons → less reactive. Across period: electronegativity ↑ → greater tendency to attract electrons → more reactive. |
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
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.
Multiple Choice
1. Which element has the highest electronegativity?
2. The bond between potassium (χ=0.8) and fluorine (χ=4.0) would be classified as:
3. Which correctly explains why caesium (Cs) is a more reactive metal than lithium (Li)?
4. Which pair correctly shows a metal that is more reactive than the other in the pair?
5. The electronegativity of silicon (Si, Group 14, Period 3) compared to phosphorus (P, Group 15, Period 3) is:
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).
Tick when you've finished all activities and checked your answers.