Periodic Trends: Electronegativity and Reactivity

Activity 1

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.

Activity 2

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).

❓ Multiple Choice

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).

Short Answer Model Answers

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).